Difference between revisions of "1959 AHSME Problems/Problem 22"
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| − | + | == Problem == | |
| − | + | 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 1 == | |
| − | + | <asy> | |
| − | Thus, 91. | + | 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 <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 13:20, 21 July 2024
Contents
Problem
The line joining the midpoints of the diagonals of a trapezoid has length 
. If the longer base is 
then the shorter base is: 
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]](http://latex.artofproblemsolving.com/6/8/0/68067696a74abf0cbeb277913e43d5b23f47b9d6.png)
Let 
be the length of the shorter base. Then:
Thus, our answer is 
.
Solution 2
Let the trapezoid be 
with 
, with 
as the midpoint of 
, and 
as the midpoint of 
, as in the diagram. As in the first solution, let the shorter base of the trapezoid (
) have length . Because , we can imagine shifting along 
by distance such that 
is at 
, 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]](http://latex.artofproblemsolving.com/e/e/f/eef0e9f72322e5c93062c1d316001b2a1b5c8e54.png)
Because 
is a midpoint connector of 
, 
, and so we have the equation 
. Solving for yields 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
