# 1982 AHSME Problems/Problem 22

## Problem 22

In a narrow alley of width $w$ a ladder of length $a$ is placed with its foot at point P between the walls. Resting against one wall at $Q$, the distance k above the ground makes a $45^\circ$ angle with the ground. Resting against the other wall at $R$, a distance h above the ground, the ladder makes a $75^\circ$ angle with the ground. The width $w$ is equal to

$\text{(A)}a\qquad \text{(B)}RQ\qquad \text{(C)}k\qquad \text{(D)}\frac{h+k}{2}\qquad \text{(E)}h$

## Solution

$[asy] import olympiad; unitsize(40); pair T,P,Q,M,L,R; T=(0,0); P=(2.5,0); Q=(10,7.5); M=(0,7.5); L=(10,0); R=(0,10); draw(R--T--L--Q); draw(P--Q--R--cycle); draw(Q--M); label("P", P, S); dot(P); label("Q", Q, E); dot(Q); label("R", R, W); dot(R); label("S", M, W); dot(M); label("T", T, S); dot(T); label("L", L, S); dot(L); label("w",(5,-1),N); label("h",(-1,5),E); markscalefactor=0.03; draw(anglemark(L,P,Q)); draw(anglemark(R,P,T)); draw(rightanglemark(P,T,M)); draw(rightanglemark(Q,L,P)); [/asy]$

We know that $m\angle QPL=45^{\circ}$ and $m\angle RPT=75^{\circ}.$ Therefore, $m\angle QPR=60^{\circ}.$ $$\qquad$$ Because the two ladders are the same length, we know that $$RP=PQ=a.$$ Since $\triangle QPR$ is isosceles with vertex angle $60^{\circ},$ we can conclude that it must be equilateral.$$\qquad$$ Now, since $\triangle PTR$ is a right triangle and $m\angle TPR=75^{\circ},$ we can conclude that $m\angle PRT=15^{\circ}.$ Because $\triangle QPR$ is equilateral, we know that $m\angle QRP = 60^{\circ}.$ It then follows that $$m\angle QRS= m\angle QRP + m\angle PRT = 60^{\circ} + 15^{\circ} = 75^{\circ}.$$ Because of ASA, $\triangle QRS\cong\triangle RPT.$ From there, it follows that $QS=TR=h.$ Since $QS$ is the width of the alley, the answer is $\boxed{\text{E) }h}.$

~ Saumya Singhal