AoPSWiki

Visit the AoPS Book Store.

Personal tools

Search

Toolbox

2008 AMC 12B Problems/Problem 25

From AoPSWiki

Problem 25

Let $ABCD$ be a trapezoid with $AB||CD, AB=11, BC=5, CD=19,$ and $DA=7$ . Bisectors of $\angle A$ and $\angle D$ meet at $P$ , and bisectors of $\angle B$ and $\angle C$ meet at $Q$ . What is the area of hexagon $ABQCDP$ ?

$\textbf{(A)}\ 28\sqrt{3}\qquad \textbf{(B)}\ 30\sqrt{3}\qquad \textbf{(C)}\ 32\sqrt{3}\qquad \textbf{(D)}\ 35\sqrt{3}\qquad \...$

Solution

Drop perpendiculars to $CD$ from $A$ and $B$ , and call the intersections $X,Y$ respectively. Now, $DA^2-BC^2=(7-5)(7+5)=DX^2-CY^2$ and $DX+CY=19-11=8$ . Thus, $DX-CY=3$ . We conclude $DX=\frac{11}{2}$ and $CY=\frac{5}{2}$ . To simplify things even more, notice that $90^{\circ}=\frac{\angle D+\angle A}{2}=180^{\circ}-\angle APD$ , so $\angle P=\angle Q=90^{\circ}$ .

Also, $\sin(\angle PDA)=\sin(\frac12\angle XDA)=\sqrt{\frac{1-\cos(\angle XDA)}{2}}=\sqrt{\frac{3}{28}}$ So the area of $\triangle APD$ is: $R\cdot c\sin a\sin b =\frac{7\cdot7}{2}\sqrt{\frac{3}{28}}\sqrt{1-\frac{3}{28}}=\frac{35}{8}\sqrt{3}$

Over to the other side: $\triangle BCY$ is $30-60-90$ , and is therefore congruent to $\triangle BCQ$ . So $[BCQ]=\frac{5\cdot5\sqrt{3}}{8}$ .

The area of the hexagon is clearly $[ABCD]-([BCQ]+[APD])$ $=\frac{15\cdot5\sqrt{3}}{2}-\frac{60\sqrt{3}}{8}=30\sqrt{3},\qquad\boxed{B}$

See Also

2008 AMC 12B (Problems • Resources)
Preceded by Problem 24	Followed by Last Question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2008_AMC_12B_Problems/Problem_25"

Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.

© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us