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Difference between revisions of "2000 USAMO Problems/Problem 2"

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as desired. Equality comes when <math>z = 4</math>; since <math>(*)</math> is symmetric in <math>y</math> and <math>z</math>, it follows that <math>y = 4</math> is also necessary for equality. Reversing our scaling, it follows that <math>x:y:z = 1:4:4</math>. <math>\blacksquare</math>
 
as desired. Equality comes when <math>z = 4</math>; since <math>(*)</math> is symmetric in <math>y</math> and <math>z</math>, it follows that <math>y = 4</math> is also necessary for equality. Reversing our scaling, it follows that <math>x:y:z = 1:4:4</math>. <math>\blacksquare</math>
  
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== See also ==
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== See Also ==
 
{{USAMO newbox|year=2000|num-b=1|num-a=3}}
 
{{USAMO newbox|year=2000|num-b=1|num-a=3}}
  
 
[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]

Revision as of 07:24, 16 September 2012

Problem

Let $S$ be the set of all triangles $ABC$ for which

\[5 \left( \dfrac{1}{AP} + \dfrac{1}{BQ} + \dfrac{1}{CR} \right) - \dfrac{3}{\min\{ AP, BQ, CR \}} = \dfrac{6}{r},\]

where $r$ is the inradius and $P, Q, R$ are the points of tangency of the incircle with sides $AB, BC, CA,$ respectively. Prove that all triangles in $S$ are isosceles and similar to one another.

Solution

We let $x = AP = s - a, y = BQ = s-b, z = CR = s-c$, and without loss of generality let $x \le y \le z$. Then $x + y + z = 3s - (a+b+c) = s$, so $r = \frac {A}{s} = \frac{\sqrt{(x+y+z)xyz}}{x+y+z} = \sqrt{\frac{xyz}{x+y+z}}$. Thus,

\[6\sqrt{\frac{xyz}{x+y+z}} = \frac{2yz + 5xy + 5xz}{xyz}\]

Squaring yields

\[0 = 4y^2z^2 + 25x^2y^2 + 25x^2z^2 - 16xy^2z - 16xyz^2 + 14x^2yz\]

We claim that the inequality

\begin{align}0 \le 4y^2z^2 + 25x^2y^2 + 25x^2z^2 - 16xy^2z - 16xyz^2 + 14x^2yz \tag{*} \end{align}

holds true, with equality iff $4x = y = z$. Then $s = x+y+z = 9x$, and $s-a = x, s-b = 4x, s-c = 4x$ yields $\boxed{a:b:c = 5:8:8}$.


Note that $(*)$ is homogeneous in $x,y,z$, so without loss of generality, scale so that $x=1$. Then

\[0 \le y^2 (4z^2 - 16z + 25) + y(14z - 16z^2) + 25z^2\]

which is a quadratic in $y$. As $4z^2 - 16z + 25 = 4(z-4)^2 + 9 \ge 0$, it suffices to show that the quadratic cannot have more than one root, or the discriminant $\Delta \le 0$. Then,

\[\Delta = z^2(14-16z)^2 - 4(4z^2 - 16z + 25)(25z^2) = -144(z-4)^2 \le 0\]

as desired. Equality comes when $z = 4$; since $(*)$ is symmetric in $y$ and $z$, it follows that $y = 4$ is also necessary for equality. Reversing our scaling, it follows that $x:y:z = 1:4:4$. $\blacksquare$

See Also

2000 USAMO (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions
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