2002 AMC 12A Problems/Problem 23

Problem

In triangle $ABC$ , side $AC$ and the perpendicular bisector of $BC$ meet in point $D$ , and $BD$ bisects $\angle ABC$ . If $AD=9$ and $DC=7$ , what is the area of triangle $ABD$ ?

$\text{(A)}\ 14 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 14\sqrt5 \qquad \text{(E)}\ 28\sqrt5$

Solution 1

$[asy] unitsize(0.25 cm); pair A, B, C, D, M; A = (0,0); B = (88/9, 28*sqrt(5)/9); C = (16,0); D = 9/16*C; M = (B + C)/2; draw(A--B--C--cycle); draw(B--D--M); label("$A$", A, SW); label("$B$", B, N); label("$C$", C, SE); label("$D$", D, S); [/asy]$ Looking at the triangle $BCD$ , we see that its perpendicular bisector reaches the vertex, therefore implying it is isosceles. Let $x = \angle C$ , so that $B=2x$ from given and the previous deducted. Then $\angle ABD=x, \angle ADB=2x$ because any exterior angle of a triangle has a measure that is the sum of the two interior angles that are not adjacent to the exterior angle. That means $\triangle ABD$ and $\triangle ACB$ are similar, so $\frac {16}{AB}=\frac {AB}{9} \Longrightarrow AB=12$ .

Then by using Heron's Formula on $ABD$ (with sides $12,7,9$ ), we have $[\triangle ABD]= \sqrt{14(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}$ .

Solution 3

Let M be the point of the perpendicular bisector on BC. By the perpendicular bisector theorem, $BD = DC = 7$ and $BM = MC$ . Also, by the angle bisector theorem, $\frac {AB}{BC} = \frac{9}{7}$ . Thus, let $AB = 9x$ and $BC = 7x$ . In addition, $BM = 3.5x$ .

Thus, $\cos\angle CBD = \frac {3.5x}{7} = \frac {x}{2}$ . Additionally, using the Law of Cosines and the fact that $\angle CBD = \angle ABD$ , $81 = 49 + 81x^2 - 2(9x)(7)\cos\angle CBD$

Substituting and simplifying, we get $x = 4/3$

Thus, $AB = 12$ . We now know all sides of $\triangle ABD$ . Using Heron's Formula on $\triangle ABD$ , $\sqrt{(14)(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}$

Solution 4

Note that because the perpendicular bisector and angle bisector meet at side $AC$ and $CD = BD$ as triangle $BDC$ is isosceles, so $BD = 7$ . By the angle bisector theorem, we can express $AB$ and $BC$ as $9x$ and $7x$ respectively. We try to find $x$ through Stewart's Theorem. So

$16(7^2+9\cdot7) = (7x)^2 \cdot 9 + (9x)^2 \cdot 7$

$16(49+63) = (49 \cdot 9 + 81 \cdot 7)x^2$

$16(49+63) = 9(49+63) \cdot x^2$

$x^2 = \frac{16}{9}$

$x=\frac{4}{3}$

We plug this to find that the sides of $\triangle ABD$ are $12,7,9$ . By Heron's formula, the area is $\sqrt{(14)(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}$ . ~skyscraper

Solution 5 (Trigonometry)

Let $\angle ACB = \theta$ , $\angle DBC = \theta$ , $\angle ABD = \theta$ , $\angle ADB = 2 \theta$ , $\angle BAC = 180^\circ - 3 \theta$

$[ABD] = \frac12 \cdot AD \cdot BD \cdot \sin \angle ADB = \frac12 \cdot 9 \cdot 7 \cdot \sin 2\theta$

By the Law of Sines we have $\frac{ \sin(180^\circ - 3 \theta) }{7} = \frac{ \sin \theta }{ 9 }$

$\frac{ 3 \theta }{7} = \frac{ \sin \theta }{ 9 }$

By the Triple-angle Identities, $\sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta$

$3 - 4 \sin^2 \theta = \frac79$ , $36 \sin^2 \theta = 20$ , $\sin^2 \theta = \frac59$

$\sin \theta = \frac{\sqrt{5}}{3}$ , $\cos \theta = \frac{2}{3}$

By the Double Angle Identity $\sin 2 \theta = 2 \sin \theta \cos \theta = 2 \cdot \frac{\sqrt{5}}{3} \cdot \frac{2}{3} = \frac{ 4 \sqrt{5} }{9}$

$[ABD] = \frac12 \cdot 9 \cdot 7 \cdot \frac{ 4 \sqrt{5} }{9} = \boxed{\textbf{(D) } 14 \sqrt{5} }$

~isabelchen

2002 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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