Difference between revisions of "2009 AMC 12A Problems/Problem 20"
(solutions by (1) worthawholebean, (2) e^pi*i = -1) |
(No difference)
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Revision as of 18:53, 11 February 2009
Problem
Convex quadrilateral has
and
. Diagonals
and
intersect at
,
, and
and
have equal areas. What is
?
Contents
[hide]Solution
Solution 1
Let denote the area of triangle
.
, so
. Since triangles
and
share a base, they also have the same height and thus
and
with a ratio of
.
.
![[asy]pathpen = linewidth(0.7);pointpen = black; pair D=MP("D",(0,0)),C=MP("C",(12,0)),A=MP("A",C+14*expi(145*pi/180),N),B=MP("B",A+(9,0),N),E=IP(A--C,B--D);MP("9",(A+B)/2,N);MP("14",(C+D)/2); fill(A--D--E--cycle,rgb(0.8,0.8,0.8));fill(B--C--E--cycle,rgb(0.8,0.8,0.8));D(A--B--C--D--cycle);D(A--C);D(B--D);D(E); [/asy]](http://latex.artofproblemsolving.com/0/2/3/023a8371cc0e18852d78718be400377b281a798f.png)
Solution 2
Using the sine area formula on triangles and
, as
, we see that
![\[(AE)(ED) = (BE)(EC)\quad \Longrightarrow\quad \frac {AE}{EC} = \frac {BE}{ED}.\]](http://latex.artofproblemsolving.com/1/1/e/11e8d0633df9bf333912e3d0c13884bc95b2114e.png)
Since , triangles
and
are similar. Their ratio is
. Since
, we must have
,
.
See also
2009 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 | |
All AMC 12 Problems and Solutions |