216. Nice ! Find the length of hypotenuse (Belarus - 2010)

by Virgil Nicula, Jan 27, 2011, 5:01 PM

Proposed problem. Let $ABC$ be an $C$-right-angled triangle with $s=15$ . Denote the altitude $CH$ of $\triangle ABC$ and the $C$-bisectors $CK$ , $CL$

of $\triangle ACH$ , $\triangle BCH$ respectively, where $\{H,K,L\}\subset (AB)$ . Denote $KL=4$ . Find the length $c$ of the hypotenuse $[AB]$ .


Proof 1 (synthetic). Very nice properties ! Prove easily that $CK\perp BI$ and $CL\perp AI$ , where $I$ is the incenter of $\triangle ABC$ . Observe that $m(\angle KCL)=45^\circ$ ,

$I$ is the circumcenter of $\triangle CKL$ and $m(\angle KIL)=90^\circ$ . It is easy to see that $\triangle AKI \sim \triangle ILB$ or $AK\cdot BL=KI^2 \ (1)$ . The inradius of $\triangle ABC$ is

half of $KL$, i.e. $r=2$ as altitude of $\triangle KIL$ and $KI=LI=2\sqrt {2}$. Therefore, $S\equiv [ABC]=r\cdot s=30$ . Hence $\boxed{a\cdot b=60}\ (2)$ . As $AL=AC$ ,

$BK=BC$ $\stackrel{(1)}{\implies}$ $(a-4)(b-4)=8$ $\iff$ $ab-4(a+b)+8=0$ $\stackrel{(2)}{\implies}$ $\boxed{a+b=17}$ $\implies$ $c^2=a^2+b^2=$ $(a+b)^2-2ab=169$ $\implies$ $c=13$ .

Proof 2 (metric). $\left\|\begin{array}{ccccc}
\frac {KA}{b}=\frac {KH}{\frac {ab}{c}} & \implies & \frac {KA}{c}=\frac {KH}{a}=\frac {\frac {b^2}{c}}{a+c} & \implies & \boxed {KA=\frac {b^2}{a+c}}\\\\
\frac {LB}{a}=\frac {LH}{\frac {ab}{c}} & \implies & \frac {LB}{c}=\frac {LH}{b}=\frac {\frac {a^2}{c}}{b+c} & \implies & \boxed {LB=\frac {a^2}{b+c}}\end{array}\right\|\ \implies$ $KL=c-\frac {b^2}{a+c}-\frac {a^2}{b+c}=$

$c-\frac {c^2-a^2}{c+a}-\frac {c^2-b^2}{c+b}=$ $c-(c-a)-(c-b)=a+b-c$ . Thus, $\left\|\begin{array}{c}
(a+b)+c=30\\\
(a+b)-c=4\end{array}\right\|\ \implies\ a+b=17$ and $c=13$ .
This post has been edited 28 times. Last edited by Virgil Nicula, Nov 22, 2015, 3:58 PM

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