Difference between revisions of "1998 AIME Problems/Problem 12"
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We want the ratio of the squares of the sides, so <math>\frac {(2)^{2}}{k^{2}} = \frac {4}{7 - 3\sqrt {5}} = 7 + 3\sqrt {5}</math> so <math>a^{2} + b^{2} + c^{2} = 7^{2} + 3^{2} + 5^{2} = \boxed{083}</math>. | We want the ratio of the squares of the sides, so <math>\frac {(2)^{2}}{k^{2}} = \frac {4}{7 - 3\sqrt {5}} = 7 + 3\sqrt {5}</math> so <math>a^{2} + b^{2} + c^{2} = 7^{2} + 3^{2} + 5^{2} = \boxed{083}</math>. | ||
− | == Solution 2 in progress == | + | == Solution 2 (in progress) == |
[[Image:1998_AIME-12.png]] | [[Image:1998_AIME-12.png]] | ||
+ | |||
+ | Without loss of generality, let <math>AB=2</math>. | ||
== See also == | == See also == |
Revision as of 14:35, 6 October 2020
Problem
Let be equilateral, and and be the midpoints of and respectively. There exist points and on and respectively, with the property that is on is on and is on The ratio of the area of triangle to the area of triangle is where and are integers, and is not divisible by the square of any prime. What is ?
Solution 1
We let , , . Since and , and .
By alternate interior angles, we have and . By vertical angles, .
Thus , so .
Since is equilateral, . Solving for and using and gives and .
Using the Law of Cosines, we get
We want the ratio of the squares of the sides, so so .
Solution 2 (in progress)
Without loss of generality, let .
See also
1998 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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