AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be a triangle. A circle passing through 
and 
intersects segments 
and 
at 
and 
, respectively. Lines 
and 
intersect at 
, while lines 
and 
intersect at 
. Prove that 
if and only if 
.
be real numbers such that 
Prove that
. Let and be the feet of the altitudes from to and from to respectively.
and 
as the reflections of across lines and , respectively. Let 
be the circumcircle of 
. Denote by 
the second intersection of line 
with , and by 
the intersection of ray 
with .
is the circumcenter of , prove that , , and are collinear if and only if quadrilateral 
can be inscribed within a circle.
of positive integers satisfying the following two conditions:
, if 
is in 
, then so are 
and 
; and contains an integer 
such that 
is not divisible by 
.
then
where 
Also, 
and 
is largest real root of the equation
be a triangle, and let 
and 
be points on segment symmetric with respect to the midpoint of . Let 
denote the circle passing through and tangent to line at . Similarly, let 
denote the circle passing through and tangent to line at 
. Let the circles and intersect again at point 
(
). Prove that 
.
and 
, let us consider the equation
[list=a]
and 
, find the least positive integer 
satisfying this equation. and , there exist infinitely many positive integers satisfying this equation.
denotes the greatest common divisor of positive integers 
and 
.)
events 
that are each continuous and last on a certain time interval. Each event has a weight 
However, one can only choose to attend activities that do not overlap with each other. The goal is to maximize the sum of weights of all activities attended. Prove or disprove that the following algorithm allows for an optimal selection:
consider 
the sum of 
over all 
such that 
and are not compatible.
If there are multiple such events, delete the event with the minimal weight.
are 
spheres centered at , with radius of 
and , which is remarked as 
and 
, respectively. Given 
points 
outside of (not including the surface of ).
as the number of sets 
such that the midpoint of 
lies entirely inside of .
be the -excenter of a scalene triangle . And let be the point symmetric to about line .
is parallel to the line through the circumcenter and the orthocenter of triangle 
.
of the marks were given to all students.
isogonal conjugates WRT the triangle 
, 
the circumcircles of 
resp. if 
are points of 
resp.such that 
are collinear and 
are collinear then 
.we have previously prove it .
the circumcenter of 
which is the midpoint of arc 
of , 
the orthocenter of which is the symmetric of 
wrt the midpoint of so are isogonal wrt and 
is cyclic then consider 
and applying the lemma yields the result.
![[asy]size(7.5cm);
pair A=(2,9),B=(0,0),C=(10,0),I,K,D,Dp,Ia,O,H,M,P;
I=incenter(A,B,C);K=(B+C)/2;D=foot(I,B,C);Dp=2*K-D;
O=extension(A,I,K,bisectorpoint(B,C));Ia=2*O-I;
H=B+C-I;M=2*Dp-Ia;P=2*I-D;
draw(circumcircle(A,B,C)^^circumcircle(Ia,B,C),green);draw(circle(I,abs(I-D)),orange);
D(MP("A",A,N)--MP("B",B,S)--MP("D",D,S)--MP("K",K,S)--MP("D' ",Dp,SE)--MP("C",C,S)--A);
D(A--MP("I",I,NE)--MP("O",O,SW)--MP("I_a",Ia,S)--MP("H",H,E)--MP("M",M,NE)--A,magenta);
D(A--MP("P",P,E)--Dp,magenta);draw(P--I--D^^I--H--B--I--C--H--O^^B--Ia--C,magenta);
dot(A^^B^^C^^I^^K^^D^^Dp^^O^^Ia^^H^^M^^P);
[/asy]](http://latex.artofproblemsolving.com/1/3/6/136190d9216ceb8b5fbdc902ee3c25f1df8606be.png)
, the midpoint of the arc not containing is the center of 
., , the antipode of the intouch point on (wrt incircle) and the extouch point on are collinear. with orthocenter 
, and 's antipode 
, 
is a parallelogram.
be the circumcenter and orthocenter of respectively. In let 
be the in- and extouch point on respectively, 
the midpoint of , the incenter. By Fact 5, is the midpoint of arc and the midpoint of 
. Also if is the antipode of in 
, we have 
, so by the Diameter of Incircle Lemma, 
. Finally by the Orthocenter-Parallelogram Lemma applied to , 
are collinear, so 
.
And this gets the job done. 
be a triangle. A triangle 
is completed which is similar to the orthic triangle 
and oriented similarly. Prove that the reflection of 
in is parallel to the Euler line of 
in be 
Let the orthocenter and circumcenter of 
be 
resp. Then the circles 
are conjugates under the 
inversion and flip in . So the reflection of in , i.e.
, is the image of , and the point is the image of . So we know that 
and are parallel, and this completes the proof.
of the marks were given to all students.
be the -excenter of a scalene triangle . And let be the point symmetric to about line . is parallel to the line through the circumcenter and the orthocenter of triangle . be the orthocenter and be the circumcenter of triangle . Note that 
hence 
are concyclic. Also 
hence lies on 
. Invert about circle with center and radius 
followed by reflection in bisector of angle 
. Then 
and 
hence 
as desired.
- radius of excircle, - center of circle circumscribed on triangle 
, - orthocenter of this triangle. Easy to proof, that 
is a quadrilateral inscribed in a circle, and by definition of we have 
, so points 
lie on one line in this order. By law of sines 
. Easy trigonometry (from definition) gives 
. 
.
hence by Tales we have 
lies on the neuberg cubic of 
leads to an almost immediate solution: it's easy to show that inversion swaps and , and and . Then 
, so we are done by similar triangles.
![[asy]size(10cm);
pair A=(2,9),B=(0,0),C=(10,0),I,K,D,Dp,Ia,O,H,M,P;
I=incenter(A,B,C);K=(B+C)/2;D=foot(I,B,C);Dp=2*K-D;
O=extension(A,I,K,bisectorpoint(B,C));Ia=2*O-I;
H=B+C-I;M=2*Dp-Ia;P=2*I-D;
draw(circumcircle(A,B,C),orange);
D(MP("A",A,N)--MP("B",B,S)--MP("D",D,S)--MP("K",K,S)--MP("D' ",Dp,SE)--MP("C",C,S)--A);
D(A--MP("I",I,NE)--MP("O",O,SW)--MP("I_a",Ia,S)--MP("H",H,E)--MP("M",M,NE)--A,heavygreen);
D(A--MP("P",P,E)--Dp,heavygreen);draw(P--I--D^^I--H--B--I--C--H--O^^B--Ia--C,red);
dot(A^^B^^C^^I^^K^^D^^Dp^^O^^Ia^^H^^M^^P);
[/asy]](http://latex.artofproblemsolving.com/a/b/1/ab1ee4a95e4c78442ab4a3c66b731502cdf4557b.png)
.
.
. .
respectively as circumcenter and orthocenter of the triangle 
.
be complex numbers lying on unit circle such that 
are vertices of the triangle and 
is the incenter. Then 
Last equation can be obtained without complex bashing by observing that 
is a parallelogram.
Finally
and
If 
we have 
implying 
. If 
#1710
be the -excenter of a scalene triangle . And let be the point symmetric to about line . is parallel to the line through the circumcenter and the orthocenter of triangle .
and 
. So, 
Inversion around followed by a reflection around the angle bisector of 
swaps 
and 
. So, 
. 
are pairs of Isogonal Conjugates WRT 
also lies on the Neuberg Cubic of . So, 
as Neuberg Cubic has its pivot at 
. 
as the unit circle. Let the coordinates of , and be 
, 
and 
so that the incenter 
is given by 
. It is well-known that the excenter is given by 
so that the coordinates of , the circumcenter of is given by 
as it is the midpoint of 
by the incenter excenter lemma. Then we calculate![\[m=\frac{(b-c)(\overline{ab+ca-bc})+\overline{b}c-b\overline{c}}{\overline{b}-\overline{c}}=\frac{(b-c)(\frac{1}{ab}+\frac{1}{ca}-\frac{1}{bc})+\frac{c}{b}-\frac{b}{c}}{\frac{1}{b}-\frac{1}{c}}\]](http://latex.artofproblemsolving.com/c/0/0/c00c8c8c0cc56aeb971dcb928575b27905295c5b.png)
As is the Euler-line of , we know that its centroid 
also lies on that line. Therefore, it suffices to show that the points 
and 
are collinear with the orgin, meaning the ratio![\[\left(\frac{m-a^2}{s-o}\right)\]](http://latex.artofproblemsolving.com/4/d/3/4d30c51a65788854c2da786b38aa8b53f99a2556.png)
should be real. From here on it is just straight calculation.
and 
Prove that 
and 
.
Prove that
such that 
.
?
then 
Equality holds iff 
and any cyclic permutations.
and the mentioned orthocenter be 
be 
,
be 
and 
and 
are collinear by simple angle chasing. Then use angle chasing again to deduct that 
and 
.
We start of with calculations after setting 
over 
.
.
.
,
.
Euler Line of 
,
,
,
,
,
.
,
,
.
,
.
,
.
,
and 
.
.
and 
.![$\overrightarrow{MA}=\frac{q}{p^2+q^2-1}[-2pq,3p^2-3+q^2]$](http://latex.artofproblemsolving.com/9/4/4/9442a0739951bc5ac1a318d18511193496e85b9b.png)
and ![$\overrightarrow{SH}=-\frac{1}{2q}[-2pq,3p^2-3+q^2]$](http://latex.artofproblemsolving.com/c/d/1/cd1a7c5bda40b0816cd42873519d052463a1046d.png)
,
.