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Difference between revisions of "1982 AHSME Problems/Problem 22"

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== Problem 22 ==
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== Problem ==
  
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In a narrow alley of width <math>w</math> a ladder of length <math>a</math> is placed with its foot at point P between the walls.  
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In a narrow alley of width <math>w</math> a ladder of length <math>a</math> is placed with its foot at point <math>P</math> between the walls.  
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Resting against one wall at <math>Q</math>, the distance k above the ground makes a <math>45^\circ</math> angle with the ground.  
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Resting against one wall at <math>Q</math>, the distance <math>k</math> above the ground makes a <math>45^\circ</math> angle with the ground.  
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Resting against the other wall at <math>R</math>, a distance h above the ground, the ladder makes a <math>75^\circ</math> angle with the ground.  
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Resting against the other wall at <math>R</math>, a distance <math>h</math> above the ground, the ladder makes a <math>75^\circ</math> angle with the ground. The width <math>w</math> is equal to  
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The width <math>w</math> is equal to  
 
  
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<math> \text{(A)}a\qquad
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<math> \textbf{(A)}\ a\qquad
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\text{(B)}RQ\qquad
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\textbf{(B)}\ RQ\qquad
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\text{(C)}k\qquad
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\textbf{(C)}\ k\qquad
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\text{(D)}\frac{h+k}{2}\qquad
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\textbf{(D)}\ \frac{h+k}{2}\qquad
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\text{(E)}h </math>
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\textbf{(E)}\ h </math>
  
 
== Solution ==
 
== Solution ==
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<asy>
 
<asy>
 
import olympiad;
 
import olympiad;
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unitsize(40);
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size(200);
 
pair T,P,Q,M,L,R;
 
pair T,P,Q,M,L,R;
 
   T=(0,0);
 
   T=(0,0);
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~ Saumya Singhal
 
~ Saumya Singhal
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== See Also ==
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{{AHSME box|year=1982|num-b=21|num-a=23}}
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[[Category:Intermediate Geometry Problems]]

Latest revision as of 21:47, 13 September 2021

Problem

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

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

Solution

[asy] import olympiad; size(200); 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 (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions
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