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

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==Problem==
 
==Problem==
In [[triangle]] <math>ABC</math> , side <math>AC</math> and the [[perpendicular bisector]] of <math>BC</math> meet in point <math>D</math>, and <math>BD</math> bisects <math>\angle ABC</math>. If <math>AD=9</math> and <math>DC=7</math>, what is the [[area]] of triangle ABD?  
+
In triangle <math>ABC</math>, side <math>AC</math> and the [[perpendicular bisector]] of <math>BC</math> meet in point <math>D</math>, and <math>BD</math> bisects <math>\angle ABC</math>. If <math>AD=9</math> and <math>DC=7</math>, what is the area of triangle ABD?  
  
 
<math>\text{(A)}\ 14 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 14\sqrt5 \qquad \text{(E)}\ 28\sqrt5</math>
 
<math>\text{(A)}\ 14 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 14\sqrt5 \qquad \text{(E)}\ 28\sqrt5</math>
  
 
==Solution==
 
==Solution==
 +
'''Solution 1'''
 
<asy>
 
<asy>
draw((0,0)--(10,0),black);
+
unitsize(0.25 cm);
draw((10,0)--(10.5,5)--(0,0),black);
+
pair A, B, C, D, M;
draw((10,0)--(5,2.4),black);
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A = (0,0);
draw((5,2.4)--(5,0),black);
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B = (88/9, 28*sqrt(5)/9);
draw((4.6,0)--(4.6,0.4)--(5.4,0.4)--(5.4,0),red);
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C = (16,0);
label("A",(10.5,5),N);
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D = 9/16*C;
label("D",(5,2.4),N);
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M = (B + C)/2;
label("C",(0,0),W);
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draw(A--B--C--cycle);
label("B",(10,0),E);
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draw(B--D--M);
label("9",(7.5,3.7),N);
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label("$A$", A, SW);
label("7",(2.5,1.6),N);
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label("$B$", B, N);
 +
label("$C$", C, SE);
 +
label("$D$", D, S);
 
</asy>
 
</asy>
 +
Looking at the triangle <math>BCD</math>, we see that its perpendicular bisector reaches the vertex, therefore implying it is isosceles. Let <math>x = \angle C</math>, so that <math>B=2x</math> from given and the previous deducted. Then <math>\angle ABD=x, \angle ADB=2x</math> 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 <math> \triangle ABD</math> and <math>\triangle ACB</math> are [[Similarity (geometry)|similar]], so <math>\frac {16}{AB}=\frac {AB}{9} \Longrightarrow AB=12</math>.
  
Looking at the triangle <math>BCD</math>, we see that its perpendicular bisector reaches the vertex, therefore hinting it is isoceles. Let <math>x = \angle C</math>, so that <math>B=2x</math> from given and the previous deducted. Then <math>\angle ABD=x, \angle ADB=2x</math> 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 <math> \triangle ABD</math> and <math>\triangle ACB</math> are [[Similar (geometry)|similar]], so <math>\frac {16}{AB}=\frac {AB}{9} \Longrightarrow AB=12</math>.
+
Then by using [[Heron's Formula]] on <math>ABD</math> (with sides <math>12,7,9</math>), we have <math>[\triangle ABD]= \sqrt{14(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}</math>.
  
Then by using Heron's Formula on <math>ABD</math> (with sides <math>12,7,9</math>), we have <math>[\triangle ABD]= \sqrt{14(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}</math>.
+
'''Solution 2'''
 +
 
 +
Let M be the point of the perpendicular bisector on BC. By the perpendicular bisector theorem, <math>BD = DC = 7</math> and <math>BM = MC</math>. Also, by the angle bisector theorem, <math>\frac {AB}{BC} = \frac{9}{7}</math>. Thus, let <math>AB = 9x</math> and <math>BC = 7x</math>. In addition, <math>BM = 3.5x</math>.
 +
 
 +
Thus, <math>\cos\angle CBD = \frac {3.5x}{7} = \frac {x}{2}</math>. Additionally, using the Law of Cosines and the fact that <math>\angle CBD = \angle ABD</math>, <math>81 = 49 + 81x^2 - 2(9x)(7)\cos\angle CBD</math>
 +
 
 +
Substituting and simplifying, we get <math>x = 4/3</math>
 +
 
 +
Thus, <math>AB = 12</math>. We now know all sides of <math> \triangle ABD</math>. Using [[Heron's Formula]] on <math>\triangle ABD</math>, <math>\sqrt{(14)(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}</math>
 +
 
 +
'''Solution 3'''
 +
 
 +
Note that because the perpendicular bisector and angle bisector meet at side <math>AC</math> and <math>CD = BD</math> as triangle <math>BDC</math> is isosceles, so <math>BD = 7</math>. By the angle bisector theorem, we can express <math>AB</math> and <math>BC</math> as <math>9x</math> and <math>7x</math> respectively. We try to find <math>x</math> through Stewart's Theorem. So
 +
 
 +
<math>16(7^2+9\cdot7) = (7x)^2 \cdot 9 + (9x)^2 \cdot 7</math>
 +
 
 +
<math>16(49+63) = (49 \cdot 9 + 81 \cdot 7)x^2</math>
 +
 
 +
<math>16(49+63) = 9(49+63) \cdot x^2</math>
 +
 
 +
<math>x^2 = \frac{16}{9}</math>
 +
 
 +
<math>x=\frac{4}{3}</math>
 +
 
 +
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
  
 
==See Also==
 
==See Also==
Line 27: Line 55:
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
[[Category:Similar Triangles Problems]]
+
 
 
[[Category:Area Problems]]
 
[[Category:Area Problems]]
 +
{{MAA Notice}}

Revision as of 22:34, 18 December 2020

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

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

See Also

2002 AMC 12A (ProblemsAnswer KeyResources)
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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