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

Revision as of 13:53, 19 August 2011

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 $\boxed{\text{B}}$ .

@@ Line 19: / Line 19: @@
 <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>
-__TOC__
 ==Solution==

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

Page

Toolbox

Search