Question

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 $a + b + c.$

Answer

Answer: 12

Concepts:equilateral triangle square (geometry)coordinate geometry

Difficulty rating: 2510

Solution:

Place $A = (0, 0),$ $B = (1, 0),$ $C = (1, 1),$ $D = (0, 1).$ By the symmetry of the equilateral triangle across diagonal $\overline{AC},$ we have $BE = DF.$ Let $BE = t,$ so $CE = CF = 1 - t.$ Then $AE^2 = 1 + t^2$ and $EF^2 = 2(1 - t)^2,$ and setting them equal gives $t^2 - 4t + 1 = 0,$ so $t = 2 - \sqrt{3}$ (taking the root less than $1$ ).

Thus $E = (1,\, 2 - \sqrt{3}),$ and line $AE$ is $y = (2 - \sqrt{3})x.$ If the smaller square has side $q,$ its vertex opposite $B$ is $(1 - q,\, q),$ which must lie on line $AE:$ $\begin{aligned} q &= (2 - \sqrt{3})(1 - q) \\ &\Longrightarrow q = \frac{2 - \sqrt{3}}{3 - \sqrt{3}} \\ &= \frac{(2 - \sqrt{3})(3 + \sqrt{3})}{6} \\ &= \frac{3 - \sqrt{3}}{6}. \end{aligned}$

So $a = 3,$ $b = 3,$ $c = 6,$ and $a + b + c = 12.$

2006 AIME II Problem 6

Answer: 12

Difficulty rating: 2510

Solution:

Problem 6 in Other Years