[/quote]
,
called ‘nice’ if and only if 
and 
are both perfect squares. How many positive integers 
such that 
are composites for all nice numbers 
?
decide whether there exist pairwise distinct positive integers 
such that for every 
,
divides 
.
of positive integers with the following property: all of the terms of the sequence 
have a greatest common divisor 
be real numbers such that 
find the largest value of 
such that 
and 
are all perfect square.
be pairwise coprime positive integers with 
being prime and 
.![$I = [0, a_1 a_2 \cdots a_n ]$](http://latex.artofproblemsolving.com/a/2/b/a2bce96b048c9b8fc94926db80aba37fd5037b4a.png)
of the real line, mark all integers that are divisible by at least one of the numbers 
. These points split 
into a number of smaller segments. Prove that the sum of the squares of the lengths of these segments is divisible by 
such that ![\[f(xy-1)+f(x)f(y)=2xy-1,\]](http://latex.artofproblemsolving.com/8/8/8/888ca39f2b7f8cec6d6426bee28d40eade40a66e.png)
for any real numbers 
and 
be a regular polygon, and let 
be its set of vertices. Each point in 
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. 
be the A-excircle. Then, 
,
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.
then has to lie on the angle bisector of 
,
,
,
,
.![[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 length of 
is the area of 
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 
.
