2000 AMC 12 Problems/Problem 19
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 1
The answer is exactly , choice . We can find the area of triangle by using the simple formula . Dropping an altitude from , we see that it has length ( we can split the large triangle into a and a triangle). Then we can apply the Angle Bisector Theorem on triangle to solve for . Solving , we get that . is the midpoint of so . Thus we get the base of triangle , to be units long. Applying the formula , we get .
Solution 2
The area of is where is the height of triangle . Using Angle Bisector Theorem, we find , which we solve to get . is the midpoint of so . Thus we get the base of triangle , to be units long. We can now use Heron's Formula on . Therefore, the answer is .
Solution 3
pair A,B,C,D,E; B=(0,0); C=(14;0); A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); draw(A--B--C--cycle); D=(7,0); E=(6.5,0); draw(A--E); draw(A--D); label("$A$",A,N); label("$B$",B,S); label("$C$",C,S); label("$E$",E,S); label("$D$",D,S); label("$13$",A--B,NW); label("$15$",A--C,NE); label("$14$",B--C,S); (Error compiling LaTeX. 32c2c358ed0750fe240d2c14d3cb08587d95c9a3.asy: 7.6: syntax error error: could not load module '32c2c358ed0750fe240d2c14d3cb08587d95c9a3.asy')
Let's find the area of by Heron,
Then,
Knowing that D is the midpoint of BC, thus the area of , which is half the area of , so .
By Angle Bisector Theorem we know that:
Also, we know that:
And, we can easily see that , so,
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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