[/quote]
,
be positive real variables. Prove that
be a set that intersects all non-finite integer arithmetic progressions, 
be the set of prime divisors of 
and 
be the set of prime divisors of 
.
.
,
be centered at 
at 
,
respectively. Let line 
again at 
again at 
.
is the other intersection of the circumcircles of triangles 
,
,
to 
and 
be the foot of the perpendicular from 
.
is on 
such that 
.
is the orthocenter of 
,
bisects 
.
,
are reals. Let 
be a circle with diameter 
and let 
meets the x-axis again at 
.
,
denote the midpoints of 
.
.
.
and 
,
(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 
?
Y by fishy15, centslordm, anantmudgal09, Bubu-Droid, buratinogigle, tree_3, HWenslawski, gadamus003, p_square, RedFlame2112, megarnie, jhu08, HamstPan38825, amar_04, son7, psi241, Ballocoma, StefanSebez, PHSH, Aryan-23, TheHU-1729, Rounak_iitr, ehuseyinyigit, Funcshun840, MS_asdfgzxcvb
Let 
be an interior point of the acute triangle 
with 
so that 
The point 
on the segment 
satisfies 
the point 
on the segment 
satisfies 
and the point 
on the line satisfies 
Let 
and 
be the circumcenters of the triangles 
and 
respectively. Prove that the lines 
and 
are concurrent.
be an interior point of the acute triangle 
with 
so that 
The point 
on the segment 
satisfies 
the point 
on the segment 
satisfies 
and the point 
on the line satisfies 
Let 
and 
be the circumcenters of the triangles 
and 
respectively. Prove that the lines 
and 
are concurrent.
intersects 
for the second time. Then 
.
it follows that 
,
.
,
and 
is cyclic. Now we prove that 
and 
,![\[\angle FED = \angle AED - \angle AEF = \angle ABP - \angle ABC = \angle PBC = \angle KDC\]](http://latex.artofproblemsolving.com/2/5/3/2535173c027dd4863ef92dfdcac6ab79e9eb4d51.png)
and some more easy angle chasing shows 
,
,
and 
,
with center 
and 
also lies on 
and let 
and 
.
.
intersect on 
,
are collinear on account of 
be a cyclic pentagon, let 
and 
,
.
.
,
.
,
we get 
collinear, implying that 
,
is the center of the Apollonian circle.
be the isogonal conjugate of 
,
is cyclic and similarly 
is cyclic. From POP we get 
therefore 
is cyclic, too. Let 
.
so 
and 
are tangent. It yields that 
is tangent to both 
and 
,
or 
.
and 
intersects again at 
,
.
or 
.
hence we need to prove that 
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.
lies on 
.
lies on 
and 
.
or 
passes through the circumcenter of 
.
.
and 
,
.
.
and 
,
So
The last equality holds by law of sines, hence the claim is proved and it yields 
.
be the second intersection of 
so 
is cyclic. On the other hand, we proved that 
and 
,
.
is the external angle bisector of 
,
.
so 
and passes through the circumcenter of 
.
and 
centred at 
at 
.
tangent to 
gives that 
are pairs of an involution on 
are coaxial (since 
be the isogonal conjugate of 
.
and 
,
hence 
hence the circles 
and 
are tangent at 
and applying radical axes, we get 
and the line through 
with 
.
passes through 
.
hence 
cyclic. Also, 
;
of 
at 
is cyclic since 
,
,
,
.
with 
and 
be the point on 
.
and 
are projective mappings, hence it suffices to check 
for three choices of 
are both the second intersection of ray 
with 
,
.
be the isogonal conjugate of 
,
,
,
implying that 
,
.
we have
This means
implying that 
is the common tangent line of ![[asy]
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,
be another intersection of 
and 
.
with center 
,
.
are concurrent at a point 
.
On the other hand,
so 
Hence, 
of ![[asy]
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in the unit circle. Let lines 
meet the unit circle again at 
Now we find the coordinate of 
And since 
as directed angles, we have
Now 
,
so we write the angle condition as
Clearing denominators,
Solving for 
gives
which we simplify to
This coordinate is a chore to deal with, but thankfully we find the very nice factorization
which we will use later. Next we find the coordinate of 
Since 
,
We have
and so
Now we find the coordinate of 
,
,
Then
where a factor of
was magically canceled from the numerator and denominator in the last step. To finish, we write
Now let 
be the intersection of lines 
are collinear. Since 
we have
And since 
![$$\frac{c(a-p)\left[d(ad+cd-ac-cp) - z(d^2-cp)\right]}{a(a-d)(c+d)(c-d)(d-p)} = \frac{d^2(a-p)\left[cp(ap+cp-ad-dp) + a(d^2-cp)(c+p-z)\right])}{cp^2(a-d)(c+d)(c-d)(d-p)}$$](http://latex.artofproblemsolving.com/2/2/5/225f85edfdd83a52aab85827f0e4a6c053480e71.png)
Thus
which we simplify to
Sure enough, when we divide, the numerators magically cancel. The result is
which is real, as desired. 
,![\[\angle D^*BF=\angle CBD = \angle ADF\]](http://latex.artofproblemsolving.com/e/7/f/e7f08be78b5e58755e32983034b696efe73ab0de.png)
and so 
are concyclic. Similarly 
are concyclic. Since 
are concurrent at ![\[AB\cdot AF = AD\cdot AD^* = AC\cdot AE\]](http://latex.artofproblemsolving.com/9/a/2/9a20ce806381b02cd27ced963a181bf5f0a77e24.png)
and so 
.
is a trapezoid with 
and 
.![\[\angle C'B'F' = \angle C'B'D+\angle DB'A'+\angle A'B'F' = \angle DCB+\angle BAD+\angle A'DF' = \angle E'DA'+\angle DAC+\angle CBD = \angle E'C'A' + \angle A'C'D + \angle DC'B' = \angle E'C'B'.\]](http://latex.artofproblemsolving.com/1/2/4/1242c1dd1a2f6f902db2827c2a85802522b2c80c.png)
Therefore we have the desired result.
.
are concyclic.![\[\angle MBX=\angle CBX-\angle CBM=\angle ACB-\angle CAM=\angle EFA-\angle CAM=\angle EMA+\angle MAE=\angle MEX\]](http://latex.artofproblemsolving.com/3/a/3/3a3b0ef61b2c3c28846b221cf68df17d29b199a3.png)
where we are using directed angles and we use Lemma 1.
and 
,
is the intersection of 
and 
,
w.r.t. 
are concyclic. It is well-known that 
is the reflection of 
is the reflection of 
are concurrent. However, since 
,
and 
are concyclic, their radical axes are concurrent. They are exactly 
and 
,
cyclic
of 
is 
is 
is isogonal w.r.t 
and 
and 
is isogonal w.r.t 
)
meets 
at 
respectively.
.
.
.
and 
are homothetic, and the homothetic center is 
.
.
.
.
colinear (육점공선 또는 Pole & Polar), which gives 
are all same by (1) and (2)
a
,
,
and 
such that 
.
is tangent to 
at 
.
(the clockwise angle between the lines 
and 
)
.
are isogonal in 
,
,
.
and 
.
.
and 
(i.e. 
is tangent to 
at the point 
for the second time at 
.
so 
.
and 
,
.
and 
,
,
.
,
,
,
and 
are cyclic. PoP at 
,
and 
.
suffices. Also, let 
.
is tangent to 
,
at 
,
themselves. By the isogonality lemma, so are 
and 
,
and 
are coaxial, because 
.
,
is cyclic. Hence, this inversion swaps 
.
is cyclic, where 
is the image of 
,
be the intersection of 
and 
,
is cyclic. Similarly, 
is cyclic. We have ![\[AF\cdot AB=AD'\cdot AD=AE\cdot AC\]](http://latex.artofproblemsolving.com/2/5/a/25a659ee92de363f22458c76d117b70df6a7d0db.png)
so 
.
.![\[\angle CDE=180^\circ-\angle CDY-\angle ADE=180^\circ-\angle CDY-\angle DCY=\angle DYC\]](http://latex.artofproblemsolving.com/5/a/e/5ae9879ea24912d868ba8bc2f5aada93f49205ea.png)
so let 
and so![\[\angle P'DE=\angle CDE-\angle P'DC = \angle DYC-\angle DBY = \angle BDY\]](http://latex.artofproblemsolving.com/9/2/4/92475216c97f40664b8eeccaaefd0128bebb9142.png)
On the other hand,![\[\angle DFZ=\angle DZE-\angle FDZ=\angle DYC-\angle DBY=\angle BDY\]](http://latex.artofproblemsolving.com/7/7/6/7765a4e68b8b1432ca8045f961c2fbda167637b9.png)
so 
,
and the following are concyclic:
so 
is cyclic. We have 
so the inversion centered at 
.![\[GA\cdot GC = GS\cdot GB=GE\cdot GX\]](http://latex.artofproblemsolving.com/8/2/5/825dfdbd47c6f629d8ee33705de729b63147b938.png)
so 
.
;
so BCEF is cyclic. Furthermore, let T be a point on the tangent to (DEF). Then 
so TD is also tangent to (CDB); in particular, the radcenter Z of (BDC),(DEF),(BCEF) means that the common tangent through D to (DEF) and (BDC) also passes through that point. Define M as the Miquel point of BCEF (so it lies on AZ, and ABCM is cyclic (!)).
.
;
,
,
be the center of (BDMX), this means 
is a line (since the circles invert to each other), but by coaxiality (DEUX is coaxial) 
There were two main steps here. The first was defining Y to prove BCEF cyclic, then using a good diagram to conjecture the tangent circles; a flurry of information was suddenly figured out. The second step was observing all the cyclic quads, and btw idt the last few concyclicities was required: for example, u can also prove BEMX cyclic!
and suppose ![\[\angle EDF + \angle BDC = 180^{\circ}\]](http://latex.artofproblemsolving.com/7/0/9/70991695b208093f9c88505a32fdbeeedb6c4419.png)
and therefore 
the point 
,
are collinear.
are both cyclic. As a corollary, ![\[\angle EDC = B+A/2, \angle FDC = C+A/2\]](http://latex.artofproblemsolving.com/f/c/4/fc480f0b581395cd808aac7128646470e73f4334.png)
and quadrilateral ![\[\angle ECD' = \angle DBC = \angle EDA = 180^{\circ}-\angle EDD'\]](http://latex.artofproblemsolving.com/2/8/d/28d1d0606876a7149140f4f8b22513f8c41090f4.png)
and therefore 
cyclic. If this circle meets 
from the isogonality condition and then ![\[\angle EDC = \angle EPC = \angle(AI, BC) = B+A/2\]](http://latex.artofproblemsolving.com/9/9/c/99c09a4139a03bb25ea3f7e38be4bc9113106c99.png)
Similarily, we obtain that 
cyclic and that 
.
and 
concur (at 
and 
are tangent to each other (at ![\[\angle BDF = \angle DEF + \angle DCB\]](http://latex.artofproblemsolving.com/d/3/b/d3b4d1db328ccd1ff5ae4b55d1cfcacdf9d15b37.png)
or that ![\[(C-\angle DCB)+A/2 = [180^{\circ}-\angle CED - \angle AEF]\]](http://latex.artofproblemsolving.com/3/6/1/36157ddc9d8cf3b3561ee32c48c34ff2dd141a8b.png)
which is equivalent to proving ![\[\angle AEF + A/2 = 180^{\circ} - [\angle ECD + \angle CED] = \angle EDC = B+A/2\]](http://latex.artofproblemsolving.com/2/e/e/2ee6d4854d839122587ae57ffaafc8d51890717d.png)
or that 
.
.
we conclude that 
is tangent to 
.
.
.
the intersection of lines 
and 
.
denote the respective reflections of 
in the perpendicular bisector of segment 
.![\[180^{\circ}-\angle LPD = \angle XPD = \angle AED = 180^{\circ}- [A/2 + \angle EDA] = 180^{\circ}-[\angle MCB+\angle DCB] = \angle LCD\]](http://latex.artofproblemsolving.com/8/4/e/84eae000680c625d76a1895c45ef27d85f1182c1.png)
proving that 
swaps 
.
.![\[\frac{PC}{AL} = \frac{XP}{XA} = \frac{XP}{XA'} = \frac{PB}{A'L}\]](http://latex.artofproblemsolving.com/6/2/3/6236361b2faf3bec322b193cfbcbd0422a057106.png)
As a result, 
.
.
and thus gives rise to the cyclic quadrilateral 
.![\[\frac{LA'}{CD''} = \frac{MA'}{MC} = \frac{MA}{MC} = \frac{LA}{CD}\]](http://latex.artofproblemsolving.com/1/0/c/10c7a9f272df369f990d1431e259110f3a1e1436.png)
As a result, 
.
Let the circle tangent to 
at 
intersect 
.
is tangent to 
whereby we conclude that 
.
is cyclic. Also note that from a relatively easy angle chase 
.![\[ \frac{PD}{PF'} = \frac{PD}{\frac{AF' \cdot PE}{AE}} = \frac{PD}{PE} \cdot \frac{AB}{AC} = \frac{CK}{DK} \cdot \frac{BK}{CK} = \frac{BK}{DK}, \]](http://latex.artofproblemsolving.com/2/6/e/26e21e2495e5325b36a3d85d6d37efa0e558c3c8.png)
and since 
we conclude that 
.
.
and 
.
.
is cyclic.
.
,
,
.
is cyclic.
.
,
are tangent.![\[ \measuredangle ADC + \measuredangle DCB = \measuredangle AKB = \measuredangle FDB \]](http://latex.artofproblemsolving.com/8/d/3/8d3f7d51af0774d2f829f6ff76a641faf7cf535b.png)
but this is very easy since we know that 
.![\[ \textbf{Invert about $T$ with radius $\sqrt{TB \cdot TC}$}. \]](http://latex.artofproblemsolving.com/a/9/5/a9528c8ecebd97b639ac81f76811feae1ff8f54d.png)
(incredible!!!) 
,
swap. Then 
and 
,
,
,
,
are tangent to each other.
.
,
.
,
.
.
.
,
.
.
be the images of 
under the inversion, respectively. Then since 
,
,
is fixed under the inversion. Let 
.
.
passes through the center of 
.
.
.
are concurrent, so 
are concurrent. Let 
.
are concurrent, so 
are concurrent, which means 
passes through 
.
is cyclic, as needed. 
and 
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where the last follows from easy trignometric manipulations
are tangent to 
,
,
,
is cyclic
is cyclic
and we're done.
and 
are cyclic. Radical axis theorem, as 
and 
tangent to 
at 
thus 
is tangent to 
and 
and the tangent to 
must contain 
collinear. Let 
are coaxial. as then their centers are collinear. First of all, all of these circles pass through 
as 
,
and line 
intersects 
intersects 
is one of the pairs of involution. Now, consider the intersection of 
in some order, 
is another pair of involution. Finally, 
,
,
are also a pair of involution.
.
with 
be the tangent to 
be the intersection of 
with ![\[X = AC \cap NM\]](http://latex.artofproblemsolving.com/f/9/2/f92dd2e2b72bce979c35b88cfc4b2bf6f8bd49c7.png)
![\[U = \ell_C \cap ZN\]](http://latex.artofproblemsolving.com/a/8/5/a85ca7e450f8311e4295e80ac4dcf4103220f10a.png)
![\[K = MC \cap AZ.\]](http://latex.artofproblemsolving.com/5/b/2/5b221215700369cd21902cbc2a5d6b8e67fb6416.png)
By Pascal's theorem on 
,
such that 
.![\[\angle ZKC = \angle AKC = \angle ADC = \angle MDC\]](http://latex.artofproblemsolving.com/6/d/5/6d5697c6f6c2fd861b45e5a9b42e00980d6b8b9c.png)
![\[\angle CZK = \angle CZA = \angle CMA = \angle CMD\]](http://latex.artofproblemsolving.com/c/9/7/c979a367589b18e36808a465f88942f0089880bf.png)
Therefore, 
.![\[\angle TCM = \angle DCM + \angle TCD = \angle KCZ + \angle CAM =\angle KCZ + \angle UCK = \angle UCZ\]](http://latex.artofproblemsolving.com/6/7/5/675bbd2706fa7e841dee013df765f2171f50ca0c.png)
![\[\angle CMT = \angle CMN = \angle CZN = \angle CZU.\]](http://latex.artofproblemsolving.com/1/c/4/1c459e3aef0f32921ee14946fafbd60d2942d785.png)
Therefore, 
.
.
with 
.![\[\angle RKC = \angle UKC = \angle TDC = \angle RDC.\]](http://latex.artofproblemsolving.com/a/5/c/a5c94100a3d35a6291a3090e0b74ab476e8fd493.png)
So, 
.
.
is tangent to ![\[TB^2 = TC^2 = TD \cdot TR.\]](http://latex.artofproblemsolving.com/a/7/f/a7f454bc9cb146b66d4b00f5e66c7283b6f24e32.png)
So, inversion at 
with 
is tangent to 
.
is isosceles.![\[\angle XRD = \angle KRD = \angle KCD = \angle MCD = \angle MAB + \angle BCD\]](http://latex.artofproblemsolving.com/6/b/7/6b75ab64f7b2557484feed0b2af67ebcecb9abf2.png)
![\[= \angle CAM + \angle ADE = \angle EAD+\angle ADE = \angle AED = \angle XED.\]](http://latex.artofproblemsolving.com/7/2/8/728835ac859e324e543a84922e6c14dd673ed25b.png)
Therefore, 
and 
.
is cyclic. Then,![\[\angle ADE = \angle BCD =\angle BD'D = \angle BD'A.\]](http://latex.artofproblemsolving.com/9/e/8/9e83673f51bd822d7b9dbd61d921bae6a29a9db2.png)
Because 
,
.
,
.
.
is cyclic.![\[\angle PDE = \angle PDD' + \angle ADE = \angle DBD' + \angle BCD = \angle DBD' + \angle BD'D = \angle BDD' = \angle BCD' = \angle DFE.\]](http://latex.artofproblemsolving.com/5/5/3/55373349cf0f48c93e41d5118b9a283b635043ae.png)
So, 
,
and ![\[\measuredangle CDB + \measuredangle FDE = \measuredangle CBD + \measuredangle DCB + \measuredangle FDA + \measuredangle ADE = \pi\]](http://latex.artofproblemsolving.com/9/3/e/93ecf5e9c44321034a7a668626fa62eca289af4f.png)
it follows that 
are isogonal with respect to 
, and 
are isogonal with respect to 
,
b
and 
,
since 
so![\[\measuredangle BCD'' = \measuredangle KCD'' = \measuredangle KED'' = \measuredangle DD''E = \measuredangle DCE\]](http://latex.artofproblemsolving.com/a/b/8/ab8d94708e6d97ec9fe2b693f8174bebc5d2ddea.png)
so lines 
are isogonal in 
.
lies on the internal 
and thus 
.
lies on 
.![\[AF \cdot AB = AD \cdot AD' = AE \cdot AC\]](http://latex.artofproblemsolving.com/3/2/5/32556d76c3e553a8c2ccc0d14910549b5d2cfc9d.png)
which concludes that 
denote the Miquel Point of cyclic quadrilateral 
also passes through 
and 
are cyclic.![\[\measuredangle EM_QB = \measuredangle EM_QA+ \measuredangle AM_QB = \measuredangle EFA + \measuredangle ACB = 2\measuredangle ACB = \measuredangle CXB = \measuredangle EXB\]](http://latex.artofproblemsolving.com/8/0/9/809437dbf8ebb31666b366e65c35743a7c28430b.png)
which implies that 
.![\[LX \cdot LE = LB \cdot LM_Q = LA \cdot LC\]](http://latex.artofproblemsolving.com/8/1/4/814c3a7e9f51f05930c51742d3211fdde6e216dc.png)
which implies that 
,
so quadrilateral 
is a common tangent to circles 
.
which we shall do as follows. Let 
.
so indeed circles 
are tangent to each other at ![\[AR^2 = RB \cdot RC = RA \cdot RM_Q\]](http://latex.artofproblemsolving.com/e/7/c/e7cb8227426a2ec63a5cac2c82d7f79cec39c174.png)
which further implies that 
is tangent to 
finishing the proof of the claim.
and 
are tangent at 
.
.![\[\measuredangle CSM_Q = \measuredangle CSD + \measuredangle DSM_Q = \measuredangle CAD + \measuredangle DBM_Q = \measuredangle CAD + \measuredangle CBM_Q + \measuredangle DBC \]](http://latex.artofproblemsolving.com/5/d/5/5d52c4dfbbbb48b2880cade6d6bb01a2b21c7c47.png)
it suffices to show that,![\[\measuredangle SAC + \measuredangle CBS = \measuredangle CAD + \measuredangle CBS\]](http://latex.artofproblemsolving.com/5/3/2/5321b682f0f87746ec1aacf94138a432f073f74e.png)
which we shall do as follows.
is tangent to 
and 
and 
.![\[\measuredangle DBC = \measuredangle M_QDR + \measuredangle RDC + \measuredangle DAM_Q = \measuredangle M_QDC + \measuredangle DAM_Q\]](http://latex.artofproblemsolving.com/1/5/2/152f44bdb3d75e92b5714425fc927a246b1646b0.png)
Further,![\[\measuredangle M_QDC + \measuredangle CBM_Q = \measuredangle M_QAD + \measuredangle CAM_Q + \measuredangle DBC\]](http://latex.artofproblemsolving.com/a/2/8/a28a1a4a8890788022b14027a7f05360719679f1.png)
which implies 
.
cyclic implies,![\[\measuredangle CAD + \measuredangle DBC = \measuredangle SDM_Q + \measuredangle M_QDC + \measuredangle M_QBS + \measuredangle CBM_Q = \measuredangle SAC + \measuredangle CBS\]](http://latex.artofproblemsolving.com/6/3/0/630a3547db669527731bc867e7ea9a28fe8868c2.png)
which is a result which we have already confirmed. Thus, circles 
it follows that the common tangent to ![\[RS^2 = RA \cdot RM_Q = RD^2\]](http://latex.artofproblemsolving.com/a/f/f/afff1fefaf8c23d1e217adc0b5d7c7ac6bb715ab.png)
so 
implying that this line indeed passes through 
and 
and so 
then 
in 
.
, both 
have degree 2. We'll now prove that 
are concyclic. By inverting through 
have degree 2 and 
has degree 1 so we need 6 special cases to check the claimed concyclicity which are the items 1,2,3,4,7 below+ the case for 
.
so the line 
where 
is the number of coincidences of 
and note that since there exists one case(no.8 below) where 
, 
.
and 
would have degree at most 5. So to prove that the recent intersection , is 
special cases which are:
, 
, 
, 
turns to be tangent to 
,
.
.
From this, it is easy to show that line 
is tangent to the circumcircle of 
are coaxial. We can show this by finding a point 
such that the power of 
such that 
.
.
,
,
and 
.
,
Using 
as we noted above, and 
which is obvious, we get
will solve the problem.
and 
,
.
.
,
has to be fixed. Considering the case 
,
.
.
and point 
We have 
,
.
We need to show 
.
.
Therefore 
,
,
,
construst a circle 
and this is equivalent to show 
are coaxial
cyclic (many of above have shown that and I’m too lazy to type again)
applying DIT to show that 
are some pairs of an involution (where 
)
such that 
so we are done
