Question

Triangle $ABC$ has $AB=90,BC=50$ and $CA=70.$ A circle is drawn with centre $P$ on $AB$ such that $CA$ and $CB$ are tangents to the circle. Find $\frac{2}{3}AP.$
(correct answer + 3, wrong answer 0)

Answer 1

Let $CA$ touches the circle at $R$ and $CB$ touches the circle at $S.$
Let $Q$ be a point on $AB$ so that $CQ$ and $AB$ are perpendicular.


Let $r$ be the radius of the circle.
From similar triangles $AQC$ and $ARP$,
$\frac{CQ}{r}=\frac{70}{AP}\cdots \left(1\right)$
From similar triangles $BQC$ and $BSP$,
$\frac{CQ}{r}=\frac{50}{BP}=\frac{50}{(90-AP)}\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right),$
$7(90-AP)=5AP$
$\Rightarrow 630=12AP$
$\Rightarrow 2AP=105$
Therefore, $\frac{2}{3}AP=\frac{105}{3}=35.$