[/quote]
,
that satisfy the equation
.
and 
be points on the side BC such that 
and 
,
be the incenter of triangles 
and 
,
is the Euler line of triangle 
.
,
,
,
,
are odd primes and![\[2^ap^b=(p+2)^c-1.\]](http://latex.artofproblemsolving.com/5/8/5/58518a8ebed14836b72af743ff100eae2efba5e2.png)
we have 
.
.
be reals such that 
and 
Prove that
be a cyclic quadrilateral. The point 
is the reflection of 
the intersection of 
is the reflection of 
.
be the reflection of 
.
concyclic.
. Prove that: 
at 
and ray 
at 
. The circumcircle of triangle 
intersects the circumcircle 
of triangle 
distinct from point 
and 
intersect 
and 
respectively.
such that 
for all real number 
and 
is an equilateral . Let 
and 
at 
and 
at 
,
on 
satisfies 
and 
.
be the point where the perpendicular from 
to line 
meets the angle bisector of 
.
holds.
be a positive integer such that 
.
.
Y by
The circle 
with center 
, is the 
-excircle of triangle 
. Which is tangent to 
at 
respectivly. Point 
is the reflection of through 
. Lines 
and 
meet at 
. Prove that ,circumcenter of 
, midpoint of 
and are collinear.
with center 
, is the 
-excircle of triangle 
. Which is tangent to 
at 
respectivly. Point 
is the reflection of through 
. Lines 
and 
meet at 
. Prove that ,circumcenter of 
, midpoint of 
and are collinear.This post has been edited 2 times. Last edited by Mr.C, Nov 18, 2020, 11:34 AM
and 
be the 
excenters. We can recast the problem in terms of the excentral triangle:
.
and define 
,
.
.
.
is on the 
.
.
,
such that 
.
.
.
,
.
,
.
is a parallelogram, and 
is defined similarly. 
that is the intersection of 
and 
.
is on the 
,
.
be the antipode of 
.
.
,
,
~
.
~
.
.
,
be the 
are the tangency points of 
e
,
so 
so 
by concurrency of lines formula.
be circumcenter of 
so
and 
be midpoint of 
and 
now it's obvious that 
.
.
are collinear.
meets the excircle at points 
such that 
is a diameter of the excircle, then note that 
becomes a right triagle and finish by noting that 
and thus 
is a tangent to the excircle
are parallel.
and 
is cyclic.
So the third lemma holds as well.
be the reflection of 
.
and 
are congruent.
by the Lemma 3. So 
and we have 
so 
is cyclic and we are done since 
is the common chord of the two circles and thus 
and DKE be T.Let the circumcentere of 
be O. we need to prove that 
are collinear.since 
is perpendicular to 
is perpendicular to 
is parallel to 
is perpendicular to EF we need to prove 
The same congruent and the fact that 
.so we are done.
,
be the circumcenter of 
to 
sends 
are concyclic. This followed by 
.
is tangent to 
be the tangent line of 
.
.
of center 
to denote the image of 
is the image of 
be the circumcenter of 
,
,
is indeed the perpendicular bisector of 
.
.
,
,
,
.
.
,
are concyclic. Hence 
.
implying that 
.
.
and 
.
.
.
(usually the incircle version appears.). But in 
,
is the orthocenter, so 
and the claim follows.
meet 
.
and 
are inversely similar. Let 
be the midpoint of 
and 
are correspondent in the two triangles. 
where the ratios are written in directed lengths. (Here we used the fact that 
are collinear. This is also known but usually the incircle version.) This means that if the perpendicular bisector of 
and 
meet 
at 
and 
,
,
.
are concyclic, this is their circumcenter, so we're done.
,
,
.
,
are concyclic. By Brocard Theorem, we are enough to prove that 
.
.
,
,
,
,
,
![[asy]
import math;
import olympiad;
import geometry;
size(10cm);
defaultpen(fontsize(11pt));
pair A = dir(130);
pair B = dir(210);
pair C = dir(330);
pair I = incenter(A,B,C);
pair Q = dir(270);
pair Ia = 2Q-I;
pair D = reflect(B,Ia)*A;
pair E = foot(Ia,A,C);
pair F = foot(Ia,A,B);
pair J = extension(B,C,Ia,E);
pair K = extension(D,Ia,E,F);
pair P = IP(circumcircle(D,K,E),Line(A,E,1),0);
pair H = (P+J)/2;
pair M = (B+C)/2;
pair N = foot(H,D,J);
pair L = foot(Ia,D,J);
draw(H--Ia,dotted+blue);
draw(circumcircle(D,K,E),dotted+red);
draw(D--P,red+blue);
draw(P--J,red+blue);
draw(rightanglemark(P,D,J));
draw(A--B--C--cycle,green+blue);
draw(circumcircle(A,B,C),blue);
draw(B--D,green);
draw(C--J,green);
draw(E--J,pink);
draw(B--F--Ia--E--C,pink);
draw(D--Ia,linewidth(0.5)+heavyred);
draw(F--E,linewidth(0.5)+heavyred);
draw(Ia--L,blue+dotted);
draw(D--A,pink);
draw(H--N,blue+dotted);
draw(P--A,pink);
dot("$N$",N,NW);
dot("$M$",M,NE);
dot("$H$",H,dir(60));
dot("$P$",P,dir(90));
dot("$K$",K,dir(90));
dot("$J$",J,dir(270));
dot("$E$",E,S);
dot("$F$",F,W);
dot("$D$",D,W);
dot("$A$",A,dir(180));
dot("$B$",B,SW);
dot("$C$",C,dir(270));
dot("$I$",I,dir(135));
dot("$I_A$",Ia,S);
dot("$L$",L,NE);
[/asy]](http://latex.artofproblemsolving.com/2/1/f/21fe1f348e3f347dc3a573f48546617b528a8784.png)
.
at 
collinear.
we have 
but 
.
which again implies 
which finally yields the collinearty of 
and 
which implies 
.
implying that the center of 
is just the midpoint of 
.
be the foot of perpendiculars from 
respectively onto 
.
to prove the collinearty of 
.
and 
.
while 
.
is just 
To prove this statement is true, we calculate parts of it separately, 
and 
Now ![\begin{align*}
\frac{c - \frac{s-b}{\cos C}}{b-c} &= \frac{(c+a) \cdot \frac{2 \cdot \Delta}{2s(s-c) - ab}}{2 \cdot \frac{\Delta}{s-a}} \\
\iff \frac{c[2s(s-c) - ab]-(s-b) \cdot ab}{b-c} &= (c+a) \cdot (s-a) \\
\iff sac+\cancel{sbc}+sc^2-ac^2-bc^2-c^3 - \cancel{abc} - sab + ab^2 &= sab - a^2b + \cancel{sbc} - \cancel{abc} - sca + a^2c - sc^2 + ac^2\\
\iff 2sac - bc^2 - c^3 &= 2sab-a^2b-ab^2+a^2c-2sc^2+2ac^2 \\
\iff a^2c + \cancel{abc} + ac^2 - bc^2 - \cancel{c^3} &= \cancel{a^2b + ab^2} + \cancel{abc} - \cancel{a^2b - ab^2} + a^2c - (a+b+\cancel{c})c^2 + 2ac^2 \\
\iff \cancel{a^2c} + ac^2 - bc^2 &= \cancel{a^2c} - ac^2 - bc^2 + 2ac^2 \\
\iff ac^2 - bc^2 &= ac^2 - bc^2 \\
\end{align*}](http://latex.artofproblemsolving.com/c/e/4/ce4869f873ea2f8c74531388a0b6add0e0c90a48.png)
which is true. This establishes 
be the in-touch , ex-touch point with 
is cyclic
such that 
)
so in order to prove that 
meet 
are center of 
is perpendicular to 
is cyclic.
is cyclic.
Note that 
.
.
and it's well known that 
so 
.
and 
are congruent.
and 
and 
.
.
and 
so 
so 
so 
.
,
,
meet the excircle at points 
,![\[\angle ADC = \angle IBC = \angle II_AC = \angle AI_AC. \ \square\]](http://latex.artofproblemsolving.com/c/4/0/c40a4905a15a0f9acc916dc4c03df2dd8c4b9379.png)
.
.
is 
.
.
,
.![\[\angle E'I_AX = \angle T_AI_AE = \angle ACB = \angle AI_AD,\]](http://latex.artofproblemsolving.com/d/a/4/da4c2a595dceecff000bf8a89dfeb940c5153fef.png)
,
,![\[\angle E'DK = \angle E'DI_A = \angle AI_AX = \angle XAI_A = \angle MI_AA.\]](http://latex.artofproblemsolving.com/0/2/9/029284a9ceea3e74b210240481b75f3240148688.png)
.![\[\angle E'DK = \angle MI_AA = \angle E'EF = \angle E'EK,\]](http://latex.artofproblemsolving.com/e/d/7/ed70cd313ceae896fdbb0456e1d86fa6e9ec784c.png)
,
be the 
.
and 
,
is a common side. Further,![\[\measuredangle I_ABD' = \measuredangle I_ABC = \measuredangle FBI_A = \measuredangle I_ABA\]](http://latex.artofproblemsolving.com/3/3/e/33e6ee5b0bb169c93a36b1a51fa89350ea28d9f4.png)
which proves that 
.
and indeed ![\[\measuredangle DI_AA =2 \measuredangle BI_AA = 2 (\measuredangle I_ABF + \measuredangle BAI_A) = \measuredangle CBF + \measuredangle BAC = \measuredangle BCA = \measuredangle DCA\]](http://latex.artofproblemsolving.com/d/f/d/dfd340025c427e5ec5631b713ea1d211583fcf3a.png)
which implies the rest of the claim.
and 
.
so 
.
maps to arc 
which implies that 
.
is clearly cyclic, we have that 
.
.
and thus, 
which implies that 
.
and 
,
.![\[\measuredangle KEL = \measuredangle FEL = \measuredangle AI_AM = \measuredangle I_AAS = \measuredangle I_ADL = \measuredangle KDL\]](http://latex.artofproblemsolving.com/2/e/4/2e49dcd658dee54a67ba6209c5be091dddb2297c.png)
which implies that 
.
(radical axis) which implies that 
as desired.
lies on 
lies on 
.
lies on 
once the problem statement is proved.
is cyclic, and 
are collinear.
. The second is easy.
as the second intersections of 
with 
Let 
in circle 
the line tangent to 
at 
Note that 
so 
We also have 
so 
is cyclic. We also have 
but 
so 
We also have 
cyclic as 
Then, note that 
so 
Then, since 
we have 
Thus, 
Similarly, 
This finishes.
is 
