and orthocenter 
,
.
to the circumcicle of 
.
be the midpoint of 
.
.
and 
be positive integers greater than 
.
grid lies a coin with its tail side up. A move consists of the following steps.
square in the grid;
for which it is possible that every coin shows head-side up after a finite number of moves.
and 
.![\[P=\frac{1}{5-2a}+\frac{1}{5-2b}+\frac{1}{5-2c}.\]](http://latex.artofproblemsolving.com/c/3/7/c3708768a84c2bbb1e4fae502a9bb1e92cb13d02.png)
be the incentre of acute-angled triangle 
,
,
respectively. Let line 
intersect the circumcircle of the triangle at 
lies between 
and 
.
so that for every 
.
such that 
.
and 
at 
be the circumcircle of acute triangle 
and 
intersect at 
and 
and 
.
are concyclic.
the centre of the circle passing through 
,
are difined similarly. Prove that 
,
,
are concurrent.
> 
that satisfy 
,
€ 
be a fixed positive integer. Given a set 
of 
points in the plane such that no three are collinear and no four concyclic, let 
be the number of circles 
that contain 
in their interior, and let ![\[m(S)=a_1+a_2+\cdots + a_n.\]](http://latex.artofproblemsolving.com/4/9/b/49b928b1f20e2d1799a1234c7f25a2bc3d62d0dd.png)
Prove that there exists a positive integer 
depending only on 
such that the points of 
are the vertices of a convex polygon if and only if 
such that 
.
and 
.
Y by buratinogigle, rightways, YadisBeles, Kezer, Davi-8191, R8450932, UpvoteFarm, Adventure10, Mango247, Rounak_iitr
We say that a triangle 
is great if the following holds: for any point 
on the side 
, if 
and 
are the feet of the perpendiculars from to the lines 
and 
, respectively, then the reflection of in the line 
lies on the circumcircle of the triangle . Prove that triangle is great if and only if 
and 
.
Senior Problems Committee of the Australian Mathematical Olympiad Committee
is great if the following holds: for any point 
on the side 
, if 
and 
are the feet of the perpendiculars from to the lines 
and 
, respectively, then the reflection of in the line 
lies on the circumcircle of the triangle . Prove that triangle is great if and only if 
and 
.Senior Problems Committee of the Australian Mathematical Olympiad Committee
This post has been edited 1 time. Last edited by MellowMelon, May 18, 2017, 3:29 AM
Reason: add proposer
Reason: add proposer
the reflection of 
.
and 
,
.
is a convex quadrilateral. Since we want this point to also lie on the circumcircle of 
,
.
,
,
.
.
.
.![\[\angle BD'A = 180^\circ-\angle ACB=180^\circ-\angle ADQ=180^\circ-\angle APQ=180^\circ-\angle D'AP,\]](http://latex.artofproblemsolving.com/1/0/e/10e3da317a40807749285cf509eacf4ae329a189.png)
which contradicts the fact that 
is non-degenerate.
or 
.
is a square, so 
as desired.
- this is because 
.
since 
- hence 
as well. This implies that 
.
.![\[\angle BD'D=\dfrac{\angle BPD}2 = \dfrac{90^\circ}2 = 45^\circ\qquad\text{and}\qquad \angle CD'D = \dfrac{\angle CQD}2 = \dfrac{90^\circ}2 = 45^\circ.\]](http://latex.artofproblemsolving.com/f/c/9/fc98d3fafd804a70a4eb0bbcebeaccca5ce6fede.png)
Hence 
and indeed 
.
by angle chasing. Now use the first observation and choose 
and 
Let 
and 
denote the reflection of 
in line 
it's easy to see that 
is the orthocenter of 
which implies that 
are symmetry over the perpendicular bisector of ![\[D^*\text{ lies on the circumcircle of }\triangle ABC\iff\angle BD^*C=\angle BE^*C=\angle A\iff E^*\equiv A.\]](http://latex.artofproblemsolving.com/c/3/6/c36b6ccdf59ee9ff74c0459f3264dcdb7f216206.png)
Therefore 
This is because 
and 
![[asy]
size(7cm); pointpen=black; pathpen=black; pointfontpen=fontsize(9pt);
void b(){
pair A=D("A",dir(110),dir(110));
pair B=D("B",dir(-150),dir(-150));
pair C=D("C",dir(-30),dir(-30));
pair D=D("D",extension(A,origin,B,C),S);
pair P=D("P",foot(D,A,B),W);
pair Q=D("Q",foot(D,A,C),E);
pair E=D("E",foot(A,B,C),S);
pair D1=D("D^*",2*foot(D,P,Q)-D,dir(120));
pair E1=D("E^*",2*foot(E,P,Q)-E,N);
D(A--B--C--cycle);
D(P--D--Q,magenta);
D(B--E1--C);
D(B--D1--C);
D(P--Q,dashed);
draw(anglemark(B,E1,C,3),deepgreen);
draw(anglemark(B,D1,C,3),deepgreen);
D(E1--E,dotted);
D(D1--D,dotted);
D(circumcircle(A,P,Q),red+linetype("4 4"));
}
b(); pathflag=false; b();
[/asy]](http://latex.artofproblemsolving.com/5/e/5/5e56a5b11abd0856f6c7de39c325411fa498a000.png)
and![\[\begin{aligned}D^* \text{ lies on the circumcircle of }\triangle ABC &\iff D^* \text{ sends $\triangle D^*PB$ to $\triangle D^*QC$}\\
&\iff \frac{DP}{PB}=\frac{D^*P}{PB}=\frac{D^*Q}{QC}=\frac{DQ}{QC}\\
&\iff \tan{\angle B}=\tan{\angle C}\\
&\iff \angle B=\angle C. \end{aligned}\]](http://latex.artofproblemsolving.com/7/2/c/72c9918701003856b554bde67c5db31bb1d94b67.png)
as desired. 
![[asy]
size(7cm); pointpen=black; pathpen=black; pointfontpen=fontsize(9pt);
void b(){
pair A=D("A",dir(70),dir(70));
pair B=D("B",dir(180),W);
pair C=D("C",dir(0),E);
pair D=D("D",foot(A,B,C),S);
pair P=D("P",foot(D,A,B),W);
pair Q=D("Q",foot(D,A,C),S);
pair D1=D("D^*",2*foot(D,P,Q)-D,dir(70));
D(P--D--Q,magenta);
D(D--D1,dotted);
D(B--P--D1--cycle,blue+dashed);
D(D1--Q--C--cycle,blue+dashed);
D(A--P--Q--cycle);
D(B--C);
}
b(); pathflag=false; b();
[/asy]](http://latex.artofproblemsolving.com/9/7/a/97aa351d7f6107d1b9e8e7cc7d748e99f92506fa.png)
by setting 
.
are the midpoints of 
respectively as 
is cyclic, thus 
.
is tangent to 
at 
,
and they are on same side of 
,
,
.
.
,
.
be projections of 
on 
is isosceles trapezoid, so 
so 
.
and 
is cyclic. 
,
.
such that ![\[f(x)+f(y)=x^2+y^2\]](http://latex.artofproblemsolving.com/a/f/9/af9cf6c9505676db971e5eec3cadadee118cb09d.png)
for all real 
and 
.
to obtain ![\[2f(x) = 2x^2\quad\implies\quad f(x) = x^2.\]](http://latex.artofproblemsolving.com/4/9/d/49d92ef58916aba960914e76165520920917af5e.png)
Now we just need to check that 
always works, which it does.
is the only possible function which satisfies the given conditions, all we need to do is to look at the case where 
.
is cyclic, so 
.
and 
gives 
,
.
is cyclic. We are done.
,
cyclic gives 
,
is a square, so 
.
is equivalent to 
,
.
as 
,
.
is cyclic, so this is equivalent to the condition![\[ \frac{D'P}{PB} = \frac{DP}{PB} = \frac{DQ}{QC} = \frac{D'Q}{QC}\]](http://latex.artofproblemsolving.com/a/0/8/a08898590da697cb79b19113e2abacca2c33563b.png)
However, we have![\[ \frac{DP}{PB} = \frac{QC}{DQ}\]](http://latex.artofproblemsolving.com/4/d/a/4da9743884d331716b747b28437cf0809f211b8d.png)
since 
,
and 
being isosceles right. But they are similar to the original, so the original condition is equivalent with 
and 
simplifies the problem quite well, and it is easier to interpret reflection in tje Cartesian plane. I will post the solution as soon as I find one.
Angle Bisector of 
lies on 
lie on opposite sides if 
lie on same side of 
which means that
.
.
.
are circumcenter of 
and 
respectively.
,
note 
,
.
.
,
and 
.
,
is cyclic, so 
,
,
is congruent to 
,
.
.
,
,
,
are the midlines of 
.
,
and 
,
.
,
let 
and 
be the feet from 
lies on 
Hence, 
and since 
is cyclic, 
and 
Let 
be the foot from 
lies on 
and 
Hence, 
is a square and 
so 
Similarly, 
so 
,
,
which means 
so 
is cyclic with diameter 
implies 
is a cyclic isosceles trapezoid. Hence, 
and 
.
which means the only possible solution is 
be the circumcenter of 
and 
are also isosceles right triangles, we have 
so 
Thus, 
follows from 
,
or 
,
.
.
,
,
and when 
,
.
,
.
are equidistant from 
is a diameter of 
so it follows that 
.
,
are isogonal in 
.
,
.
,
lie on the same side of 
,
as well. Thus 
,
.
,
is cyclic, so 
implies 
,
implies 
,
implies 
bisects 
