Difference between revisions of "2003 AIME I Problems/Problem 15"
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===Solution 3=== | ===Solution 3=== | ||
Let <math>\angle{DBM}=\theta</math> and <math>\angle{DBC}=\alpha</math>. Then because <math>BM</math> is a median we have <math>360\sin{(\alpha+\theta)}=507\sin{(\alpha-\theta)}</math>. Now we know <cmath>\sin{(\alpha+\theta)}=\sin{\alpha}\cos{\theta}+\sin{\theta}\cos{\alpha}=\dfrac{DF\cdot BD}{BF\cdot BE}+\dfrac{DE\cdot BD}{BE\cdot BF}=\dfrac{BD(DF+DE)}{BF\cdot BE}</cmath> Expressing the area of <math>\triangle{BEF}</math> in two ways we have <cmath>\dfrac{1}{2}BE\cdot BF\sin{(\alpha-\theta)}=\dfrac{1}{2}EF\cdot BD</cmath> so <cmath>\sin{(\alpha-\theta)}=\dfrac{EF\cdot BD}{BF\cdot BE}</cmath> Plugging this in we have <cmath>\dfrac{360\cdot BD(DF+DE)}{BF\cdot BE}=\dfrac{507\cdot BD\cdot EF}{BF\cdot BE}</cmath> so <math>\dfrac{DF+DE}{EF}=\dfrac{507}{360}</math>. But <math>DF=DE+EF</math>, so this simplifies to <math>1+\dfrac{2DE}{EF}=\dfrac{507}{360}=\dfrac{169}{120}</math>, and thus <math>\dfrac{DE}{EF}=\dfrac{49}{240}</math>, and <math>m+n=\boxed{289}</math>. | Let <math>\angle{DBM}=\theta</math> and <math>\angle{DBC}=\alpha</math>. Then because <math>BM</math> is a median we have <math>360\sin{(\alpha+\theta)}=507\sin{(\alpha-\theta)}</math>. Now we know <cmath>\sin{(\alpha+\theta)}=\sin{\alpha}\cos{\theta}+\sin{\theta}\cos{\alpha}=\dfrac{DF\cdot BD}{BF\cdot BE}+\dfrac{DE\cdot BD}{BE\cdot BF}=\dfrac{BD(DF+DE)}{BF\cdot BE}</cmath> Expressing the area of <math>\triangle{BEF}</math> in two ways we have <cmath>\dfrac{1}{2}BE\cdot BF\sin{(\alpha-\theta)}=\dfrac{1}{2}EF\cdot BD</cmath> so <cmath>\sin{(\alpha-\theta)}=\dfrac{EF\cdot BD}{BF\cdot BE}</cmath> Plugging this in we have <cmath>\dfrac{360\cdot BD(DF+DE)}{BF\cdot BE}=\dfrac{507\cdot BD\cdot EF}{BF\cdot BE}</cmath> so <math>\dfrac{DF+DE}{EF}=\dfrac{507}{360}</math>. But <math>DF=DE+EF</math>, so this simplifies to <math>1+\dfrac{2DE}{EF}=\dfrac{507}{360}=\dfrac{169}{120}</math>, and thus <math>\dfrac{DE}{EF}=\dfrac{49}{240}</math>, and <math>m+n=\boxed{289}</math>. | ||
+ | |||
+ | ===Solution (Overpowered Projective Geometry!!)=== | ||
+ | Firstly, angle bisector theorem yields <math>\frac{CD}{AD} = \frac{507}{360} = \frac{169}{120}</math>. We're given that <math>AM=MC</math>. Therefore, the cross ratio | ||
+ | |||
+ | <cmath> | ||
+ | (A,C;M,D) = \frac{AM(CD)}{AD(MC)} = \frac{169}{120} | ||
+ | </cmath> | ||
+ | |||
+ | We need a fourth point for this cross ratio to be useful, so let <math>K = DF \cup BA</math>. Obviously, <math>\Delta BFK</math> is isosceles and <math>BD</math> is an altitude so <math>DF = DK</math>. Therefore, | ||
+ | |||
+ | <cmath> | ||
+ | (A,C;M,D) = (F,K;D,E) \implies \frac{FD(EK)}{EF(DK)} = \frac{EK}{EF} = \frac{169}{120} | ||
+ | </cmath> | ||
+ | |||
+ | All that's left is to screw around with the ratios: | ||
+ | |||
+ | <cmath> | ||
+ | \frac{EK}{EF} = \frac{ED+DF}{EF} = \frac{EF+2DF}{EF} = 1+2(\frac{DF}{EF}) \implies \frac{DF}{EF} = \frac{49}{240} \implies \boxed{289} | ||
+ | </cmath> | ||
== See also == | == See also == |
Latest revision as of 21:08, 15 December 2019
Contents
Problem
In and Let be the midpoint of and let be the point on such that bisects angle Let be the point on such that Suppose that meets at The ratio can be written in the form where and are relatively prime positive integers. Find
Solution
In the following, let the name of a point represent the mass located there. Since we are looking for a ratio, we assume that , , and in order to simplify our computations.
First, reflect point over angle bisector to a point .
As is an angle bisector of both triangles and , we know that lies on . We can now balance triangle at point using mass points.
By the Angle Bisector Theorem, we can place mass points on of respectively. Thus, a mass of belongs at both and because BD is a median of triangle . Therefore, .
Now, we reassign mass points to determine . This setup involves and transversal . For simplicity, put masses of and at and respectively. To find the mass we should put at , we compute . Applying the Angle Bisector Theorem again and using the fact is a midpoint of , we find At this point we could find the mass at but it's unnecessary. and the answer is .
Solution 2
By the Angle Bisector Theorem, we know that . Therefore, by finding the area of triangle , we see that Solving for yields Furthermore, , so Now by the identity , we get But then , so . Thus .
Now by the Angle Bisector Theorem, , and we know that so .
We can now use mass points on triangle CBD. Assign a mass of to point . Then must have mass and must have mass . This gives a mass of . Therefore, , giving us an answer of
Solution 3
Let and . Then because is a median we have . Now we know Expressing the area of in two ways we have so Plugging this in we have so . But , so this simplifies to , and thus , and .
Solution (Overpowered Projective Geometry!!)
Firstly, angle bisector theorem yields . We're given that . Therefore, the cross ratio
We need a fourth point for this cross ratio to be useful, so let . Obviously, is isosceles and is an altitude so . Therefore,
All that's left is to screw around with the ratios:
See also
2003 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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