[/quote]
,
,
denote the midpoints of 
and 
.
be the projection of 
onto 
.
.
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 rashah76, Me.021, mathematicsy, tiendung2006, Adventure10, Mango247, home245, Rounak_iitr, deplasmanyollari, Funcshun840
In triangle 
, let 
be the excircle opposite to 
. Let 
and 
be the points where is tangent to 
, and 
, respectively. The circle 
intersects line 
at 
and 
. Let 
be the midpoint of 
. Prove that the circle 
is tangent to .
, let 
be the excircle opposite to 
. Let 
and 
be the points where is tangent to 
, and 
, respectively. The circle 
intersects line 
at 
and 
. Let 
be the midpoint of 
. Prove that the circle 
is tangent to .This post has been edited 1 time. Last edited by Amir Hossein, Jul 16, 2018, 4:01 AM
Reason: No LIME
Reason: No LIME
be the center of 
is the second intersection of 
.
,
.
is cyclic. Indeed, simply observe that
Now let 
be the point on 
is tangent to 
by La Hire's Theorem. Hence
so 
hence we are done.
,
.
is 
so 
.
is orthogonal to 
be the reflection of 
.
Moreover, 
,
and circle 
.
,
,
be the excenter opposite 
the second intersection of 
are concyclic. To this effect we define 
to be the midpoint of 
.
is perpendicular to 
,
to 
and 
is tangent to 
cut 
.
is harmonic, we have 
.
and 
be the inverses of 
is tangent to 
,
.
and 
cut 
and 
.
and 
are perpendicular to 
and 
we find that 
at 
.
is the 
.
.
is the mid point of 
excenter, and from mid point theorem. Let 
.
.
are cyclic. Also, from the well known fact that the mid point of arc lies on the chord of contact of a circle inscribed in a segment, we see that 
and 
meet 
is a harmonic quadrilateral due to the tangents at 
.
.![[asy]size(320); pair A = dir(130); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); pair Z = dir(270); pair J = 2*Z-I; pair D = foot(J, B, C); pair E = foot(J, C, A); pair F = foot(J, A, B);
path w = circumcircle(A,E,F); path l = (100*B-99*C)--(100*C-99*B); pair P = IP(l, w); pair Q = OP(l, w); pair N = foot(J, A, D); pair M = midpoint(A--D); pair T = 2*N-D;
draw(A--B--C--cycle, red); draw(w, heavycyan); draw(C--E, red); draw(F--B, red);
draw(circumcircle(D, E, F), heavygreen); pair R = extension(P, Q, E, F); draw(P--R, red); draw(F--R, heavygreen); draw(T--R, heavygreen); draw(J--R, dashed+heavygreen); draw(A--T, red); draw(circumcircle(M, P, Q), dashed+blue);
dot("$A$", A, dir(A)); dot("$B$", B, dir(120)); dot("$C$", C, dir(50)); dot("$J$", J, dir(J)); dot("$D$", D, dir(45)); dot("$E$", E, dir(350)); dot("$F$", F, dir(220)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(30)); dot("$N$", N, dir(350)); dot("$M$", M, dir(230)); dot("$T$", T, dir(T)); dot("$R$", R, dir(R));
/* TSQ Source:
A = dir 130 B = dir 210 R120 C = dir 330 R50 I := incenter A B C Z := dir 270 J = 2*Z-I D = foot J B C R45 E = foot J C A R350 F = foot J A B R220
! path w = circumcircle(A,E,F); ! path l = (100*B-99*C)--(100*C-99*B); P = IP l w Q = OP l w R30 N = foot J A D R350 M = midpoint A--D R230 T = 2*N-D
A--B--C--cycle 0.1 lightred / red w 0.1 lightcyan / heavycyan C--E red F--B red
circumcircle D E F 0.1 lightgreen / heavygreen R = extension P Q E F P--R red F--R heavygreen T--R heavygreen J--R dashed heavygreen A--T red circumcircle M P Q dashed blue
*/ [/asy]](http://latex.artofproblemsolving.com/d/6/9/d691e66ed17f21c3ef4b31a9e5755c1093cf3356.png)
.
by power of a point again, hence 
is tangent to the circumcircle of 
.
. WLOG let the points be 
with center 
; let it intersect 
about 
is swapped with 
.
is tangent to 
,
,
,
(for ![\[DM\cdot DT = (\tfrac12 AD)\cdot (2DN) = AD\cdot DN = BD\cdot DC,\]](http://latex.artofproblemsolving.com/2/d/3/2d36d3f6075a88be16d62c4a3db6765003250cad.png)
so the converse of Power of a Point establishes the concyclicity.
and 
are medians of 
and 
respectively, proving 
To compute this, use Power of a Point on 
to get 
and 
.
Huzzah. 
to be the midpoint of 
,
.
meet 
.
gives 
.
.
,
is isosceles, and the desired similarity follows since 
is also isosceles. 
,
is the 
of 
,
.
is the perpendicular bisector of 
.
and 
.
.
Hence 
,
is harmonic, hence 
is tangent to 
concur at 
,
lies on the radical axis of 
.
We claim that the 
Indeed, let 
intersect the 
we get 
Thus, by considering power at point 
we see that 
implying that 
is cyclic.
which would imply the tangency by Archimedes' Lemma. Let the perpendiculars from 
to 
be 
respectively.
WRT 
we get 
upon subtraction.
and similarly 
implying 
followed by a reflection across the 
and 
.
and line 
.![[asy]
size(7cm);
pair A = dir(130); pair B = dir(210); pair C = dir(330); draw(A--B--C--cycle, red); pair I = incenter(A, B, C); pair O = circumcenter(A, B, C);
pair M_A = -A+2*foot(O, A, I); pair M_B = -B+2*foot(O, B, I); pair M_C = -C+2*foot(O, C, I); draw(unitcircle, blue);
pair L = dir(90); pair T = -L+2*foot(O, I, L); pair F= extension(T, M_B, A, C); pair E = extension(T, M_C, A, B); draw(circumcircle(T,E,F), heavygreen); pair D = B+C - foot(I,B,C); pair Q=intersectionpoints(((E-F)*3+F)--((F-E)*3+E), circumcircle(A,B,C))[0]; pair P=intersectionpoints(((E-F)*3+F)--((F-E)*3+E), circumcircle(A,B,C))[1]; pair M = (A+D)/2; pair M_1=2*T-A; draw(circumcircle(M_1,P,Q), heavygreen);
draw(A--D^^A--M_1, orange); draw(Q--P, lightred);
dot("$A$", A, NW);
dot("$B$", B, SW);
dot("$C$", C, SE);
dot("$D$", D, SE);
dot("$F'$", E, dir(120));
dot("$E'$", F, NE);
dot("$P'$", P, NE);
dot("$Q'$", Q, dir(210));
dot("$M$", M, W);
dot("$M'$", M_1, SW);
dot("$D'$", T, SW);
[/asy]](http://latex.artofproblemsolving.com/2/f/d/2fda4d7f6a7cdff16b08506b9ff85dd88d0d5f15.png)
.
as the image of 
and 
,
and 
(call them 
and 
)
maps to the reflection of 
across 
with line 
.
is tangent to line ![[asy]
size(9cm);
pair F_1 = (-1,1);
pair E_1 = (1,1);
pair M = (0,1);
draw(circle(E_1+(0,-0.85), 0.85), red);
draw(circle(F_1+(0,-1.5), 1.5), red);
draw((-3.5,1)--(3,1), heavygreen);
pair A_1=intersectionpoints(circle(F_1+(0,-1.5), 1.5),circle(E_1+(0,-0.85), 0.85))[0];
pair D_1=intersectionpoints(circle(F_1+(0,-1.5), 1.5),circle(E_1+(0,-0.85), 0.85))[1];
draw(A_1-(3.7,0)--A_1+(2.8,0), blue);
draw(circumcircle(E_1,F_1,D_1), orange);
pair Q_1 = intersectionpoints(circumcircle(E_1,F_1,D_1),A_1-(3.7,0)--A_1+(2.8,0) )[0];
pair P_1 = intersectionpoints(circumcircle(E_1,F_1,D_1),A_1-(3.7,0)--A_1+(2.8,0) )[1];
pair M_1 = 2*D_1-A_1;
draw(circumcircle(M_1,Q_1,P_1), heavygreen);
pair X = 2*M-A_1;
draw(M_1--X, orange);
pair B_1 = intersectionpoints(circle(F_1+(0,-1.5), 1.5),A_1-(3.7,0)--A_1+(2.8,0) )[1];
pair C_1 = intersectionpoints(circle(E_1+(0,-0.85), 0.85), A_1-(3.7,0)--A_1+(2.8,0))[0];
dot("$A_1$", A_1, dir(35));
dot("$B_1$", B_1, NW);
dot("$C_1$", C_1, NE);
dot("$E_1'$", E_1, NE);
dot("$F_1'$", F_1, NW);
dot("$D'$", D_1, dir(-60));
dot("$P_1'$", P_1, SE);
dot("$Q_1'$", Q_1, SW);
dot("$X$", X, N);
dot("$M$", M, NW);
dot("$M'_1$", M_1, SE);
draw(Q_1--M_1--P_1, red);
label("$\omega_b$", circumcenter(B_1,F_1,A_1)+1.8*dir(200));
label("$\omega_c$", circumcenter(C_1,E_1,A_1)+1.2*dir(-20));
[/asy]](http://latex.artofproblemsolving.com/0/1/b/01bd611d089cdb60c5e07b1dea0b8b5836692e5b.png)
intersect 
is the radical axis of 
and 
.
over ![\[MA_1 \cdot AD' = ME_1'^2 \iff MX \cdot MD' = MF_1' \cdot MD_1' \]](http://latex.artofproblemsolving.com/0/e/9/0e946dab14da4b82006d4b8d4b50390bc3fd4ce0.png)
so 
is cyclic. By PoP again,![\[ XA_1 \cdot A_1D' = A_1Q_1' \cdot A_1P_1' \iff MA_1 \cdot A_1M_1' = A_1Q'_1 \cdot A_1P_1' \]](http://latex.artofproblemsolving.com/a/d/d/add7d3740ff78ec81be55dde72a1101c4397c618.png)
so 
is cyclic. Since 
,
is an isosceles trapezoid so 
which implies that ![\[\angle F'_1MQ'_1 = \angle MQ'_1P'_1 = \angle MP'_1Q'_1=\angle MM'_1Q'_1 \]](http://latex.artofproblemsolving.com/4/3/c/43c594c7a9d07a57458606ac76fc9e00e89f53f3.png)
so 
as desired. 
.
is cyclic. Let 
.
i.e. it is enough to prove that 
.
lies on 
.
,
,
and 
.
is the perpendicular bisector of 
is tangent to 
is tangent to 
then 
at 
the image of 
then 
then 
and 
then it sufices to prove that the center of 
is the midpoint of the centers of 
are on 
implies the bisector of 
and 
be 
.
hits 
is a right triangle and by angle chasing, 
.
and 
are similar, from which it follows that 
.
,
is cyclic as desired.
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm);
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 = -18.365249212243693, xmax = 41.64960248564837, ymin = -30.10694671047154, ymax = 5.223133080832475; /* image dimensions */
/* draw figures */
draw(circle((8.31923391956174,-16.833536832441325), 11.06046865982927), linewidth(.7));
draw((3.12,3.66)--(-2.2404698095670987,-13.543153248710096), linewidth(.7));
draw((3.12,3.66)--(16.03321684027274,-8.907097879541702), linewidth(.7));
draw(circle((5.719616959780868,-6.586768416220663), 10.57138928018629), linewidth(.7));
draw(circle((5.708997293900204,-15.197547501126968), 14.141019963430503), linewidth(.7) + linetype("4 4"));
draw((-4.819145635637421,-5.756856209628925)--(28.289404450247126,-5.797688953825286), linewidth(.7));
draw((28.289404450247126,-5.797688953825286)--(-2.2404698095670987,-13.543153248710096), linewidth(.7));
draw((3.12,3.66)--(17.69107601779835,-22.70742333634864), linewidth(.7));
draw((28.289404450247126,-5.797688953825286)--(17.69107601779835,-22.70742333634864), linewidth(.7));
/* dots and labels */
dot((3.12,3.66),dotstyle);
label("$A$", (3.3367794222982536,4.194600918057977), NE * labelscalefactor);
dot((0.18380292682926813,-5.763026341463417),dotstyle);
label("$B$", (0.40546275839092877,-5.267894979467407), NE * labelscalefactor);
dot((12.818563902439035,-5.778608780487806),dotstyle);
label("$C$", (13.004981752378553,-5.267894979467407), NE * labelscalefactor);
dot((8.31923391956174,-16.833536832441325),linewidth(4.pt) + dotstyle);
label("$I_A$", (8.530866844309479,-16.427468945570713), NE * labelscalefactor);
dot((8.332874774955208,-5.773076584236027),linewidth(4.pt) + dotstyle);
label("$D$", (8.530866844309479,-5.370748195744857), NE * labelscalefactor);
dot((-2.2404698095670987,-13.543153248710096),linewidth(4.pt) + dotstyle);
label("$E$", (-2.011587824129146,-13.136166024692319), NE * labelscalefactor);
dot((16.03321684027274,-8.907097879541702),linewidth(4.pt) + dotstyle);
label("$F$", (16.244858065118226,-8.507771292207076), NE * labelscalefactor);
dot((-4.819145635637421,-5.756856209628925),linewidth(4.pt) + dotstyle);
label("$P$", (-4.5,-5.370748195744857), NE * labelscalefactor);
dot((16.26039455257112,-5.7828535870450075),linewidth(4.pt) + dotstyle);
label("$Q$", (16.450564497673128,-5.370748195744857), NE * labelscalefactor);
dot((5.7264373874776044,-1.0565382921180135),linewidth(4.pt) + dotstyle);
label("$M$", (5.908109829234503,-0.6395002469821651), NE * labelscalefactor);
dot((17.69107601779835,-22.70742333634864),linewidth(4.pt) + dotstyle);
label("$S$", (17.890509525557427,-22.8), NE * labelscalefactor);
dot((28.289404450247126,-5.797688953825286),linewidth(4.pt) + dotstyle);
label("$T$", (28.484390802134776,-5.370748195744857), NE * labelscalefactor);
dot((13.011975396376782,-14.240249960292354),linewidth(4.pt) + dotstyle);
label("$R$", (13.210688184933453,-13.804711930495742), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/e/3/0/e30fff1c54f154949898f837911adf0080b6cd20.png)
,
,
.
,
is a harmonic quadrilateral. Thus 
,
is tangent to 
,
,
in 
so 
is cyclic.
since 
,
is a diameter in 
as the center of 
,
.
and 
.
,
must be the midpoint of 
.
is cyclic. Since 
are collinear. Then there must exist a homothety from 
.
,
is also a tangent line to 
.
and let its intersection with 
and 
,
by PoP, implying that 
,
Since 
is harmonic, 
are concurrent, say at 
Since 
we know 
is cyclic. Letting 
,
so 
Thus, 
so by PoaP 
is cyclic.
are concurrent, specifically at 
is tangent to 
we know it is also tangent to 
and 
.
,![\[DP \cdot DQ = DA \cdot DN = 2DM \cdot \frac{1}{2} DT = DM \cdot DT.\]](http://latex.artofproblemsolving.com/8/9/6/8965e059be5655531458d07b881bc8bacc90e843.png)
.![\[RQ \cdot RP = RE \cdot RF = RT^2,\]](http://latex.artofproblemsolving.com/7/a/c/7ac7d3064f62f980274d7c7b8ae6a10e71acd4a6.png)
.
.
and 
,
.
applied to 
,
,
.
,
is cyclic. It remains to show that 
is tangent to 
is tangent to both circles at 
,
as well.
)
is tangent to 
is a cyclic quadrilateral. For this, note that since 
So the claim follows.
is the perpendicular bisector of 
is right angled at 
so we are done by the converse of 
.
.
![[asy]
import graph; size(9cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.5) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -11.88233553870968, xmax = 11.830618443060125, ymin = -14.609017933970518, ymax = 5.3283339819734525; /* image dimensions */
/* draw figures */
draw(circle((1.080363026950906,-9.048627177541055), 7.048627177541055), linewidth(0.5));
draw((-2,4)--(-5.606551884284705,-6.819655652854113), linewidth(0.5));
draw((-2,4)--(6.0644951022459805,-4.0644951022459805), linewidth(0.5));
draw(circle((-0.4598185286713953,-2.524313630917376), 6.703642839471553), linewidth(0.5));
draw((-7.142925780214831,-2)--(6.2232887228720415,-2), linewidth(0.5));
draw(circle((-0.45981852867139494,-7.943987089272077), 8.943987089272078), linewidth(0.5));
/* dots and labels */
dot((-2,4),dotstyle);
label("$A$", (-1.8864969759476087,4.268992395158991), NE * labelscalefactor);
dot((-4,-2),dotstyle);
label("$B$", (-4.629920059749155,-2.6031466167399535), NE * labelscalefactor);
dot((4,-2),dotstyle);
label("$C$", (4.116438682667657,-1.7339432634562923), NE * labelscalefactor);
dot((1.0803630269509057,-2),linewidth(4pt) + dotstyle);
label("$D$", (1.18287736533531,-1.7882684730365213), NE * labelscalefactor);
dot((1.080363026950906,-9.048627177541055),linewidth(4pt) + dotstyle);
label("$I_A$", (1.4816660180265675,-9.556773443009241), NE * labelscalefactor);
dot((6.0644951022459805,-4.0644951022459805),linewidth(4pt) + dotstyle);
label("$E$", (6.262284461086689,-3.9612768562456737), NE * labelscalefactor);
dot((-5.606551884284705,-6.819655652854113),linewidth(4pt) + dotstyle);
label("$F$", (-6.1510259279955575,-7.08497640710883), NE * labelscalefactor);
dot((-7.142925780214831,-2),linewidth(4pt) + dotstyle);
label("$P$", (-7.8894326345628745,-1.9240814969870932), NE * labelscalefactor);
dot((6.2232887228720415,-2),linewidth(4pt) + dotstyle);
label("$Q$", (6.343772275457032,-1.7882684730365213), NE * labelscalefactor);
dot((-0.45981848652454715,1),linewidth(4pt) + dotstyle);
label("$M$", (-0.33822850291109224,1.2267806586661778), NE * labelscalefactor);
dot((6.705912638138943,-13.296295179999717),dotstyle);
label("$I$", (7.185813023950575,-13.332375508835144), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/9/2/1/92146219c38d78e26647078ffca83d4f4771a672.png)
inversion that takes the 
,
and the 
at 
respectively. Points 
.![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.5) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -8.981474539737476, xmax = 5.275139359946031, ymin = -9.127837480782109, ymax = 2.969682290741172; /* image dimensions */
draw((-3,2)--(-4,-2)--(1,-2)--cycle, linewidth(0.5));
/* draw figures */
draw((-3,2)--(-4,-2), linewidth(0.5));
draw((-4,-2)--(1,-2), linewidth(0.5));
draw((1,-2)--(-3,2), linewidth(0.5));
draw(circle((-1.5,-0.5), 2.91547594742265), linewidth(0.5));
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[/asy]](http://latex.artofproblemsolving.com/c/b/b/cbb3efd108dd0e5c27a55c300dbb5c82759aa9bb.png)
be the intersection of 
and Mixillinear Incircle. The tangents from 
because of radax on Mixtillinear Incircle, 
and 
.
.
,
.
.
,
,
are collinear. Now apply radax on 
to get that 
,
are actually cyclic.
.
.
,
,
.
)
,
.
,
and 
,
.
at 
that 
thus
,
which from POP implies that 
is perpendicular to 
is the midpoint of 
.
be the other tangent to 
are collinear
lies on the polar of 
by la hires we have that 
therefore 
are collinear
We know that 
since 
are collinear. We can now observe that 
are concyclic
and 
,
, which tells us that 
.
.
and 
are collinear and 
therefore there exists a homothety from 
and 
be parallel lines and let 
be points on the line 
and 
and 
be the reflection of 
across 
on the same circle and 
is tangent to 
.
which is obvious. now since 
is cyclic, this means 
indicating our desired result.
cyclic. which follows from: 
which finishes the claim.
this means 
followed by 
this indicates tangency since 
indicate by 
the image of 
, this will give: the first image.
inversion from 
to obtain our desired result
Let 
forms a harmonic quadrilateral. Thus, 
concur, and this point of concurrence is 
be the 
Since 
Next, observe that 
so 
is cyclic.
finishes.
