Difference between revisions of "2001 USAMO Problems/Problem 2"
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===Solution 1=== | ===Solution 1=== | ||
− | It is well known that the excircle opposite <math>A</math> is tangent to <math>\overline{BC}</math> at the point <math>D_2</math>. (Proof: let the points of tangency of the excircle with the lines <math>BC, AB, AC</math> be <math>D_3, F,G</math> respectively. Then <math>AB+BD_3=AB + BF=AF = AG = AC + | + | It is well known that the excircle opposite <math>A</math> is tangent to <math>\overline{BC}</math> at the point <math>D_2</math>. (Proof: let the points of tangency of the excircle with the lines <math>BC, AB, AC</math> be <math>D_3, F,G</math> respectively. Then <math>AB+BD_3=AB + BF=AF = AG = AC + CG=AC + CD_3</math>. It follows that <math>2CD_3 = AB + BC - AC</math>, and <math>CD_3 = s-b = BD_1 = CD_2</math>, so <math>D_3 \equiv D_2</math>.) |
Now consider the homothety that carries the incircle of <math>\triangle ABC</math> to its excircle. The homothety also carries <math>Q</math> to <math>D_2</math> (since <math>A,Q,D_2</math> are collinear), and carries the tangency points <math>E_1</math> to <math>G</math>. It follows that <math>\frac{AQ}{QD_2} = \frac{AE_1}{E_1G} = \frac{s-a}{E_1C + CD_2} = \frac{s-a}{CD_1 + BD_1} = \frac{s-a}{a}</math>. | Now consider the homothety that carries the incircle of <math>\triangle ABC</math> to its excircle. The homothety also carries <math>Q</math> to <math>D_2</math> (since <math>A,Q,D_2</math> are collinear), and carries the tangency points <math>E_1</math> to <math>G</math>. It follows that <math>\frac{AQ}{QD_2} = \frac{AE_1}{E_1G} = \frac{s-a}{E_1C + CD_2} = \frac{s-a}{CD_1 + BD_1} = \frac{s-a}{a}</math>. | ||
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Thus, <math>Q'</math> has normalized <math>x</math>-coordinate <math>1-\frac{s-a}{s}=\frac{a}{s}</math>. | Thus, <math>Q'</math> has normalized <math>x</math>-coordinate <math>1-\frac{s-a}{s}=\frac{a}{s}</math>. | ||
− | As the line <math>AD_2</math> has equation <math>(s-c)y=(s-b)z</math>, it can easily be found that <cmath>Q'=\left(\frac{a^2}{as}, \frac{(s-a)(s-b)}{as}, \frac{(s-a)(s-c}{as}\right)=(a^2:(s-a)(s-b):(s-a)(s-c)).</cmath> | + | As the line <math>AD_2</math> has equation <math>(s-c)y=(s-b)z</math>, it can easily be found that <cmath>Q'=\left(\frac{a^2}{as}, \frac{(s-a)(s-b)}{as}, \frac{(s-a)(s-c)}{as}\right)=(a^2:(s-a)(s-b):(s-a)(s-c)).</cmath> |
Recalling that the equation of the incircle is <cmath>a^2yz+b^2xz+c^2xy+(x+y+z)[(s-a)^2x+(s-b)^2y+(s-c)^2z]=0.</cmath> We must show that this equation is true for <math>Q'</math>'s values of <math>x,y,z</math>. | Recalling that the equation of the incircle is <cmath>a^2yz+b^2xz+c^2xy+(x+y+z)[(s-a)^2x+(s-b)^2y+(s-c)^2z]=0.</cmath> We must show that this equation is true for <math>Q'</math>'s values of <math>x,y,z</math>. | ||
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As <math>Q'</math> lies on the incircle and <math>AD_2</math>, <math>Q'=Q</math>, and our proof is complete. | As <math>Q'</math> lies on the incircle and <math>AD_2</math>, <math>Q'=Q</math>, and our proof is complete. | ||
+ | |||
+ | |||
+ | === Solution 4 === | ||
+ | We again use Barycentric coordinates. As before, let <math>A</math> be <math>(1,0,0)</math>, <math>B</math> be <math>(0,1,0)</math>, and <math>C</math> be <math>(0,0,1)</math>. Also | ||
+ | |||
+ | <cmath>D_{1}=\left( 0,\frac{s-c}{a},\frac{s-b}{a} \right), D_2=\left(0,\frac{s-b}{a},\frac{s-c}{a}\right), E_2=\left(\frac{s-a}{b},0,\frac{s-c}{b}\right), P=\left(\frac{s-a}{s},\frac{s-b}{s},\frac{s-c}{s}\right).</cmath> | ||
+ | |||
+ | Now, consider a point <math>Q'</math> for which <math>AQ'=PD_{2}</math>. Then working component-vise, we get <math>Q'+P=A+D_{2}</math> from which we can easily get the coordinates of <math>Q'</math> as; | ||
+ | |||
+ | <cmath>Q'=\left(\frac{a}{s},\frac{(s-a)(s-b)}{sa},\frac{(s-a)(s-c)}{sa}\right)</cmath> | ||
+ | |||
+ | It suffices to show that <math>Q'=Q</math>. | ||
+ | |||
+ | Let <math>I=\left(\frac{a}{2s},\frac{b}{2s},\frac{c}{2s}\right)</math> be the incenter of triangle <math>ABC</math>. We claim that <math>I</math> is the midpoint of <math>Q'D_{1}</math>. Indeed, | ||
+ | |||
+ | <cmath>\frac{1}{2}\left(0+\frac{a}{s}\right)=\frac{a}{2s}</cmath> | ||
+ | |||
+ | <cmath>\frac{1}{2}\left(\frac{(s-a)(s-b)}{sa}+\frac{s-c}{a}\right)=\frac{(s-a)(s-b)+s(s-c)}{2sa}=\frac{ab}{2sa}=\frac{b}{2s}</cmath> | ||
+ | |||
+ | <cmath>\frac{1}{2}\left(\frac{(s-a)(s-c)}{sa}+\frac{s-b}{a}\right)=\frac{c}{2s}</cmath> | ||
+ | |||
+ | Hence the claim has been proved. | ||
+ | |||
+ | Since <math>I</math> is the center of <math>\omega</math> and the midpoint of <math>Q'D_{1}</math>, thus <math>Q'</math> is the point diametrically opposite to <math>D_{1}</math>, and hence it lies on <math>\omega</math> and closer to <math>A</math> off the 2 points. Thus <math>Q=Q'</math> as desired. | ||
== See also == | == See also == |
Latest revision as of 07:35, 16 November 2017
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
.
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
.
2001usamo2-2.png
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 .
Solution 3
Here is a rather nice solution using barycentric coordinates:
Let be
,
be
, and
be
. Let the side lengths of the triangle be
and the semi-perimeter
.
Now, Thus,
Therefore, and
Clearly then, the non-normalized coordinates of
Normalizing, we have that
Now, we find the point inside the triangle on the line
such that
. It is then sufficient to show that this point lies on the incircle.
is the fraction
of the way "up" the line segment from
to
. Thus, we are looking for the point that is
of the way "down" the line segment from
to
, or, the fraction
of the way "up".
Thus, has normalized
-coordinate
.
As the line has equation
, it can easily be found that
Recalling that the equation of the incircle is We must show that this equation is true for
's values of
.
Plugging in our values, this means showing that
Dividing by
, this is just
Plugging in the value of
The first bracket is just
and the second bracket is
Dividing everything by
gives
which is
, as desired.
As lies on the incircle and
,
, and our proof is complete.
Solution 4
We again use Barycentric coordinates. As before, let be
,
be
, and
be
. Also
Now, consider a point for which
. Then working component-vise, we get
from which we can easily get the coordinates of
as;
It suffices to show that .
Let be the incenter of triangle
. We claim that
is the midpoint of
. Indeed,
Hence the claim has been proved.
Since is the center of
and the midpoint of
, thus
is the point diametrically opposite to
, and hence it lies on
and closer to
off the 2 points. Thus
as desired.
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.