1998 AIME Problems/Problem 12
Contents
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
WLOG, let have side length Then, We also notice that meaning
Let Since by congruent triangles and We can now apply Law of Cosines to and
By LoC on we get
In a similar vein, using LoC on and respectively, earns
We have and Additionally, by segment addition, Solving for and from the Law of Cosines expressions and plugging them into the segment addition gets the (admittedly-ugly) equation
Since the equation is ugly, we look at what the problem is asking for us to solve. We want We see that and since from the sine area formula. Simplifying gets us wanting to find
We see in both the denominator of what we want and under a radicand in our algebraic expression, which leads us to think the calculations may not be that bad. Isolate and square to get
Isolate the radicand and square and expand to get and moving terms to one side and dividing by we get
This can be factored into From the equation we have so plugging that value into the expression we want to find, we get
Substituting into gets an expression of so .
-PureSwag
See also
1998 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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