[/quote]
,
satisfy 
?
such that![\[f(x+yf(x))+f(xf(y)-y)=f(x)-f(y)+2xy\]](http://latex.artofproblemsolving.com/e/b/c/ebccae7e765d9fc3c8605e58a7399a70127b39b1.png)
, such that 
Prove that :
equality cases : ?
distinct positive integers such that adding the product of any two of them to 
yields a perfect square?
be a sequence of positive integers with 
,
and![\[a_n = a_{n-1}^{a_{n-1}a_{n-2}}-1\]](http://latex.artofproblemsolving.com/1/b/f/1bf1a88979e7d2b33530de979d66a4e81032953d.png)
for all 
.
is a prime less than 
for some positive integer 
,
such that 
.
for which
be an acute angled scalene triangle with circumcircle 
.
and 
be the circumcenter and orthocenter of 
respectively. Let 
and 
be points on segments 
and 
Prove that 
and 
intersect on the circumcircle of 
.
grid with 
rows and 
columns, and each square of the grid is filled with a non-negative integer. Let 
be the average of all the numbers in the grid. Prove that if 
the there exist two identical rows (meaning same numbers in the same order).
is called feirense if it is possible to write on a sheet of paper some integers such that every positive divisor of 
Y by fatant, trumpeter, anantmudgal09, solver1104, Ultroid999OCPN, popcorn1, scrabbler94, Stormersyle, ppu, Centralorbit, mathleticguyyy, Vietjung, myh2910, Bradygho, samrocksnature, megarnie, aopsuser305, Adventure10
Let 
be a triangle with 
obtuse. The 
-excircle is a circle in the exterior of 
that is tangent to side 
of the triangle and tangent to the extensions of the other two sides. Let 
, 
be the feet of the altitudes from 
and 
to lines 
and 
, respectively. Can line 
be tangent to the -excircle?
Proposed by Ankan Bhattacharya, Zack Chroman, and Anant Mudgal
be a triangle with 
obtuse. The 
-excircle is a circle in the exterior of 
that is tangent to side 
of the triangle and tangent to the extensions of the other two sides. Let 
, 
be the feet of the altitudes from 
and 
to lines 
and 
, respectively. Can line 
be tangent to the -excircle?Proposed by Ankan Bhattacharya, Zack Chroman, and Anant Mudgal
This post has been edited 3 times. Last edited by 62861, May 6, 2019, 4:09 PM
is similar to 
,
.
is isoceles, ![$$[ABC] = \frac{1}{2} \cdot AC \cdot BE = \frac{AC}{2} \cdot \sqrt{AB^2 - AE^2} = \frac{AC}{2} \cdot \sqrt{AB^2 - AB'^2} = \frac{AC}{2} \cdot \sqrt{AB^2 - AB^2} = 0,$$](http://latex.artofproblemsolving.com/9/0/7/90780c93e1f538eb3c6a7787aea933d48634bdf5.png)
contradiction. 
,
is congruent to 
.
which means that 
,![[asy]
pair A, B, C, E, F, Ep, Fp, I, J, K, L;
//A = dir(120);
//B = 2*foot(origin, A, A+dir(250)) - A;
//C = 2*foot(origin, A, A+dir(290)) - A;
A = dir(155); B = dir(230); C = dir(310);
E = foot(B, A, C);
F = foot(C, A, B);
I = incenter(A, B, C);
J = 2*circumcenter(B, I, C) - I;
K = foot(J, A, B);
L = foot(J, A, C);
Ep = reflect(A, I) * E;
Fp = reflect(A, I) * F;
draw(B--K^^C--L, rgb(1, 0.7, 0.2));
//draw(unitcircle, dotted);
draw(A--B^^A--C, rgb(0.7, 0.3, 0));
draw(B--C, heavycyan);
draw(E--F, heavycyan);
draw(Ep--Fp, heavycyan + dashed);
filldraw(circle(J, abs(J-foot(J, B, C))), opacity(0.5) + pink, red);
draw(B--E^^C--F, heavymagenta);
draw(E--Ep^^F--Fp, dotted);
dot("$A$", A, dir(110));
dot("$B$", B, dir(190));
dot("$C$", C, dir(50));
dot("$E$", E, dir(50));
dot("$F$", F, dir(190));
dot("$E'$", Ep, dir(190));
dot("$F'$", Fp, dir(50));
[/asy]](http://latex.artofproblemsolving.com/b/5/8/b58fafb70380f290482eb35f0e072fa2a17bfe12.png)
.
and 
.
must also be tangent to the 
is cyclic, we obtain 
,
.
is a chord in the circle with diameter 
,
.
then 
too, so then 
lies inside 
.
is a rectangle. In particular 
.
and 
must intersect, contradiction.
,
,
,
,
.![[asy]
size(6cm);
defaultpen(fontsize(9pt));
pen pri=royalblue;
pen sec=Cyan;
pen tri=deepgreen;
pen fil=invisible;
pair A, B, C, EE, IA, Bp, Cp, SS, F;
A=dir(140);
B=dir(230);
C=dir(310);
EE=foot(B, A, C);
IA=2*dir(270)-incenter(A, B, C);
Bp=foot(IA, A, C);
Cp=foot(IA, A, B);
SS=intersectionpoints(circle(IA, length(Bp-IA)), circle((EE+IA)/2, length(EE-IA)/2))[1];
F=extension(A, B, EE, SS);
draw(extension(A, B, IA-(0, length(IA-Bp)), IA-(1, length(IA-Bp))) -- A -- extension(A, C, IA-(0, length(IA-Bp)), IA-(1, length(IA-Bp))), pri);
draw(B -- C, pri);
filldraw(circle(IA, length(Bp-IA)), fil, pri);
fill(A--B--C--cycle,fil);
draw(B -- EE, sec); draw(C -- F, sec);
draw(EE -- F, tri); draw(Bp -- IA -- Cp, pri); draw(SS -- IA, tri);
dot("$A$", A, N);
dot("$B$", B, W);
dot("$C$", C, NE);
dot("$E$", EE, NE);
dot("$I_A$", IA, SE);
dot("$B'$", Bp, NE);
dot("$C'$", Cp, W);
dot("$S$", SS, dir(120));
dot("$F$", F, W);
[/asy]](http://latex.artofproblemsolving.com/b/9/4/b949bb7c51b611ea291b4822e860e539cc22ee5d.png)
denotes the semiperimeter, then
from which 
,
by simple angle chasing. However, 
.
by right triangle 
,
,
with 
with positive integer coefficients 
or 
all lie on one circle?
My post was just answering why 
.
is on the internal bisectors of 
and 
.
of the circumcircle of 
.
cyclic. The conclusion follows from there.
.
)
)
as the 
shares the same excenter as 
will bisect the exterior angle of 
and 
will bisect the exterior angle of 
.
and 
because quadrilateral 
,
is cyclic since 
.
.
,
and 
,
.
,
.
,
,
,
,
is cyclic. This is true because 
.
.
,
is the semiperimeter of 
.
.
.
.
,
,
.
-
is isosceles with 
,
and 
.
Then
which is impossible by the triangle inequality.
denote the 
denote the 
be the line passing through 
and 
.
It's clear that 
.
where the last inequality follows from 
.
which finishes. 
to argue that this is impossible.)
and 
be the points at which the 
was tangent to the 
.![\[ FT = FT_B = BT_B - BF = \frac{a + b - c}{2} + a\cos B,\]](http://latex.artofproblemsolving.com/8/5/0/8507680ea55854754edcd5c929cd9390a9d4fd97.png)
![\[ ET = ET_C = EC + CT_C = a\cos C + \frac{a + c - b}{2}.\]](http://latex.artofproblemsolving.com/6/3/3/633554115dfe716c70c44c4f1c5377a23f0f1cf7.png)
Also, note that 
.![\[ EF\cdot a = a\sin C\cdot a\sin B - a\cos B\cdot a\cos C = -a^2\cos (B+ C),\]](http://latex.artofproblemsolving.com/6/7/d/67ddbaaf8c8c0560a1b867ec426e16aabbf5fd54.png)
so 
.![\[ EF = FT + ET\implies -a\cos (B+C) = a + a\cos B + a\cos C\implies \cos (B+C) + \cos B + \cos C + 1 =0.\]](http://latex.artofproblemsolving.com/2/6/b/26b9c66b4d33a121ecdba774afb3ac9079102860.png)
Now, let 
and 
.![\[ xy - \sqrt{(1-x^2)(1-y)^2} + x + y + 1 = 0,\]](http://latex.artofproblemsolving.com/e/8/4/e8401fca259716e2b9f09e8111d5b2c6602b8a59.png)
![\[ (x+1)(y+1) = \sqrt{(1-x)(1-y)(1+x)(1+y},\]](http://latex.artofproblemsolving.com/9/4/5/9450899b06071a64e2658d76f2f7fa0b3aebe03d.png)
![\[ \sqrt{(x+1)(y+1)} =\sqrt{(1-x)(1-y)}\implies xy + x + y + 1 = xy - x - y + 1\implies x = -y.\]](http://latex.artofproblemsolving.com/e/e/6/ee6d67ebeaf53ae3cf6fd39e6f823e4b039ac297.png)
Hence, 
.
,
.
(since tangents are equal length, and EF can be tangent anywhere from E to F.)
from orthic triangle properties, and 
s
and factoring gives 
,
so we need 
,
which gives 
degrees which is impossible.
.
remains same after inversion, also image of excircles under inversion must be tangent to these lines. If we assume &EF& was tangent to A-excircle then it's image must be tangent to 
.
