satisfying
Prove that
and 
Prove that
be the set of integers. Determine all functions 
such that, for all integers 
and 
,
Proposed by Liam Baker, South Africa
Y by McItran, 61plus, buratinogigle, livetolove212, vsathiam, anantmudgal09, nguyendangkhoa17112003, jhu08, HWenslawski, Adventure10, Rounak_iitr, Funcshun840
Let 
and 
be points in the interiors of sides 
and 
, respectively, of a triangle 
, such that 
. Let the lines 
and 
meet at 
. Prove that the incentre 
of triangle , the orthocentre 
of triangle 
and the midpoint 
of the arc 
of the circumcircle of triangle are collinear.
and 
be points in the interiors of sides 
and 
, respectively, of a triangle 
, such that 
. Let the lines 
and 
meet at 
. Prove that the incentre 
of triangle , the orthocentre 
of triangle 
and the midpoint 
of the arc 
of the circumcircle of triangle are collinear.
and 
are obviously the reflections of 
and 
over the lines 
and 
respectively and so 
is the orthocenter of 
,
is the orthocenter of 
.
the midpoints of 
respectively. By the Lemma 
,
also becomes the perpendicular bisector of 
.
,
,
,
are congruent, and 
,
is on the perpendicular bisector of 
and 
. Note that radius is same.
, I also lies on radical axis.
are collinear(Lies on radical axis) .
meet 
at 
,
meet 
at 
,
meet 
at 
.
respectively. 
perpendicular to 
etc.)
and similarly 
,
(and similarly 
)
and 
are homothetic w.r.t. 
collinear and is equivalent to showing 
.
and similarly 
and hence we are done.
![[asy]
size(8cm);
pointpen = black;
defaultpen(fontsize(8pt));
pathpen = black;
pair A = Drawing("A", dir(110), dir(110));
pair B = Drawing("B", dir(235), dir(235));
pair C = Drawing("C", dir(-55), dir(-55));
draw(A--B--C--cycle);
pair I = Drawing("I", incenter(A,B,C), dir(-125));
pair D = Drawing("D", 2*foot(C,B,I)-C, dir(200));
pair E = Drawing("E", 2*foot(B,C,I)-B, dir(-20));
pair F = Drawing("F", extension(B,E,C,D), dir(200));
pair H = Drawing("H", orthocenter(D,E,F), dir(175));
pair M = Drawing("M", dir(90), dir(90));
draw(B--E--D--C);
draw(circumcircle(A,D,E));
draw(unitcircle);
pair X = Drawing("X", midpoint(B--D), dir(A+B));
pair Y = Drawing("Y", midpoint(C--E), dir(A+C));
pair P = Drawing("P", midpoint(B--C), dir(B+C));
draw(X--Y, black+1.5);
draw(M--I, black+1.5);
draw(M--X--I--Y--cycle, black+1.5);
draw(I--P, dashed);
[/asy]](http://latex.artofproblemsolving.com/c/5/6/c560438adb35982d9d2466375a3fc244a00c2e1c.png)
and 
the midpoints of 
the midpoint of 
.
.
of 
and 
.
,
.
;
is the perpendicular bisector of 
.
.
we have congruent triangles 
.
.
.
be a point on 
such that 
and 
.
are congruen
(
is an isosceles triangle), this means 
and we are done.
onto 
and the pedals from 
onto 
.
and 
.
,
into 
.
.
and 
therefore 
,
and similarly 
.
is isosceles, we get that 
and the conclusion follows.
and the minor arc midpoints 
as 
.
,
.
)
.
over 
.
.
,
.
.
,
![\[ k - abc \cdot \frac{ab+ac+2bc}{b+c+2a} \bar{k} = bc - \frac{1}{bc} \cdot abc \cdot \frac{ab+ac+2bc}{b+c+2a} = \frac{(b+c)(bc-a^2)}{b+c+2a} \text{(2)} \]](http://latex.artofproblemsolving.com/4/a/2/4a2e8bcb55f77d83ad1ed0d82de66ff092336482.png)
times (1) minus 
.
.
on the LHS)
,
,
,
be 
,
,
is then tangent to 
due to spiral similarity, so by easy angle chasing 
.
is on the perpendicular bisector of 
(I used angles instead of easy lengths 
). The circles with diametres 
It seems like the main difficulty of the problem is figuring out how to use the mysterious orthocenter 
intersect the sides of the triangle at points 
s.t. 
.
be the symmetric point to 
with respect to the perpendicular bisector of 
are concyclic.
and 
is syclic), then it is enough to prove that: 
.
be the external angle bisectors of 
respectively. Define 
and 
be the feet of perpendicular from 
.
and 
,
.
and 
so it holds that
implies that 
are collinear by affinity.
![[asy]
unitsize(4cm);
pair A = dir(100);
pair B = dir(250);
pair C = 2*foot(B, (0, 0), dir(270)) - B;
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
draw(A--B--C--cycle);
draw(Circle((0, 0), 1));
pair D = B + abs(C - B)*dir(A - B);
pair E = C + abs(C - B)*dir(A -C);
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
pair F = extension(B, E, C, D);
//dot("$F$", F, dir(270));
draw(B--E);
draw(C--D);
pair I = incenter(A, B, C);
dot("$I$", I, dir(270));
pair H = orthocenter(D,E, F);
dot("$H$", H, dir(90));
draw(B--I--C, heavycyan);
draw(E--H--D, heavycyan);
pair M = dir(90);
dot("$M$", M, dir(M));
draw(I--M, dotted);
[/asy]](http://latex.artofproblemsolving.com/8/4/c/84c3cd6b3549d210f799be58aa99608b44f64045.png)
and 
.
,
.
.
is 
,
and ![\[\begin{vmatrix} u & v & w \\ -b & a & 0 \\ 1 & 1 & 1 \end{vmatrix} = ua + vb - w(a + b) = 0,\]](http://latex.artofproblemsolving.com/1/f/2/1f2915b067283bbf9a4ca6d4d91e5c0f1ae77620.png)
so we can set 
,
,
,![\[(a + b)(a - c)\cdot x + (a + b)a\cdot y + a(a + b - c)\cdot z = 0,\]](http://latex.artofproblemsolving.com/7/9/9/7992d6d5ac894bd8d3c02919eb42ede4531ce6ea.png)
and similarly the equation of 
is ![\[(a + c)(a - b)\cdot x + a(a - b + c)\cdot y + a(a + c)\cdot z = 0.\]](http://latex.artofproblemsolving.com/d/2/f/d2f7fb796687f74faeb32b5bf32c95918c4eb279.png)
If we multiply the first equation by 
and the second by 
and subtract them, we get ![\[(a + b)c(a + c - 2b)\cdot y - (a + c)b(a + b - 2c)\cdot z = 0.\]](http://latex.artofproblemsolving.com/b/e/5/be5ffcffc9c0e238b8a6c3184a5223c1237159af.png)
and 
,![\[x = \frac{a((a + c)b(a +b -2c) + (a +b - c)c(a + c - 2b))}{c - a}.\]](http://latex.artofproblemsolving.com/e/a/7/ea7ba356f54f023cfaf14a0ab5d32716affdef6d.png)
,
and ![\[H = (-a(ab + ac + b^2 + c^2 - bc) : b(a + c)(a + b - 2c) : c(a + b)(a + c - 2b)).\]](http://latex.artofproblemsolving.com/9/1/b/91b6ced8e4b8d36db9d020ebf643e50cc3daf61c.png)
Now we want to calculate 
since it is on the external 
by plugging it into the equation of the circumcircle, which means ![\[M = (a^2 : b(c - b) : c(b - c)).\]](http://latex.artofproblemsolving.com/f/2/6/f2683eb53f90eb96804e12b90f5bd98f0d6a64d1.png)
Now for the final computation, we want to show ![\[\begin{vmatrix} a & b & c \\ a^2 & b(c - b) & c(b - c) \\ -a(ab + ac + b^2 + c^2 - bc) & b(a + c)(a + b - 2c) & c(a + b)(a + c - 2b)\end{vmatrix} = 0.\]](http://latex.artofproblemsolving.com/4/9/5/4952883ccfd19f1ab5a3aa45fa8ffb61495a74cb.png)
We can divide the columns by 
respectively, and expand to get ![\[\begin{vmatrix} 1 & 1 & 1 \\ a & c - b & b - c \\ -ab -ac - b^2 - c^2 + bc & a^2 + ab - ac + bc - 2c^2 & a^2 - ab + ac + bc - 2b^2 \end{vmatrix} = 0.\]](http://latex.artofproblemsolving.com/3/e/1/3e177c4806c99a07499dc10e83bd27b05332e175.png)
But the third row is 
times the first row minus 
times the second, so this is true.
be midpoint's of 
.
.
Note that 
and 
and 
.
and 
are congruent so 
.
.
.
be midpoint of 
and 
so 
.
at 
and 
lie on circles so 
and similarly 
.
.
as both are perpendicular to 
.
and 
.
.
.
.
and 
,
.
.
,
and 
are homothetic, hence 
passes through 
and 
be the midpoints of 
and 
.
.
clearly 
collinear. Done!
.
is isosceles with 
.
is perpendicular to 
and 
is perpendicular to 
,
to 
.
,
with diameter 
with diameter 
is the radical axis of 
be the foot of the altitude from 
and similarly define 
.![\[ \text{Pow}(H, \omega_B) = HH_B \cdot HD = HH_C \cdot HE = \text{Pow}(H, \omega_C) \]](http://latex.artofproblemsolving.com/8/0/4/8048e2a173dc3162cd417f2b211e13faf432e050.png)
by power of a point on 
to 
while 
is the locus of points 
.
and 
intersect at 
.
.
,
is cyclic by miquel we find 
.
.
.
and 
.
and 
,
and 
,
.
is cyclic.
,
and 
,
and 
.![\[\measuredangle ECM = \measuredangle ACM = \measuredangle ABM = \measuredangle DBM\]](http://latex.artofproblemsolving.com/f/8/5/f85d40cd2636038b7efa4553debf962367efc6aa.png)
from which it follows that 
.![\[\measuredangle AEM = \measuredangle CEM = \measuredangle BDM = \measuredangle ADM\]](http://latex.artofproblemsolving.com/e/a/2/ea24ffc3e897f9c5abdf2575027934bdb335182e.png)
from which the claim follows![\[2\measuredangle BEH = 2\measuredangle FEH = \measuredangle EMD = \measuredangle EAD = \measuredangle CAB = \measuredangle ACB + \measuredangle CBA = 2\measuredangle EBI\]](http://latex.artofproblemsolving.com/8/5/5/8556114766629afba1578468e69040fdbda03a73.png)
Thus, 
,
is isosceles, and 
.![\[\measuredangle CEQ = \measuredangle IEC = \measuredangle ECI = \measuredangle CEH\]](http://latex.artofproblemsolving.com/0/7/7/077424e97e3a1698a85ad97776a28d9d4feec9c2.png)
due to the previously proved claims. Thus, 
is a parallelogram. We can first prove the following important equality of angles. Note that,![\[2 \measuredangle FBI = 2(\measuredangle EBC + \measuredangle CBI ) = \measuredangle ECB + \measuredangle CBA = \measuredangle ACB + \measuredangle CBA = \measuredangle CAB \]](http://latex.artofproblemsolving.com/8/e/2/8e29b4b5d6b9693394cdbc4bf90dc59396bed74a.png)
we can similarly show that![\[2\measuredangle ICF = \measuredangle CAB\]](http://latex.artofproblemsolving.com/6/4/f/64f558378fbac8a77b6fc3ef31bc10c1475b8806.png)
Thus, 
.
, ![\[\measuredangle QBF = \measuredangle FBI = \measuredangle ICF = \measuredangle QCF\]](http://latex.artofproblemsolving.com/3/9/b/39b117e37171abe315c98c71b150af5c5bc3e6fe.png)
from which it is clear that 
is cyclic. We can similarly show that 
is also cyclic, which proves the claim.
and 
are the midpoints of arcs ![\[\measuredangle BLN = \measuredangle BCN = \measuredangle BCQ = \measuredangle BPQ \]](http://latex.artofproblemsolving.com/8/f/8/8f8c4bd2d379ccfc5bfe72ebc2ead25f70cda60c.png)
from which it is clear that 
.
to 
,
is a parallelogram). Thus, we must have that points 
,
we have that 
,
is the radical axis of 
and 
.
and 
.
as desired.
but 
,
,
.
is the radical axis of 
and 
also have the same radius, 
and 
.
.
![\[\angle DEM = 90 - \frac{\angle A}{2} = \angle DIQ\]](http://latex.artofproblemsolving.com/4/f/8/4f80f0555e4f47204d90232ee5a96b38c07853c1.png)
hence 
are concyclic. Similarly, 
are cyclic as well so 
is cyclic. But since 
and 
are homothetic, so 
,
and 
,
,
.
and 
are the images of the altitudes from 
and 
.
from an angle chase left as an excercise to the reader, so it suffices to show that 
.
is a parallelogram; from 
,
by a 
rotation. It thus follows that 
,
.
is the radical axis of the circles with diameters 
and we're done.
