Difference between revisions of "1990 AHSME Problems/Problem 21"
(Created page with "== Problem == Consider a pyramid <math>P-ABCD</math> whose base <math>ABCD</math> is square and whose vertex <math>P</math> is equidistant from <math>A,B,C</math> and <math>D</ma...") |
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Consider a pyramid <math>P-ABCD</math> whose base <math>ABCD</math> is square and whose vertex <math>P</math> is equidistant from <math>A,B,C</math> and <math>D</math>. If <math>AB=1</math> and <math>\angle{APB}=2\theta</math>, then the volume of the pyramid is | Consider a pyramid <math>P-ABCD</math> whose base <math>ABCD</math> is square and whose vertex <math>P</math> is equidistant from <math>A,B,C</math> and <math>D</math>. If <math>AB=1</math> and <math>\angle{APB}=2\theta</math>, then the volume of the pyramid is | ||
− | <math>\text{(A) } \frac{sin(\theta)}{6}\quad | + | <math>\text{(A) } \frac{\sin(\theta)}{6}\quad |
− | \text{(B) } \frac{cot(\theta)}{6}\quad | + | \text{(B) } \frac{\cot(\theta)}{6}\quad |
− | \text{(C) } \frac{1}{ | + | \text{(C) } \frac{1}{6\sin(\theta)}\quad |
− | \text{(D) } \frac{1-sin(2\theta)}{6}\quad | + | \text{(D) } \frac{1-\sin(2\theta)}{6}\quad |
− | \text{(E) } \frac{\sqrt{cos(2\theta)}}{ | + | \text{(E) } \frac{\sqrt{\cos(2\theta)}}{6\sin(\theta)}</math> |
== Solution == | == Solution == | ||
− | <math>\fbox{E}</math> | + | As the base has area <math>1</math>, the volume will be one third of the height. Drop a line from <math>P</math> to <math>AB</math>, bisecting it at <math>Q</math>. |
+ | <asy> | ||
+ | import three;unitsize(1cm);size(200);real h = 0.7; | ||
+ | //currentprojection=perspective(1/3,-1,1/2); | ||
+ | triple P = (.5,.5,h); | ||
+ | draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle); | ||
+ | draw((0,0,0)--P--(1,1,0)^^(1,0,0)--P--(0,1,0)); | ||
+ | draw((.5,.5,0)--P--(.5,0,0)--cycle,dotted); | ||
+ | dot((0,0,0));dot((1,0,0)); | ||
+ | label("P",P,N);label("Q",(.5,0,0),S);label("B",(1,0,0),S);label("A",(0,0,0),S); | ||
+ | </asy> | ||
+ | Then <math>\angle QPB=\theta</math>, so <math>\cot\theta=\frac{PQ}{BQ}=2PQ</math>. Therefore <math>PQ=\tfrac12\cot\theta</math>. | ||
+ | |||
+ | Now turning to the dotted triangle, by Pythagoras, the square of the pyramid's height is <cmath>PQ^2-(\tfrac12)^2=\frac{\cos^2\theta}{4\sin^2\theta}-\frac14=\frac{\cos^2\theta-\sin^2\theta}{4\sin^2\theta}=\frac{\cos 2\theta}{4\sin^2\theta}</cmath> and after taking the square root and dividing by three, the result is <math>\fbox{E}</math> | ||
== See also == | == See also == |
Latest revision as of 11:56, 4 February 2016
Problem
Consider a pyramid whose base is square and whose vertex is equidistant from and . If and , then the volume of the pyramid is
Solution
As the base has area , the volume will be one third of the height. Drop a line from to , bisecting it at . Then , so . Therefore .
Now turning to the dotted triangle, by Pythagoras, the square of the pyramid's height is and after taking the square root and dividing by three, the result is
See also
1990 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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All AHSME Problems and Solutions |
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