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Difference between revisions of "2006 AIME I Problems/Problem 6"

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<math>\angle EAF</math> is <math>60</math> degrees, so <math>\angle BAE = \frac{90 - 60}{2} = 15</math>. Thus, <math>\cos 15 = \cos (45 - 30) = \frac{\sqrt{6} + \sqrt{2}}{4} = \frac{1}{AE}</math>, so <math>AE = \frac{4}{\sqrt{6} + \sqrt{2}} \cdot \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}} = \sqrt{6} - \sqrt{2}</math>. Since <math>\triangle AEF</math> is [[equilateral]], <math>EF = AE = \sqrt{6} - \sqrt{2}</math>. <math>\triangle CEF</math> is a <math>45-45-90 \triangle</math>, so <math>CE = \frac{AE}{\sqrt{2}} = \sqrt{3} - 1</math>. Substituting back into the equation from the beginning, we get <math>s^2 = (1 - s)(2 - \sqrt{3} - s)</math>, so <math>(3 - \sqrt{3})s = 2 - \sqrt{3}</math>. Therefore, <math>s = \frac{2 - \sqrt{3}}{3 - \sqrt{3}} \cdot \frac{3 + \sqrt{3}}{3 + \sqrt{3}} = \frac{3 - \sqrt{3}}{6}</math>, and <math>a + b + c = 3 + 3 + 6 = 012</math>.
 
<math>\angle EAF</math> is <math>60</math> degrees, so <math>\angle BAE = \frac{90 - 60}{2} = 15</math>. Thus, <math>\cos 15 = \cos (45 - 30) = \frac{\sqrt{6} + \sqrt{2}}{4} = \frac{1}{AE}</math>, so <math>AE = \frac{4}{\sqrt{6} + \sqrt{2}} \cdot \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}} = \sqrt{6} - \sqrt{2}</math>. Since <math>\triangle AEF</math> is [[equilateral]], <math>EF = AE = \sqrt{6} - \sqrt{2}</math>. <math>\triangle CEF</math> is a <math>45-45-90 \triangle</math>, so <math>CE = \frac{AE}{\sqrt{2}} = \sqrt{3} - 1</math>. Substituting back into the equation from the beginning, we get <math>s^2 = (1 - s)(2 - \sqrt{3} - s)</math>, so <math>(3 - \sqrt{3})s = 2 - \sqrt{3}</math>. Therefore, <math>s = \frac{2 - \sqrt{3}}{3 - \sqrt{3}} \cdot \frac{3 + \sqrt{3}}{3 + \sqrt{3}} = \frac{3 - \sqrt{3}}{6}</math>, and <math>a + b + c = 3 + 3 + 6 = 012</math>.
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Here's an alternative geometric way to calculate <math>CE</math> (as opposed to [[trigonometry|trigonometric]]): The diagonal <math>\overline{AC}</math> is made of the [[altitude]] of the equilateral triangle and the altitude of the <math>45-45-90 \triangle</math>. The former is <math>\frac{CE\sqrt{3}}{2}</math>, and the latter is <math>\frac{CE}{2}</math>; thus <math>\frac{CE\sqrt{3} + CE}{2} = AC = \sqrt{2} \Longrightarrow CE = \sqrt{6}-\sqrt{2}</math>. The solution continues as above.
  
 
== See also ==
 
== See also ==

Revision as of 16:34, 25 September 2007

Problem

Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is equilateral. A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}.$ The length of a side of this smaller square is $\frac{a-\sqrt{b}}{c},$ where $a, b,$ and $c$ are positive integers and $b$ is not divisible by the square of any prime. Find $a+b+c.$

Solution

2006 I AIME-6.png

Call the vertices of the new square A', B', C', and D', in relation to the vertices of $ABCD$, and define $s$ to be one of the sides of that square. Since the sides are parallel, by corresponding angles and AA~ we know that triangles $AA'D'$ and $D'C'E$ are similar. Thus, the sides are proportional: $\frac{AA'}{A'D'} = \frac{D'C'}{C'E} \Longrightarrow \frac{1 - s}{s} = \frac{s}{1 - s - CE}$. Simplifying, we get that $s^2 = (1 - s)(1 - s - CE)$.

$\angle EAF$ is $60$ degrees, so $\angle BAE = \frac{90 - 60}{2} = 15$. Thus, $\cos 15 = \cos (45 - 30) = \frac{\sqrt{6} + \sqrt{2}}{4} = \frac{1}{AE}$, so $AE = \frac{4}{\sqrt{6} + \sqrt{2}} \cdot \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}} = \sqrt{6} - \sqrt{2}$. Since $\triangle AEF$ is equilateral, $EF = AE = \sqrt{6} - \sqrt{2}$. $\triangle CEF$ is a $45-45-90 \triangle$, so $CE = \frac{AE}{\sqrt{2}} = \sqrt{3} - 1$. Substituting back into the equation from the beginning, we get $s^2 = (1 - s)(2 - \sqrt{3} - s)$, so $(3 - \sqrt{3})s = 2 - \sqrt{3}$. Therefore, $s = \frac{2 - \sqrt{3}}{3 - \sqrt{3}} \cdot \frac{3 + \sqrt{3}}{3 + \sqrt{3}} = \frac{3 - \sqrt{3}}{6}$, and $a + b + c = 3 + 3 + 6 = 012$.

Here's an alternative geometric way to calculate $CE$ (as opposed to trigonometric): The diagonal $\overline{AC}$ is made of the altitude of the equilateral triangle and the altitude of the $45-45-90 \triangle$. The former is $\frac{CE\sqrt{3}}{2}$, and the latter is $\frac{CE}{2}$; thus $\frac{CE\sqrt{3} + CE}{2} = AC = \sqrt{2} \Longrightarrow CE = \sqrt{6}-\sqrt{2}$. The solution continues as above.

See also

2006 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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