1997 AHSME Problems/Problem 25


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 $A^{\prime} C^{\prime}$ and $B^{\prime}D^{\prime}$, respectively, then $MN =$

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


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 (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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