MathDB

2012 Dutch IMO TST

Part of Dutch IMO TST

Subcontests

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2

\frac{a- b}{b +c}+\frac{b- c}{c +d}+\frac{c - d}{d +a} +\frac{d - a}{a + b}>=0

Let a,b,ca, b, c and dd be positive real numbers. Prove that a−bb+c+b−cc+d+c−dd+a+d−aa+b≥0\frac{a - b}{b + c}+\frac{b - c}{c + d}+\frac{c - d}{d + a} +\frac{d - a}{a + b } \ge 0

two boxes containing balls, m and n , two actions permitted

There are two boxes containing balls. One of them contains mm balls, and the other contains nn balls, where m,n>0m, n > 0. Two actions are permitted: (i) Remove an equal number of balls from both boxes. (ii) Increase the number of balls in one of the boxes by a factor kk. Is it possible to remove all of the balls from both boxes with just these two actions, 1. if k=2k = 2? 2. if k=3k = 3?
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4
2

a_i + j = n + 1 where i = a_j, permutations of 1,2,...,n , when 4 divides n

Let nn be a positive integer divisible by 44. We consider the permutations (a1,a2,...,an)(a_1, a_2,...,a_n) of (1,2,...,n)(1,2,..., n) having the following property: for each j we have ai+j=n+1a_i + j = n + 1 where i=aji = a_j . Prove that there are exactly (12n)!(14n)!\frac{ (\frac12 n)!}{(\frac14 n)!} such permutations.

angle chasing candidate, right angle wanted, equal angles given

Let △ABC\vartriangle ABC be a triangle. The angle bisector of ∠CAB\angle CAB intersectsBC BC at LL. On the interior of line segments ACAC and ABAB, two points, MM and NN, respectively, are chosen in such a way that the lines AL,BMAL, BM and CNCN are concurrent, and such that ∠AMN=∠ALB\angle AMN = \angle ALB. Prove that ∠NML=90o\angle NML = 90^o.
5
2
1
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line passes through the incentre

A line, which passes through the incentre II of the triangle ABCABC, meets its sides ABAB and BCBC at the points MM and NN respectively. The triangle BMNBMN is acute. The points K,LK,L are chosen on the side ACAC such that ∠ILA=∠IMB\angle ILA=\angle IMB and ∠KC=∠INB\angle KC=\angle INB. Prove that AM+KL+CN=ACAM+KL+CN=AC. S. Berlov

prime power criterion with a @ b = \frac{a - b}{gcd(a, b)}

For all positive integers aa and bb, we de ne a@b=a−bgcd(a,b)a @ b = \frac{a - b}{gcd(a, b)} . Show that for every integer n>1n > 1, the following holds: nn is a prime power if and only if for all positive integers mm such that m<nm < n, it holds that gcd(n,n@m)=1gcd(n, n @m) = 1.