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.
with integer coefficients such that ![$P\left(\sqrt[3]5+\sqrt[3]{25}\right)=5+\sqrt[3]5$](http://latex.artofproblemsolving.com/5/8/e/58ec2a921cd988bd5ae017a8653544ddd758225f.png)
.
be a 
-element subset of the set 
. Prove that there exist numbers 
, 
in 
such that the sets ![\[ A_j=\{x+t_j\mid x\in A\},\qquad j=1,2,\ldots,100 \]](http://latex.artofproblemsolving.com/d/5/c/d5c2a7b943e8c7200361698f3eb6f83700ad68d6.png)
are pairwise disjoint.
positive integers 
such that for every index 
, with 
, 
is a multiple of 
. Determine the greatest possible value of the largest of the numbers.
be a triangle. Let 
and 
be the points in which the median and the angle bisector, respectively, at meet the side 
. Let 
and be the points in which the perpendicular at to 
meets 
and 
, respectively. And 
the point in which the perpendicular at to meets 
produced.
is perpendicular to .
and 
such that for all 
if 
then 
Find the smallest of sum of all elements of 
such that there exist positive integers 
for which the following holds 
for all 
. where 
is a perfect cube
of a prime number 
and a positive integer 
for which 
is an integer.
the point 
is the centre of the excircle opposite the vertex 
This excircle is tangent to the side at , and to the lines 
and 
at 
and 
, respectively. The lines 
and 
meet at 
, and the lines 
and 
meet at 
Let be the point of intersection of the lines 
and , and let 
be the point of intersection of the lines 
and 
Prove that is the midpoint of 
opposite the vertex is the circle that is tangent to the line segment , to the ray beyond 
, and to the ray beyond 
.)
such that if 
and 
are distinct rational numbers satisfying 
or 
, then 
? Justify your answer.
be a positive integer. Consider the following equation: ![\[ \{x\}+\{2x\}+ \dots + \{nx\} = \lfloor x \rfloor + \lfloor 2x \rfloor + \dots + \lfloor 2nx \rfloor\]](http://latex.artofproblemsolving.com/8/2/9/8299b3ceb1650418c0a373536cb1d2931900efd9.png)
a) For 
, solve the given equation in 
., the equation has at most 
real solutions.
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.
,![\[
\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x
\]](http://latex.artofproblemsolving.com/9/5/5/955863a0cf8747ae45b736b0631f243d3908eb84.png)
and 
.
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)
,
.