Projective Bary
by yayups, May 11, 2019, 6:06 AM
This solution shows a nontrivial usage of the projective nature of barycentric coordinates.
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
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[/asy]](//latex.artofproblemsolving.com/9/2/6/9265c88f4a81d8aca6c5ad3b7c4cd51da79b1f5f.png)
Firstly, the problem is clearly just asking us to show that
is tangent to 
. Letting 
, we see that this is equivalent to 
, but 
, so all we need to show is that
![\[TB\cdot TC=TD^2.\]](//latex.artofproblemsolving.com/2/d/8/2d8ca7873f2e762e57966529e2fb19f52d6cb0da.png)
We'll actually just compute the barycentric coordinates of 
and verify this explicitly. However, finding 
and 
is really messy, and this doesn't seem tractable immediately. However, the beauty of this solution is getting a simple way to find the coordinates of .
Let
be a variable point on 
, and define the composite map 
given by
![\[P\mapsto P'\mapsto P_1P_2\cap BC\]](//latex.artofproblemsolving.com/3/b/0/3b0a1bc34ad2accec60eebb852a299c9b94cb17e.png)
where 
is the 
inverse of and 
and 
are tangent to 
with 
. It's not easy to see that this map is projective but its true.
Now, we check that
![\[B\mapsto C\mapsto AB\mapsto B,\]](//latex.artofproblemsolving.com/b/6/9/b696c5b4d3d959c53822f3791ee22cc6e6fb3996.png)
so 
, and similarly 
. Also,
![\[\infty_{BC}\mapsto A\mapsto BC,\]](//latex.artofproblemsolving.com/7/0/5/705c34756f4cd1195dadbcd751763ef30a458a08.png)
but uh-oh, what's 
? The fix is that we actually have the tangent to at some point infinitesimally close to 
, which then intersects really close to , so we have 
.
Now, use projective coordinates for given by the last two coordinates of the barycentric coordinates of a point on . The map looks like some matrix
![\[\begin{pmatrix}w & x\\ y& z\end{pmatrix}\]](//latex.artofproblemsolving.com/9/b/9/9b96ab082094cdb16b53fba91dd088fb597fc6c8.png)
that is considered the same under scaling. We see that 
and 
map to themselves, so 
. The point 
is given by 
, and it maps to 
, so the matrix is
![\[\phi\equiv \begin{pmatrix}s-c & 0\\ 0 & -(s-b)\end{pmatrix}.\]](//latex.artofproblemsolving.com/6/8/b/68bd66cd7a38db2d1345380b628b589a960ff750.png)
Now, we see that 
since and 
are inverses (the incircle gets sent to the mixtilinear excircle), so
![\[T=\begin{pmatrix}s-c & 0\\ 0 & -(s-b)\end{pmatrix}\begin{pmatrix}s-c \\ s-b\end{pmatrix}=\begin{pmatrix}(s-c)^2 \\ -(s-b)^2\end{pmatrix},\]](//latex.artofproblemsolving.com/1/1/2/112091d2ae6406785da1a9ec592ac043583852a5.png)
so 
. The rest is a boring bash which we do below.
Homogenizing, we have
, 
, 
, and 
. By taking simply the first coordinate, we get a coordinate system on that preserves the lengths in the original setup, so
![\[TB=\frac{(s-c)^2}{(b-c)a}-1=\frac{(s-b)^2}{(b-c)a}\]](//latex.artofproblemsolving.com/d/7/0/d702095640ed1b38129962cce048d18b092682dd.png)
and 
, and
![\[TD=\frac{(s-c)^2}{(b-c)a}-\frac{s-c}{a}=\frac{s-c}{a}\cdot\frac{s-b}{b-c}=\frac{(s-b)(s-c)}{a(b-c)}.\]](//latex.artofproblemsolving.com/e/c/f/ecf4625af98721f9c35d194c39dcd73b6482258b.png)
Evidently, we have 
, as desired. As we showed before, this means 
, so is tangent to , which is tangent to at 
, so is tangent to . 
USAMO Shortlist 2014, 110 Geometery Problems #77 wrote:
Let 
be a triangle with circumcircle 
and incircle . Let be the mixtilinear excircle touch point, and let 
and 
be tangent to with 
. Let be the tangency point of with which exists by Poncelet's Porism. Show that is tangent to .
be a triangle with circumcircle 
and incircle 
. Let 
be the mixtilinear excircle touch point, and let 
and 
be tangent to with 
. Let 
be the tangency point of 
with which exists by Poncelet's Porism. Show that 
is tangent to .![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
unitsize(0.2inches);
import graph; size(0cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
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pen dotstyle = black; /* point style */
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[/asy]](http://latex.artofproblemsolving.com/9/2/6/9265c88f4a81d8aca6c5ad3b7c4cd51da79b1f5f.png)
Firstly, the problem is clearly just asking us to show that
is tangent to . Letting 
, we see that this is equivalent to 
, but 
, so all we need to show is that![\[TB\cdot TC=TD^2.\]](http://latex.artofproblemsolving.com/2/d/8/2d8ca7873f2e762e57966529e2fb19f52d6cb0da.png)
We'll actually just compute the barycentric coordinates of 
and verify this explicitly. However, finding 
and 
is really messy, and this doesn't seem tractable immediately. However, the beauty of this solution is getting a simple way to find the coordinates of .Let
be a variable point on 
, and define the composite map 
given by![\[P\mapsto P'\mapsto P_1P_2\cap BC\]](http://latex.artofproblemsolving.com/3/b/0/3b0a1bc34ad2accec60eebb852a299c9b94cb17e.png)
where 
is the 
inverse of and 
and 
are tangent to with 
. It's not easy to see that this map is projective but its true.
be the tangency point of 
to 
given by 
,
is the pole of 
,Now, we check that
![\[B\mapsto C\mapsto AB\mapsto B,\]](http://latex.artofproblemsolving.com/b/6/9/b696c5b4d3d959c53822f3791ee22cc6e6fb3996.png)
so 
, and similarly 
. Also,![\[\infty_{BC}\mapsto A\mapsto BC,\]](http://latex.artofproblemsolving.com/7/0/5/705c34756f4cd1195dadbcd751763ef30a458a08.png)
but uh-oh, what's 
? The fix is that we actually have the tangent to at some point infinitesimally close to 
, which then intersects really close to , so we have 
.Now, use projective coordinates for
given by the last two coordinates of the barycentric coordinates of a point on . The map looks like some matrix![\[\begin{pmatrix}w & x\\ y& z\end{pmatrix}\]](http://latex.artofproblemsolving.com/9/b/9/9b96ab082094cdb16b53fba91dd088fb597fc6c8.png)
that is considered the same under scaling. We see that 
and 
map to themselves, so 
. The point 
is given by 
, and it maps to 
, so the matrix is![\[\phi\equiv \begin{pmatrix}s-c & 0\\ 0 & -(s-b)\end{pmatrix}.\]](http://latex.artofproblemsolving.com/6/8/b/68bd66cd7a38db2d1345380b628b589a960ff750.png)
Now, we see that 
since and are inverses (the incircle gets sent to the mixtilinear excircle), so![\[T=\begin{pmatrix}s-c & 0\\ 0 & -(s-b)\end{pmatrix}\begin{pmatrix}s-c \\ s-b\end{pmatrix}=\begin{pmatrix}(s-c)^2 \\ -(s-b)^2\end{pmatrix},\]](http://latex.artofproblemsolving.com/1/1/2/112091d2ae6406785da1a9ec592ac043583852a5.png)
so 
. The rest is a boring bash which we do below.Homogenizing, we have
, 
, 
, and 
. By taking simply the first coordinate, we get a coordinate system on that preserves the lengths in the original setup, so![\[TB=\frac{(s-c)^2}{(b-c)a}-1=\frac{(s-b)^2}{(b-c)a}\]](http://latex.artofproblemsolving.com/d/7/0/d702095640ed1b38129962cce048d18b092682dd.png)
and 
, and![\[TD=\frac{(s-c)^2}{(b-c)a}-\frac{s-c}{a}=\frac{s-c}{a}\cdot\frac{s-b}{b-c}=\frac{(s-b)(s-c)}{a(b-c)}.\]](http://latex.artofproblemsolving.com/e/c/f/ecf4625af98721f9c35d194c39dcd73b6482258b.png)
Evidently, we have 
, as desired. As we showed before, this means 
, so is tangent to , which is tangent to at , so is tangent to . 
This post has been edited 3 times. Last edited by yayups, May 11, 2019, 10:35 PM
.
.
.
under 
which concludes the problem
April 2021
September 2020