Difference between revisions of "2001 AIME I Problems/Problem 7"
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== Solution 6 == | == Solution 6 == | ||
Let <math>A'</math> be the foot of the altitude from <math>A</math> to <math>\overline {BC}</math> and <math>K</math> be the foot of the altitude from <math>A</math> to <math>\overline{DE}</math>. Evidently, <cmath>\frac{AK}{AA'} = 1- \frac{r}{AA'} = 1 - \frac{T/s}{T/BC}</cmath> where <math>r</math> is the inradius, <math>T = [ABC]</math>, and <math>s</math> is the semiperimeter. So, <cmath>\frac{AK}{AA'} = 1 - \frac{BC}{s} = 1 - \frac{20}{63}= \frac{43}{63}</cmath> Therefore, by similar triangles, we have <math>DE = BC * \frac{AK}{AA'} = 20 * \frac{AK}{AA'}= \boxed{\frac{860}{63}}</math>. | Let <math>A'</math> be the foot of the altitude from <math>A</math> to <math>\overline {BC}</math> and <math>K</math> be the foot of the altitude from <math>A</math> to <math>\overline{DE}</math>. Evidently, <cmath>\frac{AK}{AA'} = 1- \frac{r}{AA'} = 1 - \frac{T/s}{T/BC}</cmath> where <math>r</math> is the inradius, <math>T = [ABC]</math>, and <math>s</math> is the semiperimeter. So, <cmath>\frac{AK}{AA'} = 1 - \frac{BC}{s} = 1 - \frac{20}{63}= \frac{43}{63}</cmath> Therefore, by similar triangles, we have <math>DE = BC * \frac{AK}{AA'} = 20 * \frac{AK}{AA'}= \boxed{\frac{860}{63}}</math>. | ||
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+ | == Solution 7 == | ||
+ | Label <math>P</math> the point the angle bisector of <math>A</math> intersects <math>{BC}</math>. First we find <math>{BP}</math> and <math>{PC}</math>. By the Angle Bisector Theorem, <math>\frac{BP}{PC} = \frac{21}{22}</math> and solving for each using the fact that <math>{BC} = 20</math>, we see that <math>{BP} = \frac{420}{43}</math> and <math>PC = \frac{440}{43}</math>. | ||
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+ | Because <math>{AP}</math> is the angle bisector of <math><A</math>, we can simply calculate it using Stewarts, | ||
+ | <cmath> {AP} = 21*22 - \frac{440}{43}\cdot\frac{420}{43}</cmath> | ||
+ | <cmath> {AP} = 21*22 - \frac{440\cdot420}{43^2}</cmath> | ||
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+ | 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>. | ||
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+ | 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> | ||
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+ | ~YBSuburbanTea | ||
== See also == | == See also == |
Revision as of 07:30, 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 |
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