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 $x^2y,xy^2,y^z,yz^2,z^x,zx^2$.

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 $7$.

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 $EDACB$

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
$S.sin(\frac{\pi}{7})=\frac{sin\frac{2\pi}{7}}{2}+\frac{sin\frac{4\pi}{7}-sin\frac{2\pi}{7}}{2}+\frac{sin\frac{6\pi}{7}-sin\frac{4\pi}{7}}{2}=\frac{sin\frac{6\pi}{7}}{2}\implies S=\frac{1}{2}$.

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 $x_2,x_3,x_4$.

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.