Problem 3

On the sides of an arbitrary triangle

ABC

, triangles

ABR

BCP

CAQ

are constructed externally with

\angle CBP = \angle CAQ = 45^\circ

\angle BCP = \angle ACQ = 30^\circ

\angle ABR = \angle BAR = 15^\circ

. Prove that

\angle QRP = 90^\circ

and

QR = RP

Step 1 of 6: Set up the three small triangles and the target claim

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In plain words

Each of $P$ , $Q$ , $R$ is the apex of a thin isosceles-or-not triangle glued onto one side of $ABC$ , pointing outward. Because every base angle involved ( $45^\circ, 30^\circ, 15^\circ$ ) is a "nice" angle, all the lengths $PB, PC, QA, QC, RA, RB$ can be written exactly using the law of sines, and the angles that $RP$ and $RQ$ make at $R$ can be tracked by simply adding up the known $45^\circ/30^\circ/15^\circ$ pieces around each vertex of $ABC$ . The whole problem then becomes bookkeeping with the law of cosines.

a=BC,\ b=CA,\ c=AB;\quad \angle PBC=45^\circ,\angle PCB=30^\circ;\ \angle QAC=45^\circ,\angle QCA=30^\circ;\ \angle RAB=\angle RBA=15^\circ

Detailed analysis

Let $a=BC$ , $b=CA$ , $c=AB$ and let $A,B,C$ also denote the angles of $\triangle ABC$ at those vertices. Triangle $BCP$ has $\angle PBC=45^\circ$ , $\angle PCB=30^\circ$ , hence $\angle BPC=105^\circ$ ; triangle $CAQ$ has $\angle QAC=45^\circ$ , $\angle QCA=30^\circ$ , hence $\angle AQC=105^\circ$ ; triangle $ABR$ has $\angle RAB=\angle RBA=15^\circ$ , hence $\angle ARB=150^\circ$ . The goal is $\angle PRQ = 90^\circ$ and $RP=RQ$ .