Difference between revisions of "2001 AIME I Problems/Problem 7"
(→Solution 4 (Faster)) |
(→Solution 1) |
||
Line 14: | Line 14: | ||
</asy></center> | </asy></center> | ||
− | Let <math>I</math> be the [[incenter]] of <math>\triangle ABC</math>, so that <math>BI</math> and <math>CI</math> are [[angle bisector]]s of <math>\angle ABC</math> and <math>\angle ACB</math> respectively. Then, <math>\angle BID = \angle CBI = \angle DBI,</math> so <math>\triangle BDI</math> is [[Isosceles triangle|isosceles]], and similarly <math>\triangle CEI</math> is isosceles. It follows that <math>DE = DB + EC</math>, so the perimeter of <math>\triangle ADE</math> is <math>AD + AE + DE = AB + AC = 43</math>. Hence, the ratio of the perimeters of <math>\triangle ADE</math> and <math>\triangle ABC</math> is <math>\frac{43}{63}</math>, which is the scale factor between the two similar triangles, and thus <math>DE = \frac{43}{63} \times 20 = \frac{860}{63}</math>. Thus, <math>m + n = \boxed{923}</math>. | + | Let <math>I</math> be the [[incenter]] of <math>\triangle ABC</math>, so that <math>BI</math> and <math>CI</math> are [[angle bisector]]s of <math>\angle ABC</math> and <math>\angle ACB</math> respectively. Then, <math>\angle BID = \angle CBI = \angle DBI,</math> so <math>\triangle BDI</math> is [[Isosceles triangle|isosceles]], and similarly <math>\triangle CEI</math> is isosceles. It follows that <math>DE = DB + EC</math>, so the perimeter of <math>\triangle ADE</math> is <math>AD + AE + DE = AB + AC = 43</math>. Hence, the ratio of the perimeters of <math>\triangle ADE</math> and <math>\triangle ABC</math> is <math>\frac{43}{63}</math>, which is the scale factor between the two similar triangles, and thus <math>DE = \frac{43}{63} \times 20 = \frac{860}{63}</math>. Thus, <math>m + n = \boxed{923}</math>. |
== Solution 2 == | == Solution 2 == |
Revision as of 10:09, 4 January 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
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.