,
are reals. Let 
be a circle with diameter 
and let 
be any other point on 
meets the x-axis again at 
.
.
,
and 
denote the midpoints of 
.
be the projection of 
onto 
be the reflection of 
.
.
and 
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.
Y by ac_math, tastymath75025, Carpemath, MathPassionForever, Purple_Planet, microsoft_office_word, itslumi, mathematicsy, MathLuis, centslordm, Adventure10, ike.chen, HoRI_DA_GRe8, PHSH, rama1728, Quidditch, ImSh95, Mango247, Funcshun840, and 6 other users
Points 
, 
, 
are chosen on the sides 
, 
, 
of a triangle 
respectively. The circumcircles of triangles 
, 
, 
intersect the circumcircle of triangle again at points 
, 
, 
respectively (
). Points 
, 
, 
are symmetric to , , with respect to the midpoints of the sides , , respectively. Prove that the triangles 
and 
are similar.
, 
, 
are chosen on the sides 
, 
, 
of a triangle 
respectively. The circumcircles of triangles 
, 
, 
intersect the circumcircle of triangle again at points 
, 
, 
respectively (
). Points 
, 
, 
are symmetric to , , with respect to the midpoints of the sides , , respectively. Prove that the triangles 
and 
are similar.
s.t. a quadrilateral 
be a parallelogram.
and 
be a parallelogram, too.
Lines 
and 
meet on the circumcircle of triangle 
.
is cyclic.
.
.
,
.
by 
.
on 
.
,
and 
are similar. Working with angles mod 180° we get 
,
because 
so if 
Or I missed some thing?
on 
.
to 
to 
and 
to 
(correspondingly). The lines 
meet the circumcircle of 
.
.
as 
is cyclic. Analogously 
as 
is cyclic. Therefore 
.
.
be the circumcircle of 
be diametrally opposite to 
.
be the intersection of the perpendicular from 
.
wrt 
where 
is the midpoint of 
(the circumcenter of 
(
with the circumcircle of 
then clearly 
be symmetric to 
to 
meets the perpendicular from 
to 
in a point which is symmetric to 
on its sides, as 
is congruent to 
be a Miquel point of four line 
.
(
where 
is orthocenter of four triangle that formed by four line 
are points on circumcircle of triangle 
be Simson line of 
.
(
are on circumcircle of triangle 
.
.
.
and 
be Simson line of 
wrt 
.
be midpoint of 
.
and 
.
and 
are similar and we obtain 
.
be Simson line of 
wrt 
.
.
.
and we are done.
lie on 
through midpoint of 
is Miquel points of triangle 
and 
then 
and 
.
.
be Miquel points of triangle 
.
intersect each other and make triangle 
be Miquel points of triangle 
then 
is cyclic quadrilateral. Similarly we get 
is Miquel point of triangle 
and 
be Miquel point of triangle 
.
we obtain 
and 
and 
so 
is Miquel point of triangle 
is cyclic thus 
.
is true. We are done!
is the center of spiral similarity sending 
,
.
.
gives the desired.
sends 
.
.
are similar. Also, 
,
are similar.
.![\[ k = \frac{a_2b_3-a_3b_2}{a_2-b_2-(a_3-b_3)} = \frac{\frac{bb_1-cc_1}{b+b_1-c-c_1}(a+c-b_1)-\frac{cc_1-aa_1}{c+c_1-a-a_1}(b+c-a_1)}{\frac{bb_1-cc_1}{b+b_1-c-c_1}-\frac{cc_1-aa_1}{c+c_1-a-a_1}-(b+c-a_1)+(a+c-b_1)} = \frac{\frac{(bb_1-cc_1)(a+c-b_1)(c+c_1-a-a_1)-(cc_1-aa_1)(b+c-a_1)(b+b_1-c-c_1)}{(b+b_1-c-c_1)(c+c_1-a-a_1)}}{\frac{(bb_1-cc_1)(c+c_1-a-a_1)-(cc_1-aa_1)(b+b_1-c-c_1)-(a+a_1-b-b_1)(b+b_1-c-c_1)(c+c_1-a-a_1)}{(b+b_1-c-c_1)(c+c_1-a-a_1)}} = \frac{(bb_1-cc_1)(a+c-b_1)(c+c_1-a-a_1)-(cc_1-aa_1)(b+c-a_1)(b+b_1-c-c_1)}{(bb_1-cc_1)(c+c_1-a-a_1)-(cc_1-aa_1)(b+b_1-c-c_1)-(a+a_1-b-b_1)(b+b_1-c-c_1)(c+c_1-a-a_1)}\]](http://latex.artofproblemsolving.com/8/9/a/89a15dbf67d6cf73aa606aa2ad469bc1a6d2674d.png)
![\[ (bb_1-cc_1)(a+c-b_1)(c+c_1-a-a_1)-(cc_1-aa_1)(b+c-a_1)(b+b_1-c-c_1) = \]](http://latex.artofproblemsolving.com/7/f/8/7f86a1189b14f218ff652162fe3616450418226e.png)
![\[(bb_1-cc_1)(c+c_1-a-a_1)-(cc_1-aa_1)(b+b_1-c-c_1)-(a+a_1-b-b_1)(b+b_1-c-c_1)(c+c_1-a-a_1) =\]](http://latex.artofproblemsolving.com/f/7/3/f730e79ff0a14ef38e6040c10319fa2c7d0774b7.png)
.
and 
,
.
lied on 
completely, only for the spiral similarity part.) The stronger statement follows. Let 
be points in the plane. Let 
be the centers of spiral similarity sending 
be the reflections of 
.
.![\[\frac{a_3-b_3}{a_3-c_3} = \frac{a_2-b_2}{a_2-c_2}\]](http://latex.artofproblemsolving.com/7/2/4/724fc7a3a692af5cfad9ed1cb2b11bf9ef753fd3.png)
, and this is much easier to compute than what I did.
is the symmetric of 
hours later I then see its not that... 5 minutes later I come up with this solution.
.
.
.
so we get 
.
.
.
.
are tangent to the circumcircles of 
and 
respectively.Let 
.
.
.
.
as desired.
to 
,
and 
are directly similar iff![\[
p(y-z)+q(z-x)+r(x-y)=0
\]](http://latex.artofproblemsolving.com/3/6/9/3693c92119a6dc6ac7c8498b2819ec5f07813723.png)
![\[
b_3-c_3=(a+c-b_1)-(a+b-c_1)=c+c_1-b-b_1
\]](http://latex.artofproblemsolving.com/6/1/2/612c927c05e5197276418958bc3347df626c1167.png)
and 
yields![\[
cc_1-bb_1+aa_1-cc_1+bb_1-aa_1=0
\]](http://latex.artofproblemsolving.com/b/3/6/b36d2cb2d086fdec5ee0d295224103c15bb3fb9b.png)
,
.
But 
so 
by SAS. Analogously, we conclude 
and 
.
By symmetry, we conclude 
as desired. 
meet 
again at 
.
.
denote the tangent to 
at 
,
.
in order to take advantage of symmetry. But the second ratio is connected to 
,literally took me no hints if I didn't stupidly rolled over the pages.
,
.
,
or our determinant is equal to 
of spiral similarity 
and of symmetry wrt midpoint of 
and it has same rotation and ratio as 
thus the conclusion 
This and analogous equalities yield the desired result.
on the complex plane, with 
Also, by complex midpoint, 
and one can verify using a lot of algebra that 
and 
are directly similar
Someone please call bro up to do the heavenly task. 
denote the circumcenters of 
respectively. Then
Then we have two equations for each of 
which gives a system we can solve to find 
:
This is written in a more standard form as
Using Cramer's Rule (don't laugh),
The numerator is![$$a_2^2b_2c_2\left[-(b-c)(c_2-a)(a-b_2) + (b_2-c)(c_2-a)(a-b) + (b-c_2)(c-a)(a-b_2)\right]$$](http://latex.artofproblemsolving.com/c/9/3/c9399f84938afefb9a7aea5196abe2dcabdd3dd3.png)
and the denominator is 
,
Thus by symmetry, we have
and
So we find the vector
Similarly,
and so
Thus the triangles are similar and oriented the same way.
;
and 
by spiral similarity. Then the rest is a quick angle chase to find 
as requested.
to 
.
Analogously, 
,
,
and analogous results finishes it. 
,
,
.
.
By spiral similarity. We can also repeat this same process for the cyclic variants. By SAS, Lemma is proved. 
We can repeat the same process for the cyclic variants of 
,
. (Bro headsolved the problem ![[asy]
/*
Converted from GeoGebra by User:Azjps using Evan's magic cleaner
https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py
*/
/*
A few re-additions are done using bubu-asy.py.
This adds the dps, xmin, linewidth, fontsize and directions.
*/
pair A = (-91.53138,70.81505);
pair B = (-110.35458,-31.32483);
pair C = (7.06804,-31.32483);
pair A_1 = (-79.20200,-31.32483);
pair B_1 = (-54.54733,32.50300);
pair C_1 = (-99.88684,25.47601);
pair A_3 = (-24.08453,-31.32483);
pair B_3 = (-29.91600,6.98722);
pair C_3 = (-101.99912,14.01421);
pair A_2 = (-102.39840,61.97874);
pair B_2 = (-123.65121,5.70195);
pair C_2 = (-31.45935,-58.63911);
import graph;
size(12cm);
pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps);
real xmin = -5, xmax = 5, ymin = -5, ymax = 5;
draw(A_2--B_2--C_2--cycle, linewidth(0.5) + blue);
draw(A_3--B_3--C_3--cycle, linewidth(0.5) + blue);
draw(A--B, linewidth(0.5));
draw(B--C, linewidth(0.5));
draw(C--A, linewidth(0.5));
draw(circle((-51.64327,10.65974), 72.17841), linewidth(0.5));
draw(circle((-79.72959,45.20069), 28.20244), linewidth(0.5));
draw(circle((-94.77829,-4.83039), 30.73395), linewidth(0.5));
draw(circle((-36.06698,-11.31095), 47.55192), linewidth(0.5));
draw(A_2--B_2, linewidth(0.5) + blue);
draw(B_2--C_2, linewidth(0.5) + blue);
draw(C_2--A_2, linewidth(0.5) + blue);
draw(A_3--B_3, linewidth(0.5) + blue);
draw(B_3--C_3, linewidth(0.5) + blue);
draw(C_3--A_3, linewidth(0.5) + blue);
dot("$A$", A, NW);
dot("$B$", B, SW);
dot("$C$", C, SE);
dot("$A_1$", A_1, S);
dot("$B_1$", B_1, NE);
dot("$C_1$", C_1, W);
dot("$A_3$", A_3, SE);
dot("$B_3$", B_3, NE);
dot("$C_3$", C_3, NW);
dot("$A_2$", A_2, NW);
dot("$B_2$", B_2, W);
dot("$C_2$", C_2, SE);
[/asy]](http://latex.artofproblemsolving.com/c/3/c/c3cc9040eb0b0c9410d42a65216ed88d7b7a708b.png)
and thus also the center of spiral similarity mapping 
.
.
which gives us that 
.
.
.
and similarly we also get that 
.
and we are done.
be the complex unit circle. Then if we let the lowercase letters represent their respective points translated onto the complex plane, we obtain that as ![\[a_2 = \frac{bb_1 - cc_1}{b + b_1 - c - c_1}; \phantom{e} b_2 = \frac{cc_1 - aa_1}{c + c_1 - a - a_1}; \phantom{e} c_2 = \frac{aa_1 - bb_1}{a + a_1 - b - b_1},\]](http://latex.artofproblemsolving.com/5/1/f/51f76ad7ae6704dfa1c841724f40b24e47703684.png)
and also![\[a_3 = b + c - a_1; \phantom{e} b_3 = c + a - b_1; \phantom{e} c_3 = a + b - c_1.\]](http://latex.artofproblemsolving.com/c/7/7/c777a37e28befd500029c442158ccf5f2a1637ea.png)
Now in order to show 
:![\[\frac{a_2 - b_2}{a_2 - c_2} = \frac{\frac{bb_1 - cc_1}{b + b_1 - c - c_1} - \frac{cc_1 - aa_1}{c + c_1 - a - a_1}}{\frac{bb_1 - cc_1}{b + b_1 - c - c_1} - \frac{aa_1 - bb_1}{a + a_1 - b - b_1}} = \frac{\frac{aa_1(b + b_1 - c - c_1) + bb_1(c + c_1 - a - a_1) + cc_1(a + a_1 - b - b_1)}{(c + c_1 - a - a_1)}}{\frac{aa_1(c + c_1 - b - b_1) + bb_1(a + a_1 - c - c_1) + cc_1(b + b_1 - a - a_1)}{(a + a_1 - b - b_1)}} = \frac{b + b_1 - a - a_1}{c + c_1 - a - a_1}, \]](http://latex.artofproblemsolving.com/3/6/5/365393be35f8abcc8216ab085aa14e6c406f6b0e.png)
which is also precisely 
,
.
and the similarity ratio![\[\frac{CA_3}{CB_3} = \frac{BA_1}{AB_1} = \frac{C_2A}{C_2B}.\]](http://latex.artofproblemsolving.com/9/f/4/9f4282ab55795d0c9b034ebfa47105d4ddb5f57a.png)
and symmetric variants. Note that clearly 
as 
is cyclic. Furthermore, since 
,
as desired.
and 
.
,
.
is inscribed in arc 
,
,
"onto the triangle".
(the former using spiral center and the latter just by reflecting).![\[\frac{A_2C}{A_2B}=\frac{B_1C}{C_1B}=\frac{AB_3}{AC_3}\]](http://latex.artofproblemsolving.com/8/5/1/851e7e3178cb8a497ed8f7243365099732f1639f.png)
which implies 
.
.
and so 
.
