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Difference between revisions of "2002 AMC 12A Problems/Problem 23"
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We plug this to find that the sides of <math>\triangle ABD</math> are <math>12,7,9</math>. By Heron's formula, the area is  <math>\sqrt{(14)(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}</math>. ~skyscraper
 
We plug this to find that the sides of <math>\triangle ABD</math> are <math>12,7,9</math>. By Heron's formula, the area is  <math>\sqrt{(14)(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}</math>. ~skyscraper
 +
 +
==Solution 5==
 +
 +
<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>
 +
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
 
==See Also==
 
==See Also==

Revision as of 07:41, 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 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

[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]

~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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