Difference between revisions of "2018 AMC 8 Problems/Problem 20"
(→Problem 20) |
(→Solution) |
||
Line 17: | Line 17: | ||
<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> | ||
− | |||
− | |||
− | |||
− | |||
− | |||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:53, 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
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.