[/quote]
,
,
and 
denote the midpoints of 
and 
.
be the projection of 
be the reflection of 
.
.
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.
,
?
of positive integers, with 
prime, such that 
is a perfect square.
be an integer. Two players, Alice and Bob, play the following game on the complete graph 
:
,![\[
\left( \frac{1}{4} - \varepsilon \right) n^2 + n.
\]](http://latex.artofproblemsolving.com/a/9/c/a9cca00e87c643af32eb75a2501b59060f3504fb.png)
,
(where 
)
,
with 
vertices that contains a Hamiltonian cycle 
and has more than 
edges. For which lengths 
must the graph 
?
.
with 
and 
in graph 
.
satisfying the inequality 
,
?
and graph 
?
,
(i.e., a Hamiltonian path)?
be a set of 2025 positive real numbers. For a subset 
,
as the median of 
when all elements of 
.![\[
P(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ odd}}} M_T, \quad Q(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ even}}} M_T.
\]](http://latex.artofproblemsolving.com/1/d/f/1df2ea8487289111939a53dcc49576fbe6fdef37.png)
Find the smallest real number 
such that for any set ![\[
P(A) - Q(A) \leq C \cdot \max(A),
\]](http://latex.artofproblemsolving.com/c/d/e/cdea343d11c722a423e857aeb01202863405773e.png)
where 
denotes the largest element in 
,
,
number. If a positive integer leaves a remainder that belongs to 
number.
be the Euler's totient function.
,
such that for any 
,
and 
,
is a non-negative power of 
?
,
and 
satisfying:![\[
\varphi(k_i) \in \left ( \frac{x}{a} ,x \right ], i = 1,2; \quad \varphi(k_1) \neq \varphi(k_2).
\]](http://latex.artofproblemsolving.com/6/c/b/6cbe59f15b51439d1202267f521b573a52f1fe67.png)
And these two different 
correspond to the same 
and 
as described in (1), yet 
.
as the number of elements in set 
.
and 
.
with 
.
such that for all 
,
.
Y by Math_CYCR, doxuanlong15052000, mathmaths, Davi-8191, Purple_Planet, Me.021, HWenslawski, Adventure10, Mango247, Rounak_iitr
Let 
be a triangle with 
, and let 
be the foot of the altitude from 
. A point 
is chosen inside the triangle 
so that 
bisects 
. Let 
be the intersection point of the lines 
and . Let 
be the semicircle with diameter that meets the segment 
at an interior point. A line through is tangent to at 
. Prove that the lines 
and meet on .
be a triangle with 
, and let 
be the foot of the altitude from 
. A point 
is chosen inside the triangle 
so that 
bisects 
. Let 
be the intersection point of the lines 
and . Let 
be the semicircle with diameter that meets the segment 
at an interior point. A line through is tangent to at 
. Prove that the lines 
and meet on .
meet at 
;
collinear. Let 
meet 
;
is tangent. Call 
.
at 
.
through a spiral similarity at 
,
onto 
hence 
is tangent to 
![[asy]
unitsize(3cm);
pointfontpen=fontsize(10);
pointpen=black; pathpen=rgb(0.4,0.6,0.8);
pair B=dir(0), A=-B, C=dir(110), H=foot(C,A,B), N=WP(D(C--H),0.85), D=2*N-A, P=extension(B,D,C,H), Q=tangent(P,(B+D)/2,length(B-D)/2,2), K=extension(A,D,C,Q), M=(B+D)/2;
D(CR((0,0),1,0,180),heavygreen); D(arc(M,length(K-M),degrees(B-M),degrees(D-M),CCW),red);
D(A--B--C--cycle,linewidth(1)+pathpen); DPA(D--B--Q--cycle^^Q--P--D^^H--C); D(C--K--A,dashed+pathpen);
/*Angle marks and pathticks*/
markscalefactor=0.01;
DPA(rightanglemark(B,C,A)^^rightanglemark(B,H,C)^^rightanglemark(B,Q,D));
/* Dot and label points */
D("A",A,W);
D("B",B,E);
D("C",C,dir(C));
D("D",D,SSE);
D("H",H);
D("K",K,dir(K));
D("P",P,dir(P));
D("Q",Q,dir(Q));
[/asy]](http://latex.artofproblemsolving.com/1/b/a/1baf5745016f9b7de3d358a7f88ac1408383b2e4.png)
.
.
and since 
and 
,
and our claim holds. Thus 
,
,
lies on 
bisects 
.
.
and 
,
and 
are similar. The spiral similarity sending 
to 
is centered at 
lies on the circumcircle of 
,
.
,
to 
.
and 
.
so we are done.
and 
.
and let 
.
,
,
and 
.
.
and 
.
.
and 
.
meets 
respectively, tangent to 
.
is a parallelogram 
collinear. 
.
are cyclic with diameter 
collinear. Pascal's on hexagon 
collinear thus 
.
meet at a point 
such that 
to 
.
are homothetic at 
gives 
so 
.
.
implies 
.
so 
i.e 
,
be a point on 
.
So this implies that 
is tangent to 
.
,
lies on the circle with diameter 
,
.
,
is tangent to 
,
,
.
But then 
,
be reflection of 
about 
,
colinear. Meanwhile, if 
is such that 
,
because 
.
thus 
are colinear. Also observe 
,
colinear.
,
,
,
because 
and 
.
,
is tangent to 
.
and 
.
.
bisects 
bisects 
.
meet at 
and let 
be the reflection of 
upone 
and 
are collinear
are collinear . 
are cyclic
and 
are cyclic
are cyclic and 
is tangent to 
.
,
,
collinear.
,
.
we get 
.
is right angled, 
.
,
,
.
thereby 
since then it follows that 
.
be the foot of altitude from 
(right angled at 
.
and we conclude.
at 
.
and 
.
and 
where 
,
would be on the circumcircle 
,
.
,
,
.
we obtain 
.
.
and the line 
,
(note that 
by the problem conditions).
so 
as well.
and 
,
and let 
with the base triangle. We have 
but also 
so 
as desired.
is perpendicular to 
and 
is a right triangle, 
and 
from PQ being a tangent, which leads to 
but 
and 
and 
,
,
.
AKDC is parallelogram so 
,
,
,
and CE = EK, it follows that EC = EF. Denote 
,
.
,
.
and 
,
.
.
,
.
,
and let 
meet 
We wish to show that 
is clearly cyclic since 
Since 
intersect 
and let 
by the spiral similarity, or equivalently, since 
However, since 
we see that 
Therefore, 
Since 
this implies that 
so 
since 
,
.
.
then spiral similarity at 
.
.
such that 
,
.
But as 
we have 
.
![[asy]
import geometry; size(8cm);
point A = dir(180); point C = dir(110); point B = dir(0);
triangle t = triangle(A,B,C);
point H = foot(t.VC); point M = (-0.342, 0.1); // cos(110) was in radians for some reason
point D = scale(2,A)*M; point P = intersectionpoint(line(B, D), line(C, H)); point F = intersectionpoint(parallel(H, line(A,D)), line(B,D)); point X = intersectionpoint(perpendicular(B, line(A,D)), line(A,D)); point Q = intersectionpoint(perpendicular(F, line(B,D)), line(C, X));
label("$A$", A, dir(180)); label("$B$", B, dir(0)); label("$C$", C, dir(110)); label("$H$", H, dir(225)); label("$M$", M, dir(120)); label("$D$", D, dir(225)); label("$P$", P, dir(180)); label("$F$", F, dir(270)); label("$X$", X, dir(0)); label("$Q$", Q, dir(90)); draw(A--B--C--cycle); draw(A--X); draw(C--H); draw(H--F); draw(B--P); draw(F--Q); draw(P--Q); draw(C--X, dashed+blue); draw(circle(B, D), red); draw(circle(A, B), red);
[/asy]](http://latex.artofproblemsolving.com/6/5/7/657bc57d7ad0b84f56396e777ff86f6e7cd4f26e.png)
.
and 
.
and 
are cyclic and 
is in fact 
so ![\[ \angle PDQ = 180 - \angle BDQ = 180 - \angle BAC = \angle CEB \]](http://latex.artofproblemsolving.com/3/7/0/370a1e0aceda3feb3c4fd8e027086e346612a801.png)
Also, ![\[ DQ = AC \cdot \frac{BD}{AB} \implies \frac{DQ}{EB} = \frac{AC \cdot BD}{EB\cdot AB} \]](http://latex.artofproblemsolving.com/e/b/f/ebfea1253b8fc8132717c571da469b0d1d43f24b.png)
But then, ![\[ \frac{AC\cdot BD}{EB\cdot AB} = \frac{AC\cdot PD}{HE\cdot AB} = \frac{AC \cdot PD}{AH \cdot AB} = \frac{PD}{AC} = \frac{PD}{CE} \]](http://latex.artofproblemsolving.com/2/2/a/22ad195564e4a76b40f2c47e852406d745d6e40f.png)
this readily implies 
,
which implies the conclusion. 
and let 
,
,
,
.
implying that 
hence 
.
.
.
since 
,
since both angles are acute.
we have 
.
.
.
which is sufficient as this implies that 
is the polar of 
.
so the spiral similiarity sending 
also sends 
,
and let 
.
.
,
is tangent to 
.
,
.
.
.
,
,
such that 
is center of spiral similarity taking 
to 
under the same transformation
is center of spiral sim taking 
to 
.
,
.![\[-1=(AD;M\infty)\stackrel{H}{=}(BD;PZ),\]](http://latex.artofproblemsolving.com/8/7/7/8771bf0ce68faeb5879e9a326e5cff6c5a484e60.png)
so indeed we get the desired tangency. 
![[asy]
import graph; size(8cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -7.285601963776682, xmax = 6.892842353800001, ymin = -4.759777562678323, ymax = 8.25284545671791; /* image dimensions */
pen red = rgb(0.8,0,0);
draw((-5,5)--(4,-1)--(-5,-1)--cycle, linewidth(0.7));
draw((-4.636311700472756,-1)--(-4.636311700472756,-0.6363117004727565)--(-5,-0.6363117004727565)--(-5,-1)--cycle, linewidth(0.7));
draw((-5,5)--(4,-1), linewidth(0.7) + red);
draw((4,-1)--(-5,-1), linewidth(0.7));
draw((-5,-1)--(-5,5), linewidth(0.7));
draw((-4.03076923076923,0.4538461538461567)--(0.5384615384615381,-2.5), linewidth(0.7));
draw((-0.4615384615384608,-0.19230769230769074)--(4,-1), linewidth(0.7) + red);
draw((-0.4615384615384608,-0.19230769230769074)--(0.5384615384615381,-2.5), linewidth(0.7));
draw((0.5384615384615381,-2.5)--(4,-1), linewidth(0.7));
draw(circle((1.76923076923077,-0.596153846153845), 2.267029571250247), linewidth(0.7));
draw((-0.4615384615384608,-0.19230769230769074)--(-0.46153846153846095,-1), linewidth(0.7));
draw((-0.46153846153846095,-1)--(0.5384615384615381,-2.5), linewidth(0.7));
draw(circle((-0.5,2), 5.408326913195984), linewidth(0.7));
draw((-4.03076923076923,0.4538461538461567)--(-0.4615384615384608,-0.19230769230769074), linewidth(0.7) + red);
draw((-5,-1)--(1.871006874241814,-2.8608976951071563), linewidth(0.7));
draw((-5,5)--(-0.4615384615384608,-0.19230769230769074), linewidth(0.7) + red);
draw((-0.4615384615384608,-0.19230769230769074)--(1.871006874241814,-2.8608976951071563), linewidth(0.7));
draw((-4.03076923076923,0.4538461538461567)--(-2.230769230769231,3.1538461538461537), linewidth(0.7) + blue);
draw((-5,-1)--(-4.03076923076923,0.4538461538461567), linewidth(0.7));
/* dots and labels */
dot((-5,5),linewidth(3pt) + dotstyle);
label("$A$", (-5.50283,4.926825870732575), NE * labelscalefactor);
dot((4,-1),linewidth(3pt) + dotstyle);
label("$B$", (4.184022278616064,-1.1937359447146654), NE * labelscalefactor);
dot((-5,-1),linewidth(3pt) + dotstyle);
label("$C$", (-5.70665,-1.3137469607038268), NE * labelscalefactor);
dot((-2.230769230769231,3.1538461538461537),linewidth(3pt) + dotstyle);
label("$H$", (-2.1422727070983205,3.315249370306691), NE * labelscalefactor);
dot((-2.7307692307692304,2.4038461538461546),linewidth(3pt) + dotstyle);
label("$D_{1}$", (-3.61,2.28658351897102), NE * labelscalefactor);
dot((-0.4615384615384608,-0.19230769230769074),linewidth(3pt) + dotstyle);
label("$D$", (-0.97301,-0.593), NE * labelscalefactor);
dot((-4.03076923076923,0.4538461538461567),linewidth(3pt) + dotstyle);
label("$P$", (-4.336759856614422,0.5721404334115693), NE * labelscalefactor);
dot((0.5384615384615381,-2.5),linewidth(3pt) + dotstyle);
label("$Q$", (0.2825336107293,-3.13461129711), NE * labelscalefactor);
dot((1.871006874241814,-2.8608976951071563),linewidth(3pt) + dotstyle);
label("$S$", (1.852379682255207,-3.2510676473860065), NE * labelscalefactor);
dot((-0.46153846153846095,-1),linewidth(3pt) + dotstyle);
label("$F$", (-0.8535738077951629,-1.5035822415016), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/d/3/3/d337a7b1af888859ede809e7681be006cf07919e.png)
,
and 
.
.
as 
to get
This gives 
.
is collinear and we are done. 
,
,
,
too. Thus 
.
,![\[-1 = (AD:M\infty) \overset{H}{=} (BD,PE),\]](http://latex.artofproblemsolving.com/d/c/0/dc06303c027b655222a8a8b5bd8805f2774f5295.png)
.![\[\frac{DE}{EB} = \frac{AH}{HB} \implies \angle DEQ' = \angle AHC = 90,\]](http://latex.artofproblemsolving.com/9/8/f/98f37181d6b8ad796ab222effe66f0f85a1d413e.png)
.
,
