Difference between revisions of "2002 AMC 12A Problems/Problem 23"
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draw(rightanglemark(A, N, D), linewidth(.5)); | draw(rightanglemark(A, N, D), linewidth(.5)); | ||
</asy> | </asy> | ||
+ | |||
+ | Draw <math>DN</math> such that <math>DN \bot AB</math>. As stated in Solution 1, <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] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] |
Revision as of 12:59, 26 August 2022
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
,
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
. As stated in Solution 1,
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.