Difference between revisions of "2001 USAMO Problems/Problem 2"
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'''Lemma''': Segment <math>D_1Q</math> is a diameter of circle <math>\omega</math>. | '''Lemma''': Segment <math>D_1Q</math> is a diameter of circle <math>\omega</math>. | ||
− | + | <center>[[File:2001usamo2-1.png]]</center> | |
''Proof'': Let <math>I</math> be the center of circle <math>\omega</math>, i.e., <math>I</math> is the incenter of triangle <math>ABC</math>. Extend segment <math>D_1I</math> through <math>I</math> to intersect circle <math>\omega</math> again at <math>Q'</math>, and extend segment <math>AQ'</math> through <math>Q'</math> to intersect segment <math>BC</math> at <math>D'</math>. We show that <math>D_2 = D'</math>, which in turn implies that <math>Q = Q'</math>, that is, <math>D_1Q</math> is a diameter of <math>\omega</math>. | ''Proof'': Let <math>I</math> be the center of circle <math>\omega</math>, i.e., <math>I</math> is the incenter of triangle <math>ABC</math>. Extend segment <math>D_1I</math> through <math>I</math> to intersect circle <math>\omega</math> again at <math>Q'</math>, and extend segment <math>AQ'</math> through <math>Q'</math> to intersect segment <math>BC</math> at <math>D'</math>. We show that <math>D_2 = D'</math>, which in turn implies that <math>Q = Q'</math>, that is, <math>D_1Q</math> is a diameter of <math>\omega</math>. | ||
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that is, <math>D' = D_2</math>, as desired. <math>\blacksquare</math> | that is, <math>D' = D_2</math>, as desired. <math>\blacksquare</math> | ||
− | |||
Now we prove our main result. Let <math>M_1</math> and <math>M_2</math> be the respective midpoints of segments <math>BC</math> and <math>CA</math>. Then <math>M_1</math> is also the midpoint of segment <math>D_1D_2</math>, from which it follows that <math>IM_1</math> is the midline of triangle <math>D_1QD_2</math>. Hence | Now we prove our main result. Let <math>M_1</math> and <math>M_2</math> be the respective midpoints of segments <math>BC</math> and <math>CA</math>. Then <math>M_1</math> is also the midpoint of segment <math>D_1D_2</math>, from which it follows that <math>IM_1</math> is the midline of triangle <math>D_1QD_2</math>. Hence | ||
<cmath>QD_2 = 2IM_1</cmath> | <cmath>QD_2 = 2IM_1</cmath> | ||
and <math>AD_2\parallel M_1I</math>. Similarly, we can prove that <math>BE_2\parallel M_2I</math>. | and <math>AD_2\parallel M_1I</math>. Similarly, we can prove that <math>BE_2\parallel M_2I</math>. | ||
− | + | <center>[[File:2001usamo2-2.png]]</center> | |
+ | |||
Let <math>G</math> be the centroid of triangle <math>ABC</math>. Thus segments <math>AM_1</math> and <math>BM_2</math> intersect at <math>G</math>. Define transformation <math>\mathbf{H}_2</math> as the dilation with its center at <math>G</math> and ratio <math>-1/2</math>. Then <math>\mathbf{H}_2(A) = M_1</math> and <math>\mathbf{H}_2(B) = M_2</math>. Under the dilation, parallel lines go to parallel lines and the intersection of two lines goes to the intersection of their images. Since <math>AD_2\parallel M_1I</math> and <math>BE_2\parallel M_2I</math>, <math>\mathbf{H}_2</math> maps lines <math>AD_2</math> and <math>BE_2</math> to lines <math>M_1I</math> and <math>M_2I</math>, respectively. It also follows that <math>\mathbf{H}_2(I) = P</math> and | Let <math>G</math> be the centroid of triangle <math>ABC</math>. Thus segments <math>AM_1</math> and <math>BM_2</math> intersect at <math>G</math>. Define transformation <math>\mathbf{H}_2</math> as the dilation with its center at <math>G</math> and ratio <math>-1/2</math>. Then <math>\mathbf{H}_2(A) = M_1</math> and <math>\mathbf{H}_2(B) = M_2</math>. Under the dilation, parallel lines go to parallel lines and the intersection of two lines goes to the intersection of their images. Since <math>AD_2\parallel M_1I</math> and <math>BE_2\parallel M_2I</math>, <math>\mathbf{H}_2</math> maps lines <math>AD_2</math> and <math>BE_2</math> to lines <math>M_1I</math> and <math>M_2I</math>, respectively. It also follows that <math>\mathbf{H}_2(I) = P</math> and | ||
<cmath>\frac{IM_1}{AP} = \frac{GM_1}{AG} = \frac{1}{2}</cmath> | <cmath>\frac{IM_1}{AP} = \frac{GM_1}{AG} = \frac{1}{2}</cmath> |
Revision as of 20:59, 6 July 2014
Contents
[hide]Problem
Let be a triangle and let
be its incircle. Denote by
and
the points where
is tangent to sides
and
, respectively. Denote by
and
the points on sides
and
, respectively, such that
and
, and denote by
the point of intersection of segments
and
. Circle
intersects segment
at two points, the closer of which to the vertex
is denoted by
. Prove that
.
Solution
Solution 1
It is well known that the excircle opposite is tangent to
at the point
. (Proof: let the points of tangency of the excircle with the lines
be
respectively. Then
. It follows that
, and
, so
.)
Now consider the homothety that carries the incircle of to its excircle. The homothety also carries
to
(since
are collinear), and carries the tangency points
to
. It follows that
.
![[asy] pathpen = linewidth(0.7); size(300); pen d = linetype("4 4") + linewidth(0.6); pair B=(0,0), C=(10,0), A=7*expi(1),O=D(incenter(A,B,C)),D1 = D(MP("D_1",foot(O,B,C))),E1 = D(MP("E_1",foot(O,A,C),NE)),E2 = D(MP("E_2",C+A-E1,NE)); /* arbitrary points */ /* ugly construction for OA */ pair Ca = 2C-A, Cb = bisectorpoint(Ca,C,B), OA = IP(A--A+10*(O-A),C--C+50*(Cb-C)), D2 = D(MP("D_2",foot(OA,B,C))), Fa=2B-A, Ga=2C-A, F=MP("F",D(foot(OA,B,Fa)),NW), G=MP("G",D(foot(OA,C,Ga)),NE); D(OA); D(MP("A",A,N)--MP("B",B,NW)--MP("C",C,NE)--cycle); D(incircle(A,B,C)); D(CP(OA,D2),d); D(B--Fa,linewidth(0.6)); D(C--Ga,linewidth(0.6)); D(MP("P",IP(D(A--D2),D(B--E2)),NNE)); D(MP("Q",IP(incircle(A,B,C),A--D2),SW)); clip((-20,-10)--(-20,20)--(20,20)--(20,-10)--cycle); [/asy]](http://latex.artofproblemsolving.com/d/b/6/db6875d6724835163f466dad6bbadb2278bef967.png)
By Menelaus' Theorem on with segment
, it follows that
. It easily follows that
.
Solution 2
The key observation is the following lemma.
Lemma: Segment is a diameter of circle
.

Proof: Let be the center of circle
, i.e.,
is the incenter of triangle
. Extend segment
through
to intersect circle
again at
, and extend segment
through
to intersect segment
at
. We show that
, which in turn implies that
, that is,
is a diameter of
.
Let be the line tangent to circle
at
, and let
intersect the segments
and
at
and
, respectively. Then
is an excircle of triangle
. Let
denote the dilation with its center at
and ratio
. Since
and
,
. Hence
. Thus
,
, and
. It also follows that an excircle
of triangle
is tangent to the side
at
.
It is well known that
We compute
. Let
and
denote the points of tangency of circle
with rays
and
, respectively. Then by equal tangents,
,
, and
. Hence
It follows that
Combining these two equations yields
. Thus
that is,
, as desired.
Now we prove our main result. Let and
be the respective midpoints of segments
and
. Then
is also the midpoint of segment
, from which it follows that
is the midline of triangle
. Hence
and
. Similarly, we can prove that
.

Let be the centroid of triangle
. Thus segments
and
intersect at
. Define transformation
as the dilation with its center at
and ratio
. Then
and
. Under the dilation, parallel lines go to parallel lines and the intersection of two lines goes to the intersection of their images. Since
and
,
maps lines
and
to lines
and
, respectively. It also follows that
and
or
This yields
as desired.
Note: We used directed lengths in our calculations to avoid possible complications caused by the different shapes of triangle .
See also
2001 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.