be the set of positive real numbers. Determine all functions 
and 
![\[f(x+f(xy))+y=f(x)f(y)+1\]](http://latex.artofproblemsolving.com/c/6/a/c6a968a17f6fd5fd637c1ee5d855bdff11e5423a.png)
and 
such that![\[
1^n + 2^n + 3^n + \cdots + n^n = x!
\]](http://latex.artofproblemsolving.com/6/4/b/64b5f5919fd3710cc15e96724918eae216c8e4f0.png)
.
meet at point 
.
on line 
be 
.
meet 
respectively. Let 
lie on a circle.
red cards, 
white cards, and 
blue cards. A player plays all the cards one at a time. With each play he accumulates a penalty. If he plays a blue card, then he is charged a penalty which is the number of white cards still in his hand. If he plays a white card, then he is charged a penalty which is twice the number of red cards still in his hand. If he plays a red card, then he is charged a penalty which is three times the number of blue cards still in his hand. Find, as a function of 
and 
the minimal total penalty a player can amass and all the ways in which this minimum can be achieved.
and 
Prove that
Let 
Prove that
Let 
Prove that
Let 
Prove that
with 
.
lie on 
respectively, such that 
.
to meet the circumcircle of 
respectively. Let the circumcircles of 
meet at points 
.
to 
,![$[ABC]$](http://latex.artofproblemsolving.com/d/3/3/d33cc80fa8f093e155c5be46d2e5d9da3d7e1ef5.png)
.
,
.
to meet the circumcircle of 
at 
,
.
and 
.
Equality holds when 
Equality holds when 
Equality holds when 
Y by Davi-8191, baladin, anantmudgal09, Tumon2001, doxuanlong15052000, tenplusten, Durjoy1729, AlastorMoody, snoofmath, HamstPan38825, Adventure10, Mango247, ehuseyinyigit, and 3 other users
Triangle 
is inscribed in circle 
. The interior angle bisector of angle 
intersects side 
and at 
and 
(other than ), respectively. Let 
be the midpoint of side . The circumcircle of triangle 
intersects sides 
and 
again at 
and 
(other than ), respectively. Let 
be the midpoint of segment 
, and let 
be the foot of the perpendicular from to line 
. Prove that line 
is tangent to the circumcircle of triangle 
.
is inscribed in circle 
. The interior angle bisector of angle 
intersects side 
and at 
and 
(other than ), respectively. Let 
be the midpoint of side . The circumcircle of triangle 
intersects sides 
and 
again at 
and 
(other than ), respectively. Let 
be the midpoint of segment 
, and let 
be the foot of the perpendicular from to line 
. Prove that line 
is tangent to the circumcircle of triangle 
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.
.
and 
.
is half of 
,
be the midpoint of arc 
.
,
is cyclic with diameter 
.
are collinear, so 
,
to 
to 
,
.
,
is cyclic. Then![\[
\angle NMX=\angle PCX=\angle ACX=\angle ALX=\angle DLM=\angle DHM=\angle NHM
\]](http://latex.artofproblemsolving.com/8/d/7/8d780473ff81e3e002f9cf54c5bb08b61de8cf71.png)
so 
if 
are midpoints of 
and 
resp. than 
is a rhombus because 
is the bisector of 
because 
and 
we have 
and since 
is cyclic so
so 
.
.
,
.
which passes through 
.
obviously lies on (i).
![[asy]
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dot((0.2656345589186853,3.608099001743732),dotstyle);
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dot((1.948454401984952,2.863004442880835),dotstyle);
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![\[BQ \times AB = BD \times BM \implies BQ = \dfrac{\dfrac{ac}{b+c} \times \dfrac{a}{2}}{c} = \dfrac{a^2}{2(b+c)}.\]](http://latex.artofproblemsolving.com/c/f/d/cfd43458f57d50d9d836abd4d86083b31d01ac74.png)
Note that this expression is symmetric in 
Similarly we get that 
so 
Let 
be the midpoints of 
respectively. Since ![\[XN= MY = \dfrac{1}{2} BQ = \dfrac{1}{2} CP = XM = NY,\]](http://latex.artofproblemsolving.com/1/9/1/19102c5684b6cc400effe9d44df6f659cf8797aa.png)
Consequently, 
However, since 
and 
Since 
which in turn implies quadrilateral 
is cyclic. It follows that 
so we're done. 
.
and 
are similar(cause 
is similar to 
.
.
.
and 
and equating these we get 
and 
gives 
and 
.
we get 
.
.
.
is cyclic 
.
meet 
so 
.
.
and we are through.
,
be a convex quadrilateral, and 
,
and 
.
and 
are parallel.
and 
are parallelogramm ...
,
.
.
. Thank you for it 
are aligned.
are aligned.
by Thales theorem.
.
.
.
,
bisects 
,
,
.
,
and 
.
we get that the equation of the circle passing through these points is 
We let 
,
.
Exploiting symmetry, we find that 
Now we have that 
,
We now easily compute
implying that 
and 
are two unit vectors then 
is parallel to the angle bisector of the two vectors (at the origin).
be the antipode of 
,
.
to 
,
,
.
,
,
and 
,
and 
are diameter in their respective circles). So, 
to 
,
.
is cyclic since 
and 
,
and 
is indeed tangent to 
.
,
,
.
.
,
goes through the antipode of 
,
As such, it suffices to show that 
since then we would have 
by cyclic 
are collinear.
is a terrible line to angle chase with, so we turn to barycentric coordinates to prove that 
.
.
.
and 
Solving this system of equations yields 
Thus, intersecting this circle with 
and symmetrically 
Let 
and 
and 
Thus, we have 
Thus, 
If we multiply this by 
this becomes 
However, expanding out the definition and dividing by 
,
Hence, 
is a real multiple of 
.
is also a real multiple of 
and we are done.
,
is cyclic. Thus![\[
\measuredangle LMH = \measuredangle LDH.
\]](http://latex.artofproblemsolving.com/6/1/a/61a27dd26824d5ada5156301e64472f0c926fea9.png)
We wish to prove that this is equal to 
so it suffices to show that 
By Power of a Point on ![\[
BQ \cdot AB = BD \cdot BM,
\]](http://latex.artofproblemsolving.com/f/d/b/fdb8f5039e752a767006e9d93913b6c95ef97042.png)
so![\[
BQ = \frac{BD \cdot BM}{AB}.
\]](http://latex.artofproblemsolving.com/7/9/a/79aaac2fb5deaa465c5dff500dbf2300d52d08bf.png)
Similarly,![\[
CP = \frac{CD \cdot CM}{CA}.
\]](http://latex.artofproblemsolving.com/5/c/2/5c282e9614b7c8b828d2a946429dec8d445ac59b.png)
However, by the Angle Bisector Theorem, 
and since 
this implies the claim.
(or any point on that matter) to be the origin, then the vector 
is parallel to the angle bisector of 
But this implies that the vector 
is parallel to 
so we are done.
denote the midpoint of arc 
are collinear and thus ![\[\angle ASM = \angle ASL = \angle ACL = \angle ADB\]](http://latex.artofproblemsolving.com/4/2/8/42851b39323fe3954b81c8323b8fd0640b871f6e.png)
implying that points 
are concylic. In particular, we note that 
.
,
are collinear. Since points 
are concylic comparing power wrt ![\[SD \cdot SH = SM \cdot SL\]](http://latex.artofproblemsolving.com/6/2/a/62a57756c39dacb670ebad7f7dfcd14ac15a3208.png)
Next, since the aforementioned spiral sim also maps 
and 
we have ![\[SM \cdot SD = SN \cdot SL\]](http://latex.artofproblemsolving.com/8/8/0/8806142e879b03ed1e60a07e9f7e580888c3a466.png)
From these two length conditions one deduces that 
,
is tangent to 
,
.
,
,
.
,
,
.
.![\[\angle KMN = \angle KPC = \angle KLA = \angle MHN. \quad \blacksquare\]](http://latex.artofproblemsolving.com/a/0/1/a01f92c5b7968ccac45f5c41b5d68913f5aecd1a.png)
We have 
so 
is cyclic. Furthermore it is the miquel point of 
and the spiral similarity taking 
Also notice 
is cyclic with diameter 
![\[\measuredangle HNM=\measuredangle RNM=\measuredangle RPC=\measuredangle RPA=\measuredangle RDA=\measuredangle HDL=\measuredangle HML.\]](http://latex.artofproblemsolving.com/5/e/e/5eeaf55835f4081a454e53e3a653de5b7ec819f4.png)
.
be the antipode of 
since 
and 
with 
.
.
by the gliding lemma(mean geometry theorem?), from which we get 
which gives our tangency as desired.
,
is isosceles. This is however also equivalent to showing 
,
,
.
.
,![\[\angle X'AD = 90^{\circ} = \angle X'MD \implies AX'MD \text{ is cyclic, so that } X = X'. \phantom{c} \square\]](http://latex.artofproblemsolving.com/9/8/2/982db9addce5134061059b4f8c5aec88cd954259.png)
,
.
and 
under 
,
,
,
,
.
,
and 
,
and 
,
.
.![\[ L'N \cdot L'H = L'M^2,\]](http://latex.artofproblemsolving.com/3/b/9/3b959f406838b07db33cf5128d85e6bb0a45e785.png)
so 
is diameter of circle 
as 
is tangent to 
so ![\[\angle HML=\angle HNM\]](http://latex.artofproblemsolving.com/a/c/f/acf72a09cb694809bf09b48c18663ba1466f7b5e.png)
![\[\angle HDL=\angle DNM\]](http://latex.artofproblemsolving.com/8/a/1/8a152cd26c5660816cc01dd12637b0974ba6de14.png)
![\[\angle ADN=\angle DNM,\]](http://latex.artofproblemsolving.com/f/3/8/f381f041f4b2cf94eb4547464d1d045705001e8b.png)
so we want to prove ![\[(BQ)(BA)=(BD)(BM)\Longrightarrow BQ=(BM)\cdot\frac{BD}{BA}=(CM)\cdot\frac{CD}{CA}=CP.\]](http://latex.artofproblemsolving.com/2/8/6/286ef2578303fd0a497962b17d1443ae656d7b0b.png)
Now we claim that the concyclicity is actually irrelevant: for any 
.
)
,
and 
.
,
,
has barycentric coordinates 
.![\[\overrightarrow{ZM}=\left(-\frac{b+c}{2b},\frac12,\frac{c}{2b}\right)\]](http://latex.artofproblemsolving.com/7/b/a/7ba30741acb7c9f0dfc4054d35a9895e62a6ef5c.png)
will be parallel to![\[\frac{1}{b+c}(-b-c,b,c)\]](http://latex.artofproblemsolving.com/6/9/b/69b1254a47292d7fa89b229ff68a6f4d66502d03.png)
which is exactly the displacement vector 
.
is the antipode of 
and the antipode of 
.
be the antipode of 
,
is cyclic and 
.
,
is a diameter of 
is also a diameter of 
,
,
,![\[ \frac{DX}{DN} = \frac{LX}{LM} \implies \frac{XN}{XD} = \frac{XM}{XL}. \]](http://latex.artofproblemsolving.com/2/a/2/2a2a12c797577a3d6440dcba23e8559b7310abd0.png)
This then implies that 
(by SAS) since 
and 
are collinear, so 
with 
.
,
with 
.
.
.
is given by: 
.
.
and we are done.
