1996 AHSME Problems/Problem 9

Revision as of 20:35, 18 August 2011 by Talkinaway (talk | contribs) (Solution)

Problem

Triangle $PAB$ and square $ABCD$ are in perpendicular planes. Given that $PA = 3, PB = 4$ and $AB = 5$, what is $PD$?

$\text{(A)}\ 5\qquad\text{(B)}\ \sqrt{34} \qquad\text{(C)}\ \sqrt{41}\qquad\text{(D)}\ 2\sqrt{13}\qquad\text{(E)}\ 8$

Solution

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 $\boxed{B}$

See also

1996 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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