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1959 AHSME Problems/Problem 22

Problem

The line joining the midpoints of the diagonals of a trapezoid has length $3$. If the longer base is $97,$ then the shorter base is: $\textbf{(A)}\ 94 \qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 89$

Solution 1

[asy] import geometry; point B = (0,0); point A = (3,5); point D = (13,5); point C = (15,0); point M,N; // Trapezoid draw(A--B--C--D--A); dot(A); label("A",A,NW); dot(B); label("B",B,SW); dot(C); label("C",C,SE); dot(D); label("D",D,NE); // Diagonals and their midpoints draw(A--C); draw(B--D); M = midpoint(A--C); dot(M); label("M",M,ENE); N = midpoint(B--D); dot(N); label("N",N,WNW); draw(M--N); // Length Labels label("$x$",midpoint(A--D),(0,1)); label("$97$",midpoint(B--C),S); label("$3$",midpoint(M--N),S); [/asy]

Let $x$ be the length of the shorter base. Then:

$3 = \frac{97-x}{2}$

$6 = 97-x$

$x = 91$

Thus, our answer is $\boxed{\textbf{(C) }91}$.

Solution 2

Let the trapezoid be $ABCD$ with $\overline{AD} \parallel \overline{BC}$, with $M$ as the midpoint of $\overline{AC}$, and $N$ as the midpoint of $\overline{BD}$, as in the diagram. As in the first solution, let the shorter base of the trapezoid ($\overline{AD}$) have length $x$. Because $\overline{AD} \parallel \overline{BC}$, we can imagine shifting $\overline{AC}$ along $\overleftrightarrow{BC}$ by distance $x$ such that $A$ is at $D$, at which point we get the following triangle:

[asy] import geometry; size(10cm); point B = (0,0); point C = (25,0); point A = (13,5); point M,N; // Triangle ABC draw(A--B--C--A); dot(A); label("A",A,(0,1)); dot(B); label("B",B,SW); dot(C); label("C",C,SE); // Midpoint connector M = midpoint(A--C); dot(M); label("M",M,NE); N = midpoint(B--A); dot(N); label("N",N,NW); draw(M--N); // Length Labels label("$97+x$",midpoint(B--C),S); label("$3+x$",midpoint(M--N),S); [/asy]

Because $\overline{MN}$ is a midpoint connector of $\triangle ABC$, $MN=\frac{1}{2}BC$, and so we have the equation $3+x=\frac{97+x}{2}$. Solving for $x$ yields $x=\boxed{\textbf{(C) }91}$.

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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All AHSME Problems and Solutions

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