Question

A paper equilateral triangle $ABC$ has side length $12.$ The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B.$ The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n},$ where $m,$ $n,$ and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m + n + p.$

Answer

Solution:

Let $A'$ be the landing point, with $BA' = 9$ and $CA' = 3,$ and let the crease meet $\overline{AB}$ at $P$ and $\overline{AC}$ at $Q.$ Folding preserves distances, so $PA' = PA = x$ and $QA' = QA = y.$ In triangle $PBA',$ with $PB = 12 - x$ and $\angle B = 60^\circ,$ the law of cosines gives $x^2 = (12 - x)^2 + 81 - 9(12 - x),$ which simplifies to $15x = 117,$ so $x = \frac{39}{5}.$ Similarly, in triangle $QCA',$ $y^2 = (12 - y)^2 + 9 - 3(12 - y)$ gives $21y = 117,$ so $y = \frac{39}{7}.$

Finally, in triangle $APQ$ with $\angle A = 60^\circ,$ $PQ^2 = x^2 + y^2 - xy = 39^2\left(\frac{1}{25} + \frac{1}{49} - \frac{1}{35}\right) = 39^2 \cdot \frac{49 + 25 - 35}{1225} = \frac{39^3}{1225},$ so $PQ = \frac{39\sqrt{39}}{35}.$ Thus $m + n + p = 39 + 35 + 39 = 113.$

2013 AIME I Problem 9

Solution:

Problem 9 in Other Years