and 
be given as: 
,
(where 
can be 0), and let 
be the count of numbers 
such that 
and 
is considered the same as 
)
is 
,
,
,
are positive integers, and 
are odd primes.
be positive integers, and let 
all be 
-![\[
\alpha_1^j + \alpha_2^j + \cdots + \alpha_n^j = 0 \quad \text{for any } j (0 < j < k).
\]](http://latex.artofproblemsolving.com/3/3/b/33be8eef29cd506bf1132a86def9992ec8d21cb4.png)
Prove that among 
is to be examined. The plan requires that all station pairs have the same number of jointly directly reachable stations.
,
?
Y by Ya_pank, ImSh95
Let 
be a point inside triangle 
. Let 
meet 
at 
, let 
meet 
at 
, and let 
meet 
at 
. Let 
be the point such that is the midpoint of 
, let 
be the point such that is the midpoint of 
, and let 
be the point such that is the midpoint of 
. Prove that points 
, and cannot all lie strictly inside the circumcircle of triangle .
(Australia)
be a point inside triangle 
. Let 
meet 
at 
, let 
meet 
at 
, and let 
meet 
at 
. Let 
be the point such that is the midpoint of 
, let 
be the point such that is the midpoint of 
, and let 
be the point such that is the midpoint of 
. Prove that points 
, and cannot all lie strictly inside the circumcircle of triangle .(Australia)
and 
,
is the orthocenter of 
.
,
,
,
,
,
are concyclic, so 
,![[asy]
size(200);
pair A = dir(120); dot("$A$", A, dir(A));
pair B = dir(210); dot("$B$", B, dir(B));
pair C = dir(-30); dot("$C$", C, dir(C));
pair P = 0.3A + 0.3B + 0.4C; dot("$P$", P, dir(P));
pair A1 = extension(A, P, B, C);
pair B1 = extension(B, P, C, A);
pair C1 = extension(C, P, A, B);
pair A2 = 2*A1 - P; dot("$A_2$", A2, dir(A2));
pair B2 = 2*B1 - P; dot("$B_2$", B2, dir(B2));
pair C2 = 2*C1 - P; dot("$C_2$", C2, dir(C2));
draw(A -- B -- C -- cycle);
draw(A -- A2 ^^ B -- B2 ^^ C -- C2);
draw(circumcircle(A, B, C));
add(pathticks(P -- A1, s=2));
add(pathticks(A1 -- A2, s=2));
add(pathticks(P -- B1, s=2, n=2, spacing=1));
add(pathticks(B1 -- B2, s=2, n=2, spacing=1));
add(pathticks(P -- C1, s=2, n=3, spacing=1));
add(pathticks(C1 -- C2, s=2, n=3, spacing=1));
// finds a sixth point P5 on an ellipse given five points P1, ... P5 such that P1P2 || P5P6 using Pascal's theorem
pair sixthPoint(pair P1, pair P2, pair P3, pair P4, pair P5) {
pair X = extension(P3, P5, P1, P4);
pair Y = extension(P2, P3, X, X+(P2-P1));
return extension(Y, P4, P5, P5+(P2-P1));
}
// gets conjugate diameters
guide ellipseThroughFivePoints(pair P1, pair P2, pair P3, pair P4, pair P5) {
pair O = extension((P1+P2)/2, (P5+sixthPoint(P1,P2,P3,P4,P5))/2,(P2+P3)/2, (P1+sixthPoint(P2,P3,P4,P5,P1))/2);
pair P6 = sixthPoint(P1, P2, P3, P4, P5);
pair M1 = (P1+P2)/2;
pair M2 = (P5 + P6)/2;
pair X = extension(P1, P2, P4, P5);
pair Y = extension(P1, P5, P2, P3);
pair Z = extension(X, Y, P3, P4);
pair Q1 = extension(M1, M2, P1, Z);
pair T = sqrt(abs(O - M1) * abs(O - Q1)) * dir(Q1 - O) + O;
pair M3 = extension(O, O+(P1-P2), P1, P1+(T-O));
pair Q2 = extension(Q1, P1, O, M3);
pair R = sqrt(abs(O - M3) * abs(O - Q2)) * dir(M3-O) + O;
// OR and OT are conjugate diameters
transform t = (O.x, O.y, R.x-O.x, T.x-O.x, R.y-O.y, T.y-O.y);
return (t * circle((0, 0), 1));
}
draw(ellipseThroughFivePoints(A, B2, C, A2, B), dotted);
[/asy]](http://latex.artofproblemsolving.com/2/3/9/239a77f52a2c48198a0479df2e5b90f49e297464.png)
with 
so 
.
if and only if ![\[ 0 < a^2 \cdot yz(2x+y+z) - b^2 \cdot xz(y+z) - c^2xy(y+z). \]](http://latex.artofproblemsolving.com/2/9/d/29d843bfe75982f17fd849cc3fcc5c0f2b32d789.png)
Proof. The condition we want is that ![\begin{align*} 0 &< -\operatorname{Pow}(A_2, (ABC)) \\ &= a^2 \cdot y(2x+y+z) \cdot z(2x+y+z) \\ &\qquad - b^2x(y+z) \cdot z(2x+y+z) - c^2x(y+z) \cdot y(2x+y+z) \\ &= (2x+y+z) \left[ a^2 \cdot yz(2x+y+z) - b^2 \cdot xz(y+z) - c^2 \cdot xy(y+z) \right]. \end{align*}](http://latex.artofproblemsolving.com/2/7/7/2772a22640c02d85a8298156538f3fd8c06ce4e2.png)
From 
we conclude. 
.
of 
lie on 
,
is the circumcenter of 
to meet 
.
.
is the reflection of 
so 
.
hence 
,
.
and 
are isogonal w.r.t. 
.
thus 
all lie inside 
and 
are colinear in this order so 
are colinear in this order.
is the isogonal conjugate of 
are colinear in this order. But 
so 
,
is a cubic curve.
;
of 
;
of 
,
,
and 
so translating pencils gives
so 
lie on a conic. Finally by converse of Pascal's theorem on hexagon 
gives 
lie on this conic. 
,
lie on a conic. Hence isogonal conjugate 
beyond 
similarly. This is enough to force that 
are not colinear, contradiction.
lie on circumcircle then 
with 
is 
and also we have that
Multiplying stuff from these equations yields the following few equations.
but then this must imply that 
and cyclic inequalities. But notice that summing this inequalities yields that 
which is clearly wrong. With this we achieve the desired contradicition 
the problem states that not all three inequalities 
etc. can hold simultaneously.
,
,
,
,
cyclically.
and 
.
we get a contradiction.
,
and 
is midpoint of 
.
and of 
.
.
is 
and along 
be 
.
.
.
and 
of the circles 
and 
!![[asy]
pair A = dir(110), B = dir(152), C = dir(28);
pair O = (0,0);
draw(A--B--C--cycle, royalblue);
draw(circumcircle(A,B,C),springgreen);
draw(circle((A+O)/2,0.5), fuchsia);
draw(circle((C+O)/2,0.5), fuchsia);
draw(circle((B+O)/2,0.5), fuchsia);
pair Q = (0.02, 0.2);
dot(Q, 3+orange);
label("$\underline{\underline{Q}}$",Q,dir(0),orange);
dot("$A$",A,dir(A),3+purple);
dot("$C$",C,dir(C),3+purple);
dot("$B$",B,dir(B),3+purple);
dot("$O$",O,dir(270),4+magenta);
[/asy]](http://latex.artofproblemsolving.com/b/e/2/be28bd216f3933a4a93d591da20aaa60df97b451.png)
and 
,
,
.![\[\left\{\frac{PA_1}{XA_1}, \frac{PB_1}{YB_1}, \frac{PC_1}{ZC_1}\right\}\]](http://latex.artofproblemsolving.com/d/7/0/d7097b452a917f39cff33420515827b614313839.png)
are all 
.
in two ways; using ratio lemma on 
and 
:![\[\frac{PA_1}{XA_1} = \frac{PB\sin{\angle PBC}}{XB\sin{\angle XAC}} = \frac{\sin{\angle C}\sin{\angle PBC}}{\sin{\angle APB}\sin{\angle PAC}} \]](http://latex.artofproblemsolving.com/9/7/7/9772894a4965bf364da16c5ef6e05a40d2f963f5.png)
![\[\frac{PA_1}{XA_1} = \frac{PC\sin{\angle PCB}}{XC\sin{\angle XAB}} = \frac{\sin{\angle B}\sin{\angle PCB}}{\sin{\angle APC}\sin{\angle PAB}} \]](http://latex.artofproblemsolving.com/3/4/f/34f48196ee60c9c530e5bca7971465aab6be8c62.png)
and similarly for the other two![\[\frac{PB_1}{YB_1} = \frac{\sin{\angle A} \sin{\angle PCA}}{\sin{\angle BPC}\sin{\angle PBA}}\]](http://latex.artofproblemsolving.com/0/3/a/03a54cda7d8b53a61ece50f58ddd31cc464c11d2.png)
![\[\frac{PB_1}{YB_1} = \frac{\sin{\angle C} \sin{\angle PAC}}{\sin{\angle BPA}\sin{\angle PBC}}\]](http://latex.artofproblemsolving.com/2/4/8/248cf15da23e014ae9690eb439a998ca6cc86022.png)
and![\[\frac{PC_1}{ZC_1} = \frac{\sin{\angle B}\sin{\angle PAB}}{\sin{\angle BPA} \sin{\angle PCB}}\]](http://latex.artofproblemsolving.com/9/4/f/94fe9fd8f7bfc17b309ca7375d013605c27e7234.png)
![\[\frac{PC_1}{ZC_1} = \frac{\sin{\angle A}\sin{\angle PBA}}{\sin{\angle APB} \sin{\angle PCA}}\]](http://latex.artofproblemsolving.com/a/a/2/aa29a18df5ee021eaddd96f652f3dd523956dffc.png)
Notice that multiplying 
,
,
,![\[\left(\frac{\sin{\angle C}}{\sin{\angle APB}}\right)^2, \left(\frac{\sin{\angle A}}{\sin{\angle BPC}}\right)^2, \left(\frac{\sin{\angle B}}{\sin{\angle CPA}}\right)^2 < 1\]](http://latex.artofproblemsolving.com/c/1/f/c1fd7b91a28100e5773e5b85b88df3a48a77d288.png)
which tells us that 
,
and 
.
,
,
.![\[\cos{(\tfrac{\angle APB + \angle C}{2})} \sin{(\tfrac{\angle APB - \angle C}{2})} > 0\]](http://latex.artofproblemsolving.com/a/6/5/a65e93ce1368b6b0a0a251c367984c5f536c7a83.png)
![\[\cos{(\tfrac{\angle BPC + \angle A}{2})} \sin{(\tfrac{\angle BPC - \angle A}{2})} > 0\]](http://latex.artofproblemsolving.com/f/6/a/f6a208922cb7dddc392615f7cdbad17089651b3f.png)
![\[\cos{(\tfrac{\angle CPA + \angle B}{2})} \sin{(\tfrac{\angle CPA - \angle B}{2})} > 0\]](http://latex.artofproblemsolving.com/3/a/0/3a027b8c18526cb47495f2fd18f4d412eae71fe7.png)
from which it follows that 
.![\[\angle APB + \angle C, \angle BPC + \angle A, \angle CPA + \angle B\]](http://latex.artofproblemsolving.com/6/c/6/6c6966f5b7f7103ecc7b043a4ebdabc10fa9a1dc.png)
is bounded between 
,
and summing yields 
,
be the circumcircle. Consider an affine transformation that sends 
to an orthocentric system; then the images of 
,
,
,
,
,
lie on a circle. Undoing the affine transformation, points 
.![[asy] import olympiad; import geometry;
pen pri=blue; pen sec=purple+pink; pen tri=lightblue; pen qua=lightred; pen fil=invisible; pen tfil=invisible; pen qfil=invisible;
size(7cm); defaultpen(fontsize(10pt)); pair A,B,C,P,A1,B1,C1,A2,B2,C2,D; A=dir(110); B=dir(210); C=dir(330); P=0.3*dir(250); A1=extension(A,P,B,C); B1=extension(B,P,C,A); C1=extension(C,P,A,B); A2=2A1-P; B2=2B1-P; C2=2C1-P; path e=(path)conic(A,B,C,A2,B2); for(int i=0;i<4;i+=1){ D=intersectionpoints(unitcircle,e)[i]; if(D==A||D==B||D==C)continue; break; }
filldraw(e,qfil,qua+dashed); draw(A--A2,sec); draw(B--B2,sec); draw(C--C2,sec); filldraw(unitcircle,tfil,tri); filldraw(A--B--C--cycle,fil,pri);
dot("\(A\)",A,N); dot("\(B\)",B,SW); dot("\(C\)",C,SE); dot("\(D\)",D,NE); dot("\(P\)",P,dir(60)); dot("\(A_1\)",A1,SW); dot("\(B_1\)",B1,dir(75)); dot("\(C_1\)",C1,dir(120)); dot("\(A_2\)",A2,S); dot("\(B_2\)",B2,E); dot("\(C_2\)",C2,W); [/asy]](http://latex.artofproblemsolving.com/e/c/0/ec0d672ad93c810287ae1dcb75a42ee34b09eb76.png)
of 
,
.
.
or 
![[asy]
size(400);
pair A = dir(120); dot("$A$", A, dir(A));
pair B = dir(210); dot("$B$", B, dir(B));
pair C = dir(-30); dot("$C$", C, dir(C));
pair P = 0.45A + 0.15B + 0.4C; dot("$P$", P, dir(P));
pair A1 = extension(A, P, B, C);
pair B1 = extension(B, P, C, A);
pair C1 = extension(C, P, A, B);
pair A2 = 2*A1 - P; dot("$A_1$", A1, dir(A1));
pair B2 = 2*B1 - P; dot("$B_1$", B1, dir(B1));
pair C2 = 2*C1 - P; dot("$C_1$", C1, dir(C1));
draw(A -- B -- C -- cycle);
draw(A -- A1 ^^ B -- B1 ^^ C -- C1);
path circ1 = circumcircle(A,P,B);
draw(circ1,dotted);
pair T = 2*B - C;
pair S = 2*C - B;
path lineBC = T -- S;
draw(lineBC);
pair [ ] x = intersectionpoints (lineBC , circ1);
dot("$K$", x[1] ,dir(x[0]));
[/asy]](http://latex.artofproblemsolving.com/f/6/6/f66681be9e19871355af1d0a9aa5c1cd5cd16f19.png)
be max among 
,
intersect line 
and let 
intersect 
we have 
,
and other analogous results.
.
.
be the reflection of 
.
.
cut line 
,
so 
.
.
![[asy]
pair A=dir(130),B=dir(-150),C=dir(-30),A3=dir(-85),A1=extension(A,A3,B,C),P=1.6*A1-0.6*A3,A3p=2*foot(A3,B,C)-A3,Pp=2*A1-A3,A0=foot(A3,B,C);
draw(unitcircle,brown);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$A_1$",A1,dir(A1));
dot("$A_3$",A3,dir(-40));
dot("$P$",P,dir(P));
dot("$P'$",Pp,dir(160));
dot("$A_3'$",A3p,dir(-50));
dot("$A_0$",A0,dir(-45));
draw(B--A--C,magenta);
draw(B--C^^Pp--A3p,red);
draw(A--A3^^A3--A3p,green);
[/asy]](http://latex.artofproblemsolving.com/d/9/b/d9b0c24f469741a4bc9032dd87fbee29c750aa0d.png)
then 
.
In particular, we have that
Now let 
be the orthocenter of 
as usual. The conditions we obtain imply that 
But it is easy to see that these three triangle do not have a common interior point. So we have a contradiction!
.
so Assume not.
so 
.
.
.
and 
but then we should have 
and 
and 
which is obviously not true so contradiction so at least one of 
.
.
.
If all the points lie inside the circumcircle of 
onto orthocentric quadruple, and so maps 
onto set of concyclic points. This yields that points from 
lie on one ellipse 
is obvious, so assume the opposite. Note that 
lie on 
in that order; WLOG arc 
of ellipse intersects 
again at 
One of arc 
of 
hence either 
,
be the intersections of 
,
,
,
,
,
.
Suppose all those are less than one, then the following are also less than one:
Therefore,
Which is absurd.
mapping 
to an orthocentric system; note that 
to a set of concyclic points by reflecting the orthocenter. Taking 
of the transformed image yields that 
all lie on an ellipse.
.
and 
,
,
.
so
Then
so
Similarly,
so![\[
\frac{(1-p)^2}{p+1}\mathrm{Pow}_{(ABC)}(A_2)+\frac{(1-q)^2}{q+1}\mathrm{Pow}_{(ABC)}(B_2)+\frac{(1-r)^2}{r+1}\mathrm{Pow}_{(ABC)}(C_2)=0
\]](http://latex.artofproblemsolving.com/2/d/3/2d369e8aca225c3051e4748fd98f8baaa9a347e7.png)
(note that 
)
and thus 
,
must be positive. Thus at least one of 
and its cyclic versions. Call 
we get
is the smallest among the three. Then from the condition above 
and 
.
be the intersection of 
,
,
,
and 
.
and similarly 
,
