Trisectors in a right triangle

Geometry Level 5

Triangle $ABC$ has angles $\angle A=3\alpha, \angle B=90^\circ$ and $\angle C=3\gamma$ . Let $P$ be a point in segment $BC$ such that $\angle PAB=\alpha$ , and $Q$ be a point in segment $BA$ such that $\angle QCB=\gamma$ . Let $R$ be a point within the triangle such that $\angle RAC=\alpha$ and $\angle RCA=\gamma$ . If $\angle QRC=142^\circ$ , what is the measure (in degrees) of $\alpha$ ?

The answer is 22.

1 solution

Calvin Lin Staff
May 13, 2014

First note that $\alpha+\gamma=30^\circ$ . Let $T$ be the intersection point of $CQ$ and $AP$ , then $\angle ATC=180^\circ-(2\alpha+2\gamma)=120^\circ$ and $R$ is the incenter of triangle $ATC$ . Since $TR$ is bisector of $\angle ATC$ we have $\angle ATR=\angle RTC=\angle ATQ=60^\circ$ . Now we see that triangles $AQT$ and $ART$ are congruent, thus $QR$ is perpendicular to $AT$ and $\angle QRT=30^\circ$ . Similarly, we have $\angle PRT=30^\circ$ , therefore $\angle QRP=60^\circ.$

Finally, $\angle QRC=\angle QRP+\angle PRC=60^\circ+(90^\circ-\gamma)=142^\circ$ , that is $\gamma=8^\circ$ and $\alpha=30^\circ-8^\circ=22 ^\circ.$

Trisectors in a right triangle

The answer is 22.

1 solution

0 pending reports