2000 USAMO Problems/Problem 2

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 $\boxed{a:b:c = 8:5:5}$ .

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 $x:y:z = 1:4:4$ . $\blacksquare$

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

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