Difference between revisions of "2001 AIME I Problems/Problem 7"
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pointpen = black; pathpen = black+linewidth(0.7); pen d = linewidth(0.7) + linetype("4 4"); | pointpen = black; pathpen = black+linewidth(0.7); pen d = linewidth(0.7) + linetype("4 4"); | ||
pair B=(0,0), C=(20,0), A=IP(CR(B,21),CR(C,22)), I=incenter(A,B,C), D=IP((0,I.y)--(20,I.y),A--B), E=IP((0,I.y)--(20,I.y),A--C); | pair B=(0,0), C=(20,0), A=IP(CR(B,21),CR(C,22)), I=incenter(A,B,C), D=IP((0,I.y)--(20,I.y),A--B), E=IP((0,I.y)--(20,I.y),A--C); | ||
− | D(MP("A",A,N)--MP("B",B)--MP("C",C)--cycle); D(incircle(A,B,C)); D(MP("P",I,(1,2))); D(MP("E",E,NE)--MP("D",D,NW | + | D(MP("A",A,N)--MP("B",B)--MP("C",C)--cycle); D(incircle(A,B,C)); D(MP("P",I,(1,2))); D(MP("E",E,NE)--MP("D",D,NW)); |
MP("20",(B+C)/2); MP("21",(A+B)/2,NW); MP("22",(A+C)/2,NE); | MP("20",(B+C)/2); MP("21",(A+B)/2,NW); MP("22",(A+C)/2,NE); | ||
Revision as of 08:58, 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
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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