Difference between revisions of "1957 AHSME Problems/Problem 35"
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== Solution == | == Solution == | ||
+ | |||
+ | <asy> | ||
+ | |||
+ | import geometry; | ||
+ | |||
+ | point B = (0,0); | ||
+ | point A = (0,16); | ||
+ | point C = (10,0); | ||
+ | |||
+ | // Triangle ABC | ||
+ | draw(triangle(A,B,C)); | ||
+ | dot(A); | ||
+ | label("A",A,NW); | ||
+ | dot(B); | ||
+ | label("B",B,SW); | ||
+ | dot(C); | ||
+ | label("C",C,SE); | ||
+ | |||
+ | // Parallel Lines | ||
+ | for (real x=0; x<length(segment(A,B)); x += length(segment(A,B))/8) { | ||
+ | pair[] y = intersectionpoints(parallel((0,x),line(B,C)),A--C); | ||
+ | draw((0,x)--y[0]); | ||
+ | } | ||
+ | |||
+ | // Length Label | ||
+ | label("$10$", B/2+C/2, S); | ||
+ | |||
+ | </asy> | ||
+ | |||
<math>\boxed{\textbf{(D)} \text{ is } 35}</math>. | <math>\boxed{\textbf{(D)} \text{ is } 35}</math>. | ||
Revision as of 07:55, 26 July 2024
Problem
Side of right triangle is divided into equal parts. Seven line segments parallel to are drawn to from the points of division. If , then the sum of the lengths of the seven line segments:
Solution
.
See Also
1957 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 34 |
Followed by Problem 36 | |
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.