Difference between revisions of "1959 AHSME Problems/Problem 10"
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== Solution == | == Solution == | ||
− | Note that <math>\frac{1}{2}AB * AC *\sin\angle BAC = \frac{1}{2}AD * AE *\sin\angle DAE</math>. Since <math>\angle BAC = \angle DAE</math>, we have <math>AB*AC = AD*AE</math>, so that <math>3.6*3.6 = 1.2*AE</math>. Therefore, <math>AE = \frac{3.6^2}{1.2} = | + | <asy> |
+ | |||
+ | import geometry; | ||
+ | |||
+ | point B = (0,0); | ||
+ | point C = (10,0); | ||
+ | point A = (5,12); | ||
+ | point D = 2*A/3; | ||
+ | point E = 3(C-A)+A; | ||
+ | |||
+ | // Triangle ABC | ||
+ | draw(A--B--C--A); | ||
+ | dot(A); | ||
+ | label("A",A,N); | ||
+ | dot(B); | ||
+ | label("B",B,SW); | ||
+ | dot(C); | ||
+ | label("C",C,SE); | ||
+ | |||
+ | // Triangle ADE | ||
+ | draw(A--D--E--A); | ||
+ | dot(D); | ||
+ | label("D",D,NW); | ||
+ | dot(E); | ||
+ | label("E",E,SE); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | Note that <math>[\triangle ABC]=[\triangle ADE]</math>, so <math>\frac{1}{2}AB * AC *\sin\angle BAC = \frac{1}{2}AD * AE *\sin\angle DAE</math>. Since <math>\angle BAC = \angle DAE</math>, we have <math>AB*AC = AD*AE</math>, so that <math>3.6*3.6 = 1.2*AE</math>. Therefore, <math>AE = \frac{3.6^2}{1.2}=\boxed{\textbf{(D) }10.8}</math>. | ||
+ | |||
+ | ==See also== | ||
+ | {{AHSME 50p box|year=1959|num-b=9|num-a=11}} | ||
+ | {{MAA Notice}} | ||
+ | [[Category:Introductory Geometry Problems]] |
Latest revision as of 11:20, 21 July 2024
Problem
In with , a point is taken on at a distance from . Point is joined to in the prolongation of so that is equal in area to . Then is:
Solution
Note that , so . Since , we have , so that . Therefore, .
See also
1959 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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All AHSME Problems and Solutions |
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