Difference between revisions of "2000 AMC 12 Problems/Problem 19"
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(→Solution 3) |
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Line 19: | Line 19: | ||
<asy> | <asy> | ||
+ | unitsize(0.5cm); | ||
pair A,B,C,D,E; | pair A,B,C,D,E; | ||
B=(0,0); | B=(0,0); | ||
− | C=(14 | + | C=(14,0); |
A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); | A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); | ||
draw(A--B--C--cycle); | draw(A--B--C--cycle); | ||
Line 31: | Line 32: | ||
label("$B$",B,S); | label("$B$",B,S); | ||
label("$C$",C,S); | label("$C$",C,S); | ||
− | label("$E$",E, | + | label("$E$",E,NW); |
− | label("$D$",D, | + | label("$D$",D,NE); |
label("$13$",A--B,NW); | label("$13$",A--B,NW); | ||
label("$15$",A--C,NE); | label("$15$",A--C,NE); |
Revision as of 17:18, 7 August 2017
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
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 |
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.