Difference between revisions of "2009 AMC 12A Problems/Problem 20"
m (→Solution 1) |
m (→Solution 1) |
||
Line 11: | Line 11: | ||
<center><asy>pathpen = linewidth(0.7);pointpen = black; | <center><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= | + | 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("12",(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); | 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></center> | </asy></center> |
Revision as of 17:43, 15 July 2021
- The following problem is from both the 2009 AMC 12A #20 and 2009 AMC 10A #23, so both problems redirect to this page.
Contents
Problem
Convex quadrilateral has and . Diagonals and intersect at , , and and have equal areas. What is ?
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 . , so .
Solution 2
Using the sine area formula on triangles and , as , we see that
Since , triangles and are similar. Their ratio is . Since , we must have , so .
Solution 3 (which won't work when justification is required)
The easiest way for the areas of the triangles to be equal would be if they were congruent. A way for that to work would be if were simply an isosceles trapezoid! Since and (look at the side lengths and you'll know why!),
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 |
2009 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.