Difference between revisions of "1998 AIME Problems/Problem 12"
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== Problem == | == Problem == | ||
− | Let <math>ABC</math> be [[equilateral triangle|equilateral]], and <math>D, E,</math> and <math>F</math> be the [[midpoint]]s of <math>\overline{ | + | Let <math>ABC</math> be [[equilateral triangle|equilateral]], and <math>D, E,</math> and <math>F</math> be the [[midpoint]]s of <math>\overline{BC}, \overline{CA},</math> and <math>\overline{AB},</math> respectively. There exist [[point]]s <math>P, Q,</math> and <math>R</math> on <math>\overline{DE}, \overline{EF},</math> and <math>\overline{FD},</math> respectively, with the property that <math>P</math> is on <math>\overline{CQ}, Q</math> is on <math>\overline{AR},</math> and <math>R</math> is on <math>\overline{BP}.</math> The [[ratio]] of the area of triangle <math>ABC</math> to the area of triangle <math>PQR</math> is <math>a + b\sqrt {c},</math> where <math>a, b</math> and <math>c</math> are integers, and <math>c</math> is not divisible by the square of any [[prime]]. What is <math>a^{2} + b^{2} + c^{2}</math>? |
== Solution == | == Solution == |
Revision as of 23:02, 31 October 2019
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
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
.
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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