such that 
is a perfect square.
be all the positive integers smaller than 
such that
for all 
.
such that we can choose 
natural numbers 
,
Y by mathematicsy, CoolJupiter, Adventure10, Mango247
Let the incircle of a nonisosceles triangle 
touch 
, 
and 
at points 
, 
and 
respectively. The corresponding excircle touches the side at point 
. Let 
be the common point of 
and the incircle, closest to , and 
be the common point of 
and 
. Prove that 
.
touch 
, 
and 
at points 
, 
and 
respectively. The corresponding excircle touches the side at point 
. Let 
be the common point of 
and the incircle, closest to , and 
be the common point of 
and 
. Prove that 
.
be the incircle of 
.
,
is the point diametrically opposite 
is 
.
,
and 
be the line through 
parallel to 
median then the rest is trivial
refers to directed angles.
be the incentre and ![[asy]
import geometry; import olympiad;
unitsize(4cm);pair A = dir(135), B = dir(210), C = -1/B;
pair I = incenter(A, B, C);
pair D = foot(I, B, C), E = foot(I, C, A), F = foot(I, A, B);
pair T_ = 2I - D, N = B + C - D, T = intersectionpoints(incircle(A,B,C), A -- N)[1];
pair K = extension(E, F, D, T);
pair L = extension(D, I, A, K);
draw(circumcircle(A,E,F)^^circumcircle(D, T, A), linetype("0 2")); draw(D--L, linetype("2 4"));
draw(A--B--C--cycle); draw(incircle(A,B,C)); draw(A--N); draw(D--K--F); draw(A--K, dashed);
draw(rightanglemark(A, L, D, 1.5)^^rightanglemark(I, D, B, 1.5));
draw(rightanglemark(A, F, I, 1.5)^^rightanglemark(A, E, I, 1.5));
draw(rightanglemark(A, T, D, 1.5));
draw(I--F^^I--E, linetype("2 5"));
dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(T); dot(T_); dot(N); dot(K); dot(I); dot(L);
label("A", A, dir(90));
label("B", B, dir(-90));
label("C", C, dir(-90));
label("D", D, dir(-90));
label("E", E, dir(90));
label("F", F, -dir(30));
label("K", K, dir(90));
label("T$'$", T_, dir(-135));
label("T", T, dir(-7));
label("I", I, dir(-45));
label("N", N, dir(-90));
label("L", L, dir(90));[/asy]](http://latex.artofproblemsolving.com/1/7/6/17683b1851b292373700bf0479d24b6665b21dbd.png)
are concyclic with diameter 
.![\[\measuredangle AEI = 90^{\circ} = \measuredangle AFI\]](http://latex.artofproblemsolving.com/8/7/7/877120eed02179c1a9c7530890b852803c843be4.png)
![\[\measuredangle ATD = 90^{\circ}\]](http://latex.artofproblemsolving.com/9/b/3/9b3bf32dba5f84cd72b1d481b6e00e45c5f75712.png)
,
.
is a diameter of 
.
are collinear and 
.
and 
,
is the radical axis of 
is the radical axis of 
and 
.
and 
,
.
meets 
,
and ![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(6cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -8.176743075134857, xmax = 11.428752031839883, ymin = -3.7969421960496423, ymax = 7.034290346656071; /* image dimensions */
pair A = (0.,5.), B = (-2.,0.), C = (5.,0.), D = (0.6570484595545034,0.), X = (2.3429513665770023,0.);
draw(A--B--C--cycle, linewidth(1.));
/* draw figures */
draw(A--B, linewidth(1.));
draw(B--C, linewidth(1.));
draw(C--A, linewidth(1.));
draw(circle((0.6570484976345142,1.798909413008393), 1.7989094130083934), linewidth(1.));
draw((0.6570484595545034,3.597818826016786)--D, linewidth(1.));
draw(X--A, linewidth(1.));
/* dots and labels */
dot(A,dotstyle);
label("$A$", A,dir(90));
dot(B,dotstyle);
label("$B$", B,dir(180));
dot(C,dotstyle);
label("$C$", C,dir(0));
dot(D,linewidth(4.pt) + dotstyle);
label("$D$", D,dir(-90));
dot((0.6570484595545034,3.597818826016786),linewidth(4.pt) + dotstyle);
label("$E$", (0.6570484595545034,3.597818826016786),dir(-135));
dot(X,linewidth(4.pt) + dotstyle);
label("$X$", X,dir(-90));
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/5/b/d/5bd0babd042ab4b6bb32a68205f401ed7e5f4d37.png)
and 
respectively. As 
.
centered at 
to 
which is the incircle to the excircle of ![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -7.98, xmax = 16.42, ymin = -4.14, ymax = 9.34; /* image dimensions */
pair A = (0.,5.), B = (-2.,0.), C = (5.,0.), D = (-1.0131969613578624,2.4670075966053435), F = (0.6570484595545034,0.), T = (2.03941567143766,0.6477638648687565);
draw(A--B--C--cycle, linewidth(1.));
/* draw figures */
draw(A--B, linewidth(1.));
draw(B--C, linewidth(1.));
draw(C--A, linewidth(1.));
draw(circle((0.6570484976345142,1.798909413008393), 1.7989094130083934), linewidth(1.));
draw(A--(2.3429515404454966,0.), linewidth(1.));
draw(A--(11.327350454273194,5.), linewidth(1.));
draw(F--(11.327350454273194,5.), linewidth(1.));
draw((11.327350454273194,5.)--D, linewidth(1.));
/* dots and labels */
dot(A,dotstyle);
label("$A$", (0.08,5.2), NE * labelscalefactor);
dot(B,dotstyle);
label("$B$", (-2.3,-0.38), NE * labelscalefactor);
dot(C,dotstyle);
label("$C$", (5.08,0.2), NE * labelscalefactor);
dot(D,linewidth(4.pt) + dotstyle);
label("$D$", (-1.46,2.3), NE * labelscalefactor);
dot((1.9290695423130608,3.070930457686939),linewidth(4.pt) + dotstyle);
label("$E$", (2.,3.24), NE * labelscalefactor);
dot(F,linewidth(4.pt) + dotstyle);
label("$F$", (0.6,-0.38), NE * labelscalefactor);
dot((2.3429515404454966,0.),linewidth(4.pt) + dotstyle);
label("$N$", (2.34,-0.42), NE * labelscalefactor);
dot(T,linewidth(4.pt) + dotstyle);
label("$T$", (2.2,0.42), NE * labelscalefactor);
dot((11.327350454273194,5.),linewidth(4.pt) + dotstyle);
label("$K'$", (11.4,5.16), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/9/c/4/9c453ff80361494d0162a7a20827c6cbfaf8c915.png)
.
.
and 
.
and using the lemma, we'll be done.
,![\[\frac{s-b}{b}=\frac{\sin(\alpha - C)}{\sin \alpha}=\cos C - \sin C\cot \alpha.\]](http://latex.artofproblemsolving.com/8/0/1/801e922637cb8d60401a8e05f86e97ec011838bb.png)
So, 
.
,![\[\frac{AK'}{AD}=\frac{\sin\left(\frac{B+C}{2}\right)}{\sin\left(\frac{B-C}{2}\right)}\cdot \frac{\cos\left(\frac{B+C}{2}\right)}{\cos\left(\frac{B+C}{2}\right)}=\frac{\sin(B+C)}{\sin B-\sin C}=\frac{\sin A}{\sin B-\sin C}.\]](http://latex.artofproblemsolving.com/c/1/c/c1cf85916facfe3707b0fdf468dff3e358395cd2.png)
So, 
.
at 
,![\[AX=\frac{AC\cdot AK}{FC+AK}=\frac{b\cdot AK}{FC+AK}.\]](http://latex.artofproblemsolving.com/d/9/7/d97debfb0c900fdc9251f96dca83cbf5f4e50050.png)
.![\[\cot \beta = \frac{\frac{AK'}{AX}-\cos C}{\sin C}=\frac{FC+AK}{b\sin C}-\cot C = \frac{s-c+\frac{(s-a)a}{b-c}}{b\sin C}-\cot C.\]](http://latex.artofproblemsolving.com/4/e/4/4e4c308c54c51f5c748ebaaa9739e7e85b64344a.png)
or 
.
or ![\[b^2=ab\cos C - (s-b)(s-c)+\left(\frac{(s-a)a}{b-c}\right)\left(b\cos C - (s-b)\right).\]](http://latex.artofproblemsolving.com/e/5/e/e5ec295ee4a8ffe7761bf7f332f926f84293bad8.png)
,![\[0=\frac{a^2-b^2-c^2}{2}-\frac{(a+c-b)(a+b-c)}{4}+\left(\frac{a(s-a)}{b-c}\right)\left(\frac{a^2+b^2-c^2}{2a}-(s-b)\right).\]](http://latex.artofproblemsolving.com/6/0/0/600f7ae3623af9273b49851f7df96457f4449190.png)
.
or 
.
or 
which is clearly true.
be the point diametrically opposite 
.
be the midpoint of 
colinear.
Also:
So 
as desired.
so 
and project from A we get thd desired result.
meet on the incircle at the antipode of 
are collinear. Now, let 
meet the line parallel to 
,
.
.
is cyclic from 
.
from 
is cyclic and 
,
are concurrent. Hence 
,
.
is diametrically op. to 
is a harmonic quad. Since 
collinear
.Where 
.
are harmonic pencils ! where 
is parallel to 
be second intersections of 
with the incircle of 
at 
,
are harmonic quadrilaterals, we have: 
= 
,
= 
= 
or 
meet the incircle again at a point 
,
and 
meet at a point 
be the intersection of lines 
is collinear, so 
,
,
.
collinear. Since 
is a diameter, 
.
,
at 
respectively.
be the line through 
be the antipode of 
.
,
are concurrent.
passes through 
by parallel lines.
and 
are cyclic.
are collinear, 
proving the first claim.
which proves our second claim. 
,
,
,
.
circles are collinear, then the 
be a point such that 
so 
lies on the polar of 
.
so we are done.
concur ,at 
,
let 
and let 
It is well-known that 
Notice 
so 
is the point at infinity along 
,
,
then its known that if u let 
are colinear, now taking polars w.r.t. 
concurrent and that means 
and the line through 
where 
Projecting (with perspective at 
it suffices to show that 
where 
However, it is well-known that 
where 
where 
and 
.
.
which along with the fact that 
and as 
is tangent to the incircle, we have that 
is tangent to the incircle.
and 
.
lies on the polar of 
,
becomes that polar of 
is the line 
.
and we are done.
-
such that 
.
are collinear, so 
and 
can easily be seen to be cyclic, and by radical center on the incircle, 
,
,
,
,
concur, hence 
.
passes through 
It is well-known that 
is just the midpoint of 
which implies the desired result. 
and 
and the incircle. Then from EGMO chapter 4, we know that 
.
.
is tangent to the incircle, this is enough to imply that line 
But since 
,
,
,
.
.
onto 
and 
are concurrent. Let 
so that 
.
.
concur at some point 
be the other intersection point of 
is a diameter of the incircle.![\[ (DE;ST) \overset{F}= (DE;PK) \overset{A}= (B,C;M,\overline{AK} \cap \overline{BC}) = -1 \]](http://latex.artofproblemsolving.com/0/9/e/09ec0b588794b3e7e9863549622b8de0845af24c.png)
which implies that 
intersect at 
,
denote the incircle. Define 
.
is a diameter of 
.
.
Now projecting through 
so we indeed have 
through Diameter of the Incircle lemma.
is the pole of 
.
,
concur by an incircle concurrence lemma.
,
.
,
,
with 
,
.
.
is the polar of 
is tangent to 
,
is a right triangle.
are collinear.
,
,
,
.![\[(-a+b+c)x+(-a+b-c)y+(-a-b+c)z=0,\]](http://latex.artofproblemsolving.com/d/f/f/dfff949c49cae5f346f134ca66b3e5c21d134c08.png)
line ![\[(a-b-c)(c-b)x+a(a-b+c)y+a(-a-b+c)z=0,\]](http://latex.artofproblemsolving.com/c/a/6/ca674a37cc0b1d9d6c655bb8418be5358d7d485f.png)
and line ![\[(0)x+(1)y+(-1)z=0.\]](http://latex.artofproblemsolving.com/c/f/a/cfacd23ed3ee2b980cfeeb8bca9a79df33839a60.png)
Now we plug everything into the concurrence lemma:
as desired.![\[
-1=(Z,Y;P,T)\stackrel{X}{=}(Z,Y;Q,K).
\]](http://latex.artofproblemsolving.com/9/6/6/966390ec9e48635548387e793d367b9078d44b00.png)
However, since![\[
-1=(B,C;M,\infty_{BC})\stackrel{A}{=}(Z,Y;Q,(A\infty_{BC}\cap ZY)),
\]](http://latex.artofproblemsolving.com/f/f/8/ff89b604fe7859cb57c73454a1f4bb57a32bd84d.png)
we indeed have 
,
and 
.
and also 
.
.
.
and since FM = MN, then MT = MF and as MF is tangent to the incircle, we get that MT is also tangent to the incircle. Now its obvious that the polar of A is DE 
lies on the polar of A. Now by La Hire, A should lie on the polar of 
and since D and E lie on a circle, we have that X lies on the polar of 
is the line 
so 
,
so 
is cyclic. Thus 
is also cyclic, so 
(follows by taking inverses of lines 
concur.
Since (where 
and 
and 
we know 
so 
are collinear, and we are done.
denote a harmonic bundle.
.
.
be a point on 
and 
.
On the other hand, using the ratio property of symmedians, we obtain 
We have therefore proven 
and thus 
is the antipode of 
.
.
is harmonic, we have![\[-1 = (Z,Y;T',T) \overset{X}{=} (Z,Y;P,K) \overset{A}{=} (B,C;M,Q).\]](http://latex.artofproblemsolving.com/6/f/e/6fea4d27d814f01286214cd0efe86b1d0dab7b26.png)
,
,
,
lies on the polar of 
,
is the midpoint of 
and 
be 
.
as per the OTIS problem. Let 
are collinear (proof by a homothety sending the incircle to the 
be the 
implies 
.
implies 
collinear.
.
says 
are collinear.
Clearly, 
We have that 
so 
so 
so 
so 
lie on the polar of 
Thus, 
which finishes
,
intersects 
at 
and 
intersects 
and 
implies that 
collinear. It is well-known that 
passes through 
implies that 
and since 
