[/quote]
,
,
and 
are the points of tangency of incircle with the center of 
to 
and 
.
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 
.
onto 
.
.
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 
?
.
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.
Y by mathmaths, tenplusten, Davi-8191, anantmudgal09, mathematicsy, Jalil_Huseynov, selenium_e, Adventure10, Mango247, lian_the_noob12, Rounak_iitr, Funcshun840
Let 
be an acute triangle and let 
be the midpoint of 
. A circle 
passing through 
and meets the sides 
and 
at points 
and 
respectively. Let 
be the point such that 
is a parallelogram. Suppose that lies on the circumcircle of . Determine all possible values of 
.
be an acute triangle and let 
be the midpoint of 
. A circle 
passing through 
and meets the sides 
and 
at points 
and 
respectively. Let 
be the point such that 
is a parallelogram. Suppose that lies on the circumcircle of . Determine all possible values of 
.This post has been edited 1 time. Last edited by v_Enhance, Jul 7, 2016, 9:23 PM
Reason: improve title
Reason: improve title
the second intersection of 
with 
.
thus 
.
.
.![\[ AX = \frac{AB \cdot AP}{AM} = \frac{2c^2}{b} (v+w) \quad\text{and}\quad CX = \frac{CB \cdot CQ}{CM} = \frac{2a^2}{b} (v+u). \]](http://latex.artofproblemsolving.com/7/8/b/78b1a465a308587bc764dd763554c00879841322.png)
Adding these implies that 
.![\[ BT^2 = -a^2(v-1)w-b^2wv-c^2u(v-1) = a^2w+c^2u + \underbrace{-a^2vw-b^2wu-c^2uv}_{=0}. \]](http://latex.artofproblemsolving.com/8/6/2/862c65a254dbdc612aa8e8b6d8654e8cb5ff9d5b.png)
Thus, adding gives 
,
,
is the only possible value.
be the midpoint of 
and let 
Since 
so 
at 
hence we conclude that 
.
and let 
.
be Miquel point of 
,
.
,
.
.
.
touches 
.
be two segments, and let lines 
meet at 
and 
meet again at 
.
and 
are cyclic. we have 
and 
.
and 
are similar, thus the result.
.
,
,
.
.
lies on 
.
to 
,
.
to 
.
.
,
,
lies on 
,
is tangent to circle 
,
,
with 
to 
,
gives 
.
so 
.
,
,
are concyclic, the midpoint 
of 
vary on 
respectively. Indeed, this follows since triangle 
has a fixed shape and spiral similarity moves each linear combination uniformly.
meet 
meet 
again at 
be the midpoints of 
.
.
be the midpoints of 
and 
.
.
.
from the Euclidean plane to the set of real numbers as follows: 
.
.
.
suffices, since 
shall follow analogously.
.
.
and 
.
.
and 
and hence the claim holds.
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draw(circle((5.431616963447202,5.47556856463058), 1.8064126275319419), linetype("4 4"));
draw((7.858191167907094,4.191045751226126)--(6.747139816608675,0.320286204767108));
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/* end of picture */[/asy]](http://latex.artofproblemsolving.com/3/c/f/3cf018b9c168cca84bd4aaa4cfe113f04e750a1e.png)
so that 
to be the midpts of 
resp. Similar angle chasing gives 
so that 
as desired 
cuts cicumcircle 
of 
and let 
.
,
and
Say 
meets 
at 
.
to 
we have 
,
,
.
and this means 
is tangent on circumcircle 
,
.
and why 
![[asy]
unitsize(1cm);
pair A, B, C, M, P, Q, T;
A=(2,2.5sqrt(5));
B=(0,0);
C=(8,0);
M=(B+C)/2;
pair X;
X=(1,0);
P=intersectionpoints(A--B,circumcircle(A,M,X))[1];
Q=intersectionpoints(A--C,circumcircle(A,M,X))[1];
T=P+Q-A;
draw(A--B--C--A--P--T--Q--A);
draw(circumcircle(A,M,X));
draw(circumcircle(A,B,C));
label("$B$", A, N);
label("$A$", B, SW);
label("$C$", C, SE);
label("$M$", M, S);
label("$P$", P, W);
label("$Q$", Q, E);
label("$T$", T, S);
label("$X$", X, NE);
[/asy]](http://latex.artofproblemsolving.com/b/d/4/bd45429b3352fc99ebc3f4fdb10512037b3daf6e.png)
where 
.
and 
Let 
and 
.
,
,
.
Let 
,
,
.
,
We have
.
.
.
so 
.
.
.
,
.
,
.
so 
.
is characterized by 
So after plugging in 
Since length is positive, the only possible value is 
,
be 
be 
cyclic 
cyclic, 
cyclic
to 
to 
tangent to 
.
and 
.
By distance formula, we get 
,
,
is the only value.
for which 
Then, from the parallelogram, 
Consider the equation of 
.
.
.
Therefore, the equation of the circle is 
Plugging in 
for 
We will return to the starred equation later.
so 
Let's expand this: 
Since 
lies on the circumcircle, the bolded things sum to 0, so this is just 
Let's now go back to the starred equation: 
Expanding, this becomes 
It is well known that 
so our answer is 
.
and 
is:
Let 
By plugging coordinates of 
a
Also since 
we have 
.
On the other hand 
Hence 
intersect 
.
to 
.
is the angle of the spin about 
is taken to 
.
is taken to 
,
is equal to that. Thus, 
and so the answer is just 
.
.
Since 
lies on 
.
Since 
is on the circle, it follows that 
so 
Now, suppose that 
for some 
.
so since 
so 
Thus, 
Similarly, we may obtain that 
Therefore, it follows that 
Since 
Now, note that 
so
Now, by Stewart's,
Therefore, we have that
so we must have 
as claimed. 
,
.
and 
denote the images of 
.
and 
.
.
amd 
swap. Hence line 
,
and etc. Then, 
.
for real numbers 
such that 
.
,
for some real number 
which then implies 
so 
.
.
.
.![\[BT^2 = a^2(r+t)t-b^2rt+c^2(r+t)t = a^2t+c^2r\]](http://latex.artofproblemsolving.com/e/1/b/e1be8b3ca4353197e5c69f8d8020552a24279d0e.png)
where the second equality follows from the fact that 
since 
is cyclic, we consider the equation of this circle![\[-a^2yz-b^2xz-c^2xy+(ux+vy+wz)(x+y+z)=0\]](http://latex.artofproblemsolving.com/6/9/3/69317adbb0a4cf4bbc2c020f6f869d4693ddb3f0.png)
where 
similarly, we also have that since 
.![\[-a^2yz-b^2xz-c^2xy+(c^2(s+t)x+a^2(r+s)z)(x+y+z)=0\]](http://latex.artofproblemsolving.com/d/e/3/de392805cb346c0769d837c5f810d375767f5037.png)
which since 
But this is precisely the length of 
!![\[BT^2 = a^2t + c^2 r = \frac{2a^2 - b^2 + 2c^2}{2} = 2\left(\frac{2a^2-b^2+2c^2}{4}\right) = 2BM^2\]](http://latex.artofproblemsolving.com/9/1/1/911bb7ba6270eae457dd1780e4f72666b3e059d6.png)
Thus, 
which finishes the problem.
be the circle with centre 
.
as the power of a point 
.
.
Now let 
be the second intersection of 
It follows that 
.
.
,
![\[-a^2yz-b^2xz-c^2xy+(ux+wz)(x+y+z)=0\]](http://latex.artofproblemsolving.com/e/a/6/ea68000b85fe6b1742682b18fd14bad93fa611f4.png)
where 
.
and 
.![\[-c^2p(1-p)+up=0\Longrightarrow c^2(1-p)=u\]](http://latex.artofproblemsolving.com/4/6/a/46aeecfb4c73e81f1acbc9479216701217722ecb.png)
![\[-a^2q(1-q)+wq=0\Longrightarrow a^2(1-p)=w\]](http://latex.artofproblemsolving.com/1/c/a/1ca664530aa0da7b8e6f34c0fa4baea30ff651af.png)
Therefore,![\[c^2(1-p)+a^2(1-q)=\frac{b^2}{2}\Longrightarrow c^2p+a^2q=c^2+a^2-\frac{b^2}{2}.\]](http://latex.artofproblemsolving.com/d/5/e/d5edbbdd6690ef74c3f74f5649ad0e9675f42b6b.png)
Also, note that 
,![\[a^2(1-p-q)q+b^2pq+c^2(1-p-q)p=0\]](http://latex.artofproblemsolving.com/a/1/1/a11602e3468f7b84f67512cabe0167160a9f6148.png)
![\[-a^2(-p-q)q-b^2pq-c^2(-p-q)p=a^2q+c^2p=c^2+a^2-\frac{b^2}{2}.\]](http://latex.artofproblemsolving.com/2/d/0/2d0581f5815440d6faeece54b511872ea05f92ba.png)
That looks suspiciously like the distance formula! Indeed, we get that![\[BT=\sqrt{c^2+a^2-\frac{b^2}{2}}.\]](http://latex.artofproblemsolving.com/d/e/c/decc9a5cfae884f2ecfdca137df3e21d3665362b.png)
Also, it is well known that![\[BM=\frac{1}{\sqrt2}\cdot\sqrt{c^2+a^2-\frac{b^2}{2}}.\]](http://latex.artofproblemsolving.com/0/6/b/06ba76a080f373de7b29da707d6a5957d3363966.png)
Therefore, 
![\[AC\cdot AM=AB\cdot AP+BC\cdot CQ\iff b^2/2=AP\cdot c+CQ\cdot a\]](http://latex.artofproblemsolving.com/0/3/f/03f05fd9bd2fa22f8079c70919622ce61edd04af.png)
and 
.
,
and 
.
.![\[-a^2(p+q-1)(1-q)-b^2(1-p)(1-q)-c^2(p+q-1)(1-p)=0\]](http://latex.artofproblemsolving.com/f/4/0/f404edb560c3a0f8d3132fd02c3353a6e6992373.png)
The displacement vector of 
so from the distance formula we have
for 
.
,
.
.
.
