Difference between revisions of "1971 AHSME Problems/Problem 28"
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== Solution == | == Solution == | ||
+ | |||
+ | <asy> | ||
+ | |||
+ | import geometry; | ||
+ | |||
+ | point B = origin; | ||
+ | point A = (3,5); | ||
+ | point C = (7,0); | ||
+ | triangle t = triangle(A,B,C); | ||
+ | |||
+ | point D = B*9/10 + A/10; | ||
+ | point E; | ||
+ | |||
+ | // Defining point E | ||
+ | pair[] e = intersectionpoints(parallel(D,line(B,C)),A--C); | ||
+ | E = e[0]; | ||
+ | |||
+ | // Triangle ABC and Parallel Segment | ||
+ | draw(t); | ||
+ | draw(D--E); | ||
+ | |||
+ | // Point Labels | ||
+ | dot(A); | ||
+ | label("A",A,NW); | ||
+ | dot(B); | ||
+ | label("B",B,SW); | ||
+ | dot(C); | ||
+ | label("C",C,SE); | ||
+ | dot(D); | ||
+ | label("D",D,NW); | ||
+ | dot(E); | ||
+ | label("E",E,NE); | ||
+ | |||
+ | </asy> | ||
<math>\boxed{\textbf{(C) }200}</math>. | <math>\boxed{\textbf{(C) }200}</math>. |
Revision as of 10:51, 7 August 2024
Problem
Nine lines parallel to the base of a triangle divide the other sides each into equal segments and the area into distinct parts. If the area of the largest of these parts is , then the area of the original triangle is
Solution
.
See Also
1971 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.