Difference between revisions of "2001 AIME I Problems/Problem 7"
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Applying the [[Angle Bisector Theorem]] on <math>\angle{A}</math> we have | Applying the [[Angle Bisector Theorem]] on <math>\angle{A}</math> we have | ||
<cmath>\frac{AB}{BF}=\frac{AC}{CF}</cmath> | <cmath>\frac{AB}{BF}=\frac{AC}{CF}</cmath> | ||
− | <cmath>BF=\frac{(BC-BF)\cdot AB}{ | + | <cmath>BF=\frac{(BC-BF)\cdot AB}{AC}</cmath> |
− | < | + | <cmath>BF=\frac{(20-BF)\cdot21}{22}</cmath> |
+ | <cmath>BF=\frac{420}{43}</cmath> | ||
== See also == | == See also == |
Revision as of 09:41, 12 December 2019
Problem
Triangle has , and . Points and are located on and , respectively, such that is parallel to and contains the center of the inscribed circle of triangle . Then , where and are relatively prime positive integers. Find .
Contents
[hide]Solution 1
Let be the incenter of , so that and are angle bisectors of and respectively. Then, so is isosceles, and similarly is isosceles. It follows that , so the perimeter of is . Hence, the ratio of the perimeters of and is , which is the scale factor between the two similar triangles, and thus . Thus, .
Solution 2
The semiperimeter of is . By Heron's formula, the area of the whole triangle is . Using the formula , we find that the inradius is . Since , the ratio of the heights of triangles and is equal to the ratio between sides and . From , we find . Thus, we have
Solving for gives so the answer is .
Or we have the area of the triangle as . Using the ratio of heights to ratio of bases of and from that it is easy to deduce that .
Solution 3 (mass points)
Let be the incircle; then it is be the intersection of all three angle bisectors. Draw the bisector to where it intersects , and name the intersection .
Using the angle bisector theorem, we know the ratio is , thus we shall assign a weight of to point and a weight of to point , giving a weight of . In the same manner, using another bisector, we find that has a weight of . So, now we know has a weight of , and the ratio of is . Therefore, the smaller similar triangle is the height of the original triangle . So, is the size of . Multiplying this ratio by the length of , we find is . Therefore, .
Solution 4 (Faster)
More directly than Solution 2, we have
Solution 5
Diagram borrowed from Solution 4.
Let the angle bisector of intersects at .
Applying the Angle Bisector Theorem on we have
See also
2001 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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