Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1998 AIME Problems/Problem 12"
(→‎Problem)
Line 1: Line 1:
 
== 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>?
+
Let <math>ABC</math> be [[equilateral triangle|equilateral]], and <math>D, E,</math> and <math>F</math> be the [[midpoint]]s of <math>\overline{AC}, \overline{AB},</math> and <math>\overline{BC},</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 22:59, 31 October 2019

Problem

Let $ABC$ be equilateral, and $D, E,$ and $F$ be the midpoints of $\overline{AC}, \overline{AB},$ and $\overline{BC},$ respectively. There exist points $P, Q,$ and $R$ on $\overline{DE}, \overline{EF},$ and $\overline{FD},$ respectively, with the property that $P$ is on $\overline{CQ}, Q$ is on $\overline{AR},$ and $R$ is on $\overline{BP}.$ The ratio of the area of triangle $ABC$ to the area of triangle $PQR$ is $a + b\sqrt {c},$ where $a, b$ and $c$ are integers, and $c$ is not divisible by the square of any prime. What is $a^{2} + b^{2} + c^{2}$?

Solution

1998 AIME-12.png

We let $x = EP = FQ$, $y = EQ$, $k = PQ$. Since $AE = \frac {1}{2}AB$ and $AD = \frac {1}{2}AC$, $\triangle AED \sim \triangle ABC$ and $ED \parallel BC$.

By alternate interior angles, we have $\angle PEQ = \angle BFQ$ and $\angle EPQ = \angle FBQ$. By vertical angles, $\angle EQP = \angle FQB$.

Thus $\triangle EQP \sim \triangle FQB$, so $\frac {EP}{EQ} = \frac {FB}{FQ}\Longrightarrow\frac {x}{y} = \frac {1}{x}\Longrightarrow x^{2} = y$.

Since $\triangle EDF$ is equilateral, $EQ + FQ = EF = BF = 1\Longrightarrow x + y = 1$. Solving for $x$ and $y$ using $x^{2} = y$ and $x + y = 1$ gives $x = \frac {\sqrt {5} - 1}{2}$ and $y = \frac {3 - \sqrt {5}}{2}$.

Using the Law of Cosines, we get

$k^{2} = x^{2} + y^{2} - 2xy\cos{\frac {\pi}{3}}$
$= \left(\frac {\sqrt {5} - 1}{2}\right)^{2} + \left(\frac {3 - \sqrt {5}}{2}\right)^{2} - 2\left(\frac {\sqrt {5} - 1}{2}\right)\left(\frac {3 - \sqrt {5}}{2}\right)\cos{\frac {\pi}{3}}$
$= 7 - 3\sqrt {5}$

We want the ratio of the squares of the sides, so $\frac {(2)^{2}}{k^{2}} = \frac {4}{7 - 3\sqrt {5}} = 7 + 3\sqrt {5}$ so $a^{2} + b^{2} + c^{2} = 7^{2} + 3^{2} + 5^{2} = \boxed{083}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1998_AIME_Problems/Problem_12&oldid=110745"