Difference between revisions of "2001 IMO Problems"

Revision as of 02:54, 6 October 2014

Problem 1

Consider an acute triangle $\triangle ABC$ . Let $P$ be the foot of the altitude of triangle $\triangle ABC$ issuing from the vertex $A$ , and let $O$ be the circumcenter of triangle $\triangle ABC$ . Assume that $\angle C \geq \angle B+30^{\circ}$ . Prove that $\angle A+\angle COP < 90^{\circ}$ .

Problem 2

Let $a,b,c$ be positive real numbers. Prove that $\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1$ .

Problem 3

Twenty-one girls and twenty-one boys took part in a mathematical competition. It turned out that each contestant solved at most six problems, and for each pair of a girl and a boy, there was at least one problem that was solved by both the girl and the boy. Show that there is a problem that was solved by at least three girls and at least three boys.

Problem 4

Let $n_1, n_2, \dots , n_m$ be integers where $m>1$ is odd. Let $x = (x_1, \dots , x_m)$ denote a permutation of the integers $1, 2, \cdots , m$ . Let $f(x) = x_1n_1 + x_2n_2 + ... + x_mn_m$ . Show that for some distinct permutations $a$ , $b$ the difference $f(a) - f(b)$ is a multiple of $m!$ .

Problem 5

$ABC$ is a triangle. $X$ lies on $BC$ and $AX$ bisects angle $A$ . $Y$ lies on $CA$ and $BY$ bisects angle $B$ . Angle $A$ is $60^{\circ}$ . $AB + BX = AY + YB$ . Find all possible values for angle $B$ .

@@ Line 18: / Line 18: @@
 ==Problem 5==
+<math>ABC</math> is a [[triangle]]. <math>X</math> lies on <math>BC</math> and <math>AX</math> bisects [[angle]] <math>A</math>. <math>Y</math> lies on <math>CA</math> and <math>BY</math> bisects angle <math>B</math>. Angle <math>A</math> is <math>60^{\circ}</math>. <math>AB + BX = AY + YB</math>. Find all possible values for angle <math>B</math>.
 ==Problem 6==

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