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

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The line joining the midpoints of the diagonals of a trapezoid has length <math>3</math>. If the longer base is <math>97,</math> then the shorter base is: <math>\textbf{(A)}\ 94 \qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 89</math>
 
The line joining the midpoints of the diagonals of a trapezoid has length <math>3</math>. If the longer base is <math>97,</math> then the shorter base is: <math>\textbf{(A)}\ 94 \qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 89</math>
  
== Solution ==
+
== Solution 1 ==
Let x be the length of the shorter base.
 
3 = (97 - x)/2
 
  
6 = 97 - x
+
<asy>
  
x = 91
+
import geometry;
  
Thus, 91.
+
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 <math>x</math> be the length of the shorter base. Then:
 +
 
 +
<math>3 = \frac{97-x}{2}</math>
 +
 
 +
<math>6 = 97-x</math>
 +
 
 +
<math>x = 91</math>
 +
 
 +
Thus, our answer is <math>\boxed{\textbf{(C) }91}</math>.
 +
 
 +
== Solution 2 ==
 +
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>
 +
 
 +
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 <math>\overline{MN}</math> is a [[Midpoint#Midsegments|midpoint connector]] of <math>\triangle ABC</math>, <math>MN=\frac{1}{2}BC</math>, and so we have the equation <math>3+x=\frac{97+x}{2}</math>. Solving for <math>x</math> yields <math>x=\boxed{\textbf{(C) }91}</math>.
 +
 
 +
==See also==
 +
{{AHSME 50p box|year=1959|num-b=21|num-a=23}}
 +
{{MAA Notice}}
 +
[[Category:Introductory Geometry Problems]]

Latest revision as of 12:20, 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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