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Difference between revisions of "1997 AHSME Problems/Problem 25"

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Revision as of 13:14, 5 July 2013

Problem

Let $ABCD$ be a parallelogram and let $\overrightarrow{AA^\prime}$, $\overrightarrow{BB^\prime}$, $\overrightarrow{CC^\prime}$, and $\overrightarrow{DD^\prime}$ be parallel rays in space on the same side of the plane determined by $ABCD$. If $AA^\prime = 10$, $BB^\prime = 8$, $CC^\prime = 18$, and $DD^\prime = 22$ and $M$ and $N$ are the midpoints of $A^\primeC^\prime$ (Error compiling LaTeX. Unknown error_msg) and $B^\primeD^\prime$ (Error compiling LaTeX. Unknown error_msg), respectively, then $MN =$

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution

Let $ABCD$ be a unit square with $A(0,0,0)$, $B(0,1,0)$, $C(1,1,0)$, and $D(1,0,0)$. Assume that the rays go in the +z direction. In this case, $A^\prime(0,0,10)$, $B^\prime(0,1,8)$, $C^\prime(1,1,18)$, and $D^\prime(1,0,22)$. Finding the midpoints of $A^\prime C^\prime$ and $B^\prime D^\prime$ gives $M(\frac{1}{2}, \frac{1}{2}, 14)$ and $N(\frac{1}{2}, \frac{1}{2}, 15)$. The distance $MN$ is $15 - 14 = 1$, and the answer is $\boxed{B}$.


While this solution is not a proof, the problem asks for a value $MN$ that is supposed to be an invariant. Additional assumptions that are consistient with the problem's premise should not change the answer. The assumptions made in this solution are that $ABCD$ is a square, and that all rays $XX^\prime$ are perpendicular to the plane of the square. These assumptions are consistient with all facts of the problem. Additionally, the assumptions allow for a quick solution to the problem.

See also

1997 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Problem 26
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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