IMO 2007

IMO 2007 — 6/6 solved.

6 items

IMO 2007

Official IMO 2007 problems  ·  6/6 solved.

# Status Time
1 solved 13m25s
2 solved 3m29s
3 solved 5m06s
4 solved 24m42s
5 solved 7m16s
6 solved 3m48s

Problem 1   solved · 13m25s · Solution →


Real numbers $a_1, a_2, \dots , a_n$ are given. For each $i$ ($1\le i\le n$) define $$ d_i=\max{a_j:1\le j\le i}-\min{a_j:i\le j\le n} $$ and let $$ d=\max{d_i:1\le i\le n}. $$

(a) Prove that, for any real numbers $x_1\le x_2\le \cdots\le x_n$, $$ \max{|x_i-a_i|:1\le i\le n}\ge \dfrac{d}{2} \qquad (*) $$

(b) Show that there are real numbers $x_1\le x_2\le x_n$ such that equality holds in (*)

Problem 2   solved · 3m29s · Solution →

Consider five points $A,B,C,D$, and $E$ such that $ABCD$ is a parallelogram and $BCED$ is a cyclic quadrilateral. Let $\ell$ be a line passing through $A$. Suppose that $\ell$ intersects the interior of the segment $DC$ at $F$ and intersects line $BC$ at $G$. Suppose also that $EF=EG=EC$. Prove that $\ell$ is the bisector of $\angle DAB$.

Problem 3   solved · 5m06s · Solution →

In a mathematical competition some competitors are friends. Friendship is always mutual. Call a group of competitors a clique if each two of them are friends. (In particular, any group of fewer than two competitors is a clique.) The number of members of a clique is called its size. Given that, in this competition, the largest size of a clique is even, prove that the competitors can be arranged in two rooms such that the largest size of a clique contained in one room is the same as the largest size of a clique contained in the other room.

Problem 4   solved · 24m42s · Solution →

In $\triangle ABC$ the bisector of $\angle{BCA}$ intersects the circumcircle again at $R$, the perpendicular bisector of $BC$ at $P$, and the perpendicular bisector of $AC$ at $Q$. The midpoint of $BC$ is $K$ and the midpoint of $AC$ is $L$. Prove that the triangles $RPK$ and $RQL$ have the same area.

Problem 5   solved · 7m16s · Solution →

(Kevin Buzzard and Edward Crane, United Kingdom) Let $a$ and $b$ be positive integers. Show that if $4ab-1$ divides $(4a^2-1)^2$, then $a=b$.

Problem 6   solved · 3m48s · Solution →

Let $n$ be a positive integer. Consider $$ S={(x,y,z)~:~x,y,z\in {0,1,\ldots,n },~x+y+z>0} $$ as a set of $(n+1)^3-1$ points in three-dimensional space. Determine the smallest possible number of planes, the union of which contain $S$ but does not include $(0,0,0)$.

2007 IMO (Problems) • Resources
Preceded by 2006 IMO Problems 1 2 3 4 5 6 Followed by 2008 IMO Problems
All IMO Problems and Solutions