Difference between revisions of "2012 AMC 12A Problems/Problem 18"
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-Solution by '''arowaaron''' | -Solution by '''arowaaron''' | ||
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+ | ==Solution 3 == | ||
+ | |||
+ | By heron's formula, | ||
+ | <math>sr=\sqrt{s(s-a)(s-b)(s-c)}</math> so we have <math>r^2=\frac{14\cdot 13 \cdot 12}{39}</math>. Hence <math>r=\sqrt{56}</math>. | ||
+ | |||
+ | By the law of cosines on <math>\triangle ABC</math>, | ||
+ | <math>\cos B = \frac{25^2+27^2-26^2}{2\cdot25\cdot27}=\frac{113}{225}</math> | ||
+ | Also <math>\sin^2 \frac{B}{2}=\frac{1-\cosB}{2}</math>. So, <math>\sin \frac{B}{2}=\frac{2\sqrt{14}}{\sqrt{15}}</math>. | ||
+ | |||
+ | The area of <math>\triangle BIC</math> can be calculated in 2 ways, | ||
+ | <math>\frac{1}{2} \cdot 25 r=\frac{1}{2}BI \cdot 25 \cdot \sin \frac{B}{2}</math>. Solving this equation yields <math>BI=15</math> | ||
== See Also == | == See Also == |
Revision as of 22:00, 11 February 2018
Contents
[hide]Problem
Triangle has
,
, and
. Let
denote the intersection of the internal angle bisectors of
. What is
?
Solution 1
Inscribe circle of radius
inside triangle
so that it meets
at
,
at
, and
at
. Note that angle bisectors of triangle
are concurrent at the center
(also
) of circle
. Let
,
and
. Note that
,
and
. Hence
,
, and
. Subtracting the last 2 equations we have
and adding this to the first equation we have
.
By Heron's formula for the area of a triangle we have that the area of triangle is
. On the other hand the area is given by
. Then
so that
.
Since the radius of circle is perpendicular to
at
, we have by the pythagorean theorem
so that
.
Solution 2
We can use mass points and Stewart's to solve this problem. Because we are looking at the Incenter we then label with a mass of
,
with
, and
with
. We also label where the angle bisectors intersect the opposite side
,
, and
correspondingly. It follows then that point
has mass 52. Which means that
is split into a
ratio. We can then use Stewart's to find
. So we have
. Solving we get
. Plugging it in we get
. Therefore the answer is
-Solution by arowaaron
Solution 3
By heron's formula,
so we have
. Hence
.
By the law of cosines on ,
Also $\sin^2 \frac{B}{2}=\frac{1-\cosB}{2}$ (Error compiling LaTeX. Unknown error_msg). So,
.
The area of can be calculated in 2 ways,
. Solving this equation yields
See Also
2012 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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.