Difference between revisions of "1997 JBMO Problems"

Revision as of 15:54, 15 September 2017

Problem 1

Show that given any 9 points inside a square of side length 1 we can always find 3 that form a triangle with area less than $\frac{1}{8}$

Bulgaria

Problem 2

Let $\frac{x^2+y^2}{x^2-y^2} + \frac{x^2-y^2}{x^2+y^2} = k$ . Compute the following expression in terms of $k$ : $E(x,y) = \frac{x^8 + y^8}{x^8-y^8} - \frac{ x^8-y^8}{x^8+y^8}.$ Cyprus

Problem 3

Let $ABC$ be a triangle and let $I$ be the incenter. Let $N$ , $M$ be the midpoints of the sides $AB$ and $CA$ respectively. The lines $BI$ and $CI$ meet $MN$ at $K$ and $L$ respectively. Prove that $AI+BI+CI>BC+KL$ .

Greece

Problem 4

Determine the triangle with sides $a,b,c$ and circumradius $R$ for which $R(b+c) = a\sqrt{bc}$ .

Romania

Problem 5

Let $n_1$ , $n_2$ , $\ldots$ , $n_{1998}$ be positive integers such that $n_1^2 + n_2^2 + \cdots + n_{1997}^2 = n_{1998}^2.$ Show that at least two of the numbers are even

@@ Line 9: / Line 9: @@
 Let <math>\frac{x^2+y^2}{x^2-y^2} + \frac{x^2-y^2}{x^2+y^2} = k</math>. Compute the following expression in terms of <math>k</math>:
 <cmath> E(x,y) = \frac{x^8 + y^8}{x^8-y^8} - \frac{ x^8-y^8}{x^8+y^8}.  </cmath>
-''Ciprus ''
+''Cyprus ''
 ==Problem 3==

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