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}{6sin(\theta)}\quad
+
\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)}}{6sin(\theta)}</math>
+
\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 $P-ABCD$ whose base $ABCD$ is square and whose vertex $P$ is equidistant from $A,B,C$ and $D$. If $AB=1$ and $\angle{APB}=2\theta$, then the volume of the pyramid is

$\text{(A) } \frac{\sin(\theta)}{6}\quad \text{(B) } \frac{\cot(\theta)}{6}\quad \text{(C) } \frac{1}{6\sin(\theta)}\quad \text{(D) } \frac{1-\sin(2\theta)}{6}\quad \text{(E) } \frac{\sqrt{\cos(2\theta)}}{6\sin(\theta)}$

Solution

As the base has area $1$, the volume will be one third of the height. Drop a line from $P$ to $AB$, bisecting it at $Q$. [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 $\angle QPB=\theta$, so $\cot\theta=\frac{PQ}{BQ}=2PQ$. Therefore $PQ=\tfrac12\cot\theta$.

Now turning to the dotted triangle, by Pythagoras, the square of the pyramid's height is \[PQ^2-(\tfrac12)^2=\frac{\cos^2\theta}{4\sin^2\theta}-\frac14=\frac{\cos^2\theta-\sin^2\theta}{4\sin^2\theta}=\frac{\cos 2\theta}{4\sin^2\theta}\] and after taking the square root and dividing by three, the result is $\fbox{E}$

See also

1990 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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