Problem: Given a tetrahedron ABCD, let D1 be the centroid ò the triangle ABC and let A1,B1,C1 be the intersection points of the lines parallel to DD1 and passing through the points A,B,C with the opposite faces of the tetrahedron. Prove that the volume of the tetrahedron ABCD is one-third the volume of the tetrahedron A1B1C1D1. Does the result remain true if the point D1 is replaced with any point inside the triangle ABC?
Solution: We shall prove that the statement is valid in the general case, for an arbitrary point D1 inside $\triangle{ABC}$. Since D1 belongs to the plane ABC, there are real numbers a,b,c such that $(a+b+c)\vec{DD_1}=a\vec{DA}+b\vec{DB}+c\vec{DC}$. Since $AA_\parallel DD_1$, it holds that $\vec{AA_1}=k\vec{DD_1}$ for some $k\in \mathbb{R}$. Now it is easy to get $\vec{DA_1}=-\frac{b\vec{DA}+c\vec{DC}}{a},\vec{DB_1}=-\frac{a\vec{DA}+c\vec{DC}}{b}$, and $\vec{DC_1}=-\frac{a\vec{DA}+b\vec{DB}}{c}$. This implies: $\vec{D_1A_1}=-\frac{a^2\vec{DA}+b(a+2b+c)\vec{DB}+c(a+b+2c)\vec{DC}}{a(a+b+c)},\vec{D_1B_1}=-\frac{a(2a+b+c)\vec{DA}+b^2\vec{DB}+c(a+b+2c)\vec{DC}}{b(a+b+c)}$, and $\vec{D_1C_1}=-\frac{a(2a+b+c)\vec{DA}+b(a+2b+c)\vec{DB}+c^2\vec{DC}}{c(a+b+c)}$.
By using $6V_{D_1A_1B_1C_1}=\big |[\vec{D_1A_1},\vec{D_1B_1},\vec{D_1C_1}]\big |$ and $6V_{DABC}=\big |[\vec{DA},\vec{DB},\vec{DC}]\big |$ we get
$V_{D_1A_1B_1C_1}=\frac{S}{6abc(a+b+c)^3}=3V_{DABC}$, where $S=\begin{align*}\begin{vmatrix} a^2&b(a+2b+c)&c(a+b+2c)\\ a(2a+b+c)&b^2&c(a+b+2c)\\ a(2a+b+c)& b(a+2b+c)&c^2\end{vmatrix}\end{align*}$.
Thứ Bảy, 23 tháng 3, 2019
IMO 1964 - Problem 5
Problem: Five points are given in the plane. Among the lines that connect these five points, no two coincide and no two are parallel or perpendicular. Through each point we construct an altitude to each of the other lines. What is the maximal number of intersection points of these altitudes (excluding the initial five points)>
Solution: Let us first compute the number of intersection points of the perpendiculars passing through two distinct points B and C. The perpendiculars from B to the lines through C other than BC meet all perpendiculars from C, which counts to 3.6=18 intersection points. Each perpendicular from B to the 3 lines not containing C can intersect at most 5 of the perpendicular passing through C, which counts 3.5=15 intersection points. Thus there are 18+15=33 intersection points corresponding to B,C.
It follows that the required total number is at most 10.33=330. But some of these points, namely the orthcenters of the triangles with vertices at the given points, namely the orthocenters of the triangles with vertices at the given points, are counted thrice. There are 10 such points. Hence the maximal number of intersection points is 330-2.10=310.
Remark. The jury considered only the combinaorial part of the problem and didn't require an example in which 310 points appear. However, it is "easily" verified that, for instance, the set of points A(1,1),B(e,$\pi$),C($e^2,\pi^2$,D($e^3,\pi^3$),E($e^4,\pi^4$)) works.
Solution: Let us first compute the number of intersection points of the perpendiculars passing through two distinct points B and C. The perpendiculars from B to the lines through C other than BC meet all perpendiculars from C, which counts to 3.6=18 intersection points. Each perpendicular from B to the 3 lines not containing C can intersect at most 5 of the perpendicular passing through C, which counts 3.5=15 intersection points. Thus there are 18+15=33 intersection points corresponding to B,C.
It follows that the required total number is at most 10.33=330. But some of these points, namely the orthcenters of the triangles with vertices at the given points, namely the orthocenters of the triangles with vertices at the given points, are counted thrice. There are 10 such points. Hence the maximal number of intersection points is 330-2.10=310.
Remark. The jury considered only the combinaorial part of the problem and didn't require an example in which 310 points appear. However, it is "easily" verified that, for instance, the set of points A(1,1),B(e,$\pi$),C($e^2,\pi^2$,D($e^3,\pi^3$),E($e^4,\pi^4$)) works.
Thứ Sáu, 22 tháng 3, 2019
IMO 1964 - Problem 4
Problem: Each of 17 students talked with every other student. They all talked about three different topics. Each pair of students talked about one topic. Prove that there are three students that talked about the same topic among themselves.
Solution: Let us call the topics T1,T2,T3. Consider an aarbitrary student A. By the pigeonhole principle there is a topic, say T3, he discussed with at least 6 other students. If two of these 6 students discussed T3, then we are done.
Suppose now that the 6 students discussed only T1 and T2 and choose one of them, say B. By the pigeonhole principle he discussed one of the topics, say T2, with three of these students. If two of these three students also discussed T2, then we are done. Otherwise, all the three students discussed only T1, which completes the task.
Solution: Let us call the topics T1,T2,T3. Consider an aarbitrary student A. By the pigeonhole principle there is a topic, say T3, he discussed with at least 6 other students. If two of these 6 students discussed T3, then we are done.
Suppose now that the 6 students discussed only T1 and T2 and choose one of them, say B. By the pigeonhole principle he discussed one of the topics, say T2, with three of these students. If two of these three students also discussed T2, then we are done. Otherwise, all the three students discussed only T1, which completes the task.
Thứ Hai, 11 tháng 2, 2019
IMO 1964 - Problem 3
Problem: The circle is inscribed in a triangle $ABC$ with sides $a,b,c$. Three tangents to the incircle are drawn, each of which is parallel to one side of the triangle $ABC$. These tangents form three smaller triangles (internal to $\triangle{ABC}$) with the sides of $\triangle{ABC}$. In each of these triangles an incircle is inscribed. Determine the sum of areas of all four incircles.
Solution: Let $r$ be the radius of the incircle of $\triangle{ABC},r_a,r_b,r_c$ the radius of the smaller circles corresponding to $A,B,C$, and $h_a,h_b,h_c$ the altitudes from $A,B,C$ respectively. The coefficient of similarity between the smaller triangle at $A$ and the triangle $ABC$ is $1-\frac{2r}{h_a}$, from which we easily obtain $r_a=\frac{(h_a-2r)r}{h_a}=\frac{(s-a)r}{s}$. Similarly, $r_b=\frac{(s-b)r}{s}$ and $r_c=\frac{(s-c)r}{s}$. Now a straightforward computation gives that the sum of areas of the four circles is given by $\sum =\frac{(b+c-a)(c+a-b)(a+b-c)(a^2+b^2+c^2)\pi}{(a+b+c)^3}$.
Solution: Let $r$ be the radius of the incircle of $\triangle{ABC},r_a,r_b,r_c$ the radius of the smaller circles corresponding to $A,B,C$, and $h_a,h_b,h_c$ the altitudes from $A,B,C$ respectively. The coefficient of similarity between the smaller triangle at $A$ and the triangle $ABC$ is $1-\frac{2r}{h_a}$, from which we easily obtain $r_a=\frac{(h_a-2r)r}{h_a}=\frac{(s-a)r}{s}$. Similarly, $r_b=\frac{(s-b)r}{s}$ and $r_c=\frac{(s-c)r}{s}$. Now a straightforward computation gives that the sum of areas of the four circles is given by $\sum =\frac{(b+c-a)(c+a-b)(a+b-c)(a^2+b^2+c^2)\pi}{(a+b+c)^3}$.
Thứ Sáu, 25 tháng 1, 2019
IMO 1964 - Problem 2
Problem: Denote by $a,b,c$ the lengths of the sides of a triangle. Prove that $a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$.
Solution: By substituting $a=x+y,b=y+z,c=z+x(x,y,z>0)$ the given inequality becomes $6xyz\le x^2y+xy^2+y^2z+yz^2+z^2x+zx^2$, which follows immediately by the AM-GM inequality applied to $x^2y,xy^2,y^z,yz^2,z^x,zx^2$.
Solution: By substituting $a=x+y,b=y+z,c=z+x(x,y,z>0)$ the given inequality becomes $6xyz\le x^2y+xy^2+y^2z+yz^2+z^2x+zx^2$, which follows immediately by the AM-GM inequality applied to $x^2y,xy^2,y^z,yz^2,z^x,zx^2$.
IMO 1964 - Problem 1
Problem: (a) Find all natural numbers $n$ such that the number $2^n-1$ is divisible by $7$.
(b) Prove that for all natural numbers $n$ the number $2^n+1$ is not divisible by $7$.
Solution: Let $n=3k+r$, where $0\le r<2$. Then $2^{n}=2^{3k+r}=8^{k}.2^{r}\equiv 2^{r}(\text{ mod }7)$.
Thus the remainder of $2^n$ modulo $7$ is $1,2,4$ if $n\equiv 0,1,2(\text{ mod }3)$. Hence $2^{n}-1$ is divisible by $7$ if and only if $3|n$, while $2^{n}+1$ is never divisible by $7$.
(b) Prove that for all natural numbers $n$ the number $2^n+1$ is not divisible by $7$.
Solution: Let $n=3k+r$, where $0\le r<2$. Then $2^{n}=2^{3k+r}=8^{k}.2^{r}\equiv 2^{r}(\text{ mod }7)$.
Thus the remainder of $2^n$ modulo $7$ is $1,2,4$ if $n\equiv 0,1,2(\text{ mod }3)$. Hence $2^{n}-1$ is divisible by $7$ if and only if $3|n$, while $2^{n}+1$ is never divisible by $7$.
IMO 1963 - Problem 6
Problem: Five students $A,B,C,D$ and $E$ have taken part in a certain competition. Before the competition, two persons $X$ and $Y$ tried to guess the rankings. $X$ thought that the ranking would be $A,B,C,D,E$ and $Y$ thought that the ranking would be $D,A,E,C,B$. At the end, it was revealed that $X$ didn't guess correctly any rankings of the participants , and moreover, didn't guess any of the ordering of pairs of consecutive participants. On the other hand , $Y$ guessed the correct rankings of two participants and the correct ordering of two pairs of consecutive participants. Determine the rankings of the competition.
Solution: The result is $EDACB$
Solution: The result is $EDACB$
IMO 1963 - Problem 5
Problem: Prove that $cos(\frac{\pi}{7})-cos(\frac{2\pi}{7})+cos(\frac{3\pi}{7})=\frac{1}{2}$.
Solution: The LHS of the desired idetity equals $S=cos(\frac{\pi}{7})+cos(\frac{3\pi}{7})+\frac{5\pi}{7}$. Now
$S.sin(\frac{\pi}{7})=\frac{sin\frac{2\pi}{7}}{2}+\frac{sin\frac{4\pi}{7}-sin\frac{2\pi}{7}}{2}+\frac{sin\frac{6\pi}{7}-sin\frac{4\pi}{7}}{2}=\frac{sin\frac{6\pi}{7}}{2}\implies S=\frac{1}{2}$.
Solution: The LHS of the desired idetity equals $S=cos(\frac{\pi}{7})+cos(\frac{3\pi}{7})+\frac{5\pi}{7}$. Now
$S.sin(\frac{\pi}{7})=\frac{sin\frac{2\pi}{7}}{2}+\frac{sin\frac{4\pi}{7}-sin\frac{2\pi}{7}}{2}+\frac{sin\frac{6\pi}{7}-sin\frac{4\pi}{7}}{2}=\frac{sin\frac{6\pi}{7}}{2}\implies S=\frac{1}{2}$.
IMO 1963 - Problem 4
Problem: Find all solutions $x_1,...,x_5$ to the system of equations:
$\left\{\begin{array}{I} x_5+x_2=yx_1,\\ x_1+x_3=yx_2,\\x_2+x_4=yx_3,\\ x_3+x_5=yx_4,\\x_4+x_1=yx_5, \end{array}\right.$ where $y$ is a real parameter.
Solution: Summing up all the equations yields $2(x_1+x_2+x_3+x_4+x_5)=y(x_1+x_2+x_3+x_4+x_5)$. If $y=2$, then the given equations imply $x_1-x_2=x_2-x_3=...=x_5-x_1$; hence $x_1=x_2=...=x_5$, which is clearly a solution. If $y\ne 2$, then $x_1+...+x_5=0$, and summing the first three equalities gives $x_2=y(x_1+x_2+x_3)$. Using that $x_1+x_3=yx_2$ we obtain $x_2=(y^2+y)x_2,i.e,(y^2+y-1)x_2=0$. If $y^2+y-1\ne 0$ then $x_2=0$ and similarly $x_1=...=x_5=0$. If $y^2+y-1=0$, it is easy to prove that the last two equations are the consequence of the first three. Thus choosing any values for $x_1$ and $x_5$ will give exactly one solution for $x_2,x_3,x_4$.
$\left\{\begin{array}{I} x_5+x_2=yx_1,\\ x_1+x_3=yx_2,\\x_2+x_4=yx_3,\\ x_3+x_5=yx_4,\\x_4+x_1=yx_5, \end{array}\right.$ where $y$ is a real parameter.
Solution: Summing up all the equations yields $2(x_1+x_2+x_3+x_4+x_5)=y(x_1+x_2+x_3+x_4+x_5)$. If $y=2$, then the given equations imply $x_1-x_2=x_2-x_3=...=x_5-x_1$; hence $x_1=x_2=...=x_5$, which is clearly a solution. If $y\ne 2$, then $x_1+...+x_5=0$, and summing the first three equalities gives $x_2=y(x_1+x_2+x_3)$. Using that $x_1+x_3=yx_2$ we obtain $x_2=(y^2+y)x_2,i.e,(y^2+y-1)x_2=0$. If $y^2+y-1\ne 0$ then $x_2=0$ and similarly $x_1=...=x_5=0$. If $y^2+y-1=0$, it is easy to prove that the last two equations are the consequence of the first three. Thus choosing any values for $x_1$ and $x_5$ will give exactly one solution for $x_2,x_3,x_4$.
IMO 1963 - Problem 3
Problem: Prove that if all the angles of a convex $n-gon$ are equal and the lengths of consecutive edges $a_1,...,a_n$ satisfy $a_1\ge a_2\ge ...\ge a_n$, then $a_1=a_2=...=a_n$.
Solution: Let $\vec{OA_1},\vec{OA_2},...,\vec{OA_n}$ be the vectors corresponding respectively to the edges $a_1,a_2,...,a_n$ of the polygon. By the conditions of the problem, these vectors satisfy $\vec{OA_1}+...+\vec{OA_n}=\vec{0},\angle{A_1OA_2}=\angle{A_2OA_3}=...=\angle{A_nOA_1}=\frac{2\pi}{n}$ and $OA_1\ge OA_2\ge ...\ge OA_n$. Our task is to prove that $OA_1=...=OA_n$.
Let $l$ be the line through $O$ perpendicular to $OA_n$ and $B_1,...,B_{n-1}$ the projections of $A_1,...,A_{n-1}$ onto $l$ respectively. By the assumptions, the sum of the $\vec{OB_i}'s$ is $\vec{0}$. On the other hand, since $OB_i\le OB_{n-i}$ for all $i\le \frac{n}{2}$, all the sums $\vec{OB_i}+\vec{OB_{n-i}}$ lie on the same side of the point $O$. Hence all these sums must be equal to $\vec{0}$. Consequently, $OA_i=OA_{n-i},$ from which the result immediately follows.
Solution: Let $\vec{OA_1},\vec{OA_2},...,\vec{OA_n}$ be the vectors corresponding respectively to the edges $a_1,a_2,...,a_n$ of the polygon. By the conditions of the problem, these vectors satisfy $\vec{OA_1}+...+\vec{OA_n}=\vec{0},\angle{A_1OA_2}=\angle{A_2OA_3}=...=\angle{A_nOA_1}=\frac{2\pi}{n}$ and $OA_1\ge OA_2\ge ...\ge OA_n$. Our task is to prove that $OA_1=...=OA_n$.
Let $l$ be the line through $O$ perpendicular to $OA_n$ and $B_1,...,B_{n-1}$ the projections of $A_1,...,A_{n-1}$ onto $l$ respectively. By the assumptions, the sum of the $\vec{OB_i}'s$ is $\vec{0}$. On the other hand, since $OB_i\le OB_{n-i}$ for all $i\le \frac{n}{2}$, all the sums $\vec{OB_i}+\vec{OB_{n-i}}$ lie on the same side of the point $O$. Hence all these sums must be equal to $\vec{0}$. Consequently, $OA_i=OA_{n-i},$ from which the result immediately follows.
Thứ Bảy, 19 tháng 1, 2019
IMO 1963 - Problem 2
Problem: Find the locus of points in space that are vertices of right angles of which one ray passes through a given point and the other intersects a given segment.
Solution: Let $A$ be the given point, $BC$ the given segment, and $\mathscr{B_1},\mathscr{B_2}$ the closed balls with the diameters $AB$ and $AC$ respectively. Consider one right angle $\angle{AOK}$ with $K\in [BC]$. If $B',C'$ are the feet of the perpendiculars from $B,C$ to $AO$ respectively, then $O$ lies on the segment $B'C'$, which implies that it lies on exactly one of the segment $AB',AC'$. Hence $O$ belongs to exactly one of the balls $\mathscr{B_1},\mathscr{B_2},i.e, O\in \mathscr{B_1}\Delta \mathscr{B_2}$. This is obviously the required locus.
Solution: Let $A$ be the given point, $BC$ the given segment, and $\mathscr{B_1},\mathscr{B_2}$ the closed balls with the diameters $AB$ and $AC$ respectively. Consider one right angle $\angle{AOK}$ with $K\in [BC]$. If $B',C'$ are the feet of the perpendiculars from $B,C$ to $AO$ respectively, then $O$ lies on the segment $B'C'$, which implies that it lies on exactly one of the segment $AB',AC'$. Hence $O$ belongs to exactly one of the balls $\mathscr{B_1},\mathscr{B_2},i.e, O\in \mathscr{B_1}\Delta \mathscr{B_2}$. This is obviously the required locus.
IMO 1963 - Problem 1
Problem: Determine all real solutions of the equation $\sqrt{x^2-p}+2\sqrt{x^2-1}=x$, where $p$ is a real number.
Solution: Obviously, $x\ge 0$; hence squaring the given equation yields an equivalent equation $5x^2-p-4+4\sqrt{(x^2-1)(x^2-p)}=x^2$, i.e, $4\sqrt{(x^2-1)(x^2-p)}=(p+4)-4x^2$. If 4x^2\le (p+4), we may square the equation once again to get $-16(p+1)x^2+16p=-8(p+4)x^2+(p+4)^2$, which is equivalent to $x^2=\frac{(4-p)^2}{4(4-2p)}$,i.e $x=\frac{4-p}{2\sqrt{4-2p}}$. For this to be a solution we must have $p\le 2$ and $\frac{(4-p)^2}{4-2p}=4x^2\le p+4$. Hence $\frac{4}{3}\le p\le 2$. Otherwise there is no solution.
Solution: Obviously, $x\ge 0$; hence squaring the given equation yields an equivalent equation $5x^2-p-4+4\sqrt{(x^2-1)(x^2-p)}=x^2$, i.e, $4\sqrt{(x^2-1)(x^2-p)}=(p+4)-4x^2$. If 4x^2\le (p+4), we may square the equation once again to get $-16(p+1)x^2+16p=-8(p+4)x^2+(p+4)^2$, which is equivalent to $x^2=\frac{(4-p)^2}{4(4-2p)}$,i.e $x=\frac{4-p}{2\sqrt{4-2p}}$. For this to be a solution we must have $p\le 2$ and $\frac{(4-p)^2}{4-2p}=4x^2\le p+4$. Hence $\frac{4}{3}\le p\le 2$. Otherwise there is no solution.
IMO 1963 - Problem 7
Problem: Prove that a tetrahedron $SABC$ has five different spheres that touch all six lines determined by its edges if and only if it is regular.
Solution: The sphere are arranged in a similar manner as in the planar case where we have one incircle and three excircles. Here we have one "insphere" and four "exspheres" corresponding to each of the four sides. Each vertex of the tetrahedron effectively has three tangent lines drawn from it to each of the five spheres. Repeatedly using the equality of the three tangent segments from a vertex (in the same vein as for tangent planar quadrilaterals ) we obtain $SA+BC=SB+CA=SC+AB$ from the insphere. From the hence $SA=SB=SC$ and $AB=BC=CA$. By symmetry, we also have $AB=AC=AS$. Hence indeed, all the edges of the tetraghedron are equal in length and thus we have shown that the tetrahedron is regular.
Solution: The sphere are arranged in a similar manner as in the planar case where we have one incircle and three excircles. Here we have one "insphere" and four "exspheres" corresponding to each of the four sides. Each vertex of the tetrahedron effectively has three tangent lines drawn from it to each of the five spheres. Repeatedly using the equality of the three tangent segments from a vertex (in the same vein as for tangent planar quadrilaterals ) we obtain $SA+BC=SB+CA=SC+AB$ from the insphere. From the hence $SA=SB=SC$ and $AB=BC=CA$. By symmetry, we also have $AB=AC=AS$. Hence indeed, all the edges of the tetraghedron are equal in length and thus we have shown that the tetrahedron is regular.
IMO 1962 - Problem 6
Problem: Let $ABC$ be an isosceles triangle with circumradius $r$ and inradius $p$. Prove that the distance $d$ between the circumcenter and incenter is given by $d=\sqrt{r(r-2p)}$.
Solution: This problem is a special case, when the triangle is isosceles, of Euler's formula, which holds for all triangles.
Solution: This problem is a special case, when the triangle is isosceles, of Euler's formula, which holds for all triangles.
IMO 1962 - Problem 5
Problem: On the circle $k$ three point $A,B$ and $C$ are given. Construct the fourth point on the circle $D$ such that one can inscribe in a circle in $ABCD$.
Solution: Analysis. Let $ABCD$ be the desired quadrilateral. Let us assume w.l.o.g that $AB>BC$( for $AB=BC$ the construction is trivial). For a tangent quadrilateral we have $AD-DC=AB-BC$ and $\angle{AXC}=\angle{ADC}+\angle{CDX}=180^0-\frac{\angle{ABC}}{2}$. Constructing $X$ and hence $D$ is how obvious.
Solution: Analysis. Let $ABCD$ be the desired quadrilateral. Let us assume w.l.o.g that $AB>BC$( for $AB=BC$ the construction is trivial). For a tangent quadrilateral we have $AD-DC=AB-BC$ and $\angle{AXC}=\angle{ADC}+\angle{CDX}=180^0-\frac{\angle{ABC}}{2}$. Constructing $X$ and hence $D$ is how obvious.
IMO 1962 - Problem 4
Problem: Solve the equation: $cos^2(x)+cos^2(2x)+cos^2(3x)=1$.
Solution: Since $cos(2x)=1+cos^2(x)$ and $cos(\alpha)+cos(\beta)=2cos(\frac{\alpha+\beta}{2})cos(\frac{\alpha-\beta}{2}),$ we have $cos^2(x)+cos^2(2x)+cos^2(3x)=1\iff cos(2x)+cos(4x)+2cos^2(3x)=2cos(3x)(cos(x)+cos(3x))=0\iff 4cos(3x)cos(2x)cos(x)=0$. Hence the solutions are $x\in \left\{\frac{\pi}{2}+m\pi,\frac{\pi}{4}+\frac{m\pi}{2},\frac{\pi}{6}+\frac{m\pi}{3}|m\in \mathbb{Z}\right\}$.
Solution: Since $cos(2x)=1+cos^2(x)$ and $cos(\alpha)+cos(\beta)=2cos(\frac{\alpha+\beta}{2})cos(\frac{\alpha-\beta}{2}),$ we have $cos^2(x)+cos^2(2x)+cos^2(3x)=1\iff cos(2x)+cos(4x)+2cos^2(3x)=2cos(3x)(cos(x)+cos(3x))=0\iff 4cos(3x)cos(2x)cos(x)=0$. Hence the solutions are $x\in \left\{\frac{\pi}{2}+m\pi,\frac{\pi}{4}+\frac{m\pi}{2},\frac{\pi}{6}+\frac{m\pi}{3}|m\in \mathbb{Z}\right\}$.
IMO 1962 - Problem 3
Problem: A cube $ABCDA'B'C'D'$ is given. The point $X$ is moving at a constant speed along the square $ABCD$ in the direction from $A$ to $B$. The point $Y$ is moving with the same constant speed along the square $BCC'B'$ in the direction from $B'$ to $C'$. Initially, $X$ and $Y$ start out from $A$ and $B'$ respectively. Find the locus of all the midpoints of $XY$.
Solution: By inspecting the four different stages of this periodic motion we easily obtain that the locus of the midpoints of $XY$ is the edges of $MNCQ$, where $M,N$ and $Q$ are the centers of $ABB'A'$, $BCC'B'$ and $ABCD$, respectively.
Solution: By inspecting the four different stages of this periodic motion we easily obtain that the locus of the midpoints of $XY$ is the edges of $MNCQ$, where $M,N$ and $Q$ are the centers of $ABB'A'$, $BCC'B'$ and $ABCD$, respectively.
IMO 1962 - Problem 2
Problem: Find all real number $x$ for which $\sqrt{3-x}-\sqrt{x+1}>\frac{1}{2}$.
Solution: We note that $f(x)=\sqrt{3-x}-\sqrt{x+1}$ is well-definded only for $-1\le x\le 3$ and is dereasing (and obviously continous) on this interval. We also note that $f(-1)=2>\frac{1}{2}$ and $f(1-\frac{\sqrt{31}}{8})=\sqrt{(\frac{1}{4}+\frac{\sqrt{31}}{4})^2}-\sqrt{(\frac{1}{4}-\frac{\sqrt{31}}{4})^2}=\frac{1}{2}$. Hence the inequality is satisfied for $-1\le x<1-\frac{\sqrt{31}}{8}$.
Solution: We note that $f(x)=\sqrt{3-x}-\sqrt{x+1}$ is well-definded only for $-1\le x\le 3$ and is dereasing (and obviously continous) on this interval. We also note that $f(-1)=2>\frac{1}{2}$ and $f(1-\frac{\sqrt{31}}{8})=\sqrt{(\frac{1}{4}+\frac{\sqrt{31}}{4})^2}-\sqrt{(\frac{1}{4}-\frac{\sqrt{31}}{4})^2}=\frac{1}{2}$. Hence the inequality is satisfied for $-1\le x<1-\frac{\sqrt{31}}{8}$.
IMO 1962 - Problem 1
Problem: Find the smallest natural number $n$ with the following properties:
(a) In decimal representation it, ends with 6.
(b) If we move this digit to the front of the number, we get a number $4$ times larger.
Solution: From the conditions of the problem we have $n=10x+6$ and $4n=6.10^{m}+x$ for some integer $x$. Eliminating $x$ from these two equations, we get $40n=6.10^{m+1}+n-6\implies n=\frac{2(10^{m+1})-1}{13}$. Hence we must find the smallest $m$ such that this fraction is an integer. By inspection, this happens for $m=6$, and for this $m$ we obtain $n=153846$, which indeed satisfies the conditions of the problem.
(a) In decimal representation it, ends with 6.
(b) If we move this digit to the front of the number, we get a number $4$ times larger.
Solution: From the conditions of the problem we have $n=10x+6$ and $4n=6.10^{m}+x$ for some integer $x$. Eliminating $x$ from these two equations, we get $40n=6.10^{m+1}+n-6\implies n=\frac{2(10^{m+1})-1}{13}$. Hence we must find the smallest $m$ such that this fraction is an integer. By inspection, this happens for $m=6$, and for this $m$ we obtain $n=153846$, which indeed satisfies the conditions of the problem.
Chủ Nhật, 13 tháng 1, 2019
IMO 1961 - Problem 6
Problem: A plane $\epsilon$ is given and on one side of the plane three nocollinear points $A,B$ and $C$ such that the plane determined by them is not parallel to $\epsilon$. Three arbitrary points $A',B'$ and $C'$ in $\epsilon$ are selected. Let $L,M$ and $N$ be the midpoints of $AA',BB'$ and $CC'$ and $G$ the centroid of $\triangle{LMN}$. Find the locus of all points obtained for $G$ as $A',B'$ and $C'$ are varied (independently of each other) across $\epsilon$.
Solution: Let $h(X)$ denote the distance of a point $X$ from $\epsilon$, $X$ restricted to being on the same side of $\epsilon$ as $A,B$ and $C$. Let $G_1$ be the (fixed) centroid of $\triangle{ABC}$ and $G_1'$ the centroid of $\triangle{A'B'C'}$. It is trivial to prove that $G$ is the midpoint of $G_1G_1'$. Hence varying $G_1'$ across $\epsilon$, we get that the locus of $G$ is the plane $\alpha$ parallel to $\epsilon $ such that:
$X\in \alpha\iff h(X)=\frac{h(G_1)}{2}=\frac{h(A)+h(B)+h(C)}{6}$.
Solution: Let $h(X)$ denote the distance of a point $X$ from $\epsilon$, $X$ restricted to being on the same side of $\epsilon$ as $A,B$ and $C$. Let $G_1$ be the (fixed) centroid of $\triangle{ABC}$ and $G_1'$ the centroid of $\triangle{A'B'C'}$. It is trivial to prove that $G$ is the midpoint of $G_1G_1'$. Hence varying $G_1'$ across $\epsilon$, we get that the locus of $G$ is the plane $\alpha$ parallel to $\epsilon $ such that:
$X\in \alpha\iff h(X)=\frac{h(G_1)}{2}=\frac{h(A)+h(B)+h(C)}{6}$.
IMO 1961 - Problem 5
Problem: Construct a triangle $ABC$ if the following elements are given: $AC=b,AB=c$ and $\angle{AMB}=\omega(\omega<90^0)$, where $M$ is the midpoint of $BC$. Prove that the construction has a solution if and only if $btan(\frac{\omega}{2})\le c<b$.
In what case does equality hold>
Solution: Analysis. Let $C_1$ be the midpoint of $AB$. In $\triangle{AMB}$ we have $MC_1=\frac{b}{2},$ $AB=c$ and $\angle{AMB}=\omega$. Thus, given $AB=c$, the point $M$ is at the intersection of the triangle $k(C';\frac{b}{2})$ and the set of points $e$ that view $AB$ at an angle of $\omega$. The construction of $ABC$ is now obvious.
Dicussion. It suffices to establish the conditions for which $k$ and $e$ intersect. Let $E$ be the midpoint of one of the arcs that make up $e$. A necessary and sufficient condition for $k$ to intersect $e$ is
$\frac{c}{2}=C'A\le \frac{b}{2}\le C'E=\frac{c}{2}cot\frac{w}{2}\iff btan\frac{\omega}{2}\le c<b$.
In what case does equality hold>
Solution: Analysis. Let $C_1$ be the midpoint of $AB$. In $\triangle{AMB}$ we have $MC_1=\frac{b}{2},$ $AB=c$ and $\angle{AMB}=\omega$. Thus, given $AB=c$, the point $M$ is at the intersection of the triangle $k(C';\frac{b}{2})$ and the set of points $e$ that view $AB$ at an angle of $\omega$. The construction of $ABC$ is now obvious.
Dicussion. It suffices to establish the conditions for which $k$ and $e$ intersect. Let $E$ be the midpoint of one of the arcs that make up $e$. A necessary and sufficient condition for $k$ to intersect $e$ is
$\frac{c}{2}=C'A\le \frac{b}{2}\le C'E=\frac{c}{2}cot\frac{w}{2}\iff btan\frac{\omega}{2}\le c<b$.
IMO 1961 - Problem 4
Problem: In the interior of $\triangle{P_1P_2P_3}$ a point $P$ is given. Let $Q_1,Q_2$ and $Q_3$ respectively be the intersections of $PP_1,PP_2,PP_3$ with the opposing edges of $\triangle{P_1P_2P_3}$. Prove that among the ratios $\frac{PP_1}{PQ_1},\frac{PP_2}{PQ_2}$ and $\frac{PP_3}{PQ_3}$ there exists at least one not larger than $2$ and at least one not smaller than $2$.
Solution: Let $x_i=\frac{PP_i}{PQ_i}$ for $i=1,2,3$. For all $i$ we have
$\frac{1}{x_i+1}=\frac{PQ_i}{P_iQ_i}=\frac{S_{PP_jP_k}}{S_{P_1P_2P_3}}$
where the indices $j$ and $k$ are distinct and different from $i$. Hence we have $f(x_1,x_2,x_3)=\frac{1}{x_1+1}+\frac{1}{x_2+1}+\frac{1}{x_3+1}=\frac{S_{PP_2P_3}+S_{PP_1P_3}+S_{PP_2P_1}}{S_{P_1P_2P_3}}=1$.
It follows that $\frac{1}{x_{i}+1}\ge \frac{1}{3}$ for some $i$ and $\frac{1}{x_{j}+1}\le \frac{1}{3}$ for some $j$. Consequently, $x_i\le 2$ and $x_j\ge 2$.
Solution: Let $x_i=\frac{PP_i}{PQ_i}$ for $i=1,2,3$. For all $i$ we have
$\frac{1}{x_i+1}=\frac{PQ_i}{P_iQ_i}=\frac{S_{PP_jP_k}}{S_{P_1P_2P_3}}$
where the indices $j$ and $k$ are distinct and different from $i$. Hence we have $f(x_1,x_2,x_3)=\frac{1}{x_1+1}+\frac{1}{x_2+1}+\frac{1}{x_3+1}=\frac{S_{PP_2P_3}+S_{PP_1P_3}+S_{PP_2P_1}}{S_{P_1P_2P_3}}=1$.
It follows that $\frac{1}{x_{i}+1}\ge \frac{1}{3}$ for some $i$ and $\frac{1}{x_{j}+1}\le \frac{1}{3}$ for some $j$. Consequently, $x_i\le 2$ and $x_j\ge 2$.
IMO 1961 - Problem 3
Problem: Solve the equation $cos^{n}(x)-sin^{n}(x)=1$, where $n$ is a given positive integer.
Solution: For $n\ge 2$ we have:
$1=cos^{n}(x)-sin^{n}(x)\le |cos^{n}(x)-sin^{n}(x)|\le |cos^{n}(x)|+|sin^{n}(x)|\le cos^2(x)+sin^2(x)=1$.
Hence $sin^2(x)=|sin^{n}(x)|$ and $cos^2(x)=|cos^{n}(x)|$, from which it follows that $sin(x),cos(x)\in \left\{1,0,-1\right\}\implies x\in \frac{\pi\mathbb{Z}}{2}$. By inspection one obtains the set of solutions
$\left\{m\pi |m\in \mathbb{Z}\right\}$ for even $n$ and $\left\{2m\pi,2m\pi-\frac{\pi}{2}|m\in \mathbb{Z}\right\}$ for odd $n$.
For $n=1$ we have $1=cos(x)-sin(x)=-\sqrt{2}sin(x-\frac{\pi}{4})$, which yields the set of solutions $\left\{2m\pi, 2m\pi- \frac{\pi}{2}|m\in \mathbb{Z}\right\}$.
Solution: For $n\ge 2$ we have:
$1=cos^{n}(x)-sin^{n}(x)\le |cos^{n}(x)-sin^{n}(x)|\le |cos^{n}(x)|+|sin^{n}(x)|\le cos^2(x)+sin^2(x)=1$.
Hence $sin^2(x)=|sin^{n}(x)|$ and $cos^2(x)=|cos^{n}(x)|$, from which it follows that $sin(x),cos(x)\in \left\{1,0,-1\right\}\implies x\in \frac{\pi\mathbb{Z}}{2}$. By inspection one obtains the set of solutions
$\left\{m\pi |m\in \mathbb{Z}\right\}$ for even $n$ and $\left\{2m\pi,2m\pi-\frac{\pi}{2}|m\in \mathbb{Z}\right\}$ for odd $n$.
For $n=1$ we have $1=cos(x)-sin(x)=-\sqrt{2}sin(x-\frac{\pi}{4})$, which yields the set of solutions $\left\{2m\pi, 2m\pi- \frac{\pi}{2}|m\in \mathbb{Z}\right\}$.
IMO 1961 - Problem 2
Problem: Let $a,b$ and $c$ be the lengths of a triangle whose area is $S$. Prove that $a^2+b^2+c^2\ge 4S\sqrt{3}$
Solution: Using $S=\frac{bcsin(\alpha)}{2},a^2=b^2+c^2-2bccos(\alpha)$ and $\frac{\sqrt{3}sin(\alpha)+cos(\alpha)}{2}=cos(\alpha-60^0)$ we have
$a^2+b^2+c^2\ge 4S\sqrt{3}\iff b^2+c^2\ge bc(\sqrt{3}sin(\alpha)+cos(\alpha))\iff (b-c)^2+2bc(1-cos(\alpha-60^0))\ge 0$,
where equality holds if and only if $b=c$ and $\alpha=60^0$, i.e, if the triangle is equilateral.
Solution: Using $S=\frac{bcsin(\alpha)}{2},a^2=b^2+c^2-2bccos(\alpha)$ and $\frac{\sqrt{3}sin(\alpha)+cos(\alpha)}{2}=cos(\alpha-60^0)$ we have
$a^2+b^2+c^2\ge 4S\sqrt{3}\iff b^2+c^2\ge bc(\sqrt{3}sin(\alpha)+cos(\alpha))\iff (b-c)^2+2bc(1-cos(\alpha-60^0))\ge 0$,
where equality holds if and only if $b=c$ and $\alpha=60^0$, i.e, if the triangle is equilateral.
IMO 1961 - Problem 1
Problem: Solve the following system of eqautions: $x+y+z=a,x^2+y^2+z^2=b^2,xy=z^2$.
where $a$ and $b$ are given real numbers. What conditions must hold on $a$ and $b$ for the solution to be positive and distinct?
Solution: This is a problem solvable using elementary manipulations, so we shall state only the final solutions. For $a=0$ we get $(x,y,z)=(0,0,0)$. For $a\ne 0$ we get $(x,y,z)\in \left\{(t_1,t_2,z_0),(t_2,t_1,z_0)\right\}$, where
$z_0=\frac{a^2-b^2}{2a}$ and $t_{1,2}=\frac{a^2+b^2\pm \sqrt{(3a^2-b^2)(3b^2-a^2)}}{4a}$.
For the solutions to be positive and distinct the following conditions are necessary and sifficient: $3b^2>a^2>b^2$ and $a>0$.
where $a$ and $b$ are given real numbers. What conditions must hold on $a$ and $b$ for the solution to be positive and distinct?
Solution: This is a problem solvable using elementary manipulations, so we shall state only the final solutions. For $a=0$ we get $(x,y,z)=(0,0,0)$. For $a\ne 0$ we get $(x,y,z)\in \left\{(t_1,t_2,z_0),(t_2,t_1,z_0)\right\}$, where
$z_0=\frac{a^2-b^2}{2a}$ and $t_{1,2}=\frac{a^2+b^2\pm \sqrt{(3a^2-b^2)(3b^2-a^2)}}{4a}$.
For the solutions to be positive and distinct the following conditions are necessary and sifficient: $3b^2>a^2>b^2$ and $a>0$.
Thứ Bảy, 12 tháng 1, 2019
IMO 1960 - Problem 7
Problem: A sphere is inscribed in a regular cone. Around the sphere a cylinder is circumscribed so that its base is in the same plane as the base of the cone. Let $V_1$ be the volume of the cone and $V_2$ the volume of the cylinder.
(a) Prove that $V_1=V_2$ is impossible.
(b) Find the smallest $k$ for which $V_1=kV_2$, and in this case construct the angle at the vertex of the cone.
Solution: Let $A$ be the vertex of the cone, $O$ the center of the sphere, $S$ the center of the base of the cone, $B$ a point on the base circle, and $r$ the radius of the sphere. Let $\angle{SAB}=\alpha$. We easily obtain $AS=\frac{r(1+sin(\alpha))}{sin(\alpha)}$ and $SB=\frac{r(1+sin(\alpha)tan(\alpha)}{sin(\alpha)}$ and hence $V_1=\pi.SB^2.\frac{SA}{3}=\pi.r^3.\frac{(1+sin(\alpha))^2}{3sin(\alpha)(1-sin(\alpha))}$. We also have $V_2=2\pi.r^3$ and hence $k=\frac{(1+sin(\alpha))^2}{6sin(\alpha)(1-sin(\alpha))}\implies (1+6k)sin^2(\alpha)+2(1-3k)sin(\alpha)+1=0$.
The discriminant of this quadratic must be nonnegative: $(1-3k)^2-(1+6k)\ge 0\implies k\ge \frac{4}{3}$. Hence we cannot have $k=1$. For $k=\frac{4}{3}$ we have $sin(\alpha)=\frac{1}{3}$, whose construction is elementary.
(a) Prove that $V_1=V_2$ is impossible.
(b) Find the smallest $k$ for which $V_1=kV_2$, and in this case construct the angle at the vertex of the cone.
Solution: Let $A$ be the vertex of the cone, $O$ the center of the sphere, $S$ the center of the base of the cone, $B$ a point on the base circle, and $r$ the radius of the sphere. Let $\angle{SAB}=\alpha$. We easily obtain $AS=\frac{r(1+sin(\alpha))}{sin(\alpha)}$ and $SB=\frac{r(1+sin(\alpha)tan(\alpha)}{sin(\alpha)}$ and hence $V_1=\pi.SB^2.\frac{SA}{3}=\pi.r^3.\frac{(1+sin(\alpha))^2}{3sin(\alpha)(1-sin(\alpha))}$. We also have $V_2=2\pi.r^3$ and hence $k=\frac{(1+sin(\alpha))^2}{6sin(\alpha)(1-sin(\alpha))}\implies (1+6k)sin^2(\alpha)+2(1-3k)sin(\alpha)+1=0$.
The discriminant of this quadratic must be nonnegative: $(1-3k)^2-(1+6k)\ge 0\implies k\ge \frac{4}{3}$. Hence we cannot have $k=1$. For $k=\frac{4}{3}$ we have $sin(\alpha)=\frac{1}{3}$, whose construction is elementary.
IMO 1960 - Problem 6
Problem: An isosceles trapezoid with bases $a$ and $b$ and height $h$ is given.
(a) On the line of symmetry construct the point $P$ such that both (non-base) sides are seen form $P$ with an angle of $90^0$.
(b) Find the distance of $P$ from one of the bases of the trapezoid .
(c) Under what conditions for $a,b$ and $h$ can be the point $P$ be constructed (analyze all possible cases) ?
Solution: Let $E,F$ respectively be the midpoints of the bases $AB,CD$ of the isosceles trapezoid $ABCD$.
(a) The point $P$ is on the intersection of $EF$ and the circle with diameter $BC$.
(b) Let $x=EP$. Since $\triangle{BEP}\sim \triangle{PFC}$, we have $x(h-x)=\frac{ab}{4}\implies x_{1,2}=\frac{h\pm \sqrt{h^2-ab}}{2}$.
(c) If $h^2>ab$ there are two solutions, if $h^2=ab$ there is only one solution, and if $h^2<ab$ there are no solutions.
(a) On the line of symmetry construct the point $P$ such that both (non-base) sides are seen form $P$ with an angle of $90^0$.
(b) Find the distance of $P$ from one of the bases of the trapezoid .
(c) Under what conditions for $a,b$ and $h$ can be the point $P$ be constructed (analyze all possible cases) ?
Solution: Let $E,F$ respectively be the midpoints of the bases $AB,CD$ of the isosceles trapezoid $ABCD$.
(a) The point $P$ is on the intersection of $EF$ and the circle with diameter $BC$.
(b) Let $x=EP$. Since $\triangle{BEP}\sim \triangle{PFC}$, we have $x(h-x)=\frac{ab}{4}\implies x_{1,2}=\frac{h\pm \sqrt{h^2-ab}}{2}$.
(c) If $h^2>ab$ there are two solutions, if $h^2=ab$ there is only one solution, and if $h^2<ab$ there are no solutions.
IMO 1960 - Problem 5
Problem: A cube $ABCDA'B'C'D'$ is given.
(a) Find the locus of all midpoints of segments $XY$, where $X$ is any point on segment $AC$ and $Y$ any point on segment $B'D'$.
(b) Find the locus of all point $Z$ on segments $XY$ such that $\vec{ZY}=2\vec{XZ}$.
Solution:
(a) The locus of the points is the square $EFGH$ where these four points are the centers of the faces $ABB'A',BCC'B',CDD'C'$ and $DAA'D'$.
(b) The locus of the points is the rectangle $IJKL$ where these points are on $AB',CB',CD'$ and $AD'$ at a distance of $\frac{AA'}{3}$ with respect to the plane $ABCD$.
(a) Find the locus of all midpoints of segments $XY$, where $X$ is any point on segment $AC$ and $Y$ any point on segment $B'D'$.
(b) Find the locus of all point $Z$ on segments $XY$ such that $\vec{ZY}=2\vec{XZ}$.
Solution:
(a) The locus of the points is the square $EFGH$ where these four points are the centers of the faces $ABB'A',BCC'B',CDD'C'$ and $DAA'D'$.
(b) The locus of the points is the rectangle $IJKL$ where these points are on $AB',CB',CD'$ and $AD'$ at a distance of $\frac{AA'}{3}$ with respect to the plane $ABCD$.
IMO 1960 - Problem 4
Problem: Construct a triangle $ABC$ whose lengths of heights $h_a$ and $h_b$ (from $A$ and $B$, respectively) and length of median $m_a$ (from $A$) are given.
Solution: Analysis. Let $A'$ and $B'$ be the feet of the perpendiculars from $A$ and $B$, respectively, to the opposite sides, $A_1$ the midpoint of $BC$, and let $D'$ be the foot of the perpendicular from $A_1$ to $AC$. We then have $AA_1=m_a$, $AA'=h_a,\angle{AA'A_1}=90^0,A_1D'=\frac{h_b}{2}$ and $\angle{AD'A_1}=90^0$.
Construction. We construct the quadrilateral $AD'A_1A'$( starting from the circle with diameter $AA_1$). Then $C$ is the intersection of $A'A_1$ and $AD'$, and $B$ is on the line $A_1C$ such that $CA_1=A_1B$ and $\mathscr{B}(B,A_1,C)$.
Dicussion. We must have $m_a\ge h_a$ and $m_a\ge \frac{h_b}{2}$. The number of solutions is $0$ if $m_a=h_a=\frac{h_b}{2},1$ if two of $m_a,h_a,\frac{h_b}{2}$ are equal, and $2$ otherwise.
Solution: Analysis. Let $A'$ and $B'$ be the feet of the perpendiculars from $A$ and $B$, respectively, to the opposite sides, $A_1$ the midpoint of $BC$, and let $D'$ be the foot of the perpendicular from $A_1$ to $AC$. We then have $AA_1=m_a$, $AA'=h_a,\angle{AA'A_1}=90^0,A_1D'=\frac{h_b}{2}$ and $\angle{AD'A_1}=90^0$.
Construction. We construct the quadrilateral $AD'A_1A'$( starting from the circle with diameter $AA_1$). Then $C$ is the intersection of $A'A_1$ and $AD'$, and $B$ is on the line $A_1C$ such that $CA_1=A_1B$ and $\mathscr{B}(B,A_1,C)$.
Dicussion. We must have $m_a\ge h_a$ and $m_a\ge \frac{h_b}{2}$. The number of solutions is $0$ if $m_a=h_a=\frac{h_b}{2},1$ if two of $m_a,h_a,\frac{h_b}{2}$ are equal, and $2$ otherwise.
IMO 1960 - Problem 3
Problem: A right - angled triangle $ABC$ is given for which the hypotenuse $BC$ has length $a$ and is divided into $n$ equal segments, where $n$ is odd. Let $\alpha$ be the angle with which the point $A$ sees segment containing the middle of the hypotenuse. Prove that: $tan(\alpha)=\frac{4nh}{(n^2-1)a}$.
Solution: Let $B'C'$ be the middle of the $n=2k+1$ segments and let $D$ be the foot of the perpendicular from $A$ to the hypotenuse. Let us assume $\mathscr{B}(C,D,C',B',B)$. Then from $CD<BD,CD+BD=a$, and $CD.BD=h^2$ we have $CD^2-a.CD+h^2=0\implies CD=\frac{a-\sqrt{a^2-4h^2}}{2}$. Let us define $\angle{DAC'}=\gamma$ and $\angle{DAB'}=\beta$; then $tan(\beta)=\frac{DB'}{h}$ and $tan(\gamma)=\frac{DC'}{h}$. Since $DB'=CB'-CD=\frac{(k+1)a}{2k+1}-\frac{c-\sqrt{c^2-4h^2}}{2}$ and $DC'=\frac{ka}{2k+1}-\frac{c-\sqrt{c^2-4h^2}}{2}$, we have : $tan(\alpha)=tan(\beta-\gamma)=\frac{tan(\beta)-tan(\gamma)}{1+tan(\beta).tan(\gamma)}=\frac{\frac{a}{(2k+1)h}}{1+\frac{a^2-4h^2}{4h^2}-\frac{a^2}{4h^2(2k+1)^2}}=\frac{4h(2k+1)}{4ak(k+1)}=\frac{4nh}{(n^2-1)a}$.
Solution: Let $B'C'$ be the middle of the $n=2k+1$ segments and let $D$ be the foot of the perpendicular from $A$ to the hypotenuse. Let us assume $\mathscr{B}(C,D,C',B',B)$. Then from $CD<BD,CD+BD=a$, and $CD.BD=h^2$ we have $CD^2-a.CD+h^2=0\implies CD=\frac{a-\sqrt{a^2-4h^2}}{2}$. Let us define $\angle{DAC'}=\gamma$ and $\angle{DAB'}=\beta$; then $tan(\beta)=\frac{DB'}{h}$ and $tan(\gamma)=\frac{DC'}{h}$. Since $DB'=CB'-CD=\frac{(k+1)a}{2k+1}-\frac{c-\sqrt{c^2-4h^2}}{2}$ and $DC'=\frac{ka}{2k+1}-\frac{c-\sqrt{c^2-4h^2}}{2}$, we have : $tan(\alpha)=tan(\beta-\gamma)=\frac{tan(\beta)-tan(\gamma)}{1+tan(\beta).tan(\gamma)}=\frac{\frac{a}{(2k+1)h}}{1+\frac{a^2-4h^2}{4h^2}-\frac{a^2}{4h^2(2k+1)^2}}=\frac{4h(2k+1)}{4ak(k+1)}=\frac{4nh}{(n^2-1)a}$.
IMO 1960 - Problem 2
Problem: For which real numbers $x$ does the following inequality hold: $\frac{4x^2}{(1-\sqrt{1+2x})^2}<2x+9?$
Solution: The LHS term is well-defined for $x\ge \frac{-1}{2}$ and $x\ne 0$. Furthermore, $\frac{4x^2}{(1-\sqrt{1+2x})^2}=(1+\sqrt{1+2x})^2$. Since $f(x)=(1+\sqrt{1+2x})^2-2x-9=2\sqrt{1+2x}-7$ is increasing and since $f(\frac{45}{8})=0$, it follows that the inequality holds precisely for $\frac{-1}{2}\le x<\frac{45}{8}$ and $x\ne 0$.
Solution: The LHS term is well-defined for $x\ge \frac{-1}{2}$ and $x\ne 0$. Furthermore, $\frac{4x^2}{(1-\sqrt{1+2x})^2}=(1+\sqrt{1+2x})^2$. Since $f(x)=(1+\sqrt{1+2x})^2-2x-9=2\sqrt{1+2x}-7$ is increasing and since $f(\frac{45}{8})=0$, it follows that the inequality holds precisely for $\frac{-1}{2}\le x<\frac{45}{8}$ and $x\ne 0$.
IMO 1960 - Problem 1
Problem: Find all the three-digit numbers for which one obtains, when dividing the number by $11$, the sum of squares of the digits of the initial number.
Solution: Given the number $\overline{acb},$ since $11|\overline{acb}$, it follows that $c=a+b$ or $c=a+b-11$. In the first case, $a^2+b^2+(a+b)^2=10a+b$, and in the second case, $a^2+b^2+(a+b-11)^2=10(a-1)+b$. In the first case the LHS is even, and hence $b\in \left\{0,2,4,6,8\right\}$, while in the second case it is odd, and hence $b\in \left\{1,3,5,7,9\right\}$. Analyzing the $10$ quadratic equation for $a$ we obtains that the only valid solutions are $550$ and $803$.
Solution: Given the number $\overline{acb},$ since $11|\overline{acb}$, it follows that $c=a+b$ or $c=a+b-11$. In the first case, $a^2+b^2+(a+b)^2=10a+b$, and in the second case, $a^2+b^2+(a+b-11)^2=10(a-1)+b$. In the first case the LHS is even, and hence $b\in \left\{0,2,4,6,8\right\}$, while in the second case it is odd, and hence $b\in \left\{1,3,5,7,9\right\}$. Analyzing the $10$ quadratic equation for $a$ we obtains that the only valid solutions are $550$ and $803$.
Putnam 2018 - B3
Problem: Find all positive integers $n<10^{100}$ for which simultaneously $n$ divides $2^n$, $n-1$ divides $2^n-1$ and $n-2$ divides $2^n-2$.
Solution: Since $n|2^{n}\implies n=2^{k}$ for some $k\ge 0$. Substituting into the second divisibility, $2^{k}-1|2^{2^{k}}-1$. Now we use the result $gcd(2^{a}-1,2^{b}-1)=2^{gcd(a,b)}-1$ to see $gcd(2^{2^{k}}-1,2^{k}-1)=2^{gcd(2^{k},k)}-1=2^{k}-1$, therefore $gcd(2^{k},k)=k$ and $k|2^{k}$, so $k=2^{m}$. Therefore, $n=2^{2^{m}}$. Substituting into the final divisibility,
$2^{2^{m}}-2|2^{2^{2^{m}}}-2\implies 2^{2^{m}-1}-1|2^{2^{2^{m}-1}}-1$. Using the exponential divisible sequence property twice, $gcd(2^{2^{2^{m}-1}}-1,2^{2^{m}-1}-1)=2^{gcd(2^{2^{m}-1},2^{m}-1)}-1=2^{2^{gcd(2^{m},m)}-1}-1=2^{2^{m}-1}-1,$
therefore $gcd(2^{m},m)=m$ and $m=2^{l}$ for some $l\ge 0$. Therefore $n=2^{2^{2^{l}}}$ for $l=0,1,2,3$ giving the solutions $n=4,16,65536,2^{256}$. Note that $2^{256}<2^{300}=8^{100}<10^{100}$, therefore these all satisfy $n<10^{100}$. Clearly for $l\ge 4,n=2^{2^{2^{l}}}\ge 2^{2^{16}}=2^{65536}>>10^{100}$.
Solution: Since $n|2^{n}\implies n=2^{k}$ for some $k\ge 0$. Substituting into the second divisibility, $2^{k}-1|2^{2^{k}}-1$. Now we use the result $gcd(2^{a}-1,2^{b}-1)=2^{gcd(a,b)}-1$ to see $gcd(2^{2^{k}}-1,2^{k}-1)=2^{gcd(2^{k},k)}-1=2^{k}-1$, therefore $gcd(2^{k},k)=k$ and $k|2^{k}$, so $k=2^{m}$. Therefore, $n=2^{2^{m}}$. Substituting into the final divisibility,
$2^{2^{m}}-2|2^{2^{2^{m}}}-2\implies 2^{2^{m}-1}-1|2^{2^{2^{m}-1}}-1$. Using the exponential divisible sequence property twice, $gcd(2^{2^{2^{m}-1}}-1,2^{2^{m}-1}-1)=2^{gcd(2^{2^{m}-1},2^{m}-1)}-1=2^{2^{gcd(2^{m},m)}-1}-1=2^{2^{m}-1}-1,$
therefore $gcd(2^{m},m)=m$ and $m=2^{l}$ for some $l\ge 0$. Therefore $n=2^{2^{2^{l}}}$ for $l=0,1,2,3$ giving the solutions $n=4,16,65536,2^{256}$. Note that $2^{256}<2^{300}=8^{100}<10^{100}$, therefore these all satisfy $n<10^{100}$. Clearly for $l\ge 4,n=2^{2^{2^{l}}}\ge 2^{2^{16}}=2^{65536}>>10^{100}$.
Thứ Sáu, 11 tháng 1, 2019
IMO 1959 - Problem 6
Problem: Let $\alpha$ and $\beta$ be two planes intersenting at a line $p$. In $\alpha$ a point $A$ is given and in $\beta$ a point $C$ is given, neither of which lies on $p$. Construct $B$ in $\alpha$ and $D$ in $\beta$ such that $ABCD$ is an equilateral trapezoid, $AB\parallel CD,$ in which a circle can be inscribed.
Solution:
Analysis. For $AB\parallel CD$ to hold evidently neither must intersect $p$ and hence constructing lines $r$ in $\alpha$ through $A$ and $s$ in $\beta$ through $C$, both being parallel to $p$, we get that $B\in r$ and $D\in s$. Hence the problem reduces to a planar problem in $\gamma$, determined by $r$ and $s$. Denote by $A'$ the foot of the perpendicular from $A$ to $s$. Since $ABCD$ is isosceles and has an incircle, it follows $AD=BC=\frac{AB+CD}{2}=A'C$. The remaining parts of the problem are now obvious.
Solution:
Analysis. For $AB\parallel CD$ to hold evidently neither must intersect $p$ and hence constructing lines $r$ in $\alpha$ through $A$ and $s$ in $\beta$ through $C$, both being parallel to $p$, we get that $B\in r$ and $D\in s$. Hence the problem reduces to a planar problem in $\gamma$, determined by $r$ and $s$. Denote by $A'$ the foot of the perpendicular from $A$ to $s$. Since $ABCD$ is isosceles and has an incircle, it follows $AD=BC=\frac{AB+CD}{2}=A'C$. The remaining parts of the problem are now obvious.
IMO 1959 - Problem 5
Problem: A segment $AB$ is given and on it a point $M$. On the same side of $AB$ squares $AMCD$ and $BMFE$ are constructed. The circumcircles of the two squares, whose centers are $P$ and $Q$, intersects in $M$ and another point $N$.
(a) Prove that lines $FA$ and $BC$ intersect at $N$.
(b) Prove that all such constructed lines $MN$ pass through the same point $S$, regardless of the selention of $M$.
(c) Find the locus of the midpoints of all segments $PQ$, as $M$ varies along the segment $AB$.
Solution:
(a) It suffices to prove that $AF\bot BC$, since then for the intersection point $X$ we have $\angle{AXC}=\angle{BXF}=90^0$, implying that $X$ belongs to the circumcircles of both squares and thus that $X=N$. The relation $AF\bot BC$ holds because from $MA=MC,MF=MB$, and $\angle{AMC}=\angle{FMB}$ it follows that $\triangle{AMF}$ is obtained by rotating $\triangle{BMC}$ by $90^0$ around $M$.
(b) Since $N$ is on the circumcircle of $BMFE$, it follows that $\angle{ANM}=\angle{MNB}=45^0$. Hence $MN$ is the bisector of $\angle{ANB}$. It follows that $MN$ passes through the midpoint of the arc $AB$ of the circle with diameter $AB$(i.e, the circumcircle of $\triangle{ABN}$) not containing $N$.
(c) Let us introduce a coordinate system such that $A=(0,0),B=(b,0)$ and $M=(m,0)$. Setting in general $W=(x_W,y_W)$ for an arbitrary point $W$ and denoting by $R$ the midpoint of $PQ$, we have $y_R=\frac{y_P+y_Q}{2}=\frac{m+b-m}{4}=\frac{b}{4}$ and $x_R=\frac{x_P+x_Q}{2}=\frac{m+m+b}{4}=\frac{2m+b}{4}$, the parameter $m$ varying from $0$ to $b$. Thus the locus of all points $R$ is the closed segment $R_1R_2$ where $R_1=(\frac{b}{4},\frac{b}{4})$ and $R_2=(\frac{b}{4},\frac{3b}{4})$.
(a) Prove that lines $FA$ and $BC$ intersect at $N$.
(b) Prove that all such constructed lines $MN$ pass through the same point $S$, regardless of the selention of $M$.
(c) Find the locus of the midpoints of all segments $PQ$, as $M$ varies along the segment $AB$.
Solution:
(a) It suffices to prove that $AF\bot BC$, since then for the intersection point $X$ we have $\angle{AXC}=\angle{BXF}=90^0$, implying that $X$ belongs to the circumcircles of both squares and thus that $X=N$. The relation $AF\bot BC$ holds because from $MA=MC,MF=MB$, and $\angle{AMC}=\angle{FMB}$ it follows that $\triangle{AMF}$ is obtained by rotating $\triangle{BMC}$ by $90^0$ around $M$.
(b) Since $N$ is on the circumcircle of $BMFE$, it follows that $\angle{ANM}=\angle{MNB}=45^0$. Hence $MN$ is the bisector of $\angle{ANB}$. It follows that $MN$ passes through the midpoint of the arc $AB$ of the circle with diameter $AB$(i.e, the circumcircle of $\triangle{ABN}$) not containing $N$.
(c) Let us introduce a coordinate system such that $A=(0,0),B=(b,0)$ and $M=(m,0)$. Setting in general $W=(x_W,y_W)$ for an arbitrary point $W$ and denoting by $R$ the midpoint of $PQ$, we have $y_R=\frac{y_P+y_Q}{2}=\frac{m+b-m}{4}=\frac{b}{4}$ and $x_R=\frac{x_P+x_Q}{2}=\frac{m+m+b}{4}=\frac{2m+b}{4}$, the parameter $m$ varying from $0$ to $b$. Thus the locus of all points $R$ is the closed segment $R_1R_2$ where $R_1=(\frac{b}{4},\frac{b}{4})$ and $R_2=(\frac{b}{4},\frac{3b}{4})$.
IMO 1959 - Problem 4
Problem: Construct a right-angled triangle whose hypotenuse $c$ is given if it is known that the median from the right angle equals the geometric mean of the remaining two sides of the triangle.
Solution: Analysis. Let $a$ and $b$ be the other two sides of the triangle. From the conditions of the problem we have $c^2=a^2+b^2$ and $\frac{c}{2}=\sqrt{ab}\iff \frac{3}{2}c^2=a^2+b^2+2ab=(a+b)^2\iff \sqrt{\frac{3}{2}}c=a+b$. Given a desired $\triangle{ABC}$ let $D$ be a point on ($AC$ such that $CD=CB$). In that case, $AD=a+b=\sqrt{\frac{3}{2}}c,$ and also, since $BC=CD$, it follows that $\angle{ADB}=45^0$.
Construction. From a segment of length $c$ we elementarily construct a segment $AD$ of length $\sqrt{\frac{3}{2}}c$. We then construct a ray $DX$ such that $\angle{ADX}=45^0$ and a circle $k(A,c)$ that intersects the ray at point $B$.
Finally, we construct the perpendicular from $B$ to $AD$; point $C$ is the foot of that perpendicular.
Proof. It holds that $AB=c$ and since $CB=CD$, it also holds that $AC+CB=AC+CD=AD=\sqrt{\frac{3}{2}}c$. From this it follows that $\sqrt{AC.CB}=\frac{c}{2}$. Since $BC$ is perpendicular to $AD$, it follows that $\angle{BCA}=90^0$. Thus $ABC$ is the desired triangle.
Dicussion. Since $AB\sqrt{2}=\sqrt{2}c>\sqrt{\frac{3}{2}}c=AD>AB$, the circle $k$ intersects the ray $DX$ in exactly two points, which correspond to two symmetric solutions.
Solution: Analysis. Let $a$ and $b$ be the other two sides of the triangle. From the conditions of the problem we have $c^2=a^2+b^2$ and $\frac{c}{2}=\sqrt{ab}\iff \frac{3}{2}c^2=a^2+b^2+2ab=(a+b)^2\iff \sqrt{\frac{3}{2}}c=a+b$. Given a desired $\triangle{ABC}$ let $D$ be a point on ($AC$ such that $CD=CB$). In that case, $AD=a+b=\sqrt{\frac{3}{2}}c,$ and also, since $BC=CD$, it follows that $\angle{ADB}=45^0$.
Construction. From a segment of length $c$ we elementarily construct a segment $AD$ of length $\sqrt{\frac{3}{2}}c$. We then construct a ray $DX$ such that $\angle{ADX}=45^0$ and a circle $k(A,c)$ that intersects the ray at point $B$.
Finally, we construct the perpendicular from $B$ to $AD$; point $C$ is the foot of that perpendicular.
Proof. It holds that $AB=c$ and since $CB=CD$, it also holds that $AC+CB=AC+CD=AD=\sqrt{\frac{3}{2}}c$. From this it follows that $\sqrt{AC.CB}=\frac{c}{2}$. Since $BC$ is perpendicular to $AD$, it follows that $\angle{BCA}=90^0$. Thus $ABC$ is the desired triangle.
Dicussion. Since $AB\sqrt{2}=\sqrt{2}c>\sqrt{\frac{3}{2}}c=AD>AB$, the circle $k$ intersects the ray $DX$ in exactly two points, which correspond to two symmetric solutions.
IMO 1959 - Problem 3
Problem: Let $x$ be an angle and let the real number $a,b,c,cos(x)$ satisfy the following equation: $acos^2(x)+bcos(x)+c=0$.
Write the analogous quadratic equation for $a,b,c,cos(2x)$. Compare the given and the obtained equality for $a=4,b=2,c=-1$.
Solution: Multiplying the equality by $4(acos^2(x)-bcos(x)+c)$, we obtain $4a^2cos^4x+2(4ac-2b^2)cos^2(x)+4c^2=0$. Plugging in $2cos^2(x)=1+cos(2x)$ we obtain (after quite a bit of manipulation):
$a^2cos^2(2x)+(2a^2+4ac-2b^2)cos(2x)+(a^2+4ac-2b^2+4c^2)=0$.
For $a=4,b=2$ and $c=-1$ we get $4cos^2(x)+2cos(x)-1=0$ and $16cos^(2x)+8cos(2x)-4=0\implies 4cos^2(2x)+2cos(2x)-1=0$
Write the analogous quadratic equation for $a,b,c,cos(2x)$. Compare the given and the obtained equality for $a=4,b=2,c=-1$.
Solution: Multiplying the equality by $4(acos^2(x)-bcos(x)+c)$, we obtain $4a^2cos^4x+2(4ac-2b^2)cos^2(x)+4c^2=0$. Plugging in $2cos^2(x)=1+cos(2x)$ we obtain (after quite a bit of manipulation):
$a^2cos^2(2x)+(2a^2+4ac-2b^2)cos(2x)+(a^2+4ac-2b^2+4c^2)=0$.
For $a=4,b=2$ and $c=-1$ we get $4cos^2(x)+2cos(x)-1=0$ and $16cos^(2x)+8cos(2x)-4=0\implies 4cos^2(2x)+2cos(2x)-1=0$
IMO 1959 - Problem 2
Problem 2: For which real number $x$ do the following equations hold:
(a) $\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2},$
(b) $\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=1,$
(c) $\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=2,$
Solution: For the square roots to be real we must have $2x-1\ge 0\implies x\ge \frac{1}{2}$ and $x\ge \sqrt{2x-1}\implies (x-1)^2\ge 0,$ which always holds. Then we have $\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=c\iff c^2=2x+2\sqrt{(x-1)^2}=2x+2|x-1|=\left\{\begin{array}{I} 2,\frac{1}{2}\le x\le 1,\\ 4x-2,x\ge 1\end{array}\right.$
(a) $c^2=2$. The equation holds for $\frac{1}{2}\le x\le 1$.
(b) $c^2=1$. The equation has no solution.
(c) $c^2=4$. The equation holds for $4x-2=4\implies x=\frac{3}{2}$.
(a) $\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2},$
(b) $\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=1,$
(c) $\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=2,$
Solution: For the square roots to be real we must have $2x-1\ge 0\implies x\ge \frac{1}{2}$ and $x\ge \sqrt{2x-1}\implies (x-1)^2\ge 0,$ which always holds. Then we have $\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=c\iff c^2=2x+2\sqrt{(x-1)^2}=2x+2|x-1|=\left\{\begin{array}{I} 2,\frac{1}{2}\le x\le 1,\\ 4x-2,x\ge 1\end{array}\right.$
(a) $c^2=2$. The equation holds for $\frac{1}{2}\le x\le 1$.
(b) $c^2=1$. The equation has no solution.
(c) $c^2=4$. The equation holds for $4x-2=4\implies x=\frac{3}{2}$.
Thứ Hai, 7 tháng 1, 2019
IMO 1959- Problem 1
Problem: For every integer $n$ prove that the fraction $\frac{21n+4}{14n+3}$ cannot be reduced any further.
Solution: The desired result $(14n+3,21n+4)=1$ follows from $3(14n+3)-2(21n+4)=1$.
Solution: The desired result $(14n+3,21n+4)=1$ follows from $3(14n+3)-2(21n+4)=1$.
IMO 1994- Problem 6
Problem: Show that there exists a set $A$ of positive integers with the following property: For any infinite set $S$ of primes there exists two positive integers $m\in A$ and $n\notin A$ each of which is a product of $k$ distinct elements of $S$ for some $k\ge 2$.
Solution: Let $A$ be the of all positive integers of the form $q_1q_2...q_{q_1}$ where $q_1<q_2<...<q_{q_1}$ are primes. For any infinite set $S=\left\{p_1,p_2,...,\right\}$ of primes with $p_1<p_2,...$, we can satisfy he requirement of the problem by taking $k=p_1,m=p_1p_2...p_k$ and $n=p_2p_3...p_{k+1}$.
Source: http://sms.math.nus.edu.sg/simo/IMO_Problems/94.pdf
Solution: Let $A$ be the of all positive integers of the form $q_1q_2...q_{q_1}$ where $q_1<q_2<...<q_{q_1}$ are primes. For any infinite set $S=\left\{p_1,p_2,...,\right\}$ of primes with $p_1<p_2,...$, we can satisfy he requirement of the problem by taking $k=p_1,m=p_1p_2...p_k$ and $n=p_2p_3...p_{k+1}$.
Source: http://sms.math.nus.edu.sg/simo/IMO_Problems/94.pdf
IMO 1994- Problem 5
Problem: Let $\mathbb{S}$ be the set of real numbers strictly greater than $-1$. Find all functions $f:\mathbb{S}\to \mathbb{S}$ satisfying the two conditions:
(i) $f(x)+f(y)+xf(y)=y+f(x)+yf(x)$ for all $x$ and $y$ in $\mathbb{S}$.
(ii) $\frac{f(x)}{x}$ is strictly increasing on each of the intervals $-1<x<0$ and $0<x$.
Solution: Condition (ii) implies that $f(x)=x$ has at most three solutions, one in $(-1,0),$ one equal to $0$ and the third in $(0,\infty)$.
Suppose $f(u)=u$ for some $u\in (-1,0)$. Setting $x=y=u$ in (i), we get $f(u^2+2u)=u^2+2u\in (-1,0)$.
This means $u^2+2u=u$. But then $u\notin (-1,0)$. The case $f(v)=v$ for some $v>0$ leads to a similar contradiction.
However, $f(x+(1+x)f(x))=x+(1+x)f(x)$ for all $x\in \mathbb{S}$. So we have $x+(1+x)f(x)=0$ which gives $f(x)=-\frac{x}{1+x}$.
It's routine to check that $f(x)=-\frac{x}{1+x}$ satisfies the desired property.
Source:http://sms.math.nus.edu.sg/simo/IMO_Problems/94.pdf
(i) $f(x)+f(y)+xf(y)=y+f(x)+yf(x)$ for all $x$ and $y$ in $\mathbb{S}$.
(ii) $\frac{f(x)}{x}$ is strictly increasing on each of the intervals $-1<x<0$ and $0<x$.
Solution: Condition (ii) implies that $f(x)=x$ has at most three solutions, one in $(-1,0),$ one equal to $0$ and the third in $(0,\infty)$.
Suppose $f(u)=u$ for some $u\in (-1,0)$. Setting $x=y=u$ in (i), we get $f(u^2+2u)=u^2+2u\in (-1,0)$.
This means $u^2+2u=u$. But then $u\notin (-1,0)$. The case $f(v)=v$ for some $v>0$ leads to a similar contradiction.
However, $f(x+(1+x)f(x))=x+(1+x)f(x)$ for all $x\in \mathbb{S}$. So we have $x+(1+x)f(x)=0$ which gives $f(x)=-\frac{x}{1+x}$.
It's routine to check that $f(x)=-\frac{x}{1+x}$ satisfies the desired property.
Source:http://sms.math.nus.edu.sg/simo/IMO_Problems/94.pdf
IMO 1994- Problem 4
Problem: Determine all ordered pairs $(m,n)$ of positive integers such that $\frac{n^3+1}{mn-1}$ is an integer.
Solution: Note that $mn-1$ and $m^3$ are relative prime. That $mn-1$ dividing $n^3+1$ is therefore equivalent to $mn-1$ dividing $m^3(n^3+1)=m^3n^3-1+m^3+1$, which is in turn equivalent to $mn-1$ dividing $m^3+1$. If $m=n$, we have $\frac{n^3+1}{n^2-1}=n+\frac{1}{n-1}$. This is an integer if and only if $n=2$. We now consider the case $m>n$. If $n=1,\frac{2}{m-1}$ is an integer. This is so if and only if $m=2,3$. Suppose $n\ge 2$. Note that $n^3+1\equiv 1(\text{ mod }n)$ while $mn-1\equiv -1(\text{ mod }n)$. Hence $\frac{n^3+1}{mn-1}=kn-1$ for some positive integer $k$. Now $kn-1<\frac{n^3+1}{n^2-1}=n+\frac{1}{n-1}$ or $(k-1)n<1+\frac{1}{n-1}$. Hence $k=1$, so that $n^3+1=(mn-1)(n-1)$. This yields $m=\frac{n^2+1}{n+1}=n+1+\frac{2}{n-1}$, which is an integer if and only if $n=2,3$. In each case, we have $m=5$. In summary, there are $9$ solutions, namely $(2,2),(2,1),(3,1),(5,2),(5,3),(1,2),(1,3),(2,5),(3,5)$
the last $4$ obtained by symmetry.
Source: http://sms.math.nus.edu.sg/simo/IMO_Problems/94.pdf
Solution: Note that $mn-1$ and $m^3$ are relative prime. That $mn-1$ dividing $n^3+1$ is therefore equivalent to $mn-1$ dividing $m^3(n^3+1)=m^3n^3-1+m^3+1$, which is in turn equivalent to $mn-1$ dividing $m^3+1$. If $m=n$, we have $\frac{n^3+1}{n^2-1}=n+\frac{1}{n-1}$. This is an integer if and only if $n=2$. We now consider the case $m>n$. If $n=1,\frac{2}{m-1}$ is an integer. This is so if and only if $m=2,3$. Suppose $n\ge 2$. Note that $n^3+1\equiv 1(\text{ mod }n)$ while $mn-1\equiv -1(\text{ mod }n)$. Hence $\frac{n^3+1}{mn-1}=kn-1$ for some positive integer $k$. Now $kn-1<\frac{n^3+1}{n^2-1}=n+\frac{1}{n-1}$ or $(k-1)n<1+\frac{1}{n-1}$. Hence $k=1$, so that $n^3+1=(mn-1)(n-1)$. This yields $m=\frac{n^2+1}{n+1}=n+1+\frac{2}{n-1}$, which is an integer if and only if $n=2,3$. In each case, we have $m=5$. In summary, there are $9$ solutions, namely $(2,2),(2,1),(3,1),(5,2),(5,3),(1,2),(1,3),(2,5),(3,5)$
the last $4$ obtained by symmetry.
Source: http://sms.math.nus.edu.sg/simo/IMO_Problems/94.pdf
IMO 1994- Problem 3
Problem: For any positive integer $k$, let $f(k)$ be the number of elements in the set $\left\{k+1,k+2,...,2k\right\}$ whose base $2$ representation has precisely three $1s$.
(a) Prove that, for each positive integer $m$, there exists at least one positive integer $k$ such that $f(k)=m$.
(b) Determine all positive integers $m$ for which there exists exactly one $k$ with $f(k)=m$.
Solution: Let $g(k)$ denote the number of elements in the set $\left\{1,...,n\right\}$ whose binary representation has exactly three ones. Then $f(k)$ and $g(k)$ are both increasing and $f(k)=g(2k)-g(k)$. Hence
$f(k+1)-f(k)=g(2k+2)-g(k+1)-g(2k)+g(k)=g(2k+2)-g(2k)-[g(k+1)-g(k)]$
Since either both $2k+2$ is counted in $g(2k+2)$ and $k+1$ is counted in $g(k+1)$ or neither is. Thus $f(k+1)-f(k)$ is either $1$ or $0$ depending on where $2k+1$ is counted in $g(2k+1)$ or not. Since $f(2^n)=C_{n+1}^3-C_{n}^3=C_{n}^2$, the image of $f$ is $\mathbb{N}\cup \left\{0\right\}$. This proves (a)
.
Let $m$ be any positive integer for which there is only one $k$ with $f(k)=m$. Then $f(k+1)-f(k)=1=f(k)-f(k-1)$.
The former means $2k+1$ is counted in $g(2k+2)$, or equivalently, the binary representation of $k$ has exactly two ones. The same holds for $k-1$. This happens only when the last two digits of $k-1$ are $01$. In other words, $k=2^n+2$. But
$f(2^n+2)=g(2^{n+1}+4)-g(2^n+2)=1+g(2^{n+1})-g(2^n)=1+C_{n}^2$
Thus the answer is any number of the form $1+C_{n}^2,n\ge 2$.
Source: http://sms.math.nus.edu.sg/simo/IMO_Problems/94.pdf
(a) Prove that, for each positive integer $m$, there exists at least one positive integer $k$ such that $f(k)=m$.
(b) Determine all positive integers $m$ for which there exists exactly one $k$ with $f(k)=m$.
Solution: Let $g(k)$ denote the number of elements in the set $\left\{1,...,n\right\}$ whose binary representation has exactly three ones. Then $f(k)$ and $g(k)$ are both increasing and $f(k)=g(2k)-g(k)$. Hence
$f(k+1)-f(k)=g(2k+2)-g(k+1)-g(2k)+g(k)=g(2k+2)-g(2k)-[g(k+1)-g(k)]$
Since either both $2k+2$ is counted in $g(2k+2)$ and $k+1$ is counted in $g(k+1)$ or neither is. Thus $f(k+1)-f(k)$ is either $1$ or $0$ depending on where $2k+1$ is counted in $g(2k+1)$ or not. Since $f(2^n)=C_{n+1}^3-C_{n}^3=C_{n}^2$, the image of $f$ is $\mathbb{N}\cup \left\{0\right\}$. This proves (a)
.
Let $m$ be any positive integer for which there is only one $k$ with $f(k)=m$. Then $f(k+1)-f(k)=1=f(k)-f(k-1)$.
The former means $2k+1$ is counted in $g(2k+2)$, or equivalently, the binary representation of $k$ has exactly two ones. The same holds for $k-1$. This happens only when the last two digits of $k-1$ are $01$. In other words, $k=2^n+2$. But
$f(2^n+2)=g(2^{n+1}+4)-g(2^n+2)=1+g(2^{n+1})-g(2^n)=1+C_{n}^2$
Thus the answer is any number of the form $1+C_{n}^2,n\ge 2$.
Source: http://sms.math.nus.edu.sg/simo/IMO_Problems/94.pdf
IMO 1994- Problem 2
Problem: $ABC$ is an isosceles triangle with $AB=AC$. Suppose that:
(i) $M$ is the midpoint of $BC$ and $O$ is the point on the line $AM$ such that $OB$ is perpendicular to $AB$.
(ii) $Q$ is an arbitrary point on the segment $BC$ different from $B$ and $C$.
(iii) $E$ lies on the line $AB$ and $F$ lies on the line $AC$ such that $E,Q$ and $F$ are distinct and collinear.
Prove that $OQ$ is perpendicular to $EF$ if and only if $QE=QF$.
Solution: First assume that $OQ$ is perpendicular to $EF$. Now $OEBQ$ and $OCFQ$ are cyclic.
Hence $\angle{OEQ}=\angle{OBQ}=\angle{OCQ}=\angle{OFQ}$. It follows that $QE=QF$.
Suppose now that $QE=QF$ and that the perpendicular through $O$ to $EF$ meet $BC$ at $Q'\ne Q$. Draw the line through $Q'$ parallel to $EF$, meeting the lines $AB$ and $AC$ at $E'$ and $F'$,respectively. Then $Q'E'=Q'F'$ as before. Let $AQ'$ meet $EF$ at $N$. Then $N\ne Q$ and $NE=NF$, so that $QE\ne QF$, a contradiction. So $Q'=Q$.
Source: http://sms.math.nus.edu.sg/simo/IMO_Problems/94.pdf
(i) $M$ is the midpoint of $BC$ and $O$ is the point on the line $AM$ such that $OB$ is perpendicular to $AB$.
(ii) $Q$ is an arbitrary point on the segment $BC$ different from $B$ and $C$.
(iii) $E$ lies on the line $AB$ and $F$ lies on the line $AC$ such that $E,Q$ and $F$ are distinct and collinear.
Prove that $OQ$ is perpendicular to $EF$ if and only if $QE=QF$.
Solution: First assume that $OQ$ is perpendicular to $EF$. Now $OEBQ$ and $OCFQ$ are cyclic.
Hence $\angle{OEQ}=\angle{OBQ}=\angle{OCQ}=\angle{OFQ}$. It follows that $QE=QF$.
Suppose now that $QE=QF$ and that the perpendicular through $O$ to $EF$ meet $BC$ at $Q'\ne Q$. Draw the line through $Q'$ parallel to $EF$, meeting the lines $AB$ and $AC$ at $E'$ and $F'$,respectively. Then $Q'E'=Q'F'$ as before. Let $AQ'$ meet $EF$ at $N$. Then $N\ne Q$ and $NE=NF$, so that $QE\ne QF$, a contradiction. So $Q'=Q$.
Source: http://sms.math.nus.edu.sg/simo/IMO_Problems/94.pdf
IMO 1994- Problem 1
Problem: Let $m$ and $n$ be positive integers. Let $a_1,a_2,...,a_m$ be distinct elements of $\left\{1,2,...,n\right\}$ such that whenever $a_i+a_j\le n$ for some $i,j,1\le i\le j\le m$, there exists $k$, $1\le k\le m,$ with $a_i+a_j=a_k$. Prove that: $\frac{a_1+a_2+...+a_m}{m}\ge \frac{n+1}{2}$.
Solution: Without loss of generality, we may assume that $a_1>a_2>...>a_m$. We claim that $a_i+a_{m+1-i}\ge n+1$ for $i=1,...,m$. The result then follows readily. To prove the claim ,we assume that on the contrary that it's false. Thus there exists $i$ such that $a_i+a_{m+1-i}<n+1$. Then $a_i<a_i+a_m<a_i+a_{m-1}<...<a_i+a_{m+1-i}\le n$. Thus $\left\{a_i+a_m,a_i+a_{m-1},...,a_{i}+a_{m+1-i}\right\}\subseteq \left\{a_1,a_2,...,a_{i-1}\right\}$
which is impossible. Thus the claim follows.
Source: http://sms.math.nus.edu.sg/simo/IMO_Problems/94.pdf
Solution: Without loss of generality, we may assume that $a_1>a_2>...>a_m$. We claim that $a_i+a_{m+1-i}\ge n+1$ for $i=1,...,m$. The result then follows readily. To prove the claim ,we assume that on the contrary that it's false. Thus there exists $i$ such that $a_i+a_{m+1-i}<n+1$. Then $a_i<a_i+a_m<a_i+a_{m-1}<...<a_i+a_{m+1-i}\le n$. Thus $\left\{a_i+a_m,a_i+a_{m-1},...,a_{i}+a_{m+1-i}\right\}\subseteq \left\{a_1,a_2,...,a_{i-1}\right\}$
which is impossible. Thus the claim follows.
Source: http://sms.math.nus.edu.sg/simo/IMO_Problems/94.pdf
IMO 1995- Problem 6
Problem: Let $p$ be an odd prime. Find the number of $p$- element subsets $A$ of $\left\{1,2,...,2p\right\}$ such that the sum if all elements of $A$ is divisible by $p$.
Solution: For any $p$-element subset $A$ of $\left\{1,2,...,2p\right\}$, denote by $s(A)$ the sum of the elements of $A$. Of the $C_{2p}^{p}$ such subsets, $B=\left\{1,2,...,p\right\}$ and $C=\left\{p+1,p+2,...,2p\right\}$ satisfy $s(B)\equiv s(C)\equiv 0(\text{ mod }p)$. For $A\ne B,C$, we have $A\cap B\ne \emptyset \ne A\cap C$. Partition the $C_{2p}^{p}-2p-$ element subsets other than $b$ and $C$ into groups of size $p$ as follows. Two subsets $A$ and $A'$ are in the same group if and only if $A'\cap C=A\cap C$ and $A'\cap B$ is a cyclic permutation of $A\cap B$ within $B$. Suppose $A\cap B$ has $n$ elements, $0\le n<p$. For some $m$ such that $0<m<p$,
$A'\cap B=\left\{x+m:x\in A\cap B,x+m\le p\right\}\cup \left\{x+m-p:x\in A\cap B,x\le p<x+m\right\}$.
Hence $s(A')-s(A)\equiv mn(\text{ mod }p)$, but $mn$ is not divisible by $p$. It follows that exactly one subset $A$ in each group satisfies $s(A)\equiv 0(\text{ mod }p)$, and the total number of such subsets is $p^-1(C_{2p}^{p}-2)+2$.
Second solution: Let $\omega$ be a primitive $pth$ root of unity. Then
$a_0+a_1\omega+...+a_{p-1}\omega^{p-1}=0$ iff $a_0=a_1=...=a_{p-1}$
Also $\prod\limits_{k=1}^{2p}(x-\omega^k)=(x^{p}-1)^2=2^{2p}-2x^{p}+1$.
If $t(\omega)=\sum a_i\omega^{i}$, then $a_i$ is the number of combinations with $i_1+...+i_p\equiv i(\text{ mod }p)$.
Therefore :$(a_0-2)+a_1\omega+...+a_{p-1}^{p-1}\equiv 0$
and $a_0-2=a_1=...=a_p$. So $|S|=\frac{C_{2p}^{p}-2}{p}+2$
Source: https://sms.math.nus.edu.sg/Simo/IMO_Problems/95.pdf
Solution: For any $p$-element subset $A$ of $\left\{1,2,...,2p\right\}$, denote by $s(A)$ the sum of the elements of $A$. Of the $C_{2p}^{p}$ such subsets, $B=\left\{1,2,...,p\right\}$ and $C=\left\{p+1,p+2,...,2p\right\}$ satisfy $s(B)\equiv s(C)\equiv 0(\text{ mod }p)$. For $A\ne B,C$, we have $A\cap B\ne \emptyset \ne A\cap C$. Partition the $C_{2p}^{p}-2p-$ element subsets other than $b$ and $C$ into groups of size $p$ as follows. Two subsets $A$ and $A'$ are in the same group if and only if $A'\cap C=A\cap C$ and $A'\cap B$ is a cyclic permutation of $A\cap B$ within $B$. Suppose $A\cap B$ has $n$ elements, $0\le n<p$. For some $m$ such that $0<m<p$,
$A'\cap B=\left\{x+m:x\in A\cap B,x+m\le p\right\}\cup \left\{x+m-p:x\in A\cap B,x\le p<x+m\right\}$.
Hence $s(A')-s(A)\equiv mn(\text{ mod }p)$, but $mn$ is not divisible by $p$. It follows that exactly one subset $A$ in each group satisfies $s(A)\equiv 0(\text{ mod }p)$, and the total number of such subsets is $p^-1(C_{2p}^{p}-2)+2$.
Second solution: Let $\omega$ be a primitive $pth$ root of unity. Then
$a_0+a_1\omega+...+a_{p-1}\omega^{p-1}=0$ iff $a_0=a_1=...=a_{p-1}$
Also $\prod\limits_{k=1}^{2p}(x-\omega^k)=(x^{p}-1)^2=2^{2p}-2x^{p}+1$.
If $t(\omega)=\sum a_i\omega^{i}$, then $a_i$ is the number of combinations with $i_1+...+i_p\equiv i(\text{ mod }p)$.
Therefore :$(a_0-2)+a_1\omega+...+a_{p-1}^{p-1}\equiv 0$
and $a_0-2=a_1=...=a_p$. So $|S|=\frac{C_{2p}^{p}-2}{p}+2$
Source: https://sms.math.nus.edu.sg/Simo/IMO_Problems/95.pdf
Chủ Nhật, 6 tháng 1, 2019
IMO 1995 - Problem 5
Problem: Let $ABCDEF$ be a convex hexagon with $AB=BC=CD,DE=EF=FA$ and $\angle{BCD}=\angle{EFA}=\frac{\pi}{3}$. Let $G$ and $H$ be two points interior to the hexagon such that angles $AGB$ and $DHE$ are both $\frac{2\pi}{3}$. Prove that: $AG+GB+GH+DH+HE\ge CF$.
Solution: Note that $BCD$ and $EFA$ are equilateral triangles. Hence $BE$ is an axis of symmetry of $ABDE$. Reflect $BCD$ and $EFA$ about $BE$ to $BC'A$ and $EF'D$ respectively. Since $\angle{BGA}=180^0-\angle{AC'B}$, $G$ lies on the circumcircle of $ABC'$. By Ptolemy's Theorem, $AG+GB=C'G$. Similarly, $DH+HE=HF'$. It follows that:
$CF=C'F'\le C'G+GH+HF'=AG+GB+GH+DH+HE,$ with equality if and only if $G$ and $H$ both lie on $C'F'$
Source: https://sms.math.nus.edu.sg/Simo/IMO_Problems/95.pdf
Solution: Note that $BCD$ and $EFA$ are equilateral triangles. Hence $BE$ is an axis of symmetry of $ABDE$. Reflect $BCD$ and $EFA$ about $BE$ to $BC'A$ and $EF'D$ respectively. Since $\angle{BGA}=180^0-\angle{AC'B}$, $G$ lies on the circumcircle of $ABC'$. By Ptolemy's Theorem, $AG+GB=C'G$. Similarly, $DH+HE=HF'$. It follows that:
$CF=C'F'\le C'G+GH+HF'=AG+GB+GH+DH+HE,$ with equality if and only if $G$ and $H$ both lie on $C'F'$
Source: https://sms.math.nus.edu.sg/Simo/IMO_Problems/95.pdf
IMO 1995- Problem 4
Problem: The positive real numbers $x_0,x_1,...,x_{1995}$ satisfy $x_0=x_{1995}$ and $x_{i-1}+\frac{2}{x_{i-1}}=2x_i+\frac{1}{x_i}$ for $i=1,2,...,1995$. Find the maximum value that $x_0$ can have.
Solution: The given condition is equivalent to $(2x_i^2-x_{i-1})(x_i-\frac{1}{x_{i-1}})=0$
which yield $x_i=\frac{x_{i-1}}{2}$ and $x_i=\frac{1}{x_{i-1}}$. We shall prove by induction that $x_i=2^{k_i}x_0^{\epsilon_{i}}$, where $|k_i|\le i$ and $\epsilon_i=(-1)^{k_i+i}$. This is true for $i=0$, with $k_0=0,\epsilon_0=1$. Assume that it is true for $i-1$. Then for $x_i=\frac{1}{2}x_{i-1}$, we have $k_i=k_{i-1}-1$ and $\epsilon_i=\epsilon_{i-1}$. For $x_i=\frac{1}{x_{i-1}}$, we have $k_i=-k_{i-1}$ and $\epsilon_{i}=-\epsilon_{i-1}$. In each case, it is immediate that $|k_i|\le i$ and $\epsilon_i=(-1)^{k_i+i}$. Thus $x_0=x_{1995}=2^{k}x_0^{\epsilon}$ where $k=k_{1995}$ and $\epsilon=\epsilon_{1995}$. Thus $|k|\le 1995$ and $\epsilon=(-1)^{1995+k}$. If $k$ is odd then $\epsilon=1$ and this is impossible. Thus $k$ is even and $\epsilon=-1$ and $x_0^2=2^k$. Since $|k|\le 1995$, we have $k\le 1994$. Hence $x_0\le 2^{997}$. The value $x_0=2^{997}$ can be attained by choosing $x_i=\frac{1}{2}x_{i-1}$ for $i=1,2,...,1994$ and $x_{1995}=\frac{1}{x_{1994}}$.
Source: https://sms.math.nus.edu.sg/Simo/IMO_Problems/95.pdf
Solution: The given condition is equivalent to $(2x_i^2-x_{i-1})(x_i-\frac{1}{x_{i-1}})=0$
which yield $x_i=\frac{x_{i-1}}{2}$ and $x_i=\frac{1}{x_{i-1}}$. We shall prove by induction that $x_i=2^{k_i}x_0^{\epsilon_{i}}$, where $|k_i|\le i$ and $\epsilon_i=(-1)^{k_i+i}$. This is true for $i=0$, with $k_0=0,\epsilon_0=1$. Assume that it is true for $i-1$. Then for $x_i=\frac{1}{2}x_{i-1}$, we have $k_i=k_{i-1}-1$ and $\epsilon_i=\epsilon_{i-1}$. For $x_i=\frac{1}{x_{i-1}}$, we have $k_i=-k_{i-1}$ and $\epsilon_{i}=-\epsilon_{i-1}$. In each case, it is immediate that $|k_i|\le i$ and $\epsilon_i=(-1)^{k_i+i}$. Thus $x_0=x_{1995}=2^{k}x_0^{\epsilon}$ where $k=k_{1995}$ and $\epsilon=\epsilon_{1995}$. Thus $|k|\le 1995$ and $\epsilon=(-1)^{1995+k}$. If $k$ is odd then $\epsilon=1$ and this is impossible. Thus $k$ is even and $\epsilon=-1$ and $x_0^2=2^k$. Since $|k|\le 1995$, we have $k\le 1994$. Hence $x_0\le 2^{997}$. The value $x_0=2^{997}$ can be attained by choosing $x_i=\frac{1}{2}x_{i-1}$ for $i=1,2,...,1994$ and $x_{1995}=\frac{1}{x_{1994}}$.
Source: https://sms.math.nus.edu.sg/Simo/IMO_Problems/95.pdf
IMO 1995 - Problem 3
Problem: Determine all integers $n>3$ such that there are $n$ points $A_1,A_2,...,A_n$ in the plane which satisfy the following two conditions simultaneously.
(a) No three lie on the same line
(b) There exist real numbers $p_1,p_2,...,p_n$ such that the area of $\triangle{A_iA_jA_k}$ is equal to $p_i+p_j+p_k$ for $1\le i<j<k\le n$.
Solution: We claim $n=4$ is the only answer. For $n=4$, let $A_1A_2A_3A_4$ be a unit square and let $p_1=p_2=p_3=p_4=\frac{1}{6}$. It remains to show no solution exists for $n=5$ which implies that there no solution for any $n\ge 5$.
Suppose to the contrary that there is solution with $n=5$. Denote the area of $\triangle{A_iA_jA_k}$ by $[ijk]=p_i+p_j+p_k$. We cannot have $p_i=p_j$. If for example $p_4=p_5,$ then $[124]=[125]$ and $[234]=[235]$, which implies that $A_4A_5$ is parallel to both $A_1A_2$ and $A_2A_3$. This is impossible since $A_1,A_2,A_3$ are not collinear. We also observe that if $A_iA_jA_kA_l$ is convex, then $p_i+p_k=p_j+p_l$. This follows from $[ijk]+[kli]=[jkl]+[lij]$.
Consider the convex hull of $A_1,...,A_5$, we have $3$ cases. First, suppose that the convex hull is a pentagon $A_1A_2A_3A_4A_5$. Since $A_1A_2A_3A_4$ and $A_1A_2A_3A_5$ are convex, our observation yields $p_1+p_3=p_2+p_4$ and $p_1+p_3=p_2+p_5$. Hence $p_4=p_5$, a contradiction.
Next we suppose that the convex hull is a quadrilateral $A_1A_2A_3A_4$. We may assume that $A_5$ is within $A_3A_4A_1$. Then $A_1A_2A_3A_5$ is convex and we have the same contradiction as before.
Finally suppose that the convex hull is a triangle $A_1A_2A_3$. Since $[124]+[234]+[314]=[125]+[235]+[315]$ we have $p_4=p_5$, a contradiction.
Source: https://sms.math.nus.edu.sg/Simo/IMO_Problems/95.pdf
(a) No three lie on the same line
(b) There exist real numbers $p_1,p_2,...,p_n$ such that the area of $\triangle{A_iA_jA_k}$ is equal to $p_i+p_j+p_k$ for $1\le i<j<k\le n$.
Solution: We claim $n=4$ is the only answer. For $n=4$, let $A_1A_2A_3A_4$ be a unit square and let $p_1=p_2=p_3=p_4=\frac{1}{6}$. It remains to show no solution exists for $n=5$ which implies that there no solution for any $n\ge 5$.
Suppose to the contrary that there is solution with $n=5$. Denote the area of $\triangle{A_iA_jA_k}$ by $[ijk]=p_i+p_j+p_k$. We cannot have $p_i=p_j$. If for example $p_4=p_5,$ then $[124]=[125]$ and $[234]=[235]$, which implies that $A_4A_5$ is parallel to both $A_1A_2$ and $A_2A_3$. This is impossible since $A_1,A_2,A_3$ are not collinear. We also observe that if $A_iA_jA_kA_l$ is convex, then $p_i+p_k=p_j+p_l$. This follows from $[ijk]+[kli]=[jkl]+[lij]$.
Consider the convex hull of $A_1,...,A_5$, we have $3$ cases. First, suppose that the convex hull is a pentagon $A_1A_2A_3A_4A_5$. Since $A_1A_2A_3A_4$ and $A_1A_2A_3A_5$ are convex, our observation yields $p_1+p_3=p_2+p_4$ and $p_1+p_3=p_2+p_5$. Hence $p_4=p_5$, a contradiction.
Next we suppose that the convex hull is a quadrilateral $A_1A_2A_3A_4$. We may assume that $A_5$ is within $A_3A_4A_1$. Then $A_1A_2A_3A_5$ is convex and we have the same contradiction as before.
Finally suppose that the convex hull is a triangle $A_1A_2A_3$. Since $[124]+[234]+[314]=[125]+[235]+[315]$ we have $p_4=p_5$, a contradiction.
Source: https://sms.math.nus.edu.sg/Simo/IMO_Problems/95.pdf
IMO 1995 - Problem 2
Problem: Let $a,b$ and $c$ be positive real numbers such that $abc=1$. Prove that: $\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\ge \frac{3}{2}$.
Solution: Let $S$ be the right hand side and $T=a(b+c)+b(a+c)+c(a+b)=2(ab+bc+ca)\ge 6(abc)^{\frac{2}{3}}=6$. Then by Cauchy's inequality, we have :
$ST\ge (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2=\frac{T^2}{4}$.
Thus $S\ge \frac{T}{4}\ge \frac{3}{2}$ as desired.
Second solution: Let $x=\frac{1}{a},y=\frac{1}{b},z=\frac{1}{c}$. Then $xyz=1$ and $S=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$.
By Cauchy's inequality,
[(x+y)+(z+x)+(x+y)] S\ge (x+y+z)^2.
or $S\ge \frac{x+y+z}{2}\ge \frac{3}{2}(xyz)^{\frac{1}{3}}=\frac{3}{2}$ as desired.
Source: https://sms.math.nus.edu.sg/Simo/IMO_Problems/95.pdf
Solution: Let $S$ be the right hand side and $T=a(b+c)+b(a+c)+c(a+b)=2(ab+bc+ca)\ge 6(abc)^{\frac{2}{3}}=6$. Then by Cauchy's inequality, we have :
$ST\ge (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2=\frac{T^2}{4}$.
Thus $S\ge \frac{T}{4}\ge \frac{3}{2}$ as desired.
Second solution: Let $x=\frac{1}{a},y=\frac{1}{b},z=\frac{1}{c}$. Then $xyz=1$ and $S=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$.
By Cauchy's inequality,
[(x+y)+(z+x)+(x+y)] S\ge (x+y+z)^2.
or $S\ge \frac{x+y+z}{2}\ge \frac{3}{2}(xyz)^{\frac{1}{3}}=\frac{3}{2}$ as desired.
Source: https://sms.math.nus.edu.sg/Simo/IMO_Problems/95.pdf
IMO 1995- Problem 1
Problem: Let $A,B,C,D$ be distinct points on a line, in that order. The circles with diameters $AC$ and $BD$ intersect at $X$ and $Y$. $O$ is an arbitrary point on line $XY$ but not on $AD$. $CO$ intersects the circle with diameter $AC$ again at $M$, and $BO$ intersects the other circle again at $N$. Prove that $AM,DN$ and $XY$ are concurrent.
Solution: The result is trivial if $O$ coincides with $X$ or $Y$. Suppose it does not. From $ON.ON=OX.OY=OC.OM,$
$BCMN$ is cyclic. If $O$ is on the segment $XY$, then
$\angle{MAD}+\angle{MNB}+\angle{BND}=\angle{MAD}+\angle{MCA}+\angle{AMC}=180^0$.
If $O$ is not on the segment $XY$, then
$\angle{MAD}=180^0-\angle{AMC}-\angle{MAC}=180^0-\angle{BND}-\angle{ONM}=\angle{MND}$.
In either case, $ADNM$ is also cyclic. Let $AM$ and $DN$ intersect at $Z$. Let the line $ZX$ intersect the first circle at $Y_1$ and the second at $Y_2$. Then $ZX.ZY_1=ZA.ZM=ZD.ZN=ZX.ZY_2$.
Hence $Y_1=Y_2=Y$ and indeed $Z$ lies on $XY$.
Source: https://sms.math.nus.edu.sg/Simo/IMO_Problems/95.pdf
Solution: The result is trivial if $O$ coincides with $X$ or $Y$. Suppose it does not. From $ON.ON=OX.OY=OC.OM,$
$BCMN$ is cyclic. If $O$ is on the segment $XY$, then
$\angle{MAD}+\angle{MNB}+\angle{BND}=\angle{MAD}+\angle{MCA}+\angle{AMC}=180^0$.
If $O$ is not on the segment $XY$, then
$\angle{MAD}=180^0-\angle{AMC}-\angle{MAC}=180^0-\angle{BND}-\angle{ONM}=\angle{MND}$.
In either case, $ADNM$ is also cyclic. Let $AM$ and $DN$ intersect at $Z$. Let the line $ZX$ intersect the first circle at $Y_1$ and the second at $Y_2$. Then $ZX.ZY_1=ZA.ZM=ZD.ZN=ZX.ZY_2$.
Hence $Y_1=Y_2=Y$ and indeed $Z$ lies on $XY$.
Source: https://sms.math.nus.edu.sg/Simo/IMO_Problems/95.pdf
Taiwan TST 2014
Problem: Positive integers $x_1,x_2,...,x_n(n\ge 4)$ are arranged in a circle such that each $x_i$ divides the sum of the neighbors, that is $\frac{x_{i-1}+x_{i+1}}{x_i}=k_i$ is an integer for each $i$, where $x_0=x_n,x_{n+1}=x_1$. Prove that: $2n\le k_1+k_2+...+k_n<3n$.
Solution: We first establish the lower bound. Note that:
$\sum\limits_{i=1}^{n}k_i=\sum\limits_{i=1}^{n}(\frac{x_{i+1}}{x_i}+\frac{x_{i-1}}{x_i})\ge (2n)\sqrt[2n]{1}=2n$, where the inequality is $AM-GM$.
We claim that the upper bound holds for all $n\ge 2$. We proceed by induction on $n$. Consider the base case $n=2$. Here, $k_1=\frac{2x_2}{x_1},$ and $k_2=\frac{2x_1}{x_2}$. Note that $k_1,k_2\in \mathbb{N}$ and $k_1k_2=4$, so either $k_1=4$ and $k_2=1$ or $k_1=k_2=2$. In the first case, we get $(x,2x)$, in the second we get $(x,x)$, both of which work.
Assume the upper bound holds for $n-1$. Now consider the situation for $n$. WLOG, let $x_n$ be the maximum element. There are three cases.
Case 1: $k_n\ge 3$. In this case, $x_1+x_{n-1}\ge 3x_n$. However, this clearly violates the maximality of $x_n$, so this case does not exist.
Case 2: $k_n=2$. In this case, $x_1+x_{n-1}=2x_n$. The only way to not violate maximality of $x_n$ is for $x_{n-1}=x_1=x_n$. However, one quickly sees that this implies that the whole sequence is constant, in which case the average of the $k_is$ is $2$, so this case is resolved.
Case 3: $k_n=1$. Now, $x_n=x_1+x_{n-1}$. Note that:
$\frac{k_1+...+k_n}{n}=\frac{k_1+k_{n-1}+k_n+\sum\limits_{i=2}^{n-2}k_i}{n}$
$=\frac{3+\frac{x_{n-2}+x_1}{x_{n-1}+\frac{x_2+x_{n-1}}{x_1}}}{n}<\frac{3+3(n-1)}{n}=3$, where the inequality follows from the inductive hypothesis. Therefore, our induction, and our proof is complete.
Source: https://artofproblemsolving.com/community/c473124h1546246_more_combinatorics
USAMO 2009
Problem: Let $n$ be a positive integer. Determine the size of the largest subsets of $\left\{-n,-n+1,...,n-1,n\right\}$ which does not contain three elements $a,b,c$( not necessarily distinct) satisfying $a+b+c=0$.
Solution: We claim the answer is $n$ for $n$ even and $n+1$ for $n$ odd. In both cases, we simply take the subsets of all odd numbers as an equality case.
Let $P$ be the set of positive numbers in our subset, and $N$ be the set of negative numbers( note that $0$ is not in our subset since $0+0+0=0$). Now, the condition is equivalent to saying that $-P,-N,P+N$ are disjoint.
Note that: $P+N=\left\{p+n:p\in P,n\in N\right\}$.
We now claim that $|P+N|\ge |P|+|N|-1$. To do this, let $P=\left\{x_1<...<x_m\right\}$ and let $N=\left\{y_1<...<y_n\right\}$. Then, the following elements are in $P+N:$
$x_1+y_1<x_2+y_1<...<x_m+y_1<x_m+y_2<...<x_m+y_n$
This proves the claim. Now, since $-P,-N,P+N$ are disjoint, we have that:
$|-P|+|-N|+|P+N|\le 2n+1\implies |P|+|N|+|P|+|N|-1\le 2n+1$
$\implies |P|+|N|\le n+1$.
This completes the proof for the odd case, but this is not strong enough for the even case.
We will now show that in fact $|P|+|N|\le n$ if $n$ is even. Assume not, so then equality must hold in all the above inequalities. Therefore, $(-P)\cup (-N)\cup (P+N)=\left\{-n,...\right\},$ and $|P+N|=|P|+|N|-1$. To start, since $0\notin P,N$, we must have that $n,-n\notin P+N$, so $n\in P$ and $-n\in N$.
Say that some $x\in P$. Then, since $-n\in N$, we must have that $x-n\in P+N$. However, since $-P,-N,P+N$ are disjoint, we must have that $x-n\notin -P,$ or $n-x\in P$. Thus,
$x\in P\implies n-x\notin P$. Therefore, one seems that from the numbers $1,...,n$, at most $1+\frac{n-2}{2}=\frac{n}{2}$ of them can be in $P$(note that $\frac{n}{2}\notin P$ since $\frac{n}{2}\in P\implies n-\frac{n}{2}\notin P$). Similar, one can show that there are at most $\frac{n}{2}$ elements in $N$. However, this implies $|P|+|N|\le n,$ which is a contradiction, since we started by assuming $|P|+|N|>n$. Therefore, for $n$ even, we can have at most $n$ elements in the subset.
Source: https://artofproblemsolving.com/community/c473124h1546246_more_combinatorics
Solution: We claim the answer is $n$ for $n$ even and $n+1$ for $n$ odd. In both cases, we simply take the subsets of all odd numbers as an equality case.
Let $P$ be the set of positive numbers in our subset, and $N$ be the set of negative numbers( note that $0$ is not in our subset since $0+0+0=0$). Now, the condition is equivalent to saying that $-P,-N,P+N$ are disjoint.
Note that: $P+N=\left\{p+n:p\in P,n\in N\right\}$.
We now claim that $|P+N|\ge |P|+|N|-1$. To do this, let $P=\left\{x_1<...<x_m\right\}$ and let $N=\left\{y_1<...<y_n\right\}$. Then, the following elements are in $P+N:$
$x_1+y_1<x_2+y_1<...<x_m+y_1<x_m+y_2<...<x_m+y_n$
This proves the claim. Now, since $-P,-N,P+N$ are disjoint, we have that:
$|-P|+|-N|+|P+N|\le 2n+1\implies |P|+|N|+|P|+|N|-1\le 2n+1$
$\implies |P|+|N|\le n+1$.
This completes the proof for the odd case, but this is not strong enough for the even case.
We will now show that in fact $|P|+|N|\le n$ if $n$ is even. Assume not, so then equality must hold in all the above inequalities. Therefore, $(-P)\cup (-N)\cup (P+N)=\left\{-n,...\right\},$ and $|P+N|=|P|+|N|-1$. To start, since $0\notin P,N$, we must have that $n,-n\notin P+N$, so $n\in P$ and $-n\in N$.
Say that some $x\in P$. Then, since $-n\in N$, we must have that $x-n\in P+N$. However, since $-P,-N,P+N$ are disjoint, we must have that $x-n\notin -P,$ or $n-x\in P$. Thus,
$x\in P\implies n-x\notin P$. Therefore, one seems that from the numbers $1,...,n$, at most $1+\frac{n-2}{2}=\frac{n}{2}$ of them can be in $P$(note that $\frac{n}{2}\notin P$ since $\frac{n}{2}\in P\implies n-\frac{n}{2}\notin P$). Similar, one can show that there are at most $\frac{n}{2}$ elements in $N$. However, this implies $|P|+|N|\le n,$ which is a contradiction, since we started by assuming $|P|+|N|>n$. Therefore, for $n$ even, we can have at most $n$ elements in the subset.
Source: https://artofproblemsolving.com/community/c473124h1546246_more_combinatorics
Middle European Mathematical Olympiad T-2
Problem: Determine all functions $f:\mathbb{R}\to \mathbb{R}$ such that: $xf(xy)+xyf(x)\ge f(x^2)f(y)+x^2y$ holds for all $x,y\in \mathbb{R}$.
Solution: Let $(x,y)$ denote the assertion $xf(xy)+xyf(x)\ge f(x^2)f(y)+x^2y$.
$(x,x)\implies (x^2-f(x^2))(x-f(x))\le 0,$denote this by $(1)$
$(0,0)\implies f(0)=0$
$(1,1)\implies f(1)=1$.
$(x,1)\implies 2xf(x)\ge f(x^2)+x^2\implies f(x)^2-f(x^2)\ge (f(x)-x)^2,$denoted this by $(2)$.
Now, suppose that there exists $a\ge 0$ such that $a>f(a)$.
$(1)\implies (a^2-f(a^2))(a-f(a))\le 0\implies a^2-f(a^2)\le 0\implies -f(a^2)\le -a^2$.
$(2)\implies (f(a)-a)^2\le f(a)^2-f(a^2)\le f(a)^2-a^2=(f(a)-a)(f(a)+a)$
As $f(a)-a<0$, we get $f(a)-a\ge f(a)+a\implies 2a\le 0$, which isn't true. So $f(x)\ge x$ for every $x\ge 0$.
That means $f(x^2)\ge x^2\forall x$.
$(1)\implies x\ge f(x)\forall x\implies \boxed{f(x)=x\forall x\ge 0}$
$(x<0,y<0)\implies x(yf(x)-xf(y))\ge 0\implies yf(x)\le xf(y)$, and let that be $(3)$.
$(y,x)\rightarrow (3)\implies xf(y)\le yf(x)\implies xf(y)=yf(x)\implies f(x)=cx\forall x<0$.
$(2)\implies 2xf(x)\ge f(x^2)+x^2\ge 2x^2\implies x(f(x)-x)\ge 0\implies f(x)\le x\forall x<0$
So, $\boxed{f(x)=cx\forall x<0}$ with $c\ge 1$.
It's easy to see that:
$\boxed{f(x)=\left\{\begin{array}{I} x\text{ if }x\ge 0\\cx\text{ if }x<0 \end{array}\right.,c\ge 1}$
Source: https://artofproblemsolving.com/community/c6h607043_functional_inequality
Solution: Let $(x,y)$ denote the assertion $xf(xy)+xyf(x)\ge f(x^2)f(y)+x^2y$.
$(x,x)\implies (x^2-f(x^2))(x-f(x))\le 0,$denote this by $(1)$
$(0,0)\implies f(0)=0$
$(1,1)\implies f(1)=1$.
$(x,1)\implies 2xf(x)\ge f(x^2)+x^2\implies f(x)^2-f(x^2)\ge (f(x)-x)^2,$denoted this by $(2)$.
Now, suppose that there exists $a\ge 0$ such that $a>f(a)$.
$(1)\implies (a^2-f(a^2))(a-f(a))\le 0\implies a^2-f(a^2)\le 0\implies -f(a^2)\le -a^2$.
$(2)\implies (f(a)-a)^2\le f(a)^2-f(a^2)\le f(a)^2-a^2=(f(a)-a)(f(a)+a)$
As $f(a)-a<0$, we get $f(a)-a\ge f(a)+a\implies 2a\le 0$, which isn't true. So $f(x)\ge x$ for every $x\ge 0$.
That means $f(x^2)\ge x^2\forall x$.
$(1)\implies x\ge f(x)\forall x\implies \boxed{f(x)=x\forall x\ge 0}$
$(x<0,y<0)\implies x(yf(x)-xf(y))\ge 0\implies yf(x)\le xf(y)$, and let that be $(3)$.
$(y,x)\rightarrow (3)\implies xf(y)\le yf(x)\implies xf(y)=yf(x)\implies f(x)=cx\forall x<0$.
$(2)\implies 2xf(x)\ge f(x^2)+x^2\ge 2x^2\implies x(f(x)-x)\ge 0\implies f(x)\le x\forall x<0$
So, $\boxed{f(x)=cx\forall x<0}$ with $c\ge 1$.
It's easy to see that:
$\boxed{f(x)=\left\{\begin{array}{I} x\text{ if }x\ge 0\\cx\text{ if }x<0 \end{array}\right.,c\ge 1}$
Source: https://artofproblemsolving.com/community/c6h607043_functional_inequality
Iran TST 2014 third exam - Problem 6
Problem: The incircle of a non-isosceles triangle $ABC$ with the center $I$ touches the side $BC$ at $D$. Let $X$ is a point on arc $BC$ from circumcircle of triangle $ABC$ such that if $E,F$ are feet of perpendicular from $X$ on $BI,CI$ and $M$ is midpoint of $EF$ we have $MB=MC$.
Prove that: $\angle{BAD}=\angle{CAX}$.
Solution:
Let $\angle{CBA}=2a,\angle{BCA}=2b$.
Let $Q,W,R$ are foot of perpendicular from $E,M,F$ on $BC$.
$MB=MC\implies BW=WC$, but $M$ is midpoint of $EF\implies WQ=WR\implies BQ=CR\implies BE.cos(a)=CF.cos(b)$.
Let $I_a$ is $A-excenter$.
Let $T,S$ are foot of perpendicular from $X$ to $BI_a,CI_a\implies \frac{XT}{XS}=\frac{BE}{CF}=\frac{cos(b)}{cos(a)}=\frac{sin\angle{BCI_a}}{sin\angle{CBI_a}}=\frac{sin\angle{XI_aB}}{sin\angle{XI_aC}}\implies XI_a$ is symmedian of triangle $CI_aB$.
Let $L$ is midpoint of arc $BAC$, $\angle{LBC}=\angle{LCB}=\angle{CI_aB}\implies LB,LC$ are tangent to circumcircle of triangle $CI_aB$.
$\frac{BX}{CX}=\frac{sin\angle{BLX}}{sin\angle{CLX}}=(\frac{cos(b)}{cos(a)})^2(1)$.
$\frac{sin\angle{BAD}}{sin\angle{CAD}}=\frac{AC.BD}{AB.CD}=\frac{tan(b)}{tan(a)}.\frac{sin(2a)}{sin(2b)}=(\frac{cos(a)}{cos(b)})^2(2)$.
$(1)(2)\implies \frac{sin\angle{BAD}}{sin\angle{CAD}}=\frac{sin\angle{CLX}}{sin\angle{XLB}},(\angle{BAD}+\angle{CAD}=\angle{CLX}+\angle{BLX})\implies \angle{CAX}=\angle{CLX}=\angle{BAD}$.
So we are done.
Source: https://artofproblemsolving.com/community/c6h590555p3497371
Prove that: $\angle{BAD}=\angle{CAX}$.
Solution:
Let $\angle{CBA}=2a,\angle{BCA}=2b$.
Let $Q,W,R$ are foot of perpendicular from $E,M,F$ on $BC$.
$MB=MC\implies BW=WC$, but $M$ is midpoint of $EF\implies WQ=WR\implies BQ=CR\implies BE.cos(a)=CF.cos(b)$.
Let $I_a$ is $A-excenter$.
Let $T,S$ are foot of perpendicular from $X$ to $BI_a,CI_a\implies \frac{XT}{XS}=\frac{BE}{CF}=\frac{cos(b)}{cos(a)}=\frac{sin\angle{BCI_a}}{sin\angle{CBI_a}}=\frac{sin\angle{XI_aB}}{sin\angle{XI_aC}}\implies XI_a$ is symmedian of triangle $CI_aB$.
Let $L$ is midpoint of arc $BAC$, $\angle{LBC}=\angle{LCB}=\angle{CI_aB}\implies LB,LC$ are tangent to circumcircle of triangle $CI_aB$.
$\frac{BX}{CX}=\frac{sin\angle{BLX}}{sin\angle{CLX}}=(\frac{cos(b)}{cos(a)})^2(1)$.
$\frac{sin\angle{BAD}}{sin\angle{CAD}}=\frac{AC.BD}{AB.CD}=\frac{tan(b)}{tan(a)}.\frac{sin(2a)}{sin(2b)}=(\frac{cos(a)}{cos(b)})^2(2)$.
$(1)(2)\implies \frac{sin\angle{BAD}}{sin\angle{CAD}}=\frac{sin\angle{CLX}}{sin\angle{XLB}},(\angle{BAD}+\angle{CAD}=\angle{CLX}+\angle{BLX})\implies \angle{CAX}=\angle{CLX}=\angle{BAD}$.
So we are done.
Source: https://artofproblemsolving.com/community/c6h590555p3497371
Thứ Bảy, 5 tháng 1, 2019
Relationship between Fibonacci and Lucas numbers
Đầu tiên, ta nhắc lại các định nghĩa của dãy Fibonacci và dãy Lucas như sau:
a) Dãy Fibonacci: $F_0=0;F_1=1;F_{n+2}=F_{n+1}+F_{n},\forall n\ge 2,n\in \mathbb{N}$.
b) Dãy Lucas: $L_0=2;L_1=1;L_{n+2}=L_{n+1}+L_{n},\forall n\ge 2,n\in \mathbb{N}$.
Và dưới đây là mối liên hệ giữa dãy Fibonacci và dãy Lucas.
1) $L_{2n}=L_{n}^2+2.(-1)^{n+1}$.
2) $L_{2n}.L_{2n+2}-5F_{n+1}^2=1$.
3) $\sum\limits_{k=1}^{n}F_{2k}=F_{2n+1}-1$
4) $\sum\limits_{k=1}^{n}F_{2k-1}=F_{2n}$
5) $\sum\limits_{k=1}^{n}(-1)^{k+1}F_k=(-1)^{n+1}F_{n-1}+1$
6) $L_n=F_{n+1}+F_{n-1}$
7) $5F_n=L_{n+1}+L_{n-1}$
8) $F_{n+m}=F_{n-1}F_{m}+F_{n}F_{m+1}$
9) $F_{n+m}=\frac{L_nF_m+L_mF_n}{2}$
10) $L_{n+m}=F_mL_{n+1}+F_{m-1}L_n$
11) $F_{n-m}=(-1)^m.\frac{L_mF_n-F_mL_n}{2}$
12) $L_{n-m}=(-1)^m.\frac{L_mL_n-5F_mF_n}{2}$
13) $F_nF_m-F_{n-k}F_{m+k}=(-1)^{n-k}.F_k.F_{m+k-n}$
14) $F_{n+3}^2=2F_{n+2}^2+2F_{n+1}^2-F_n^2$
15) $F_{n+4}^3=3F_{n+3}^3+6F_{n+2}^3-3F_{n+1}^3-F_n^3$
Source: https://artofproblemsolving.com/community/c260h1565488p9589231
a) Dãy Fibonacci: $F_0=0;F_1=1;F_{n+2}=F_{n+1}+F_{n},\forall n\ge 2,n\in \mathbb{N}$.
b) Dãy Lucas: $L_0=2;L_1=1;L_{n+2}=L_{n+1}+L_{n},\forall n\ge 2,n\in \mathbb{N}$.
Và dưới đây là mối liên hệ giữa dãy Fibonacci và dãy Lucas.
1) $L_{2n}=L_{n}^2+2.(-1)^{n+1}$.
2) $L_{2n}.L_{2n+2}-5F_{n+1}^2=1$.
3) $\sum\limits_{k=1}^{n}F_{2k}=F_{2n+1}-1$
4) $\sum\limits_{k=1}^{n}F_{2k-1}=F_{2n}$
5) $\sum\limits_{k=1}^{n}(-1)^{k+1}F_k=(-1)^{n+1}F_{n-1}+1$
6) $L_n=F_{n+1}+F_{n-1}$
7) $5F_n=L_{n+1}+L_{n-1}$
8) $F_{n+m}=F_{n-1}F_{m}+F_{n}F_{m+1}$
9) $F_{n+m}=\frac{L_nF_m+L_mF_n}{2}$
10) $L_{n+m}=F_mL_{n+1}+F_{m-1}L_n$
11) $F_{n-m}=(-1)^m.\frac{L_mF_n-F_mL_n}{2}$
12) $L_{n-m}=(-1)^m.\frac{L_mL_n-5F_mF_n}{2}$
13) $F_nF_m-F_{n-k}F_{m+k}=(-1)^{n-k}.F_k.F_{m+k-n}$
14) $F_{n+3}^2=2F_{n+2}^2+2F_{n+1}^2-F_n^2$
15) $F_{n+4}^3=3F_{n+3}^3+6F_{n+2}^3-3F_{n+1}^3-F_n^3$
Source: https://artofproblemsolving.com/community/c260h1565488p9589231
Lucal sequences
Problem: Let $(L_n)$ be the Lucas sequence, that is, $L_0=2,L_1=1,L_{n+2}=L_{n+1}+L_n$. Prove that: $L_n$ is a perfect square if and only if $n=3$.
Solution:
If $n$ is even, we have $L_{2m}=L_{m}^2-2(-1)^{n-1}$ which gives a contradiction.
If $n$ is odd
Case 1: $n=4t+1$ and $n\ne 1$. We write $n=1+2.3^r.k$ so $k$ is not divisible by $3$ and $2|k$. We have $L_{m+2k}\equiv -L_m(\text{ mod }L_k)$ so $L_n\equiv -L_1\equiv -1(\text{ mod }L_k)$.
And for every even $k$ satisfy $k$ is not divisible by $3$, we can check that $L_k\equiv 3(\text{ mod }4)$ so if $L_n=x^2$, we have $x^2+1$ has a prime divisor with the form $4k+3$.
Case 2: $n=4t+3$, we write $n=3+2.3^r.k$, then do similar with the first case we obtain $x^2+4$ has a prime divisor with the form $4k+3$ with $x$ is odd. So $x$ is even, then $L_n=4$ which gives $n=3$.
Solution:
If $n$ is even, we have $L_{2m}=L_{m}^2-2(-1)^{n-1}$ which gives a contradiction.
If $n$ is odd
Case 1: $n=4t+1$ and $n\ne 1$. We write $n=1+2.3^r.k$ so $k$ is not divisible by $3$ and $2|k$. We have $L_{m+2k}\equiv -L_m(\text{ mod }L_k)$ so $L_n\equiv -L_1\equiv -1(\text{ mod }L_k)$.
And for every even $k$ satisfy $k$ is not divisible by $3$, we can check that $L_k\equiv 3(\text{ mod }4)$ so if $L_n=x^2$, we have $x^2+1$ has a prime divisor with the form $4k+3$.
Case 2: $n=4t+3$, we write $n=3+2.3^r.k$, then do similar with the first case we obtain $x^2+4$ has a prime divisor with the form $4k+3$ with $x$ is odd. So $x$ is even, then $L_n=4$ which gives $n=3$.
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