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.

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.

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.

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.

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.

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 $x\in \left\{\frac{\pi}{2}+m\pi,\frac{\pi}{4}+\frac{m\pi}{2},\frac{\pi}{6}+\frac{m\pi}{3}|m\in \mathbb{Z}\right\}$.

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.

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 $-1\le x<1-\frac{\sqrt{31}}{8}$.

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.

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:
$X\in \alpha\iff h(X)=\frac{h(G_1)}{2}=\frac{h(A)+h(B)+h(C)}{6}$.

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
$\frac{c}{2}=C'A\le \frac{b}{2}\le C'E=\frac{c}{2}cot\frac{w}{2}\iff btan\frac{\omega}{2}\le c<b$.

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 $x_j\ge 2$.

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 $\left\{2m\pi, 2m\pi- \frac{\pi}{2}|m\in \mathbb{Z}\right\}$.

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.

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 $a>0$.