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
![$k^{2} = x^{2} + y^{2} - 2xy\cos{\frac {\pi}{3}}$](http://latex.artofproblemsolving.com/2/a/3/2a37f32a1429651e899001af78ae6a6ceda7514c.png)
We want the ratio of the squares of the sides, so so
.
Solution 2 (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.