Difference between revisions of "2001 USAMO Problems/Problem 2"
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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 < | + | 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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<cmath>2a[(-a+b+c)(a-b+c)(a+b-c)+4b^2(a+b-c)+4c^2(a-b+c)]+(a+b+c)[4a^2(-a+b+c)+(a-b+c)^3+(a+b-c)^3]=0</cmath> | <cmath>2a[(-a+b+c)(a-b+c)(a+b-c)+4b^2(a+b-c)+4c^2(a-b+c)]+(a+b+c)[4a^2(-a+b+c)+(a-b+c)^3+(a+b-c)^3]=0</cmath> | ||
The first bracket is just | The first bracket is just | ||
− | <cmath>-a^3-b^3-c^3+a^2b+a^2c+b^2c+b^2a+c^2a+c^2b-2abc+4ab^2+4b^3-4b^2c+4ac^2-4bc^2+4c^3=-a^3+3b^3+3c^3+a^2b+a^2c-3b^2c+5b^2a+5c^2a-3c^2b-2abc</cmath> | + | <cmath>-a^3-b^3-c^3+a^2b+a^2c+b^2c+b^2a+c^2a+c^2b-2abc+4ab^2+4b^3-4b^2c+4ac^2-4bc^2+4c^3</cmath> |
+ | <cmath>=-a^3+3b^3+3c^3+a^2b+a^2c-3b^2c+5b^2a+5c^2a-3c^2b-2abc</cmath> | ||
and the second bracket is <cmath>-2a^3+4a^2b+4a^2c+6ab^2+6ac^2-12abc.</cmath> | and the second bracket is <cmath>-2a^3+4a^2b+4a^2c+6ab^2+6ac^2-12abc.</cmath> | ||
Dividing everything by <math>2a</math> gives | Dividing everything by <math>2a</math> gives |
Revision as of 20:35, 10 May 2016
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.
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.