Difference between revisions of "1997 JBMO Problems"

Revision as of 17:22, 15 September 2017

Problem 1

(Bulgaria) 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 $\frac{1}{8}$

Problem 2

(Cyprus) 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 $k$ : $E(x,y) = \frac{x^8 + y^8}{x^8-y^8} - \frac{ x^8-y^8}{x^8+y^8}.$

Problem 3

(Greece) 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 $AI+BI+CI>BC+KL$ .

Problem 4

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

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 1: / Line 1: @@
 ==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 <math>\frac{1}{8}</math>
+''(Bulgaria)'' 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 <math>\frac{1}{8}</math>
-'' Bulgaria''
 ==Problem 2==
-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>:
+''(Cyprus)'' 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>
-''Cyprus ''
 ==Problem 3==
-Let <math>ABC</math> be a triangle and let <math>I</math> be the incenter. Let <math>N</math>, <math>M</math> be the midpoints of the sides <math>AB</math> and <math>CA</math> respectively. The lines <math>BI</math> and <math>CI</math> meet <math>MN</math> at <math>K</math> and <math>L</math> respectively. Prove that <math>AI+BI+CI>BC+KL</math>.
+''(Greece)'' Let <math>ABC</math> be a triangle and let <math>I</math> be the incenter. Let <math>N</math>, <math>M</math> be the midpoints of the sides <math>AB</math> and <math>CA</math> respectively. The lines <math>BI</math> and <math>CI</math> meet <math>MN</math> at <math>K</math> and <math>L</math> respectively. Prove that <math>AI+BI+CI>BC+KL</math>.
-''Greece''
 ==Problem 4==
-Determine the triangle with sides <math>a,b,c</math> and circumradius <math>R</math> for which <math>R(b+c) = a\sqrt{bc}</math>.
+''(Romania)'' Determine the triangle with sides <math>a,b,c</math> and circumradius <math>R</math> for which <math>R(b+c) = a\sqrt{bc}</math>.
-''Romania''
 ==Problem 5==
-Let <math>n_1</math>, <math>n_2</math>, <math>\ldots</math>, <math>n_{1998}</math> be positive integers such that <cmath> n_1^2 + n_2^2 + \cdots + n_{1997}^2 = n_{1998}^2. </cmath> Show that at least two of the numbers are even
+Let <math>n_1</math>, <math>n_2</math>, <math>\ldots</math>, <math>n_{1998}</math> be positive integers such that <cmath> n_1^2 + n_2^2 + \cdots + n_{1997}^2 = n_{1998}^2. </cmath> Show that at least two of the numbers are even.

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