Question

Let $ABCD$ be an isosceles trapezoid with $\overline{AD} \parallel \overline{BC}$ whose angle at the longer base $\overline{AD}$ is $\frac{\pi}{3}.$ The diagonals have length $10\sqrt{21},$ and point $E$ is at distances $10\sqrt{7}$ and $30\sqrt{7}$ from vertices $A$ and $D,$ respectively. Let $F$ be the foot of the altitude from $C$ to $\overline{AD}.$ The distance $EF$ can be expressed in the form $m\sqrt{n},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n.$

Answer

Answer: 32

Concepts:trapezoid law of sines triangle inequality bounding to limit cases

Difficulty rating: 2990

Solution:

By the triangle inequality, $30\sqrt{7} = DE$ $\le DA + AE$ $= DA + 10\sqrt{7},$ so $DA \ge 20\sqrt{7}.$ On the other hand, in triangle $ACD$ the angle at $D$ is $\frac{\pi}{3}$ and $AC = 10\sqrt{21},$ so the Law of Sines gives $\begin{aligned} DA &= \frac{AC \sin\angle DCA}{\sin\frac{\pi}{3}} \\ &= \frac{10\sqrt{21}}{\sqrt{3}/2}\sin\angle DCA \\ &= 20\sqrt{7}\sin\angle DCA \\ &\le 20\sqrt{7}. \end{aligned}$

Both bounds force $DA = 20\sqrt{7},$ so $\angle DCA = 90^\circ,$ and equality in the triangle inequality means $E$ lies on line $AD$ with $A$ between $D$ and $E.$ From the right triangle, $DC = \sqrt{DA^2 - AC^2}$ $= \sqrt{2800 - 2100}$ $= 10\sqrt{7},$ and since $\angle CDF = 60^\circ,$ the foot satisfies $DF = DC\cos 60^\circ = 5\sqrt{7}.$

Points $F$ and $E$ are on line $AD$ on the same side of $D,$ so $EF = DE - DF$ $= 30\sqrt{7} - 5\sqrt{7}$ $= 25\sqrt{7},$ and $m + n = 25 + 7 = 32.$

2008 AIME I Problem 10

Answer: 32

Difficulty rating: 2990

Solution:

Problem 10 in Other Years