,
and 
denote the midpoints of 
and 
.
be the projection of 
onto 
be the reflection of 
.
.
be given as: 
,
(where 
can be 0), and let 
be the count of numbers 
such that 
and 
is considered the same as 
)
is 
,
,
,
are positive integers, and 
are odd primes.
be positive integers, and let 
all be 
-![\[
\alpha_1^j + \alpha_2^j + \cdots + \alpha_n^j = 0 \quad \text{for any } j (0 < j < k).
\]](http://latex.artofproblemsolving.com/3/3/b/33be8eef29cd506bf1132a86def9992ec8d21cb4.png)
Prove that among 
be the orthocenter of acute triangle 
,
be the reflection of 
intersects line 
and 
.
.
Y by VulcanForge, itised, ETS1331, tree_3, centslordm, Smileyklaws, megarnie, tigerzhang, HrishiP, ike.chen, OliverA, ihatemath123, rayfish, samrocksnature, e_plus_pi, asimov, son7, ImSh95, crazyeyemoody907, rama1728, itslumi, pearlhaas, deplasmanyollari, pomodor_ap, Funcshun840
Let 
be a triangle with 
, incenter 
, and 
excenter 
. The incircle meets 
at 
. Define 
, 
. Show that the circumcircle of 
and 
are tangent to each other
be a triangle with 
, incenter 
, and 
excenter 
. The incircle meets 
at 
. Define 
, 
. Show that the circumcircle of 
and 
are tangent to each otherThis post has been edited 1 time. Last edited by Eyed, Jul 20, 2021, 8:51 PM
have orthocenter 
,
be the foot from 
.
intersects 
at 
respectively, show 
and 
are tangent.
be the 
.
all lie on the circle with diameter 
.
,
and 
.
says 
cyclic, and finally
says that 
and 
are in fact tangent at 
.
.
,
,
and 
.
.
is cyclic
,
.
We can easily find, 
and 
.
.
:
So, 
cyclic.
and 
are tangent at 
.
Hence, proven
(normally 
)
be the midpoint of 
be the inverse of 
be the midpoint of 
.
intersects 
and 
at 
is tangent to 
(at 
is the diameter of 
.
is the diameter of 
,
,
,
"
,
is tangent to 
is the midpoint of 
.![\[\angle PBD\stackrel{(1)}{=}\angle PDM_B\stackrel{(2)}{=}\angle TDM_B\stackrel{(3)}{=}\angle TM_AM_B\stackrel{(4)}{=}\angle TAB\stackrel{(5)}{=}\angle M_AAB\stackrel{(6)}{=}\angle B'BC\stackrel{(7)}{=}\angle M'BD,\]](http://latex.artofproblemsolving.com/4/6/c/46cacb9c3cd123cb8802e5fda9af04eb6d3f641a.png)
as desired.
and 
.
Finally 
,
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to 
.
,
,
,
.
Hence 
as well.
.
Therefore,
which implies 
is tangent to 
and 
be the 
and 
-
.
and 
intersect again at 
.
,
collinear, and also 
,
,
,
,
and 
meet 
again at 
and 
.
is cyclic, we have 
by Reim's. Analogously 
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dot("\(A\)",A,dir(80)); dot("\(B\)",B,dir(215)); dot("\(C\)",C,E); dot("\(I\)",I,dir(300)); dot("\(D\)",D,SE); dot("\(I_A\)",IA,SE); dot("\(E\)",EE,NE); dot("\(F\)",F,SE); dot("\(I_B\)",IB,NE); dot("\(I_C\)",IC,N); dot("\(L\)",L,dir(260)); dot("\(S\)",SS,W); dot("\(E'\)",Ep,S); dot("\(F'\)",Fp,NW); [/asy]](http://latex.artofproblemsolving.com/8/f/9/8f9293d31efea20f452b3fef886aa16c9e6e4886.png)
so 
,
.
,
.
.
,
where 
is the midpoint arc of 
.
is cyclic.
.![\[ \measuredangle A'XI_A = \measuredangle IXI_A = 90^{\circ} = \measuredangle A'AM = \measuredangle A'AI_A \]](http://latex.artofproblemsolving.com/7/5/6/7566839e9b44d7e819c4468f0bf5ff6e42df4544.png)
since 
is the diameter of 
by definition. Hence, 
is cyclic.
,
concur. Suppose they concur at 
is cyclic since 
.
as well. Therefore, 
is cyclic.
and 
are both cyclic.![\[ \measuredangle XDE = \measuredangle XDF = \measuredangle XDA = \measuredangle XIA = \measuredangle XII_A = \measuredangle XBI_A = \measuredangle XBE. \]](http://latex.artofproblemsolving.com/e/9/8/e98bcd18f2f731ae4904b4297aefae2f6e8b5442.png)
is cyclic.![\[ \measuredangle XI_A E = \measuredangle XI_A B = \measuredangle XCB = \measuredangle XCD = \measuredangle XFD = \measuredangle XFE \]](http://latex.artofproblemsolving.com/9/4/9/949b3b4e7398a99d7c9b8c666abf5e9c6a6fac0f.png)
from our previous claims. Now, draw a line 
tangent to 
at 
as well. It suffices to prove that 
,![\[ \measuredangle YAX = \measuredangle YDX = \measuredangle BDX = \measuredangle BEX = \measuredangle I_A EX = \measuredangle I_A FX \]](http://latex.artofproblemsolving.com/0/1/b/01bd39058d65d484f40c90be311c899874b21919.png)
Hence, 
be the midpoint of arc 
,
,
be the reflection of 
,
so 
,
.
.
is the 
is the orthic triangle of the excentral triangle, 
is the 
is cyclic.
and 
as desired. 
,
,
be the reflection of 
meets 
,
and 
are tangent at 
and 
,
is harmonic.
to 
.
and similarly 
,
and 
are indeed similar. I will prove that 
is the similarity ratio, which solves the problem because obviously 
as 
,
and 
.
as desired. 
be the center of 
be the center of 
and 
,
by 
respectively. Define 
.
intersect 
again at 
.
and 
,
.
,
.
intersect 
again at 
.
.
and 
,
,
.
,
are collinear. Moreover by the similarity, 
.
and 
are bases of the same angle, which implies that 
,
.
.
We use complex numbers with the assumption that 
have complex coordinates 
respectively. Since 
From 
we have 
By the definition of 
It remains to check 
Luckily, these differences have nice forms 
Hence 
is 
.
is cyclic. But by 
done. 
and 
are tangent.
observe easily that 
is cyclic. 
Consequently the desired tangency holds. 
as the Miquel point of 
Proof
It's well-known that 
is cyclic, 
are collinear, and 
is harmonic. Observe that 
so 
is cyclic and thus 
is the Miquel Point of 
![[asy]
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with 
denoting the image of a point 
Since![\[-1 = (T, I; B, C) = (\infty, I'; B', C'),\]](http://latex.artofproblemsolving.com/e/f/c/efc8605d0bd52e840ac8db50c954ba3c49ce87c5.png)
it follows that 
Thus, omitting the 
s
be a quadrilateral with 
and Miquel Point 
Suppose that 
if 
and 
and we're done. ![[asy]
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to hold true. Also, my original proof for 
so 
is cyclic.
denote the circumcircle of 
be the second intersection of 
be the intersection of 
and 
be the intersection of 
and 
because they're both perpendicular to 
.
Let 
and 
;
.
and 
.
.
is cyclic.
.
is cyclic.
.
is cyclic.
are definitively collinear (by analogous reasoning, 
would also be collinear)
,
.
is cyclic.
proving the claim. 
and 
as 
.
at ![\[\measuredangle(\ell, GI)=90^{\circ}-\measuredangle(I_AG,\ell)=90^{\circ}-\measuredangle I_AEG=90^{\circ} -\measuredangle BEG =90 ^{\circ} -\measuredangle BDG=\measuredangle GAI\]](http://latex.artofproblemsolving.com/e/1/b/e1bc301e0082980ca5cb2373f1e13c1fd1a62ebe.png)
finishing. 
.
.
Thus 
are concyclic. Now let 
.
,
is tangent to 
,
.
,
.
are tangent to 
,
are collinear, we have 
.
.
.
,
.
.
we can easily check 
since in normalized barycentric coordinates they both have 
.
,
,
.
.
is cyclic. Now angle chasing gives 
so 
is tangent to 
,
is the other tangent.
is tangent to 
.
be the unit complex number corresponding to the direction of 
.
Both of these are equal to 
,
,
where 
.
is cyclic.
Let 
.
reveals that 
,
we have 
cyclic. But then since 
we also have 
is cyclic.
.
,![\[ \measuredangle TDE' = \measuredangle TBE' = \measuredangle TIM = \measuredangle TIA = \measuredangle TDA, \]](http://latex.artofproblemsolving.com/9/0/f/90f51c6e1e17f3372621352977969e82685990e9.png)
so 
.
.
is cyclic. 
just simply note that![\[ \measuredangle SDT = \measuredangle BDT = \measuredangle BET = \measuredangle I_aET \]](http://latex.artofproblemsolving.com/4/2/3/423c9190b260a36ebd7064a83973b876eee9ac2f.png)
which shows that the arcs have the same measure, hence the tangency is proven.
meets 
again at 
,
is cyclic. Then by the Radical Center theorem, we know that 
concur at some point 
all lie on the circle with diameter 
,
is cyclic. Observe that 
.
,
.
is cyclic, and so is 
,
to 
,
meet 
again at 
.
and 
are tangent is equivalent to showing 
.
,
,
,
.
and 
.
has equation 
and line 
has equation 
.![\[\left|\begin{matrix}z&\overline z&1\\a&\overline a&1\\d&\overline d&1\end{matrix}\right|=0\]](http://latex.artofproblemsolving.com/e/6/c/e6c1866674bbd65c84c9dcc4022ed7e0de196f15.png)
,
,
.
would be a factor of each numerator because 
or 
implies 
.![\[\ell_3(z)=j(bj+cj+2bc)z-bcj^2(b+c+2j)\overline z+(j^2+bc)(-j^2+bc)=0\]](http://latex.artofproblemsolving.com/9/e/c/9ec3c22a990f5232f6a9443985119369d0a034ee.png)
is![\[C(z)=k_1\ell_2(z)\ell_3(z)+k_2\ell_3(z)\ell_1(z)+k_3\ell_1(z)\ell_2(z)=0\]](http://latex.artofproblemsolving.com/b/c/8/bc84c32234c919e49fa40c6816154fa3c4bf1ffc.png)
where 
,
,
are chosen such that the 
and 
terms vanish.
.
.
,
.
and 
,
.
,
.
,
So the other intersection is 
.
which is true by inspection.
at 
.
at 
,
.
which is true by inspection.
,
,
intersect at one point 
is a diameter of 
is perpendicular to 
is perpendicular to 
,
is perpendicular to 
,
is perpendicular to 
are collinear
,
is perpendicular to 
,
are concyclic
,
and 
are parallel and homothety yields that the circumcircles of 
are tangent to each other
.![\[\angle BTC = \angle BTD + \angle DTC = \angle BED + \angle DFC = \angle FEI_A + \angle EFI_A = \angle BI_AC\]](http://latex.artofproblemsolving.com/d/d/e/dde7e8bff79dd4217562da0d66fa463545d9a7c9.png)
![\[\angle ITD = \angle BTD - \angle BTI = \angle BEA - \angle BI_AI = \angle IAD\]](http://latex.artofproblemsolving.com/5/d/0/5d0b1b43444a3f48a32d15f8693595ce28d36b1d.png)
First we notice,![\[\angle TAI = 180 - \angle IDT = 180 - (90 + \angle BDT) = 90 - \angle BDT = 90 - \angle TFC.\]](http://latex.artofproblemsolving.com/d/8/2/d829ed186c3bed26a87a60f336c0066c3d1b5eb8.png)
we get that 
is an orthogonal system. Hence 
are collinear.
lye on the circle with diameter 
.![\[\angle BPD=\angle BPI+\angle IPD=\frac{\angle C}{2}+\angle IAD=\angle EI_AI+\angle I_AAE=\angle BEA=\angle BED\]](http://latex.artofproblemsolving.com/4/7/f/47f70099d48c250a1caadf9d40dd6e6bd8862500.png)
Thus, 
.
and 
it is the miquel point of 
.
.![\[\angle PI_AE+\angle DKP=\angle DBE=\angle DPE\]](http://latex.artofproblemsolving.com/7/b/5/7b5db6e6b3ff09931ceb1bc4af6633563e04f438.png)
If 
is tangent to 
at 
then![\[\angle DPE-\angle (PD,l)=\angle (l,PE)=\angle PI_AE=\angle DPE-\angle DKP\]](http://latex.artofproblemsolving.com/1/f/5/1f5f96332ad6ea92146ab826d6d15c0b923632d0.png)
Hence 
.
and 
are tangent to each other at 
meets line 
.
,
is cyclic, since then it follows that 
.![\[\angle PBE = \angle PII_A = 180^{\circ} - \angle PIA = 180^{\circ} - \angle PDA = \angle PDE.\]](http://latex.artofproblemsolving.com/b/e/3/be39e44bb4561878bf84871af078808cc4e3f4ef.png)
![\[ \angle KDP = \angle BEP = \angle PFI_A.\]](http://latex.artofproblemsolving.com/1/d/1/1d1e38652f52a60d5b0d7465806d0c760515c93d.png)
of 
and 
)
,
:
L
,
with 
,
)
,
,
we have that :
and 
,
.
,
be the intersection of lines 
,
,
,
,
.
,
,
,
and 
)
,
(Cuz 
,
![[asy]
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 = -9, xmax = 4, ymin = -5, ymax = 9.8; /* image dimensions */
draw((-3,6)--(-4,3)--(1,3)--cycle, linewidth(0.7));
draw((-3,6)--(-4,3), linewidth(0.7));
draw((-4,3)--(1,3), linewidth(0.7));
draw((1,3)--(-3,6), linewidth(0.7));
draw((-3,6)--(-1.162277660168379,-3.4868329805051377), linewidth(0.7));
draw((1,3)--(-1.162277660168379,-3.4868329805051377), linewidth(0.7));
draw((-4,3)--(-0.5811388300841897,-1.7434164902525695), linewidth(0.7));
draw(circle((-7.511328635098222,3.5698101949859686), 5.1242471195534), linewidth(0.7));
draw(circle((-2.7565835097474296,-1.9868329805051381), 2.189020589670635), linewidth(0.7));
draw(circle((-1.5,1.1981019498596834), 3.08169378477153), linewidth(0.7));
draw(circle((-3.209430584957906,1.2905694150420945), 1.883388681284223), linewidth(0.7)+dashed);
draw((-3,6)--(-0.5811388300841897,-1.7434164902525695), linewidth(0.7));
draw((-2.4188611699158105,4.139620389971937)--(-4.179782467020843,-0.3236084406375195), linewidth(0.7));
draw((-4.179782467020843,-0.3236084406375195)--(-0.5811388300841897,-1.7434164902525695), linewidth(0.7));
draw((-2.4188611699158105,3)--(-4.179782467020843,-0.3236084406375195), linewidth(0.7));
draw((-4.179782467020843,-0.3236084406375195)--(-1.837722339831621,0), linewidth(0.7));
draw((-3,6)--(-4.179782467020843,-0.3236084406375195), linewidth(0.7));
dot((-3,6),linewidth(3pt) + dotstyle);
label("$A$", (-2.9202501646632117,6.219693079558107), NE * labelscalefactor);
dot((-4,3),linewidth(3pt) + dotstyle);
label("$B$", (-4.450345658129821,3.09574811373045), NE * labelscalefactor);
dot((1,3),linewidth(3pt) + dotstyle);
label("$C$", (1.2875124423699638,2.9682401559415657), NE * labelscalefactor);
dot((-2.4188611699158105,4.139620389971937),linewidth(3pt) + dotstyle);
label("$I$", (-2.325213028315086,4.264571060128553), NE * labelscalefactor);
dot((-0.5811388300841897,-1.7434164902525695),linewidth(3pt) + dotstyle);
label("$I_A$", (-0.3488396825873824,-2.2595861134026767), NE * labelscalefactor);
dot((-2.4188611699158105,3),linewidth(3pt) + dotstyle);
label("$D$", (-2.325213028315086,2.4369569984878825), NE * labelscalefactor);
dot((-1.837722339831621,0),linewidth(3pt) + dotstyle);
label("$E$", (-1.5814166078799288,-0.07069950469350209), NE * labelscalefactor);
dot((-1.162277660168379,-3.4868329805051377),linewidth(3pt) + dotstyle);
label("$F$", (-1.1776414082151292,-4.087200175043347), NE * labelscalefactor);
dot((-4.179782467020843,-0.3236084406375195),linewidth(3pt) + dotstyle);
label("$R$", (-4.917874836689063,-0.4744747043583013), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]](http://latex.artofproblemsolving.com/f/b/2/fb216a478c0cf7cc398838a78267002ed8b9d182.png)
.
and 
.
all intersect at a point .(Just angle chase for this )
Observe that 
. Let 
.
and self-intersecting 
coincide at a point 
.
,
.
so 
.![\[\angle DPI = \angle CPI - \angle CPD = \frac B2 - \angle CFD = \frac A2 - \theta = \angle IAD. \ \blacksquare\]](http://latex.artofproblemsolving.com/2/c/d/2cdcd5626acbc3799ada95d444b047afc38ca20a.png)
Thus done.
we only need
Clearly all segments are easy to calculate
