 # Problem of the Week Problem C and Solution All Squared Up

## Problem

Ziibi drew a square. Starting at one corner and moving around the square, he labelled the vertices $$J$$, $$K$$, $$L$$, and $$M$$, in order. He drew points $$P$$ and $$Q$$ outside the square so that both $$\triangle JMP$$ and $$\triangle MLQ$$ are equilateral. Determine the measure, in degrees, of $$\angle MPQ$$.

## Solution

Since $$JKLM$$ is a square, $$JK=KL=LM=MJ$$.

Since $$\triangle JMP$$ is equilateral, $$MJ=JP=MP$$.

Since $$\triangle MLQ$$ is equilateral, $$LM=LQ=QM$$.

It follows that $JK=KL=LM=MJ=JP=MP=LQ=QM.$ Each angle in a square is $$90^{\circ}$$. Therefore, $$\angle JML =90^{\circ}$$.

Each angle in an equilateral triangle is $$60^{\circ}$$. Therefore, $$\angle JMP =60^{\circ}$$ and $$\angle LMQ =60^{\circ}$$.

A complete revolution is $$360^{\circ}$$. Since $$\angle PMQ$$, $$\angle JMP$$, $$\angle JML$$, and $$\angle LMQ$$ form a complete revolution, then \begin{aligned} \angle PMQ&=360^\circ -\angle JMP-\angle JML-\angle LMQ\\ &=360^\circ - 60^\circ -90^\circ-60^\circ\\ &=150^\circ\end{aligned}

In $$\triangle MPQ$$, $$MP=QM$$ and the triangle is isosceles. It follows that $$\angle MPQ=\angle MQP$$.

In a triangle, the sum of the three angles is $$180^\circ$$. Since $$\angle PMQ=150^\circ$$, then the sum of the two remaining equal angles must be $$30^\circ$$. Therefore, each of the remaining two angles must equal $$15^\circ$$ and it follows that $$\angle MPQ=15^\circ$$.