Difference between revisions of "2011 AIME I Problems/Problem 4"
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== Solution 2 == | == Solution 2 == | ||
− | Let <math>I</math> be the incenter of <math>ABC</math>. Since <math>I</math> lies on <math>BM</math> and <math>AN</math>, <math>IM \perp MC</math> and <math>IN \perp NC</math>, so <math>\angle IMC + \angle INC = 180^\circ</math>. This means that <math>CMIN</math> is a cyclic quadrilateral. By the Law of Sines, <math>\frac{MN}{\sin \angle MIN} = \frac{2R}{\sin \angle CMI} = 2R = CI</math>, where <math>R</math> is the radius of the circumcircle of <math>CMIN</math>. Since <math>\sin \angle MIN = \sin \angle BIA = \sin (90^\circ + \tfrac 12 \angle BCA) = \cos \ | + | Let <math>I</math> be the incenter of <math>ABC</math>. Since <math>I</math> lies on <math>BM</math> and <math>AN</math>, <math>IM \perp MC</math> and <math>IN \perp NC</math>, so <math>\angle IMC + \angle INC = 180^\circ</math>. This means that <math>CMIN</math> is a cyclic quadrilateral. By the Law of Sines, <math>\frac{MN}{\sin \angle MIN} = \frac{2R}{\sin \angle CMI} = 2R = CI</math>, where <math>R</math> is the radius of the circumcircle of <math>CMIN</math>. Since <math>\sin \angle MIN = \sin \angle BIA = \sin (90^\circ + \tfrac 12 \angle BCA) = \cos \tfrac 12 \angle BCA = \cos \angle BCI</math>, we have that <math>MN = CI \cdot \sin \angle MIN = CI \cdot \cos \angle BCI</math>. Letting <math>H</math> be the point of contact of the incircle of <math>ABC</math> with side <math>BC</math>, we have <math>MN = CI \cdot \cos \angle BCI = CI \cdot \frac{CH}{CI} = CH</math>. Thus, <math>MN = s - AB = \frac{117+120-125}{2}=\boxed{056}</math>. |
== Solution 3 (Bash) == | == Solution 3 (Bash) == |
Revision as of 18:52, 2 August 2023
Contents
[hide]Problem
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 intersect line
at points
and
, respectively.
Since
is the angle bisector of angle
and
is perpendicular to
,
must be an isoceles triangle, so
, and
is the midpoint of
. For the same reason,
, and
is the midpoint of
.
Hence
. Since
so
.
Solution 2
Let be the incenter of
. Since
lies on
and
,
and
, so
. This means that
is a cyclic quadrilateral. By the Law of Sines,
, where
is the radius of the circumcircle of
. 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 (LoC) formula gives
. To find
, we use LoC on
. Our formula now becomes
. After simplifying, we get
.
--lucasxia01
Solution 4
Because ,
is cyclic.
Applying Ptolemy's theorem on CMIN:
by sine angle addition formula.
.
Let be where the incircle touches
, then
.
, for a final answer of
.
Note: This is similar to Solution 2 after the first four lines
Video Solution
https://www.youtube.com/watch?v=yIUBhWiJ4Dk ~Mathematical Dexterity
Video Solution
https://www.youtube.com/watch?v=vkniYGN45F4
~Shreyas S
Alternate Solution: https://www.youtube.com/watch?v=L2OzYI0OJsc&t=12s
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 |
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