Question

In quadrilateral $ABCD,$ $BC = 8,$ $CD = 12,$ $AD = 10,$ and $m\angle A = m\angle B = 60^\circ.$ Given that $AB = p + \sqrt{q},$ where $p$ and $q$ are positive integers, find $p + q.$

Answer

Answer: 150

Concepts:law of cosines equilateral triangle

Difficulty rating: 2450

Solution:

Extend rays $AD$ and $BC$ until they meet at $P.$ Triangle $ABP$ has $60^\circ$ angles at $A$ and $B,$ so it is equilateral: $PA = PB = AB.$ Writing $x = AB,$ we get $PD = PA - AD = x - 10$ and $PC = PB - BC = x - 8.$

The Law of Cosines in triangle $PDC,$ with $\angle P = 60^\circ$ and $DC = 12,$ gives $\begin{aligned} 144 &= (x-10)^2 + (x-8)^2 \\ &\quad {}- (x-10)(x-8) \\ &= x^2 - 18x + 84, \end{aligned}$ so $x^2 - 18x - 60 = 0$ and $x = 9 + \sqrt{81 + 60} = 9 + \sqrt{141}.$

Thus $p + q = 9 + 141 = 150.$

2005 AIME I Problem 7

Answer: 150

Difficulty rating: 2450

Solution:

Problem 7 in Other Years