has no solutions in integers 
,
,
.
be a 
board. Some of its cells are colored black in such a way that every 
board of 
black cells. Find the maximum amount of black cells that the board may have.
intersect at 
,
tangent to the extensions of 
at 
respectively. Circle 
passes through points 
,
.
let 
to the side 
,
,
be the incenter, 
-
and 
the other intersection points of the circle 
with the lines 
and 
,
.
denote the sum of the digits of a positive integer 
.
,
is a positive integer?
Y by anhtaitran, aopser123, artsolver, kingofgeedorah, Tawan, BogdanB, anantmudgal09, Desimathematics, AlastorMoody, karitoshi, amar_04, itslumi, Ya_pank, centslordm, HamstPan38825, samrocksnature, Rg230403, jhu08, megarnie, OronSH, Adventure10, Mango247, aidan0626, whslovemath, ehuseyinyigit
Elmo is now learning olympiad geometry. In triangle 
with 
, let its incircle be tangent to sides 
, 
, and 
at 
, 
, and 
, respectively. The internal angle bisector of 
intersects lines 
and 
at 
and 
, respectively. Let 
and 
be distinct points on side such that 
. Finally, let 
be the circumcircle of 
.
(a) Help Elmo show that is tangent to the circumcircle of 
.
(b) Help Elmo show that is tangent to the incircle of .
James Lin
with 
, let its incircle be tangent to sides 
, 
, and 
at 
, 
, and 
, respectively. The internal angle bisector of 
intersects lines 
and 
at 
and 
, respectively. Let 
and 
be distinct points on side such that 
. Finally, let 
be the circumcircle of 
.(a) Help Elmo show that
is tangent to the circumcircle of 
.(b) Help Elmo show that
is tangent to the incircle of .James Lin
This post has been edited 1 time. Last edited by ABCDE, Jun 24, 2016, 2:07 PM
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/c/8/6/c86df53573b41e802885b75d753e1c3bad0aa6e4.png)
and 
and 
be the circumcenter of triangle 
cuts 
.
and 
so 
implies 
or 
.
are concyclic. Similarly, we have 
implies 
,
are concyclic. Also, note that 
so 
or 
.
is cyclic AND 
so 
.
.
implies 
.
implies 
.
so we follows that 
are bisectors of 
,
.
implies 
.
.
be the circumcenter of triangle 
.
be the orthocenter of triangle 
.
,
.
or 
.
.
are on the same side wrt 
are collinear, which means 
and 
are tangent to each other.
Since 
so 
of 
is the incenter, 
or 
.
.
then since 
so the inversion will send 
.
since 
.
implies 
.
is the radius of 
so we obtain that 
are orthogonal circles. This means that the image of 
,
and 
are tangent to 
,
and 
.
, so 
so we are done.
but 
is the bisector of 
is the bisector of 
is the bisector of 
i.e 
tangent to 
and 
and incircle of 
are concyclic. Let 
and incircle of triangle 
.
.
and projecting this quadruple on the line 
then 
.
and 
so, 
and so 
.
and an inversion about 
followed by reflection in 
are isogonal in angle 
and result of part a.) follows.
inversion (inversion about 
followed by reflection in the bisector of angle 
the image of point 
is tangent to the 
be the mixtilinear excenter opposite 
where 
is that for the mixtilinear excircle. We want 
.
.
.
.
at 
.
as 
.
,
.
and 
perpendicular to 
is also perpendicular to 
is the centre of homothety of two circles, and thus tangent point.
.
.
as 
are tangent to 
.
swap under an inversion about the circle with diameter 
maps to line 
,
is an angle bisector of 
,
and 
are isogonal in 
at 
and 
respectively, since the midpoint of minor arc 
,
are homothetic, as desired.
is orthogonal to 
,
is the inradius. Equivalently, we can show that 
swap under an inversion about the incircle. Of course, it is well-known that 
(as 
are tangent to the incircle), which is the circle with diameter 
.
maps to itself under this inversion, it follows that the image of 
are ugly; you can 
and keep in this way, but maybe yes, it makes use of many synthetic observations
denote the inverse of an arbitrary point 
with 
.
,
at 
concur by considering the radical center of 
.
and 
,
and 
.
and let the capital letters denote the angles. Let the 
.
and 
and 
However, by Law of Sines, we have
Similarly, 
.
and clearly 
and by homothety we're done.
is tangent to the mixtillinear excircle (oops im not sure about this name).
be the where the excircle opposite 
is where the mixtillinear excircle meets 
.
and
Thus 
.
.
,
sends the mixtillinear excircle opposite 
.
we know that 
.
,
is the angle bisector of 
.
.
at 
be the center of 
and 
,
is perpendicular to 
.
We know that 
.
,
.
This is not so hard to check. Hence 
,
and 
.
by Apollonian incircle lemma 
bisects 
and 
are isogonal, so simple angle chasing shows that the tangents to 
.
Then under inversion at 
,
by incenter-excenter lemma, 
.
First we claim the following:
and 
implies that 
and similarly for 
be the center of 
If we invert around 
then 
so if we had that 
then we get that 
showing that 
and the incircle are tangent. To do this, we let 
Now the following claim shows that 
so 
so we are left to show that 
But this is trivially true as 
For this apply Monge d'Alembert's Theorem to 
the incircle and the circumcenter. Notice how we wish to show that the tangency point of 
with 
goes through the excimillicenter of 
We wish to show 
ciclic.
followed by a reflection across the bisector of 
Now, if we let 
be the external 
we get that 
which we will call 
and 
which is the midpoint of arc 
In order to treat the inverse of 
we notice that 
We now wish to show 
But this is easy as considering the homothety centered at 
which sends 
by the fact that 
then 
so under the inversion we find 
are the points on the angle bisector with 
Then, the midpoint of 
is the point 
which is the arc midpoint of 
be the intersections of the circle with diameter 
we find that 
are isogonal with respect to 
Since 
invert to 
we may assume that 
are collinear and 
are collinear. Then since the circles with diameters 
we may check that this homothety also sends 
to each other. Then, we may check that this homothety has scale factor 
Since 
and 
we get 
However, we can also check that this is the scale factor of the homothety sending the incenter 
However, 
so 
.
and the incircle. Notice something about the two circles? If you did part a) in the way I hinted towards, maybe you'll be able to motivate something too... something that would invert your expectations perhaps? From there it shouldn't be too difficult to finish.
.
which yields that 
.
we obtain that 
is cyclic with center 
.
,
and 
,
.
and 
respectively. I claim that their exsimilicenter is 
it suffices to show that
We thus want to show by 
of 
,
,
are a distance 
away from 
lie on 
is cyclic, so 
.
,
,
as usual. By considering 
we find that 
and the inradius is 
.
which is true. 
and this follows by projecting from the point at infinity along the line perpendicular to 
However, since 
and similarly 
Thus, it suffices to show that 
However, this follows easily as 
and 
.
and similarly for 
by the above claim, we have 
which reduces to 
This means that 
so 
and 
and 
This means that 
so 
This means that 
that passes through both 
Let 
intersect the incircle again at 
.
however this is clearly true since 
is the polar of 
(aka line 
.![\[(A, BC \cap AI; X, Y) \overset{D}{=} (AD \cap (DEF), D; E, F) = -1\]](http://latex.artofproblemsolving.com/5/2/d/52d51510d9232513001fb7513b6f886e1b739941.png)
,![\[\angle IDX = \frac C2 = 90 - \frac{A+B}{2} = \angle YFB - \angle YAF = \angle DYI. \quad {\color{blue} \Box}\]](http://latex.artofproblemsolving.com/c/7/5/c7519548ff5dc675b35f43c901a47e0b4c517700.png)
.
,
.
.
.
and 
.![\[\left|\begin{matrix}1&-a&-z&az\\1&-\overline z&-a&a\overline z\\2&-x-y&-x-y&2\\0&1&d^2&-2d\end{matrix}\right|=0.\]](http://latex.artofproblemsolving.com/e/e/f/eefe89e03681add29c1803f795be93da1dbc9735.png)
and 
to 
.
to 
can be derived as a linear combination of the two circles' equations. Thus, 
vanishes. By expanding and taking the coefficients of 
,
,
,
,![\[\left|\begin{matrix}1&a&z&az\\1&\overline z&a&a\overline z\\2&x+y&x+y&2\\0&1&d^2&2d\end{matrix}\right|=0.\]](http://latex.artofproblemsolving.com/a/d/c/adc091c5c216db8a91e0bf6e1b7233eed7f43844.png)
,
in the determinant is
.
to 
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/2/8/2/28227bae529dd1f6f51f885158433747d1ee3799.png)
and 
are cyclic. Now as the incircle and 
,
until it meets the incircle again at 
where 
is the radius of circle 
:
respectively. 
.
.
and 
,![\[\frac{AX}{AY}=\frac{BX}{CY}=\frac{DX}{DY}\implies (A,D;X,Y)=-1\]](http://latex.artofproblemsolving.com/2/f/e/2fe3be6cbe5b5ed65f52584efaeb775196c0718e.png)
![\[-1=(A,D;X,Y)\overset{S}{=}(K,T;X,Y)\]](http://latex.artofproblemsolving.com/1/5/6/156a06716e3353d813b940065dd67d3e117dee48.png)
.
and 
thus 
.
hence 
:
intersects 
meet at 
then prove that 
is tangent to 
which is 
q
are collinear.
be the altitudes.![\[(N,AD_{\infty};X,Y)=-1=(EF\cap BC,D;B,C)=(AQ,AD;AB,AC)=(AQ\cap XY,AD_{\infty};X,Y)\]](http://latex.artofproblemsolving.com/d/8/3/d838b19a98de42618c767e08e6279fc92232f4c1.png)
Hence 
passes through 
.![\[\measuredangle BMQ=90-\measuredangle QNB=\measuredangle MNQ\]](http://latex.artofproblemsolving.com/f/c/b/fcb4f0a81da79107991361d010be2668a4b84a37.png)
Set 
as the antipodes of 
.
yields 
are collinear thus, 
are collinear. Similarily 
are collinear. We get that 
.
and 
are concyclic.![\[\measuredangle QBM=180-\measuredangle MYQ=\measuredangle QYX\]](http://latex.artofproblemsolving.com/6/9/c/69c25db4cf14346a754554019608d05b127a1aeb.png)
Hence 
,
.
implies 
.![\[\cot B+\cot C=\frac{DB}{DA}+\frac{DC}{DA}=\frac{BC}{DA}=\frac{BC}{AC.\sin C}=\frac{\sin A}{\sin B.\sin C}\]](http://latex.artofproblemsolving.com/4/6/c/46cfc290f11ed5928fbdbe8ac7613d823ff7455b.png)
Which completes the proof of lemma
are concyclic.
.
if 
are the orthocenter of 
and 
then 
gives that 
h
.
hence 
.
and 
.![\[\frac{MX.MY}{MO.MR}\overset{?}{=}\frac{QA.QN}{\frac{QA^2}{2}}=\frac{2QN}{QA}=2.\frac{MN}{MO}.\frac{A'O}{A'A}=\frac{MN}{MO}\]](http://latex.artofproblemsolving.com/b/a/3/ba38bdad20cef0bc915629c244493a99dd5ac064.png)
![\[MX.MY\overset{?}{=}MN.MR=(\frac{MX+MY}{2})MR\iff \frac{2}{MR}\overset{?}{=}\frac{1}{MX}+\frac{1}{MY}\]](http://latex.artofproblemsolving.com/5/f/7/5f78c220965909c8446f9f663d605af70e8e2e56.png)
Multiply both sides by 
to get![\[\frac{BC}{MR}\overset{?}{=}\cot B+\cot C=\frac{\sin A}{\sin B.\sin C}\]](http://latex.artofproblemsolving.com/d/2/4/d244bf8becd2b2a09568641fa4ea3ae59a08b958.png)
Since 
![\[\frac{BC}{MR}=\frac{BC}{AC.\frac{MY}{CY}}=\frac{\sin A}{\sin B.\sin C}\]](http://latex.artofproblemsolving.com/6/3/9/6392edd6759f239f4f1a2face8de970076a2b8a0.png)
Which gives that 
.
is tangent to 
.
be the midpoint of 
and 
.![\[\frac{VQ}{VO}=\frac{NQ}{NA}.\frac{UO}{UA}=\frac{NQ}{NA}=\frac{MQ}{MH}=\frac{MQ}{MA'}\]](http://latex.artofproblemsolving.com/7/7/4/774b943ba4a1380ab10560e30d24e71695a4e948.png)
Hence 
.
because of coaxiality lemma and it lies on the perpendicular bisector of 
.
are collinear subsequently 
are tangent to each other as desired
.
,
,
.
intersect 
.
and let 
intersect 
,
centered at 
is the tangent to 
.
and 
,
bisects 
.
,
where 
.
chapter 
.
where 
.
,
under inversion about 
,
.
be the intersection of 
Since 
we can project onto AI to get 
Since 
,
bisects 
bisects 
from iran lemma. Hence, inverting about incircle swaps 
.
Thus, 
However, we know that the incircle inverts to itself. This is finished.
by the Iran lemma. So ![\[BP\cdot BA=BS\cdot BT=BX^2,\]](http://latex.artofproblemsolving.com/1/0/3/103554e7b9ff444f00a8c77499368184e3211f52.png)
so 
.
.
.
