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Difference between revisions of "1959 AHSME Problems/Problem 22"

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== Solution 2 ==
 
== Solution 2 ==
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Let the trapezoid be <math>ABCD</math> with <math>\overline{AD} \parallel \overline{BC}</math>, with <math>M</math> as the midpoint of <math>\overline{AC}</math> and <math>N</math> as the midpoint of <math>\overline{BD}</math>, as in the diagram. Because <math>\overline{AD} \parallel \overline{BC}</math>, we can imagine shifting <math>\overline{AC}</math> along <math>\overleftrightarrow{BC}</math> until <math>A</math> is at point <math>D</math> until we get the following triangle:
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Let the trapezoid be <math>ABCD</math> with <math>\overline{AD} \parallel \overline{BC}</math>, with <math>M</math> as the midpoint of <math>\overline{AC}</math>, and <math>N</math> as the midpoint of <math>\overline{BD}</math>, as in the diagram. As in the first solution, let the shorter base of the trapezoid (<math>\overline{AD}</math>) have length <math>x</math>. Because <math>\overline{AD} \parallel \overline{BC}</math>, we can imagine shifting <math>\overline{AC}</math> along <math>\overleftrightarrow{BC}</math> by distance <math>x</math> such that <math>A</math> is at <math>D</math>, at which point we get the following triangle:
  
 
<asy>
 
<asy>

Latest revision as of 13:20, 21 July 2024

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
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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