Difference between revisions of "2011 AIME I Problems/Problem 4"
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== Solution 1 == | == Solution 1 == | ||
Extend <math>{CM}</math> and <math>{CN}</math> such that they intersect line <math>{AB}</math> at points <math>P</math> and <math>Q</math>, respectively. | Extend <math>{CM}</math> and <math>{CN}</math> such that they intersect line <math>{AB}</math> at points <math>P</math> and <math>Q</math>, respectively. | ||
+ | <asy> | ||
+ | defaultpen(fontsize(10)+0.8); size(200); | ||
+ | pen p=fontsize(9)+linewidth(3); | ||
+ | pair A,B,C,D,K,L,M,N,P,Q; | ||
+ | A=origin; B=(125,0); C=IP(CR(A,117),CR(B,120)); L=extension(B,C,A,bisectorpoint(B,A,C)); K=extension(A,C,B,bisectorpoint(C,B,A)); M=foot(C,B,K); N=foot(C,A,L); | ||
+ | draw(A--B--C--A); draw(A--L^^B--K, gray+dashed+0.5); draw(M--C--N^^N--extension(A,B,C,N)^^M--extension(A,B,C,M), gray+0.5); | ||
+ | dot("$A$",A,dir(200),p); dot("$B$",B,right,p); dot("$C$",C,up,p); dot("$L$",L,2*dir(70),p); dot("$N$",N,2*dir(-90),p); dot("$M$",M,2*dir(-90),p); dot("$P$",extension(A,B,C,M),2*down,p); dot("$Q$",extension(A,B,C,N),2*down,p); | ||
+ | label("$125$",A--B,down,fontsize(10)); label("$117$",A--C,2*dir(130),fontsize(10)); label("$120$",B--C,1.5*dir(30),fontsize(10)); | ||
+ | </asy> | ||
Since <math>{BM}</math> is the angle bisector of angle <math>B</math>, and <math>{CM}</math> is perpendicular to <math>{BM}</math>, so <math>BP=BC=120</math>, and <math>M</math> is the midpoint of <math>{CP}</math>. For the same reason, <math>AQ=AC=117</math>, and <math>N</math> is the midpoint of <math>{CQ}</math>. | Since <math>{BM}</math> is the angle bisector of angle <math>B</math>, and <math>{CM}</math> is perpendicular to <math>{BM}</math>, so <math>BP=BC=120</math>, and <math>M</math> is the midpoint of <math>{CP}</math>. For the same reason, <math>AQ=AC=117</math>, and <math>N</math> is the midpoint of <math>{CQ}</math>. | ||
Hence <math>MN=\tfrac 12 PQ</math>. Since <cmath>PQ=BP+AQ-AB=120+117-125=112,</cmath> so <math>MN=\boxed{056}</math>. | Hence <math>MN=\tfrac 12 PQ</math>. Since <cmath>PQ=BP+AQ-AB=120+117-125=112,</cmath> so <math>MN=\boxed{056}</math>. |
Revision as of 13:05, 3 August 2022
Contents
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
, 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
. 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 (LoC) 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
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.