Difference between revisions of "2001 AIME I Problems/Problem 7"
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<cmath> {AP} = 21*22 - \frac{440\cdot420}{43^2}</cmath> | <cmath> {AP} = 21*22 - \frac{440\cdot420}{43^2}</cmath> | ||
− | Now we can calculate what <math>{AO}</math> is. Using the formula to find the distance from a vertex to the incenter, <math>{AO} = \frac{43}{63} \cdot[21\cdot22 - \frac{420*440}{43^2} = \frac{43^2\cdot22 - 20\cdot440}{43\cdot3}</math>. | + | Now we can calculate what <math>{AO}</math> is. Using the formula to find the distance from a vertex to the incenter, <math>{AO} = \frac{43}{63} \cdot[21\cdot22 - \frac{420*440}{43^2}] = \frac{43^2\cdot22 - 20\cdot440}{43\cdot3}</math>. |
Now because <math>\triangle{APE} ~ \triangle{ABC}</math>, we can find <math>{DE}</math> by <math>\frac{AO}{AP} \cdot 20</math>. Dividing and simplifying, we see that <math>\frac{1}{21}\cdot\frac{43}{3}\cdot20 = \frac{860}{63}</math>. So the answer is <math>\boxed{923}</math> | Now because <math>\triangle{APE} ~ \triangle{ABC}</math>, we can find <math>{DE}</math> by <math>\frac{AO}{AP} \cdot 20</math>. Dividing and simplifying, we see that <math>\frac{1}{21}\cdot\frac{43}{3}\cdot20 = \frac{860}{63}</math>. So the answer is <math>\boxed{923}</math> |
Revision as of 07:31, 1 December 2022
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 incenter; 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 3.
Let the angle bisector of intersects at .
Applying the Angle Bisector Theorem on we have Since is the angle bisector of , we can once again apply the Angle Bisector Theorem on which gives Since we have Solving gets . Thus .
~ Nafer
Solution 6
Let be the foot of the altitude from to and be the foot of the altitude from to . Evidently, where is the inradius, , and is the semiperimeter. So, Therefore, by similar triangles, we have .
Solution 7
Label the point the angle bisector of intersects . First we find and . By the Angle Bisector Theorem, and solving for each using the fact that , we see that and .
Because is the angle bisector of , we can simply calculate it using Stewarts,
Now we can calculate what is. Using the formula to find the distance from a vertex to the incenter, .
Now because , we can find by . Dividing and simplifying, we see that . So the answer is
~YBSuburbanTea
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.