Difference between revisions of "2011 AIME I Problems/Problem 4"
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+ | == Solution 4== | ||
+ | |||
+ | Because <math>\angle CMI = \angle CNI = 90</math>, <math>CMIN</math> is cyclic. | ||
+ | |||
+ | Ptolemy on CMIN: | ||
+ | |||
+ | <math>CN*MI+CM*IN=CI*MN</math> | ||
+ | |||
+ | <math>CI^2(\cos \angle ICN \sin \angle ICM + \cos \angle ICM \sin \angle ICN) = CI * MN</math> | ||
+ | |||
+ | <math>MN = CI \sin \angle MCN</math> by angle addition formula. | ||
+ | |||
+ | <math>\angle CMN = 180 - \angle MIN = 90 - \angle BCI</math>. | ||
+ | |||
+ | Let <math>H</math> be where the incircle touches <math>BC</math>, then <math>CI \cos \angle BCI = CH = \frac{a+b-c}{2}</math>. | ||
+ | <math>a=120, b=117, c=125</math>, for a final answer of <math>\boxed{056}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2011|n=I|num-b=3|num-a=5}} | {{AIME box|year=2011|n=I|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:11, 25 February 2018
Problem 4
In triangle ,
,
and
. The angle bisector of angle
intersects
at point
, and the angle bisector of angle
intersects
at point
. Let
and
be the feet of the perpendiculars from
to
and
, respectively. Find
.
Solution 1
Extend and
such that they intersects lines
at points
and
, respectively.
Since
is the angle bisector of angle B,and
is perpendicular to
,so ,
, M is the midpoint of
.For the same reason,
,N is the midpoint of
.
Hence
.But
,so
.
Solution 2
Let be the incenter of
. Now, since
and
, we have
is a cyclic quadrilateral. Consequently,
. Since
, we have that
. Letting
be the point of contact of the incircle of
with side
, we have
thus
Solution 3 (Bash)
Project onto
and
as
and
.
and
are both in-radii of
so we get right triangles with legs
(the in-radius length) and
. Since
is the hypotenuse for the 4 triangles (
and
),
are con-cyclic on a circle we shall denote as
which is also the circumcircle of
and
. To find
, we can use the Law of Cosines on
where
is the center of
. Now, the circumradius
can be found with Pythagorean Theorem with
or
:
. To find
, we can use the formula
and by Heron's,
. To find
, we can find
since
.
. Thus,
and since
, we have
. Plugging this into our Law of Cosines formula gives
. To find
, we use LoC on
. Our formula now becomes
. After simplifying, we get
.
--lucasxia01
Solution 4
Because ,
is cyclic.
Ptolemy on CMIN:
by angle addition formula.
.
Let be where the incircle touches
, then
.
, for a final answer of
.
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.