are positive real numbers, then
and 
Prove that
such that 
is a perfect square for all integers 
and 
.
Prove that
Y by AlastorMoody, megarnie, Adventure10, Mango247, Funcshun840
Let 
be a triangle with its inceter 
, incircle 
, and let 
be a midpoint of the side 
. A line through the point 
perpendicular to the line and a line through the point perpendicular to the line 
meet at 
. Show that a circle with line segment 
as the diameter touches .
be a triangle with its inceter 
, incircle 
, and let 
be a midpoint of the side 
. A line through the point 
perpendicular to the line and a line through the point perpendicular to the line 
meet at 
. Show that a circle with line segment 
as the diameter touches .
is expected to have the property that 
is tangent to 
.
by the tangency point where 
,
,
,
,
;
.
;
is self-polar WRT 
be the antipode of 
WRT 
and 
by 
,
sends 
.
are tangent at 
.
,
,or 
lies on the polar of 
,
,
.
,
as the foot from 
,
as the 
as the antipode of 
,
as the second intersection of 
(this collinearity is well-known) with 
be the point on 
.
,
.
,
;
,
.
;
,
,
are tangent at 
intersect 
and let 
,
is a parallelogram, which means that 
,
is also a parallelogram and 
.
,
and we are done by homothety.
be the tangent from 
be the antipode of 
are collinear since if 
,
.
we obtain that 
.
as both of them are perpendicular to 
and mapping 
to 
and the midpoint of 
be the reflection of 
(which also lies on 
and 
.
is the 
.
,
is cyclic. Then applying Radical Axes Theorem on 
,
are concurrent, i.e. 
.
,
and 
.
This means that 
is the common external tangent of 
![[asy]
size(9cm,0);
defaultpen(fontsize(10pt));
pair A = (1,3.5);
pair B = (0,0);
pair C = (5,0);
pair I = incenter(A,B,C);
pair D = foot(I,B,C);
pair M = (B+C)/2;
pair D1 = 2*I-D;
pair P = 2*foot(I,A,D1)-D1;
pair K = extension(A,foot(A,B,C),D,P);
draw(A--B--C--cycle,linewidth(1.5));
draw(incircle(A,B,C));
draw(K--A--P);
draw(K--P,dashed);
draw(D--D1);
draw(circumcircle(A,P,K),dashed);
dot(A);dot(B);dot(C);dot(I);dot(D);dot(M);dot(D1);dot(P);dot(K);
label("$A$",A,N);
label("$B$",B,dir(145));
label("$C$",C,SE);
label("$D$",D,NW);
label("$M$",M,S);
label("$K$",K,S);
label("$I$",I,W);
label("$D'$",D1,dir(70));
label("$P$",P,SE);
[/asy]](http://latex.artofproblemsolving.com/a/8/9/a896e8f5e972a9ce6d64de86c13e3cbf45412e36.png)
.
and 
thus 
.
thus 
are colinear. This implies that 
or 
.
thus 
and 
+ 
= 
+ 
+ 
+ 
+ 
+ 
or 
= 
= 
be feet of perpendiculars from 
respectively on 
and 
.
.
is the nagel cevian hence 
or 
is tagent to 
.
concur.
is cyclic with diameter 
finishes the proof of the Lemma
is nothing but 
.
hence 
is the external homothety center of 
tangent as desired
![[asy]
unitsize(4cm);
pair A, B, C, D, D1, I, R, S, L, M, X, K;
A = dir(135);
B = dir(210);
C = dir(330);
M = (B + C) / 2;
I = incenter(A, B, C);
D = foot(I, B, C);
D1 = 2 * M - D;
X = foot(D, A, D1);
R = IP(Line(A, I, 10), Line(B, C, 10));
S = foot(A, B, C);
K = IP(Line(X, D, 10), Line(A, S, 10));
L = IP(Line(A, I, 10), Line(M, K, 10));
draw(unitcircle, heavyred);
draw(A--B--C--A, heavyred);
draw(incircle(A, B, C), blue);
draw(A--K^^X--K^^M--K, heavygreen);
draw(A--L, orange);
draw(CP(M, D), heavygreen);
draw(A--D1, heavygreen);
dot("$A$", A, N);
dot("$B$", B, SW);
dot("$C$", C, SE);
dot("$D$", D, S);
dot("$I$", I, W);
dot("$D'$", D1, S);
dot("$X$", X, NE);
dot("$R$", R, S);
dot("$S$", S, S);
dot("$K$", K, S);
dot("$L$", L, SE);
dot("$M$", M, S);
[/asy]](http://latex.artofproblemsolving.com/7/7/f/77fec2b138e6ee6fd9f856a44ef69f6ca1f3f15b.png)
,
denote the feet of the 
.
.
is tangent to 
and 
wrt 
,
.
.
(since 
,
.
be 
Hence, 
and 
.
,
are collinear. Therefore, our redefined point 
,
passes through the antipode of 
lies on both 
.
to 
,
passes through the antipode of 
and 
,
so in fact 
is also tangent to 
be the ![[asy] pair A = dir(135); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); pair M = (B+C)/2; filldraw(incircle(A, B, C), invisible, deepcyan); filldraw(A--B--C--cycle, invisible, deepcyan); pair D = foot(I, B, C); pair E = 2*M-D; pair H = 2*I-D; pair T = foot(D, A, E); pair F = foot(A, B, C); pair Kp = extension(A, F, T, D); pair L = foot(A, M, Kp);
filldraw(circumcircle(T, M, D), invisible, orange); draw(A--T--Kp, lightred); draw(T--E, lightred); draw(A--Kp, lightblue); draw(M--Kp, deepgreen); draw(I--L, deepgreen); draw(A--I, deepgreen+dotted);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, dir(I)); dot("$M$", M, dir(M)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot(H); dot("$T$", T, dir(10)); dot("$F$", F, dir(235)); dot("$K'$", Kp, dir(Kp)); dot("$L$", L, dir(L));
/* TSQ Source:
A = dir 135 B = dir 210 C = dir 330 I = incenter A B C M = (B+C)/2 incircle A B C 0.1 lightcyan / deepcyan A--B--C--cycle 0.1 lightcyan / deepcyan D = foot I B C E = 2*M-D H .= 2*I-D T = foot D A E R10 F = foot A B C R235 K' = extension A F T D L = foot A M Kp
circumcircle T M D 0.1 yellow / orange A--T--Kp lightred T--E lightred A--Kp lightblue M--Kp deepgreen I--L deepgreen A--I deepgreen dotted
*/ [/asy]](http://latex.artofproblemsolving.com/1/3/9/139d6959d1924c954eeaf1ab361154a74ebb2210.png)
with 
is certainly tangent to 
.
with 
,
.
is cyclic.
.
as well. Thus we conclude 
are perpendicular at 
by 
,
meet 
where 
since 
it's suffice to prove 
so 
is projection of 
on 
and therefore 
is radical axis of 
Thus 
and 
