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{BC}, \overline{CA},</math> and <math>\overline{AB},</math> respectively. There exist [[point]]s <math>P, Q,</math> and <math>R</math> on <math> | + | 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>? |
[[Image:1998_AIME-12.png]] | [[Image:1998_AIME-12.png]] | ||
== Solution == | == Solution == | ||
− | + | We let <math>x = EP = FQ</math>, <math>y = EQ</math>, <math>k = PQ</math>. Since <math>AE = \frac {1}{2}AB</math> and <math>AD = \frac {1}{2}AC</math>, <math>\triangle AED \sim \triangle ABC</math> and <math>ED \parallel BC</math>. | |
− | <math>EP = FQ | ||
− | <math>EQ | ||
− | <math>PQ | ||
− | Since <math> | ||
− | + | By alternate interior angles, we have <math>\angle PEQ = \angle BFQ</math> and <math>\angle EPQ = \angle FBQ</math>. By vertical angles, <math>\angle EQP = \angle FQB</math>. | |
− | + | Thus <math>\triangle EQP \sim \triangle FQB</math>, so <math>\frac {EP}{EQ} = \frac {FB}{FQ}\Longrightarrow\frac {x}{y} = \frac {1}{x}\Longrightarrow x^{2} = y</math>. | |
− | |||
− | Since <math> | + | Since <math>\triangle EDF</math> is equilateral, <math>EQ + FQ = EF = BF = 1\Longrightarrow x + y = 1</math>. Solving for <math>x</math> and <math>y</math> using <math>x^{2} = y</math> and <math>x + y = 1</math> gives <math>x = \frac {\sqrt {5} - 1}{2}</math> and <math>y = \frac {3 - \sqrt {5}}{2}</math>. |
Using the [[Law of Cosines]], we get | Using the [[Law of Cosines]], we get | ||
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:<math> = \left(\frac {\sqrt {5} - 1}{2}\right)^{2} + \left(\frac {3 - \sqrt {5}}{2}\right)^{2} - 2\left(\frac {\sqrt {5} - 1}{2}\right)\left(\frac {3 - \sqrt {5}}{2}\right)\cos{\frac {\pi}{3}}</math> | :<math> = \left(\frac {\sqrt {5} - 1}{2}\right)^{2} + \left(\frac {3 - \sqrt {5}}{2}\right)^{2} - 2\left(\frac {\sqrt {5} - 1}{2}\right)\left(\frac {3 - \sqrt {5}}{2}\right)\cos{\frac {\pi}{3}}</math> | ||
:<math> = 7 - 3\sqrt {5}</math></div> | :<math> = 7 - 3\sqrt {5}</math></div> | ||
− | We want the ratio of the squares of the sides, so <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>. |
== See also == | == See also == |
Revision as of 15:33, 24 April 2008
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

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 |