Difference between revisions of "1980 AHSME Problems/Problem 21"
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== Solution == | == Solution == | ||
| + | We can use the principle of same height same area to solve this problem. | ||
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Latest revision as of 18:07, 16 July 2023
Problem
![[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair B=origin, C=(15,3), D=(5,1), A=7*dir(72)*dir(B--C), E=midpoint(A--C), F=intersectionpoint(A--D, B--E); draw(E--B--A--C--B^^A--D); label("$A$", A, dir(D--A)); label("$B$", B, dir(E--B)); label("$C$", C, dir(0)); label("$D$", D, SE); label("$E$", E, N); label("$F$", F, dir(80));[/asy]](http://latex.artofproblemsolving.com/7/0/1/70161822674949b3054b3faa63150a9a6f751e97.png)
In triangle 
, 
, 
is the midpoint of side 
,
and 
is a point on side 
such that 
; 
and 
intersect at 
.
The ratio of the area of triangle 
to the area of quadrilateral 
is
Solution
We can use the principle of same height same area to solve this problem.
See also
| 1980 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
