Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution 2)
 
(36 intermediate revisions by 20 users not shown)
Line 1: Line 1:
 
==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 <math>ABD</math>?
  
−
In triangle <math>ABC</math> , side <math>AC</math> and the perpendicular bisector of <math>BC</math> meet in point <math>D</math>, and  bisects <math><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>A) 14</math> <math>B) 21</math> <math>C)28</math> <math>D)14\sqrt5</math> <math>E)28\sqrt5</math>
+
==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 <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>.
  
−
==Solution==
+
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>.
  
−
This problems needs a picture. You can help by adding it.
+
==Solution 2==
  
−
Looking at the triangle BCD, we see that its perpendicular bisector reaches the vertex, therefore hinting it is isoceles. Let angle C be x. B=2x from given and the previous deducted. <ABD=x, <ADB=2x (outer angle). That means ABD and ACB are similar.
+
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>.  
  
−
<math>\frac {16}{AB}=\frac {AB}{9}</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>
−
<math>AB=12</math>
 
  
−
Then by using Heron's Formula on ABD (12,7,9 as sides), we have
+
Substituting and simplifying, we get <math>x = 4/3</math>
−
<math>\sqrt{14(2)(7)(5)}</math>
+
 
−
<math>14\sqrt5=E</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>
 +
 
 +
"Note:-you could also drop a perpendicular from D to AB at point let say,F then BF = 3.5x by pyathgoras theorem we can find DF and (AB ×DF )÷2 is our answer"
 +
 
 +
==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
 +
 
 +
==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 <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>\frac{AB}{AD} = \frac{AC}{AB}</math>, <math>AB^2 = AD \cdot AC = 9 \cdot 16</math>, <math>AB = 12</math>
 +
 
 +
By the [[Angle Bisector Theorem]], <math>\frac{BC}{AB} = \frac{CD}{AD}</math>
 +
 
 +
<math>\frac{BC}{12} = \frac{7}{9}</math>
 +
 
 +
<math>BC = \frac{28}{3}</math>, <math>CM = \frac{14}{3}</math>, <math>DN = DM = \sqrt{CD^2 - CM^2} = \frac{7 \sqrt{5} }{3}</math>
 +
 
 +
<math>[ABD] = \frac12 \cdot AB \cdot DN = \frac12 \cdot 12 \cdot \frac{7 \sqrt{5} }{3} = \boxed{\textbf{(D) } 14 \sqrt{5} }</math>
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 +
 
 +
==Solution 5 (Trigonometry) ==
 +
 
 +
Let <math>\angle ACB = \theta</math>, <math>\angle DBC = \theta</math>, <math>\angle ABD = \theta</math>, <math>\angle ADB = 2 \theta</math>, <math>\angle BAC = 180^\circ - 3 \theta</math>
 +
 
 +
<math>[ABD] = \frac12 \cdot AD \cdot BD \cdot \sin \angle ADB = \frac12 \cdot 9 \cdot 7 \cdot \sin 2\theta</math>
 +
 
 +
By the [[Law of Sines]] we have <math>\frac{ \sin \angle BAC }{BD} = \frac{ \sin \angle ABD }{AD}</math>
 +
 
 +
<math>\frac{ \sin(180^\circ - 3 \theta) }{7} = \frac{ \sin \theta }{ 9 }</math>
 +
 
 +
<math>\frac{ \sin 3 \theta }{7} = \frac{ \sin \theta }{ 9 }</math>
 +
 
 +
By the [https://artofproblemsolving.com/wiki/index.php/Trigonometric_identities#Triple-angle_identities Triple-angle Identities], <math> \sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta</math>
 +
 
 +
<math>3 - 4 \sin^2 \theta = \frac79</math>, <math>36 \sin^2 \theta = 20</math>, <math>\sin^2 \theta = \frac59</math>
 +
 
 +
<math>\sin \theta = \frac{\sqrt{5}}{3}</math>, <math>\cos \theta = \frac{2}{3}</math>
 +
 
 +
By the [https://artofproblemsolving.com/wiki/index.php/Trigonometric_identities#Double-angle_identities Double Angle Identity] <math>\sin 2 \theta = 2 \sin \theta \cos \theta = 2 \cdot \frac{\sqrt{5}}{3} \cdot \frac{2}{3} = \frac{ 4 \sqrt{5} }{9}</math>
 +
 
 +
<math>[ABD] = \frac12 \cdot 9 \cdot 7 \cdot \frac{ 4 \sqrt{5} }{9} = \boxed{\textbf{(D) } 14 \sqrt{5} }</math>
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
 
==See Also==
 
==See Also==
 +
{{AMC12 box|year=2002|ab=A|num-b=22|num-a=24}}
 +
 +
[[Category:Introductory Geometry Problems]]
  
−
{{AMC12 box|year=2002|ab=A|num-b=24|after=Last<br>Problem}}
+
[[Category:Area Problems]]
 +
{{MAA Notice}}

Latest revision as of 00:41, 5 July 2023

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 $\triangle ABD$, $\sqrt{(14)(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}$

"Note:-you could also drop a perpendicular from D to AB at point let say,F then BF = 3.5x by pyathgoras theorem we can find DF and (AB ×DF )÷2 is our answer"

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_23&oldid=195282"