$ABC$ is a right triangle with $\angle B=90^\circ$ , $AB=5$ , $BC=12$ , and $AB$ tangent to a circle. $CP$ and $CQ$ are also tangent to the circle at $P$ and $Q$ respectively.

If the length of chord $PQ$ is $\dfrac {a\sqrt b}c$ , where $a$ , $b$ , and $c$ are positive integers, with $a$ and $c$ being coprime and $b$ square-free, find $a+b+c$ .

The answer is 54.

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

Since $AC=13$ (By Pythagorean theorem) , let $AP=x$ , therefore $PC=x+13=QC=QB+12$ which implies $QB=x+1$ .

Now let the point of contact of $AB$ with the circle be equal to $O$ . Now we know that $AP=AO=x$ and $BQ=BO=x+1$ , which implies $x+x+1=5 \implies x=2=AP$ . Therefore $BQ=3$ .

Now draw a parallel line through $A$ to $PQ$ to cut $CQ$ at $D$ . Now we can see that $ADQP$ is an isosceles trapezium as $\angle CQP= \angle CPQ$ . This implies $AP=DQ=2$ . So $BD=1$ . Now again by Pythagorean theorem, we have $AD=\sqrt{1^2+5^2}=\sqrt{26}$ .

Since $\triangle CQP$ is similar to $\triangle CDA$ , we have $\dfrac{AD}{PQ}=\dfrac{CD}{CQ}$ (because sides of similar triangles are proportional) $\implies \dfrac{\sqrt{26}}{PQ}=\dfrac{13}{15}$ . Therefore $PQ=\dfrac{15\sqrt{26}}{13}$ making the answer $\boxed{54}$ .