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Belarusian Mathematical Olympiad

Geometry Difficulty 5.2 AIME, harder Prove it Belarus

Given a right-angled triangle $ABC$ with $\angle C = 90^\circ$ . Its perimeter is equal to $30$ cm. $CH$ is the altitude of this triangle, $CK$ and $CL$ are the bisectors of the angles $ACH$ and $BCH$ respectively.
Find the length of the hypotenuse $AB$ if $KL = 4$ cm.

Solution

Let $a = BC$ , $b = AC$ , $c = AB$ , $\alpha = \angle BAC$ , $\beta = \angle ABC$ , $P = a+b+c$ . Then $\angle ACH = \beta$ , $\angle BCH = \alpha$ , thus $\angle ACK = \angle KCH = \beta/2$ , $\angle HCL = \angle LCB = \alpha/2$ . Since $\angle HKC = \angle KAC + \angle ACK$ (exterior angle of triangle $ACK$ ), we have
$\angle HKC = \alpha + \beta/2 = \angle BCH + \angle LCH = \angle BCK.$
Hence triangle $KCB$ is isosceles and $BK = BC = a$ . Similarly, $AL = AC = b$ . Therefore, $AK = AB - BK = c - a$ , so $KL = AL - AK = b - (c-a) = a+b-c$ . Hence
$c = \frac{(a+b+c) - (a+b-c)}{2} = \frac{P - KL}{2} = \frac{30-4}{2} = 13 \text{ (cm).}$

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