Difference between revisions of "1982 AHSME Problems/Problem 22"
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− | == Problem | + | == Problem == |
− | 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. | + | 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. |
− | Resting against one wall at <math>Q</math>, the distance k above the ground makes a <math>45^\circ</math> angle with the ground. | + | 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. |
− | 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. | + | 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 |
− | The width <math>w</math> is equal to | ||
− | <math> \ | + | <math> \textbf{(A)}\ a\qquad |
− | \ | + | \textbf{(B)}\ RQ\qquad |
− | \ | + | \textbf{(C)}\ k\qquad |
− | \ | + | \textbf{(D)}\ \frac{h+k}{2}\qquad |
− | \ | + | \textbf{(E)}\ h </math> |
== Solution == | == Solution == | ||
Line 16: | Line 15: | ||
<asy> | <asy> | ||
import olympiad; | import olympiad; | ||
− | + | size(200); | |
pair T,P,Q,M,L,R; | pair T,P,Q,M,L,R; | ||
T=(0,0); | T=(0,0); | ||
Line 39: | Line 38: | ||
label("$L$", L, S); | label("$L$", L, S); | ||
dot(L); | dot(L); | ||
− | label("$w$",(5,-1)); | + | label("$w$",(5,-1),N); |
− | label("$h$",(-1,5)); | + | label("$h$",(-1,5),E); |
markscalefactor=0.03; | markscalefactor=0.03; | ||
draw(anglemark(L,P,Q)); | draw(anglemark(L,P,Q)); | ||
Line 53: | Line 52: | ||
<cmath>RP=PQ=a.</cmath> | <cmath>RP=PQ=a.</cmath> | ||
Since <math>\triangle QPR</math> is isosceles with vertex angle <math>60^{\circ},</math> we can conclude that it must be equilateral.<cmath>\qquad</cmath> | Since <math>\triangle QPR</math> is isosceles with vertex angle <math>60^{\circ},</math> we can conclude that it must be equilateral.<cmath>\qquad</cmath> | ||
− | Now, since <math>\triangle PTR</math> is a right triangle and <math>m\angle TPR=75^{\circ},</math> we can conclude that <math>m\angle PRT=15^{\circ}.</math> | + | Now, since <math>\triangle PTR</math> is a right triangle and <math>m\angle TPR=75^{\circ},</math> we can conclude that <math>m\angle PRT=15^{\circ}.</math> Because <math>\triangle QPR</math> is equilateral, we know that <math>m\angle QRP = 60^{\circ}.</math> |
+ | It then follows that <cmath>m\angle QRS= m\angle QRP + m\angle PRT = 60^{\circ} + 15^{\circ} = 75^{\circ}.</cmath> | ||
Because of ASA, <math>\triangle QRS\cong\triangle RPT.</math> From there, it follows that <math>QS=TR=h.</math> Since <math>QS</math> is the width of the alley, the answer is <math>\boxed{\text{E) }h}.</math> | Because of ASA, <math>\triangle QRS\cong\triangle RPT.</math> From there, it follows that <math>QS=TR=h.</math> Since <math>QS</math> is the width of the alley, the answer is <math>\boxed{\text{E) }h}.</math> | ||
~ Saumya Singhal | ~ Saumya Singhal | ||
+ | |||
+ | == See Also == | ||
+ | {{AHSME box|year=1982|num-b=21|num-a=23}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 21:47, 13 September 2021
Problem
In a narrow alley of width a ladder of length
is placed with its foot at point
between the walls.
Resting against one wall at
, the distance
above the ground makes a
angle with the ground.
Resting against the other wall at
, a distance
above the ground, the ladder makes a
angle with the ground. The width
is equal to
Solution
We know that and
Therefore,
Because the two ladders are the same length, we know that
Since
is isosceles with vertex angle
we can conclude that it must be equilateral.
Now, since
is a right triangle and
we can conclude that
Because
is equilateral, we know that
It then follows that
Because of ASA,
From there, it follows that
Since
is the width of the alley, the answer is
~ 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 |