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Difference between revisions of "1996 AHSME Problems/Problem 9"

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Triangle <math>PAB</math> and square <math>ABCD</math> are in [[perpendicular]] planes. Given that <math>PA = 3, PB = 4</math> and <math>AB = 5</math>, what is <math>PD</math>?  
 
Triangle <math>PAB</math> and square <math>ABCD</math> are in [[perpendicular]] planes. Given that <math>PA = 3, PB = 4</math> and <math>AB = 5</math>, what is <math>PD</math>?  
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<asy>
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real r=sqrt(2)/2;
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draw(origin--(8,0)--(8,-1)--(0,-1)--cycle);
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draw(origin--(8,0)--(8+r, r)--(r,r)--cycle);
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filldraw(origin--(-6*r, -6*r)--(8-6*r, -6*r)--(8, 0)--cycle, white, black);
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filldraw(origin--(8,0)--(8,6)--(0,6)--cycle, white, black);
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pair A=(6,0), B=(2,0), C=(2,4), D=(6,4), P=B+1*dir(-65);
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draw(A--P--B--C--D--cycle);
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dot(A^^B^^C^^D^^P);
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label("$A$", A, dir((4,2)--A));
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label("$B$", B, dir((4,2)--B));
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label("$C$", C, dir((4,2)--C));
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label("$D$", D, dir((4,2)--D));
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label("$P$", P, dir((4,2)--P));
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</asy>
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<math> \text{(A)}\ 5\qquad\text{(B)}\ \sqrt{34} \qquad\text{(C)}\ \sqrt{41}\qquad\text{(D)}\ 2\sqrt{13}\qquad\text{(E)}\ 8 </math>
  
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<math> \text{(A)}\ 5\qquad\text{(B)}\ \sqrt{34} \qquad\text{(C)}\ \sqrt{41}\qquad\text{(D)}\ 2\sqrt{13}\qquad\text{(E)}\ 8 </math>
 
  
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__TOC__
 
 
==Solution==
 
==Solution==
 
=== Solution 1 ===
 
=== Solution 1 ===
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[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 
[[Category:3D Geometry Problems]]
 
[[Category:3D Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 13:07, 5 July 2013

Problem

Triangle $PAB$ and square $ABCD$ are in perpendicular planes. Given that $PA = 3, PB = 4$ and $AB = 5$, what is $PD$? [asy] real r=sqrt(2)/2; draw(origin--(8,0)--(8,-1)--(0,-1)--cycle); draw(origin--(8,0)--(8+r, r)--(r,r)--cycle); filldraw(origin--(-6*r, -6*r)--(8-6*r, -6*r)--(8, 0)--cycle, white, black); filldraw(origin--(8,0)--(8,6)--(0,6)--cycle, white, black); pair A=(6,0), B=(2,0), C=(2,4), D=(6,4), P=B+1*dir(-65); draw(A--P--B--C--D--cycle); dot(A^^B^^C^^D^^P); label("$A$", A, dir((4,2)--A)); label("$B$", B, dir((4,2)--B)); label("$C$", C, dir((4,2)--C)); label("$D$", D, dir((4,2)--D)); label("$P$", P, dir((4,2)--P)); [/asy] $\text{(A)}\ 5\qquad\text{(B)}\ \sqrt{34} \qquad\text{(C)}\ \sqrt{41}\qquad\text{(D)}\ 2\sqrt{13}\qquad\text{(E)}\ 8$


Solution

Solution 1

Since the two planes are perpendicular, it follows that $\triangle PAD$ is a right triangle. Thus, $PD = \sqrt{PA^2 + AD^2} = \sqrt{PA^2 + AB^2} = \sqrt{34}$, which is option $\boxed{\text{B}}$.

Solution 2

Place the points on a coordinate grid, and let the $xy$ plane (where $z=0$) contain triangle $PAB$. Square $ABCD$ will have sides that are vertical.

Place point $P$ at $(0,0,0)$, and place $PA$ on the x-axis so that $A(3,0,0)$, and thus $PA = 3$.

Place $PB$ on the y-axis so that $B(0,4,0)$, and thus $PB = 4$. This makes $AB = 5$, as it is the hypotenuse of a 3-4-5 right triangle (with the right angle being formed by the x and y axes). This is a clean use of the fact that $\triangle PAB$ is a right triangle.

Since $AB$ is one side of $\square ABCD$ with length $5$, $BC = 5$ as well. Since $BC \perp AB$, and $BC$ is also perpendicular to the $xy$ plane, $BC$ must run stright up and down. WLOG pick the up direction, and since $BC = 5$, we travel $5$ units up to $C(0,4,5)$. Similarly, we travel $5$ units up from $A(3,0,0)$ to reach $D(3,0,5)$.

We now have coordinates for $P$ and $D$. The distance is $\sqrt{(5-0)^2 + (0-0)^2 + (3-0)^2} = \sqrt{34}$, which is option $\boxed{\text{B}}$.

See also

1996 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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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