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 
.
![[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]](http://latex.artofproblemsolving.com/1/4/2/1421b3d3db612ff6a5d4211a2407c9b2ed7c9d54.png)
Then 
, so 
. Therefore 
.
Now turning to the dotted triangle, by Pythagoras, the square of the pyramid's height is ![\[PQ^2-(\tfrac12)^2=\frac{\cos^2\theta}{4\sin^2\theta}-\frac14=\frac{\cos^2\theta-\sin^2\theta}{4\sin^2\theta}=\frac{\cos 2\theta}{4\sin^2\theta}\]](http://latex.artofproblemsolving.com/9/2/6/926bf69880784480a2e5dec73862e4ddff2d6583.png)
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 | |
| 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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