Difference between revisions of "1994 AHSME Problems/Problem 13"
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+ | ==Problem== | ||
+ | In triangle <math>ABC</math>, <math>AB=AC</math>. If there is a point <math>P</math> strictly between <math>A</math> and <math>B</math> such that <math>AP=PC=CB</math>, then <math>\angle A =</math> | ||
+ | <asy> | ||
+ | draw((0,0)--(8,0)--(4,12)--cycle); | ||
+ | draw((8,0)--(1.6,4.8)); | ||
+ | label("A", (4,12), N); | ||
+ | label("B", (0,0), W); | ||
+ | label("C", (8,0), E); | ||
+ | label("P", (1.6,4.8), NW); | ||
+ | dot((0,0)); | ||
+ | dot((4,12)); | ||
+ | dot((8,0)); | ||
+ | dot((1.6,4.8)); | ||
+ | </asy> | ||
+ | <math> \textbf{(A)}\ 30^{\circ} \qquad\textbf{(B)}\ 36^{\circ} \qquad\textbf{(C)}\ 48^{\circ} \qquad\textbf{(D)}\ 60^{\circ} \qquad\textbf{(E)}\ 72^{\circ} </math> | ||
+ | |||
==Solution== | ==Solution== | ||
<asy> | <asy> |
Latest revision as of 12:24, 16 July 2024
Problem
In triangle , . If there is a point strictly between and such that , then
Solution
Let . Since , we have as well. Then . Since , we have .
So
--Solution by TheMaskedMagician
See Also
1994 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 | ||
All AHSME Problems and Solutions |
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