## Sunday, March 12, 2006

### Cotangible. Topic: Geometry/Inequalities. Level: Olympiad

Problem: (2002 USAMO - #2) Let $ABC$ be a triangle such that

$\left( \cot \frac{A}{2} \right)^2 + \left( 2\cot \frac{B}{2} \right)^2 + \left( 3\cot \frac{C}{2} \right)^2 = \left( \frac{6s}{7r} \right)^2$,

where $s$ and $r$ denote its semiperimeter and its inradius, respectively. Prove that triangle $ABC$ is similar to a triangle $T$ whose side lengths are all positive integers with no common divisors and determine these integers. Solution: At first glance, the condition seems extremely strange and convoluted, but remembering our basic triangle facts we can simplify it greatly.

First, recall that the intersection of the angle bisectors is the incenter, $I$. Since we have a bunch of equivalent tangents (see picture), we can write $a = x+y$, $b = x+z$, and $c = y+z$. Then $s = \frac{a+b+c}{2} = x+y+z$.

Also, because radii make right angles with tangent lines, we have right triangles with which we can evaluate the cotangent terms:

$\cot{\frac{A}{2}} = \frac{z}{r}$, $\cot{\frac{B}{2}} = \frac{y}{r}$, $\cot{\frac{C}{2}} = \frac{x}{r}$.

Notice that we have $r^2$ in the denominator of every term, so we can multiply through to get rid of it. Substituting everything in, we have

$z^2+4y^2+9x^2 = \frac{36}{49}(x+y+z)^2$.

To solve this, we use Cauchy to find

$\left(1+\frac{1}{4}+\frac{1}{9}\right)(z^2+4y^2+9x^2) \ge (x+y+z)^2$,

or

$z^2+4y^2+9x^2 \ge \frac{36}{49}(x+y+z)^2$.

That means the equality condition on Cauchy holds, that is

$z = \frac{2y}{\frac{1}{2}} = \frac{3x}{\frac{1}{3}}$.

Setting this arbitrarily to $36k$, we have

$z = 36k$, $y = 9k$, $x = 4k$.

Therefore our triangle has sides $a = 13k$, $b = 40k$, $z = 45k$. So all such triangles are similar to the triangle $T$ with side lengths $13, 40, 45$. QED.

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Comment: This is actually a really cool problem and is good practice for converting geometric quantities into algebraic ones. Knowing the incircle relations is extremely useful and comes up often in olympiads.

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Practice Problem: (2000 USAMO - #2) Let $S$ be the set of all triangles ABC for which

$5 ( \frac{1}{AP} + \frac{1}{BQ} + \frac{1}{CR} ) - \frac{3}{\min\{ AP, BQ, CR \}} = \frac{6}{r}$,

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.

#### 1 comment:

1. Wow, that was a really cool problem!