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Difference between revisions of "2002 AMC 12A Problems/Problem 23"

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Draw <math>DN</math> such that <math>DN \bot AB</math>, <math>\triangle BND \cong \triangle BMD</math>. As stated in Solution 1, <math>AB = 12</math>
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Draw <math>DN</math> such that <math>DN \bot AB</math>, <math>\triangle BND \cong \triangle BMD</math>.
  
 
<math>\angle ACB = \angle DBC = \angle ABD</math>, <math>\triangle ABD \sim \triangle ACB</math> by <math>AA</math>
 
<math>\angle ACB = \angle DBC = \angle ABD</math>, <math>\triangle ABD \sim \triangle ACB</math> by <math>AA</math>

Revision as of 12:08, 26 August 2022

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 2

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 3

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 4

[asy] size(12cm, 12cm); pair A, B, C, D, M, N; A = (0,0); B = (88/9, 28*sqrt(5)/9); C = (16,0); D = 9/16*C; M = (B + C)/2; N = (6,4.27); draw(A--B--C--cycle); draw(B--D--M); draw(D--N--B); label("$A$", A, SW); label("$B$", B, N); label("$C$", C, SE); label("$D$", D, S); label("$M$", M, NE); label("$N$", N, NW); draw(rightanglemark(B, M, D), linewidth(.5)); draw(rightanglemark(A, N, D), linewidth(.5)); [/asy]

Draw $DN$ such that $DN \bot AB$, $\triangle BND \cong \triangle BMD$.

$\angle ACB = \angle DBC = \angle ABD$, $\triangle ABD \sim \triangle ACB$ by $AA$

$\frac{AB}{AD} = \frac{AC}{AB}$, $AB^2 = AD \cdot AC = 9 \cdot 16$, $AB = 12$

By the Angle Bisector Theorem, $\frac{BC}{AB} = \frac{CD}{AD}$

$\frac{BC}{12} = \frac{7}{9}$

$BC = \frac{28}{3}$, $CM = \frac{14}{3}$, $DN = DM = \sqrt{CD^2 - CM^2} = \frac{7 \sqrt{5} }{3}$

$[ABD] = \frac12 \cdot AB \cdot DN = \frac12 \cdot 12 \cdot \frac{7 \sqrt{5} }{3} = \boxed{\textbf{(D) } 14 \sqrt{5} }$

~isabelchen

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 $\frac{ \sin \angle BAC }{BD} = \frac{ \sin \angle ABD }{AD}$

$\frac{ \sin(180^\circ - 3 \theta) }{7} = \frac{ \sin \theta }{ 9 }$

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

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

See Also

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

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