Difference between revisions of "2018 AMC 12B Problems/Problem 21"
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Finally, we apply the Shoelace Theorem to find <math>[MOI]:</math> <cmath>[MOI]=\frac12\left|\left(4\cdot\frac52+6\cdot2+2\cdot4\right)-\left(4\cdot6+\frac52\cdot2+2\cdot4\right)\right|=\boxed{\textbf{(E)}\ \frac72}.</cmath> | Finally, we apply the Shoelace Theorem to find <math>[MOI]:</math> <cmath>[MOI]=\frac12\left|\left(4\cdot\frac52+6\cdot2+2\cdot4\right)-\left(4\cdot6+\frac52\cdot2+2\cdot4\right)\right|=\boxed{\textbf{(E)}\ \frac72}.</cmath> | ||
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<u><b>Remark</b></u> | <u><b>Remark</b></u> | ||
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Therefore, we get <cmath>[MOI]=\frac12\cdot MI\cdot h_O=\frac72.</cmath> | Therefore, we get <cmath>[MOI]=\frac12\cdot MI\cdot h_O=\frac72.</cmath> | ||
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~pieater314159 ~MRENTHUSIASM | ~pieater314159 ~MRENTHUSIASM | ||
Revision as of 22:39, 20 October 2021
Contents
Problem
In with side lengths
,
, and
, let
and
denote the circumcenter and incenter, respectively. A circle with center
is tangent to the legs
and
and to the circumcircle of
. What is the area of
?
Diagram
~MRENTHUSIASM
Solution
In this solution, let the brackets denote areas.
We place the diagram in the coordinate plane: Let and
Since is a right triangle with
its circumcenter is the midpoint of
from which
Note that the circumradius of
is
Let denote the semiperimeter of
The inradius of
is
from which
Since is tangent to both coordinate axes, its center is at
and its radius is
for some positive number
Let
be the point of tangency of
and
As
and
are both perpendicular to the common tangent line at
we conclude that
and
are collinear. It follows that
or
Solving this equation, we have
Finally, we apply the Shoelace Theorem to find
Remark
Alternatively, to find we can use
as the base and the distance from
to
as the height:
- By the Distance Formula, we have
- The equation of
is
so the distance from
to
is
Therefore, we get
~pieater314159 ~MRENTHUSIASM
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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.