Difference between revisions of "2001 AIME I Problems/Problem 7"
m |
Fortenforge (talk | contribs) (→Solution 2) |
||
Line 25: | Line 25: | ||
</asy></center> | </asy></center> | ||
− | The [[semiperimeter]] of <math>ABC</math> is <math>s = \frac{20 + 21 + 22}{2} = \frac{63}{2}</math>. By [[Heron's formula]], the area of the whole triangle is <math>A = \sqrt{s(s-a)(s-b)(s-c)} = \frac{21\sqrt{1311}} | + | The [[semiperimeter]] of <math>ABC</math> is <math>s = \frac{20 + 21 + 22}{2} = \frac{63}{2}</math>. By [[Heron's formula]], the area of the whole triangle is <math>A = \sqrt{s(s-a)(s-b)(s-c)} = \frac{21\sqrt{1311}}{4}</math>. Using the formula <math>A = rs</math>, we find that the [[inradius]] is <math>r = \frac{A}{s} = \frac{\sqrt{1311}}6</math>. Since <math>\triangle ADE \sim \triangle ABC</math>, the ratio of the heights of triangles <math>ADE</math> and <math>ABC</math> is equal to the ratio between sides <math>DE</math> and <math>BC</math>. From <math>A=\frac{1}{2}bh</math>, we find <math>h_{ABC} = \frac{21\sqrt{1311}}{40}</math>. Thus, we have |
<center><math>\frac{h_{ADE}}{h_{ABC}} = \frac{h_{ABC}-r}{h_{ABC}} = \frac{21\sqrt{1311}/40-\sqrt{1311}/6}{21\sqrt{1311}/40}=\frac{DE}{20}.</math></center> Solving for <math>DE</math> gives <math>DE=\frac{860}{63},</math> so the answer is <math>m+n=\boxed{923}</math>. | <center><math>\frac{h_{ADE}}{h_{ABC}} = \frac{h_{ABC}-r}{h_{ABC}} = \frac{21\sqrt{1311}/40-\sqrt{1311}/6}{21\sqrt{1311}/40}=\frac{DE}{20}.</math></center> Solving for <math>DE</math> gives <math>DE=\frac{860}{63},</math> so the answer is <math>m+n=\boxed{923}</math>. | ||
Line 33: | Line 33: | ||
Using the ratio of heights to ratio of bases of <math>ADE</math> and <math>ABC</math> | Using the ratio of heights to ratio of bases of <math>ADE</math> and <math>ABC</math> | ||
<math>\frac {\frac {2S}{20}-\frac {2S}{63}}{\frac {2S}{20}}= \frac {DE}{BC(20)}</math> | <math>\frac {\frac {2S}{20}-\frac {2S}{63}}{\frac {2S}{20}}= \frac {DE}{BC(20)}</math> | ||
− | from that it is easy to deduce that <math>DE=\frac{860}{63}</math> | + | from that it is easy to deduce that <math>DE=\frac{860}{63}</math>. |
− | |||
=== Solution 3 ([[mass points]]) === | === Solution 3 ([[mass points]]) === |
Revision as of 14:17, 17 June 2012
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
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, .
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 |