,
,
and 
,
is the intersection point of the diagonals 
and 
.
and radius 
intersects 
and 
.
is tangent to the circles with diameters 
and 
.
be a triangle inscribed in a circle, where the tangents to the circle at points 
and 
intersect at the point 
and 
.
and 
intersect at point 
.
be the reflection of 
and 
intersect at point 
.
.
be non-negative real numbers such that 
Prove that
such that 
and 
for any real numbers 
and 
.
is written on the blackboard. Every time, Carmela can erase the number 
on the black board and replace it with a new number 
,
is a perfect square. Prove or disprove that all positive integers 
can be written exactly once on the blackboard.
and 
as ![\[\sqrt{2}=1.a_1a_2a_3\dots\quad\quad\sqrt{3}=1.b_1b_2b_3\dots\]](http://latex.artofproblemsolving.com/4/d/5/4d5049d5c3b26757a25c5dd34462c5462228b83d.png)
where each 
or 
is a digit between 0 and 9. Prove that there exist at least 1000 values of 
between 
and 
such that 
.
of positive integers with 
is called good if we can colour each integer red, blue, green or purple, in such a way that
consecutive integers at least one is coloured red;
of each 
consecutive integers at least one is coloured blue;
of each 
consecutive integers at least one is coloured green;
of each 
consecutive integers at least one is coloured purple.
.
,
points in the plane, no three of them belonging to a same line. Each of these points is colored using one of four given colors. Prove that there is a color and at least 
scalene triangles with vertices of that color.
such that :
.
such that 
and 
and 
,
be the intersection of the altitude through 
be the intersection of the altitude through 
.
be the intersection of the median through 
.
either the sum of the first 
Y by doxuanlong15052000, Tumon2001, Davi-8191, Centralorbit, thczarif, amar_04, Piano_Man123, nima.sa, Adventure10, Mango247, Rounak_iitr, NO_SQUARES, eggymath
Let 
be a triangle with orthocenter 
and let 
be the midpoint of 
Suppose that 
and 
are distinct points on the circle with diameter 
different from 
such that lies on line 
Prove that the orthocenter of 
lies on the circumcircle of 
Proposed by Michael Ren
be a triangle with orthocenter 
and let 
be the midpoint of 
Suppose that 
and 
are distinct points on the circle with diameter 
different from 
such that lies on line 
Prove that the orthocenter of 
lies on the circumcircle of 
Proposed by Michael Ren
be the center of 
;
is the antipode of 
lies on the dilation of 
at 
,
,
.
is cyclic with diameter 
Respectively we have that 
so 
and 
are tangent to 
.
(
to 
and 
to 
and 
,
at 
at 
.
.
and let 
,
under inversion wrt 
and that 
(
is cyclic), we have that 
and 
is perpendicular to 
,
is the orthic triangle of 
,
be the 
,
,
be the reflection of 
is cyclic with diameter 
,
is tangent to 
be a point on 
is tangent to 
,
,
and the image of 
.
be the image of 
must be cyclic and that it also passes through 
because 
are colinear. But 
passes through 
passes through 
and 
be the altitudes in 
are tangent to 
be midpoints of 
and 
.
hence 
as desired. 
;
be the orthocenter of 
and we will use lowercase for the complex numbers.
.
.
.
.
and 
are parallel vectors, since 
are colinear.
and 
.
and 
are parallel.
has magnitude 
as the origin. Take a point 
.
.
is just the reflection of 
,
,
and 
.
be the orthocenter of 
, 
and 
are parallelogram.
is a parallelogram !
.
is a parallelogram!
.
at 
so 
so perpendicular bisector of 
at midpoint of 
is 
in perpendicular bisector of 
in perpendicular bisector of 
![[asy]size(7cm);
pair B=(0,0),A=(7,11),C=(10,0),H,M,P,Q,X,Y,Z;
H=orthocenter(A,B,C);M=(B+C)/2;X=(A+H)/2;
Q=X+(A-X)*dir(45);Y=foot(X,Q,M);P=2*Y-Q;
Z=orthocenter(A,P,Q);
draw(circumcircle(A,P,Q)^^circumcircle(Y,X,M)^^circumcircle(A,B,C),green);
D(MP("A",A,N)--MP("B",B,S)--MP("M",M,S)--MP("C",C,SE)--cycle);
D(MP("H",H,S)--MP("X",X,NE)--A--MP("Q",Q,W)--MP("Y",Y,SSW)--MP("P",P,W)--M,magenta);
D(X--Y,magenta);D(P--A--MP("Z",Z,S),magenta);
dot(A^^B^^C^^H^^M^^X^^Q^^Y^^P^^Z);
[/asy]](http://latex.artofproblemsolving.com/5/5/5/5558ce8de2077ee24b4364f1de6944bffa074710.png)
be the circle with diameter 
,
be the homothety centered at 
;
,
.
(both are 
to 
by a well-known property of orthocenters. So 
.
since 
,
,
.
are collinear 
for some constant 
.
.
has roots 
.
lie on the circle with diameter 
.
and 
.
and 
.
.
is a parallelogram.
.
.
.
,
.
.
,
.
.
.
.
.
Simplification of coordinates of 
.
,
we get after simplifying, 
Coordinates of 
.
we get, 
Equation of circumcircle of 
be the center of circle 
,
.
.
be the midpoint of 
.
,
.
.
.
.
.
We cancel the denominators.
Simplifying and taking similar terms together we get,
So, 
be the midpoint of 
= 
.
is a straight line.
.
the midpoint of 
the midpoint of 
,
and 
we get 
is the midline of 
.
So from 
we get 
is a paralellogram 
,
in 
.
and 
.
to 
,
is the diameter of the nine-point circle, so since 
is the perpendicular bisector of 
.
altitudes. Recall that 
are tangents to the circle 
.
.
lies on the circumcircle of 
with a given point 
and 
,
.![\[ H=4O-A-B-C=4O-A-2D. \]](http://latex.artofproblemsolving.com/c/f/2/cf25dc43520b9d8c62f0a06bc2572648b3ee04f2.png)
So 
and 
,
.
.
is a parallelogram, which means 
and 
.
the unit circle, centered at 
,
.
,
.
,
is a parallelogram.
are now known to satisfy ![\[ p + q = \frac{2ef}{e+f} + \frac{2pq}{e+f} \]](http://latex.artofproblemsolving.com/b/a/8/ba8019a3d679ac1f2758ff03b5b5633d6b6aa5ed.png)
so ![\[ (a + p + q) - o = \frac{2pq}{e+f} \qquad \text{and}\qquad a - o = \frac{2ef}{e+f} \]](http://latex.artofproblemsolving.com/f/3/8/f386ef05a419829f7a6e10de25bdb84529dd47ea.png)
which clearly have the same magnitude. Hence the orthocenter of 
and 
as the antipode of 
and 
are both perpendicular to 
.
and 
.
,
(since it's the midpoint of 
antipode of 
and 
and note 
Now note that the orthocenter of 
,
and 
.
Upon expansion we have,
Now taking the conjugate of this expression we find,
and we're done.
,
.
to lie on the circle with diameter 
.
where 
,![\[\frac{p+q}{p+q-2m} = \overline{\left(\frac{p+q}{p+q-2m}\right)} = \frac{p+q}{p+q-2pq \overline{m}}\]](http://latex.artofproblemsolving.com/b/3/f/b3f5408fae532d7ccb8a5b0ba4f172c4041a3b6c.png)
and we're done since 
.
and orthocentre are collinear. Hence, 
are collinear.
and 
are tangents to the circle 
.
is the 
,
is symmedian and 
.
are collinear, so 
is a harmonic bundle. Hence, 
s
.![\[\angle FKM=\angle FKQ=\angle FPE=\angle FAE=\angle BAC=\angle FEM\]](http://latex.artofproblemsolving.com/f/8/a/f8a868d87b51ca63d6ab848ed7834eba9893ee8b.png)
and it follows that 
are concyclic. Hence, 
,
is the midpoint of 
;
.
,![\[\frac{\operatorname{pow}(J,(AH))}{\operatorname{pow}(J,(AM))} = \frac{\operatorname{pow}((B,(AH))}{\operatorname{pow}(B,(AM))} \iff \frac{-AJ \cdot 2JK}{-AJ \cdot JK} = \frac{BF \cdot BA}{BD \cdot BM} \iff 2 = 2. \quad \blacksquare\]](http://latex.artofproblemsolving.com/4/7/c/47c70d9a63331d88bc415dd2be52a08135e5d5e7.png)
Then 
by three tangents lemma, so 
Also we have 
so the antigonal conjugate of 
We can show it must be 
and using continuity.
we get the orthocenter of 
which lies on the circumcircle by homothety.
where 
due to 
is on 
.
.
and therefore 
which implies 
as normal.
-
-
,
,
,
.
and 
.
also works.
,
works since then the orthocenter is just 
Let 
.
and since ![\[-1=\frac{\text{Pow}(M,(AH))}{\text{Pow}(M,(ABC))}=\frac{\text{Pow}(T,(AH))}{\text{Pow}(T,(ABC))} \iff TA \cdot TW'=-TP \cdot TQ\]](http://latex.artofproblemsolving.com/9/4/d/94d95e9f02d2a5422d99209f941993ae3364f411.png)
And see that obviously 
and so 
(here lengths are directed).
is a parallelogram. In-order to have 
,
at 
and thus: 
which implies 
(which is the nine-point circle).
be the 
be the 
A
and that 
.
.
Observe that 
.
