2018 AMC 8 Problems/Problem 20: Difference between revisions
Created page with "==Problem 20== In <math>\triangle ABC,</math> a point <math>E</math> is on <math>\overline{AB}</math> with <math>AE=1</math> and <math>EB=2.</math> Point <math>D</math> is on..." |
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<math>\textbf{(A) } \frac{4}{9} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{5}{9} \qquad \textbf{(D) } \frac{3}{5} \qquad \textbf{(E) } \frac{2}{3}</math> | <math>\textbf{(A) } \frac{4}{9} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{5}{9} \qquad \textbf{(D) } \frac{3}{5} \qquad \textbf{(E) } \frac{2}{3}</math> | ||
{{AMC8 box|year=2018|num-b=19|num-a=21}} | |||
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Revision as of 11:05, 21 November 2018
Problem 20
In
a point
is on
with
and
Point
is on
so that
and point
is on
so that
What is the ratio of the area of
to the area of
| 2018 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
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| All AJHSME/AMC 8 Problems and Solutions | ||
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