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Difference between revisions of "1999 JBMO Problems"

(1999 JBMO Problems are up!)
 
 
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==Problem 4==
 
==Problem 4==
  
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Let <math>S</math> be a square with the side length 20 and let <math>M</math> be the set of points formed with the vertices of <math>S</math> and another 1999 points lying inside <math>S</math>. Prove that there exists a triangle with vertices in <math>M</math> and with area at most equal with <math>\frac 1{10}</math>.
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Let <math>ABC</math> be a triangle with <math>AB=AC</math>. Also, let <math>D\in[BC]</math> be a point such that <math>BC>BD>DC>0</math>, and let <math>\mathcal{C}_1,\mathcal{C}_2</math> be the circumcircles of the triangles <math>ABD</math> and <math>ADC</math> respectively. Let <math>BB'</math> and <math>CC'</math> be diameters in the two circles, and let <math>M</math> be the midpoint of <math>B'C'</math>. Prove that the area of the triangle <math>MBC</math> is constant (i.e. it does not depend on the choice of the point <math>D</math>).
  
 
[[1999 JBMO Problems/Problem 4|Solution]]
 
[[1999 JBMO Problems/Problem 4|Solution]]

Latest revision as of 12:13, 25 August 2018

Problem 1

Let $a,b,c,x,y$ be five real numbers such that $a^3 + ax + y = 0$, $b^3 + bx + y = 0$ and $c^3 + cx + y = 0$. If $a,b,c$ are all distinct numbers prove that their sum is zero.

Solution

Problem 2

For each nonnegative integer $n$ we define $A_n = 2^{3n}+3^{6n+2}+5^{6n+2}$. Find the greatest common divisor of the numbers $A_0,A_1,\ldots, A_{1999}$.

Solution

Problem 3

Let $S$ be a square with the side length 20 and let $M$ be the set of points formed with the vertices of $S$ and another 1999 points lying inside $S$. Prove that there exists a triangle with vertices in $M$ and with area at most equal with $\frac 1{10}$.

Solution

Problem 4

Let $ABC$ be a triangle with $AB=AC$. Also, let $D\in[BC]$ be a point such that $BC>BD>DC>0$, and let $\mathcal{C}_1,\mathcal{C}_2$ be the circumcircles of the triangles $ABD$ and $ADC$ respectively. Let $BB'$ and $CC'$ be diameters in the two circles, and let $M$ be the midpoint of $B'C'$. Prove that the area of the triangle $MBC$ is constant (i.e. it does not depend on the choice of the point $D$).

Solution

See Also

1999 JBMO (Problems • Resources)
Preceded by
1998 JBMO Problems 		Followed by
2000 JBMO Problems
1 • 2 • 3 • 4
All JBMO Problems and Solutions


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