2006 AIME II Problems/Problem 6

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 1

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 $AE$ (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 $\frac{AE\sqrt{3}}{2}$ , and the latter is $\frac{AE}{2}$ ; thus $\frac{AE\sqrt{3} + AE}{2} = AC = \sqrt{2} \Longrightarrow AE= \sqrt{6}-\sqrt{2}$ . The solution continues as above.

Solution 2

Since $\triangle AFE$ is equilateral, $\overline{AE} = \overline{AF}$ . It follows that $\overline{FC} = \overline{EC}$ . Let $\overline{FC} = x$ . Then, $\overline{EF} = x\sqrt{2}$ and $\overline{DF} = 1-x$ .

$\overline{AF} = \sqrt{1+(1-x)^2} = x\sqrt{2}$ .

Square both sides and combine/move terms to get $x^2+2x-2 = 0$ . Therefore $x = -1 + \sqrt{3}$ and $x = -1 - \sqrt{3}$ . The second solution is obviously extraneous, so $x = -1 + \sqrt{3}$ .

Now, consider the square ABCD to be on the Cartesian Coordinate Plane with $A = (0,0)$ . Then, the line containing $\overline{AF}$ has slope $\frac{1}{2-\sqrt{3}}$ and equation $y = \frac{1}{2-\sqrt{3}}x$ .

The distance from $\overline{DC}$ to $\overline{AF}$ is the distance from $y = 1$ to $y = \frac{1}{2-\sqrt{3}}x$ .

Similarly, the distance from $\overline{AD}$ to $\overline{AF}$ is the distance from $x = 0$ to $y = \frac{1}{2-\sqrt{3}}x$ .

For some value $x = s$ , these two distances are equal.

$(s-0) = (1 - (\frac{1}{2-\sqrt{3}})s)$

Solving for s, $s = \frac{3 - \sqrt{3}}{6}$ , and $a + b + c = 3 + 3 + 6 = 012$ .

2006 AIME II (Problems • Answer Key • Resources)
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2006 AIME II Problems/Problem 6

Contents

Problem

Solution 1

Solution 2

See also