,
and 
are both integers.
be real numbers such that 
Prove that 
Let 
Prove that 
and 
Prove that
Y by OronSH, peace09, ehuseyinyigit, pingupignu, v4913, Rijul saini, ys-lg, Rounak_iitr, Funcshun840
Let 
be an acute, scalene triangle with orthocentre 
. Let 
be the line through the reflection of 
with respect to 
and the reflection of 
with respect to 
. Lines 
and 
are defined similarly. Suppose lines , , and determine a triangle 
.
Prove that the orthocentre of, the circumcentre of , and are collinear.
Fedir Yudin, Ukraine
be an acute, scalene triangle with orthocentre 
. Let 
be the line through the reflection of 
with respect to 
and the reflection of 
with respect to 
. Lines 
and 
are defined similarly. Suppose lines , , and determine a triangle 
.Prove that the orthocentre of
, the circumcentre of , and are collinear.Fedir Yudin, Ukraine
This post has been edited 1 time. Last edited by MarkBcc168, Jul 17, 2024, 12:30 PM
Reason: add proposer
Reason: add proposer
as the unit circle and define:
the centroid of 
the circumcenter of 
the reflection of 
the reflection of 
and 
are collinear and then euler line implies the required result.
to meet 
Similarly for 
For a point 
let 
denote the image 
is just a combination of a dilation and a translation, so it suffices to prove that 
and 
are collinear.
and 
and because 
is a real number which is true since:
be the reflection of 
over the 
similarly. Let 
and define 
similarly.
so 
passes through 
and by symmetry 
Next, since 
we get 
and 
but pairs 
and 
share a midpoint, which is the foot of the altitude from 
and thus the radical axis is line 
are congruent, so 
so 
are concyclic. By definition we have 
so 
Also 
so by symmetry we have 
oppositely oriented.
over the perpendicular bisector of 
which passes thorugh 
If 
then 
similarly oriented. We also have 
and 
be midpoints of 
we get 
Now if 
then by spiral similarity we have 
cyclic, and by Gliding Principle, also 
But the right angles give 
is cyclic with diameter 
Since 
we must then have 
and since 
is cyclic we get that it is an isosceles trapezoid.
because reflecting back gives the desired result. Call 
the circumcenters of 
and 
the orthocenter of 
Now if 
we would have 
cyclic by spiral similarity. Thus it suffices to show 
are cyclic. Since 
it suffices to show 
is an isosceles trapezoid.
be the midpoint of 
and thus the circumcenter of 
There is a spiral similarity at 
to 
to 
so there is a spiral similarity at 
to 
to 
so by Gliding Principle 
Thus the perpendicular bisectors of 
are parallel and both pass through 
so we are done.![[asy]
size(9cm);
defaultpen(fontsize(10pt));
import olympiad;
import geometry;
pair A = (1,3.6);
pair B = (0,0);
pair C = (4,0);
pair H = orthocenter(A,B,C);
pair B_c = reflect(C,H) * B;
pair C_b = reflect(B,H) * C;
pair C_a = reflect(A,H) * C;
pair A_c = reflect(C,H) * A;
pair A_b = reflect(B,H) * A;
pair B_a = reflect(A,H) * B;
pair A_1 = extension(A_b,B_a,A_c,C_a);
pair B_1 = extension(B_c,C_b,A_b,B_a);
pair C_1 = extension(B_c,C_b,A_c,C_a);
pair O_1 = circumcenter(A_1,B_1,C_1);
pair H_1 = orthocenter(A_1,B_1,C_1);
pair O = circumcenter(A,B,C);
fill(A--B--C--cycle, mediumgray);
fill(A_1--B_1--C_1--cycle, mediumgray);
draw(circumcircle(A,A_1,A_b), linewidth(0.7));
draw(B_c--H--C_b, gray);
draw(B_1--C_b, dashed);
draw(B_1--A_b, dashed);
draw(A_1--C_a, dashed);
draw(C_b--A, dashed);
draw(C_a--B, dashed);
draw(A--B--C--cycle, linewidth(1));
draw(A_1--B_1--C_1--cycle, linewidth(1));
draw(H--O_1, black);
dot("$A$", A, dir(63));
dot("$B$", B, dir(-90));
dot("$C$", C, dir(-26));
dot("$H$", H, dir(-15));
dot("$B_c$", B_c, dir(170));
dot("$C_b$", C_b, dir(107));
dot("$C_a$", C_a, dir(157));
dot("$A_c$", A_c, dir(-129));
dot("$A_b$", A_b, dir(16));
dot("$B_a$", B_a, dir(-85));
dot("$A_1$", A_1, dir(-140));
dot("$B_1$", B_1, dir(-42));
dot("$C_1$", C_1, dir(-70));
dot("$O_1$", O_1, dir(135));
dot("$H_1$", H_1, dir(150));
dot("$O$", O, dir(90));
[/asy]](http://latex.artofproblemsolving.com/a/5/2/a5200e599e12959669b88175eea603dd119c59f4.png)
denote the reflection of 
,
,
,
,
similarly. Let 
be the circumcenter of 
,
and 
similarly.
and 
lies on 
.
are isogonal w.r.t. 
,
are isogonal w.r.t. 
Similarly, we get that 
,
implying the similarity. 
,
are concyclic. However, since 
,
.
and 
.
To finish, note that 
,
,
.
and hence passes through the inverse of 
w.r.t. 
and 
,
be the reflection of 
.
and define 
,
similarly. Clearly this is the triangle that we care about.
,
is antiparallel to 
,
is cyclic. More particularly this means that 
,
to 
is cyclic. Moreover since we have reflections over 
.
be the orthocenter and circumcenter respectively of 
.![\[ \frac{A'H'}{B'H'} = \frac{AH}{BH} = \frac{A'H}{B'H} \]](http://latex.artofproblemsolving.com/c/a/6/ca6f48b042ade0d47f4f51c58c6a110057c06595.png)
and similar, so 
,
,
.
by say, inversion. Particularly these three circles stay fixed, and 
etc. Let 
be the reflection of 
be the reflection of 
and 
.
and 
.
,
,![\[\begin{cases}
\left(ac \sigma_1 \right)f + (abc\sigma_2)\bar f = \left(ab+ac-c^2\right) \sigma_2 + \left(ac^2+abc-a^2b\right)\sigma_1 \\
\left(bc \sigma_1 \right)f + (abc\sigma_2)\bar f = \left(ab+bc-c^2\right) \sigma_2 + \left(bc^2+abc-ab^2\right)\sigma_1
\end{cases}\]](http://latex.artofproblemsolving.com/7/5/9/759f6c4132c833ab4ace007f92ce04d0a6d0d63f.png)
Similarly, we can get that 
and 
.
is simply 
and the orthocenter of the triangle 
.
lies on this line. So it suffices to show that ![\[\frac{\sigma_2/\sigma_1-\sigma_2^2/abc}{\sigma_2/\sigma_1} \in \mathbb R \Longleftrightarrow \frac{\sigma_1\sigma_2}{abc} \in \mathbb R \]](http://latex.artofproblemsolving.com/e/9/a/e9ad1cb5b453d7a5db0cc30d6932f8c2348f1bdd.png)
![\[\frac{\overline{(a+b+c)} \cdot \overline{(ab+ac+bc)}}{\overline{abc}} = \frac{(ab+ac+bc)(a+b+c)}{abc}\]](http://latex.artofproblemsolving.com/0/b/3/0b377a53eb9927c8d3217a1219e8ea9a0e855389.png)
the isogonal conjugate of the point of infinity along the Euler line; 
actually lies on the circumcircle of 
.
lies on 
,
,
as well (where 
are the vertices of 
,
will give you the triangle formed by the three lines 
denote the antisteiner point of euler line in 
,
be the antipode of 
and let 
,
,
are similar and orthologic with centers of orthology as antisteiner points, also 
.
,
,
and 
.
and 
.
,
,
are concyclic.
and 
,
must be the intersection of 
and 
by Miquel points.
and so 
.
we must have
as 
by aforementioned concyclicity. This expression is symmetric in 
passes through the circumcenter 
is the reflection of 
and some angle chasing we get 
concyclic
by 
then 
collinear
,
,
denote the reflection of 
.
,
.
and 
is cyclic with circumcenter 
is cyclic and that 
.
as 
and 
.
and the other two analogous triangles have the same circumradius (denote it by 
)
We will use the classic trick to deal with a line passing through a circumcenter by reflecting a point on the circumcircle in this line. Let 
;
,
.
,
as the intersection of the line through 
with 
.
and 
are isosceles trapezoids. This means that 
are congruent, which implies that 
.
via law of cosines in 
.
.
,
,
and 
,
Now we can divide by 
and we can express with the sides of 
by law of sines and cosines in 
We will again introduce the line through 
meet at 
,
We now conjecture that 
,
and 
by simple angle chasing. This motivates us to redefine 
,
.
and we want to show that 
,
and 
as the reflection of 
is parallel to 
are reflections in 
and 
,
are reflections in 
,
,
.
is a parallelogram will finish the problem. We have that 
,
.
- this can be done by showing 
,
(and this length bash is by far much better than the one in Approach 1). This finishes the problem.
,
similarly. Then 
is the reflection of 
,
is cyclic, and hence 
and 
,
.
are concyclic, and 
.
,
.
be the midpoints, and let 
,
,
(resp. 
)
and 
.![\[ \frac{M'X}{M'D'} = \frac{N'Y}{N'E'} = \frac{P'Z}{P'F'}, \]](http://latex.artofproblemsolving.com/2/a/e/2ae3db1a776d0d78484b7596ad6e73d590e549f8.png)
in the directed sense. We will show the second equality, using complex numbers to avoid dealing with config issues. Since 
,
,
,
.![\[ \frac{y - n'}{z - p'} = \frac{c_b - a'}{b_c - a'}. \]](http://latex.artofproblemsolving.com/a/e/9/ae927d89833b2ccb44c9e59b09d78694528447c3.png)
On the other hand, if we let 
,
,
,
,![\[ \frac{e' - n'}{f' - p'} = \overline{\bigg(\frac{e - n}{f - p}\bigg)} = \overline{\bigg(\frac{c_a - a}{b_a - a}\bigg)}, \]](http://latex.artofproblemsolving.com/e/b/c/ebc692fa481686f6fb1ad1995093aa9bde0b02b6.png)
and so to show our identity, it suffices to show that 
and 
are inversely similar. But this is clear since we have 
because 
is cyclic, and similarly 
because 
is cyclic.
,
and 
meet at 
,
be the orthocenter and circumcenter of 
is cyclic with center 
and 
are similar as 
and 
.
,
and 
with cyclic equalities also holding.
.
be such that 
Thus 
is the Apollonian Circle of 
with ratio 
be the triangle 
and 
and 
together implies that 
.
.
so we are done. 
and 
.
,
and 
be the circumcenter and orthocenter of 
is the reflection of 
,
.
by the main claim. Thus 
.
.
which implies 
.
and 
,
such that 
,
,
.
.
.
be the Anti-Steiner point of 
so this is true.
,
,
,
homothety at 
,
implies that 
,
be the altitudes of 
intersect 
at 
be reflections of 
in 
respectively. Since 
we have 
lie on a circle. Similarly, we have 
lie on a circle. Consider the inversion
So
This means 
lie on a circle. From this, if we let 
WRT 
such as 
then
or 
.
.
with the same way as 
we have 
But 
then 
.
we have 
with the opposite direction, hence 
.
.
which we define 
)
and 
or 
are collinear. Similarly, we have 
are collinear. Hence 
or 
.
.
and 
we have 
.
be circumcenter and orthocenter of 
WRT 
intersects 
.
we have 
are isogonal conjugate in 
or 
and 
or 
.
or 
are collinear
.
analogously.
,
similarly.
form a triangle.
So, 
Let 
Thus, 
Then, we have 
So, 
are cyclic, with the center 
.
intersect 
at 
be the point such that 
Consider the ratio 
We get that 
Thus, 
is a diameter of the circumcircle of 
collinear, as desired.
.
,
,
.
.
similarly.
.
together like that figure without rotating any of them thanks to the lenght equalities and angles among these triangles.
.
appears among the lines 
.
,
,
.
,
,
and 
.
.
.
One can easily check that this is true. 
hence 
.
.
,
.
hence 
.![$[HH_1]$](http://latex.artofproblemsolving.com/b/0/e/b0e5a95dd4804ce18dfce7c84dc0944d5302b245.png)
.
.
and let 
.
with 
,
and let 
and 
.
.
.
be the reflection of 
and let 
be the reflection of 
.
with 
and
Note that
and similarly 
.
rotation around 
rotation around 
cases. When this is applied to 
positively oriented it is also true, proving the claim.
negatively oriented.
.
,
and 
and 
,
,
.
and 
are tangent to 
.
so 
is tangent to 
and 
.
have inscribed triangles 
and 
.
and 
are inverses in 
negatively oriented.
and similarly 
,
,
and 
.
.
,
,
.
be the inverse of 
.
and 
and 
at 
negatively oriented and that 
and 
.
be the antigonal conjugate of 
with 
is the reflection of 
,
,
goes through the circumcenter 
of 
and 
and 
,
where 
follows from the fact that 
,
.
and 
,
.)![[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. https://github.com/Bubu-Droid/dotfiles/blob/master/bubu-scripts/bubu-asy.py */ pair A = (9.15982,17.71344); pair B = (2.70665,-14.29283); pair C = (45.12867,-14.68292); pair X = (-2.10282,27.85747); pair Y = (-27.39869,-14.01600); pair Bp = (5.86979,1.39564); pair Ap = (-0.58337,-30.61063); pair D = (19.20189,-42.85472); pair T = (32.14381,21.33431);
import graph; size(12cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -5, xmax = 5, ymin = -5, ymax = 5;
draw(A--B, linewidth(0.6)); draw(B--C, linewidth(0.6)); draw(C--A, linewidth(0.6)); draw(circle((24.03305,-1.93901), 24.64614), linewidth(0.6) + linetype("4 4") + red); draw(circle((-12.16657,5.35962), 24.64614), linewidth(0.6) + linetype("4 4") + blue); draw(circle((8.92903,-7.38428), 36.92808), linewidth(0.6)); draw(circle((20.74302,-18.25681), 24.64614), linewidth(0.6) + linetype("4 4") + red); draw(X--D, linewidth(0.6)); draw(D--Y, linewidth(0.6)); draw(Y--B, linewidth(0.6)); draw(X--A, linewidth(0.6));
dot("$A$", A, N); dot("$B$", B, dir(270)); dot("$C$", C, SE); dot("$X$", X, N); dot("$Y$", Y, SW); dot("$B'$", Bp, NW); dot("$A'$", Ap, SW); dot("$D$", D, dir(270)); dot("$T$", T, NE); [/asy]](http://latex.artofproblemsolving.com/7/3/d/73d6c67b50634dfd26716231d0351dbaf8da3ad7.png)
.
.
and 
.
is cyclic.![\[ CP\cdot CB = CQ\cdot CA \implies 2\cdot CP\cdot CB =2\cdot CQ\cdot CA \implies CY\cdot CB = CX \cdot CA \]](http://latex.artofproblemsolving.com/9/0/3/903c51da63d82856032a88034ce0bf204b88dc16.png)
which gives us that 
and 
,
.![\[ \measuredangle A'CB'=\measuredangle BCA =\measuredangle YCA. \]](http://latex.artofproblemsolving.com/1/d/1/1d1c5f52298ebb5a2824ea37fa20c90fc272e0b6.png)
Also note that, ![\[ \measuredangle CB'A'=\measuredangle ABC =\measuredangle ABY=\measuredangle AXY=\measuredangle CXY .\]](http://latex.artofproblemsolving.com/7/a/0/7a0974c244a3d2667f1c16ae4067cc2fb4b8b7ef.png)
.
.
and 
.
which gives us that 
on reflecting about 
.
,
and 
denote the affixes of 
.
,
and 
.![\[ \frac{ab+bc+ca}{a+b+c}+c-\frac{ab}{c},\qquad \frac{ab+bc+ca}{a+b+c} + a - \frac{bc}{a}\qquad, \frac{ab+bc+ca}{a+b+c}+b-\frac{ca}{b}. \]](http://latex.artofproblemsolving.com/4/2/d/42d09447889507cc7c5a639db4b1425d427e2c26.png)
![\[ g=\frac{ab+bc+ca}{a+b+c}+\frac{a+b+c}{3}-\frac{\frac{ab}{c} +\frac{bc}{a}+\frac{ca}{b}}{3} .\]](http://latex.artofproblemsolving.com/9/a/3/9a36c6a916dcfdaa0fd0f4f6de9ca667f6028b20.png)
.![\[ \frac{ab+bc+ca}{a+b+c}+a-\frac{bc}{a}\mapsto a-c+\frac{ab}{c} -\frac{bc}{a},\qquad \frac{ab+bc+ca}{a+b+c}+b-\frac{ca}{b} \mapsto b-c+\frac{ab}{c}-\frac{ca}{b}. \]](http://latex.artofproblemsolving.com/9/b/f/9bf396da888e1f43b8373344bb1e52b744adac5e.png)
,
and 
is 
.
![\[ \frac{ab}{c}+b+a\mapsto a+b+c+\frac{ab+bc+ca}{a+b+c} .\]](http://latex.artofproblemsolving.com/e/8/1/e817d79aaf46e12fdadbe6b47e980e42a653bb56.png)
Thus to prove that ![\[ a+b+c+\frac{ab+bc+ca}{a+b+c},\qquad a+b+c,\qquad \frac{ab +bc+ca}{a+b+c}+\frac{a+b+c}{3}-\frac{\frac{ab}{c} +\frac{bc}{a}+\frac{ca}{b}}{3} \]](http://latex.artofproblemsolving.com/c/9/f/c9f022b0f295d51a9460c561a46c76c6759a4454.png)
are collinear.
which is clearly real and we are done!
and define the other points similarly. Let 
so 
is cyclic. Similarly, 
and 
are cyclic.
.![\[\measuredangle CC_AB_A=\measuredangle ACB=\measuredangle A_BA_CA\]](http://latex.artofproblemsolving.com/9/e/5/9e55697324ed10baf309fd4a266aff21c23887e3.png)
And clearly 
so ![\[\triangle A_CAA_B\sim \triangle C_AAB_A\implies\frac{AA_C}{AC_A}=\frac{AA_B}{AB_A}\]](http://latex.artofproblemsolving.com/2/7/7/277a875de340f871f83f69457d42268f927c8c45.png)
This, together with 
gives 
.
so 
is cyclic. In conclusion, 
which implies the desired claim by symmetry. 
so 
and similarly 
,
.![\[\frac{A’H}{A'H'}=\frac{B’H}{B'H'}=\frac{C’H}{C'H'}\]](http://latex.artofproblemsolving.com/9/8/6/986376a728c269dbaa1533d2ea22ecba6ccbee39.png)
,
,
and 
.
and ![\[k=\frac{\vert h-x\vert }{\vert y+z\vert}=\frac{\vert h-y\vert}{\vert x+z\vert}=\frac{\vert h-z\vert}{\vert x+y\vert}\]](http://latex.artofproblemsolving.com/e/5/3/e53c125f30f2b1be4457cddea3e5c8147902b402.png)
Where 
and 
is a real number. Squaring gives ![\[h\overline{h}-\frac{h}{x}-\overline{h}x+1=k^2\left(2+\frac{y}{z}+\frac{z}{y}\right)\]](http://latex.artofproblemsolving.com/0/5/b/05bfe101f73625106317758713c1dba5ada1a616.png)
![\[h\overline{h}-\frac{h}{y}-\overline{h}y+1=k^2\left(2+\frac{z}{x}+\frac{x}{z}\right)\]](http://latex.artofproblemsolving.com/c/d/2/cd245a9981fb13fe870c30616ea6d7444831ae1e.png)
![\[h\overline{h}-\frac{h}{z}-\overline{h}z+1=k^2\left(2+\frac{x}{y}+\frac{y}{x}\right)\]](http://latex.artofproblemsolving.com/6/d/b/6dbc6e2ad3b6665c054f01111880d1aadbadee76.png)
Subtracting the first two gives ![\[h\frac{x-y}{xy}+\overline{h}(y-x)=k^2\left(z\frac{x-y}{xy}+\frac{1}{z}(y-x)\right)\]](http://latex.artofproblemsolving.com/c/0/f/c0f61fc7ac443ab17954a6d13edeb1c08ceeaa6c.png)
![\[\iff \frac{h}{xy}-\overline{h}=k^2\left(\frac{z}{xy}-\frac{1}{z}\right)\]](http://latex.artofproblemsolving.com/1/0/0/1003b44a2d3044ce56695bde85b757a3c2581b23.png)
From symmetry, we also have![\[\frac{h}{yz}-\overline{h}=k^2\left(\frac{x}{yz}-\frac{1}{x}\right)\]](http://latex.artofproblemsolving.com/c/7/4/c74bc16307447d3a235fcec3171c9d01904d19ea.png)
Subtracting these yields ![\[\frac{h}{y}\cdot\frac{z-x}{zx}=k^2\left(\frac{1}{y}\cdot\frac{(z-x)(z+x)}{zx}+\frac{z-x}{zx}\right)\]](http://latex.artofproblemsolving.com/e/f/3/ef39948061a2361b2ba697553b6ea68653714802.png)
![\[\iff \frac{h}{xyz}=k^2\left(\frac{z+x}{xyz}+\frac{1}{zx}\right)\]](http://latex.artofproblemsolving.com/8/5/2/852d2aae92ad77fd486abf5f7592177e5b99f04a.png)
![\[\iff h=k^2(x+y+z)\]](http://latex.artofproblemsolving.com/c/9/a/c9a4be5fcedde695f6810bb168068c4e0a7bbb89.png)
And since 
is real, we get that 
,
,
,
,
be the orthocentre of triangle 
.
,
is cyclic. In particular, the circumcentre of 
is 
;
and 
.![\[XH : YH : ZH = AH : BH : CH = XH_{\mathcal T} : YH_{\mathcal T} : ZH_{\mathcal T}\]](http://latex.artofproblemsolving.com/e/b/8/eb8f455a2dadadf3c8547c3c08f3906b37ddf7c1.png)
which implies that 
and 
be altitudes in 
and 
and 
meet at 
.
and 
.
therefore
so it remains to show that 
.
.
,
and 
(by harmonic bundles). Similarly, 
,
and 
.
and 
)
However it is well-known that 
(since 
is simply the orthic axis of 
as desired. Our proof is complete.
.
is a parallelogram and so 
.
and 
.
.
respectively. We will show that 
are collinear on a line parallel to 
.
.
or equivalently
However 
,
But 
and 
,
.
,
so
which reduces to the desired.
is a parallelogram.
.
.
Due to the evaluation of 
provided in the previous claim, we have
Because 
.
so it suffices to show that
However by the law of sines 
(as computed previously). Moreover, also by the law of sines 
so it remains to show that
But
The claim is proven.
be the reflection of 
,
and 
similarly. Let 
denote 
,
and 
similarly.
be the reflection of 
across line 
;
and ![\[AB_D \cdot AA_E' = AB \cdot AA_E = AC \cdot AA_F,\]](http://latex.artofproblemsolving.com/b/1/2/b1254566ecbd0845522560fa1bbebb9e8bffba5b.png)
we have 
cyclic. So, 
,
is cyclic. Therefore, 
.
,
,
and 
and 
,
.
is a translation of 
(so that the circumcenter of 
,
is a parallelogram. Repeating this twice more, it follows that 
,
is cyclic, but we can just cut to the chase at this point.
and 
already, so it suffices to show that 
.
and 
.
and 
,
.
is an isosceles trapezoid; combined with ![\[\angle A'O'H' = \angle AOH = \angle AO_A H = \angle A'O'H.\]](http://latex.artofproblemsolving.com/1/d/a/1da4cbe47648c3437b5fbf3764736c72a04db485.png)
with a point 
be the reflection of 
across the perpendicular line 
.
and let 
be the image of 
.
.
and 
,
.
,
.
