Question

$ABCD$ is a square of side length $\sqrt{3} + 1.$ Point $P$ is on $\overline{AC}$ such that $AP = \sqrt{2}.$ The square region bounded by $ABCD$ is rotated $90^\circ$ counterclockwise with center $P,$ sweeping out a region whose area is $\dfrac{1}{c}(a\pi + b),$ where $a, b,$ and $c$ are positive integers and $\gcd(a, b, c) = 1.$ What is $a + b + c?$

$15$

$17$

$19$

$21$

$23$

Answer: C

Concepts:transformation sector area decomposition

Difficulty rating: 2520

Solution:

Let $A', B', C', D'$ be the images of the vertices under the rotation. The swept region decomposes into four circular sectors and four triangles.

Since $AP = \sqrt{2}$ and $PC = AC - AP = \sqrt{6},$ the sectors at $A$ and $C$ have areas $\tfrac{\pi}{2}$ and $\tfrac{3\pi}{2}.$ If $H$ is the midpoint of $AA',$ then $PH=AH=1$ and $HB=\sqrt3,$ so $\triangle BPH$ is a $30$ - $60$ - $90^\circ$ triangle and $PB=2.$ Hence the two $60^\circ$ sectors along $BC$ each have area $2\pi/3.$ The two triangles with altitude $PH$ contribute $\sqrt3-1,$ and the other congruent pair has altitude $\sqrt3$ and contributes $3-\sqrt3.$ Thus the four triangles contribute $2.$

The total area is $\begin{gathered} \dfrac{\pi}{2} + \dfrac{3\pi}{2} + 2\cdot\dfrac{2\pi}{3} + 2 \\ = \dfrac{10\pi + 6}{3}, \end{gathered}$ so $a + b + c = 10 + 6 + 3 = 19.$

Thus, the correct answer is C.

2013 AMC 12A Problem 23

Answer: C

Difficulty rating: 2520

Solution:

Problem 23 in Other Years