1959 AHSME Problems/Problem 22: Difference between revisions
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== Solution == | == Solution == | ||
<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 <math>x</math> be the length of the shorter base. Then: | Let <math>x</math> be the length of the shorter base. Then: | ||
Revision as of 12:00, 21 July 2024
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
![[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 
.
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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
