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 <math>ABD</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 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>. | ||
+ | |||
+ | 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> | ||
− | <math>\text{( | + | 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> | ||
+ | |||
+ | "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>\frac { | + | <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]] | ||
− | {{ | + | [[Category:Area Problems]] |
+ | {{MAA Notice}} |
Latest revision as of 01:41, 5 July 2023
Contents
Problem
In triangle , side
and the perpendicular bisector of
meet in point
, and
bisects
. If
and
, what is the area of triangle
?
Solution 1
Looking at the triangle
, we see that its perpendicular bisector reaches the vertex, therefore implying it is isosceles. Let
, so that
from given and the previous deducted. Then
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
and
are similar, so
.
Then by using Heron's Formula on (with sides
), we have
.
Solution 2
Let M be the point of the perpendicular bisector on BC. By the perpendicular bisector theorem, and
. Also, by the angle bisector theorem,
. Thus, let
and
. In addition,
.
Thus, . Additionally, using the Law of Cosines and the fact that
,
Substituting and simplifying, we get
Thus, . We now know all sides of
. Using Heron's Formula on
,
"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 and
as triangle
is isosceles, so
. By the angle bisector theorem, we can express
and
as
and
respectively. We try to find
through Stewart's Theorem. So
We plug this to find that the sides of are
. By Heron's formula, the area is
. ~skyscraper
Solution 4
Draw such that
,
,
by
,
,
By the Angle Bisector Theorem,
,
,
Solution 5 (Trigonometry)
Let ,
,
,
,
By the Law of Sines we have
By the Triple-angle Identities,
,
,
,
By the Double Angle Identity
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.