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


See also

1982 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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