Difference between revisions of "1959 AHSME Problems/Problem 22"
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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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.