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>\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 == | ||
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+ | [[Image:1998_AIME-12.png]] | ||
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+ | 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>. | ||
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+ | 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>. | ||
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+ | 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 | ||
+ | <div style="text-align:center;"><math>k^{2} = x^{2} + y^{2} - 2xy\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> | ||
+ | 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 == | ||
− | + | {{AIME box|year=1998|num-b=11|num-a=13}} | |
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 15:04, 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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.