Difference between revisions of "2000 AMC 12 Problems/Problem 19"
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== Solution == | == Solution == | ||
| − | The answer is exactly 3, choice \mathrm{(C)} | + | The answer is exactly 3, choice <math>\mathrm{(C)}</math>. |
We can use either Heron's on the larger triangle to calculate the height of ADE or we can decompose ABC into two right triangles of 5-12-13 and 9-12-15 to notice that height is 12. Then angle bisector theorem leads directly to the base of ADE, which leads to an area of 3. | We can use either Heron's on the larger triangle to calculate the height of ADE or we can decompose ABC into two right triangles of 5-12-13 and 9-12-15 to notice that height is 12. Then angle bisector theorem leads directly to the base of ADE, which leads to an area of 3. | ||
Revision as of 04:11, 4 February 2015
Problem
In triangle 
, 
, 
, 
. Let 
denote the midpoint of 
and let 
denote the intersection of with the bisector of angle 
. Which of the following is closest to the area of the triangle 
?
Solution
The answer is exactly 3, choice 
.
We can use either Heron's on the larger triangle to calculate the height of ADE or we can decompose ABC into two right triangles of 5-12-13 and 9-12-15 to notice that height is 12. Then angle bisector theorem leads directly to the base of ADE, which leads to an area of 3.
See also
| 2000 AMC 12 (Problems • Answer Key • Resources) | |
| Preceded by Problem 18 |
Followed by Problem 20 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
