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Difference between revisions of "2015 EGMO Problems"

(Created page with "==Day 1== ===Problem 1=== Let <math>\triangle ABC</math> be an acute-angled triangle, and let <math>D</math> be the foot of the altitude from <math>C.</math> The angle bisecto...")
 
(No difference)

Latest revision as of 13:47, 24 December 2022

Day 1

Problem 1

Let $\triangle ABC$ be an acute-angled triangle, and let $D$ be the foot of the altitude from $C.$ The angle bisector of $\angle ABC$ intersects $CD$ at $E$ and meets the circumcircle $\omega$ of triangle $\triangle ADE$ again at $F.$ If $\angle ADF = 45^{\circ}$, show that $CF$ is tangent to $\omega .$

Solution

Problem 2

A domino is a $2 \times 1$ or $1 \times 2$ tile. Determine in how many ways exactly $n^2$ dominoes can be placed without overlapping on a $2n \times 2n$ chessboard so that every $2 \times 2$ square contains at least two uncovered unit squares which lie in the same row or column.

Solution

Problem 3

Let $n, m$ be integers greater than $1$, and let $a_1, a_2, \dots, a_m$ be positive integers not greater than $n^m$. Prove that there exist positive integers $b_1, b_2, \dots, b_m$ not greater than $n$, such that\[\gcd(a_1 + b_1, a_2 + b_2, \dots, a_m + b_m) < n,\]where $\gcd(x_1, x_2, \dots, x_m)$ denotes the greatest common divisor of $x_1, x_2, \dots, x_m$.

Solution

Day 2

Problem 4

Determine whether there exists an infinite sequence $a_1, a_2, a_3, \dots$ of positive integers which satisfies the equality\[a_{n+2}=a_{n+1}+\sqrt{a_{n+1}+a_{n}}\]for every positive integer $n$.

Solution

Problem 5

Let $m, n$ be positive integers with $m > 1$. Anastasia partitions the integers $1, 2, \dots , 2m$ into $m$ pairs. Boris then chooses one integer from each pair and finds the sum of these chosen integers. Prove that Anastasia can select the pairs so that Boris cannot make his sum equal to $n$.

Solution

Problem 6

Let $H$ be the orthocentre and $G$ be the centroid of acute-angled triangle $ABC$ with $AB\ne AC$. The line $AG$ intersects the circumcircle of $ABC$ at $A$ and $P$. Let $P'$ be the reflection of $P$ in the line $BC$. Prove that $\angle CAB = 60$ if and only if $HG = GP'$

Solution

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