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
![$k^{2} = x^{2} + y^{2} - 2xy\cos{\frac {\pi}{3}}$](http://latex.artofproblemsolving.com/2/a/3/2a37f32a1429651e899001af78ae6a6ceda7514c.png)
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
.
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 | ||
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