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)
be a given integer, 
.
with 
vertices satisfying the condition: for any two non-adjacent vertices 
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 
,
,
when divided by 19, then that positive integer is called an 
number. If a positive integer leaves a remainder that belongs to 
when divided by 19, then that positive integer is called a 
number.
Y by Davi-8191, Aryan-23, megarnie, Adventure10, Rounak_iitr, ehuseyinyigit, and 2 other users
Let 
be a triangle with 
. The angle bisectors of 
and 
meet the sides 
and 
at 
and 
, respectively. Let 
be the incentre of triangle 
. Suppose that 
. Find all possible values of .
Jan Vonk, Belgium, Peter Vandendriessche, Belgium and Hojoo Lee, Korea
be a triangle with 
. The angle bisectors of 
and 
meet the sides 
and 
at 
and 
, respectively. Let 
be the incentre of triangle 
. Suppose that 
. Find all possible values of .Jan Vonk, Belgium, Peter Vandendriessche, Belgium and Hojoo Lee, Korea
or 
.
.
,
and 
from angle bisector theorem and sine theorem for other triangles. Also 
.
,
or 
I don't know if it's correct, but the problem can be surely done like this
be the incenter of triangle 
.
then 
and note that 
we can easy find 
.
or 
.
;
be the incenter of 
.
meets 
at
.
;
.
and triangle 
are similar (since 
)
.
(notice that 
.
.
.
.![\[ \frac{(AD+a+1)a}{AD(a+2}=\frac{1+AD}{a}\]](http://latex.artofproblemsolving.com/8/f/c/8fca2cbfde2a42be7691d2a6e20274fd92f9e0c1.png)
![\[ \leftrightarrow AD.a^2+a^3+a^2=a.AD+a.AD^2+2.AD+2.AD^2\]](http://latex.artofproblemsolving.com/5/2/b/52be09c686d98eb049c76a84bddf3106cb2070e5.png)
![\[ \leftrightarrow AD.a^2=a.AD-a+2AD+a^2-2\]](http://latex.artofproblemsolving.com/4/0/d/40d35f3aeb82833c714e0df9632355b0c91b016c.png)
)![\[ \leftrightarrow
AD(a^2-a-2)=a^2-a-2\]](http://latex.artofproblemsolving.com/8/3/1/831e67ce38932bbbba3c389bf457f51eeec959d9.png)
or 
.
thus 
.
,
,
is the incenter of 
.
.
,![\[ \frac{EC}{\sin 45^\circ - \alpha/2} = \frac{2R \sin 2 \alpha}{\sin {45^\circ - 3\alpha/2}}.\]](http://latex.artofproblemsolving.com/c/a/a/caa923e44cd45f4ef6f73475c84f7fff605ae39f.png)
,![\[ \frac{JC}{\alpha/2} = \frac{2R \sin {90^\circ - \alpha}}{\sin 135^\circ}.\]](http://latex.artofproblemsolving.com/6/6/9/669031996a46c3c585c757578632dc00763735af.png)
,![\[ \frac{EC}{\sin{135^{\circ} - \alpha}} = \frac{JC}{\sin {3 \alpha / 2}}.\]](http://latex.artofproblemsolving.com/6/6/a/66acb4f58f08261abaaa766425beb863a5a01dfa.png)
![\[ \frac{\sin {45^\circ + \alpha}}{\sin \alpha} \cdot \frac{\sin {45^\circ + 3\alpha/2}}{\sin {3 \alpha/2}} =
\sqrt{2} \cdot \frac{\sin {45^\circ - \alpha/2}}{\sin {\alpha/2}}.\]](http://latex.artofproblemsolving.com/7/b/e/7be0a85afa86a5668ed66b27b9ca7fbd31b281e7.png)
and using formulas for double and triple angle, it follows![\[ \left(\frac{x^2-1}{2x}+1\right)\left(\frac{3x-x^3}{1-3x^2}+1\right)=2(x-1).\]](http://latex.artofproblemsolving.com/6/7/a/67a14744bdfb4591bd0c3e987e666588f2634ee1.png)
.
or 
,
or 
.
.
internally by a cevian making angle 
, along with the base angles being 
and 
respectively, 
.
be the incenter of triangle 
and by Quadrilateral Ceva on 
Let 
and note that 
and 
Hence, 
Therefore, 
and 
or 
First, note it's true when 
or 
Let 
be the incenter of 
over 
to 
Note that 
Given that 
is cyclic. Thus, 
This means that 
which implies 
so 
then 
which implies that 
so 
and 
.
.
,
,
,
,
,
.![\[\frac{\sin(45-x)}{\sin(3x)}\div \frac{\sin(90-2x)}{\sin(45)}=\frac{EK}{KC}\div \frac{EK}{IK}=\frac{IK}{KC}=\frac{ID}{CD}=\frac{\sin(45-x)}{\sin(45+x)}.\]](http://latex.artofproblemsolving.com/e/c/5/ec57821a6b7eb7cf872ddec8776496cb47c9f5e9.png)
So we obtain![\[\frac{1}{2}(\cos(x)-\cos(90+x))=\sin(45)\sin(45+x)=\sin(3x)\sin(90-2x)=\frac{1}{2}(\cos(5x-90)-\cos(90+x))\]](http://latex.artofproblemsolving.com/b/c/0/bc03c849b7d5deb56a94377a359dd30a9a07fa4c.png)
which means that![\[2\sin(3x-45)\sin(2x-45)=\cos(x)-\cos(5x-90)=0\implies x=15, \frac{45}{2}\implies \angle CAB=\boxed{60^{\circ},90^{\circ}}.\]](http://latex.artofproblemsolving.com/9/0/7/907eaca1dfe14c88b8ab2fb4c5166ec1407f7a61.png)
We are done. 
mapping 
to 
are collinear. We proceed using 
,
Also, by LoS on 
and 
,
so
It's not hard to verify that 
are the solutions. We are done. 
,
.
,
.
However, by triangle 
,
Hence, 
However, 
,
Multiplying by 2 this becomes 
This gives 
and 
as solutions, which correspond to answers of 
and 
degrees. These clearly work (e.g. by coordbashing), so we are done.
.
On the other hand, if 
and 
bisect 
,
.![\[\angle AIE=\angle BID=\angle DIC=90-\angle DCI=90-a,\]](http://latex.artofproblemsolving.com/2/4/d/24d7e6f289b4bf3d0c9531d83713b37c9a6ede26.png)
and that![\[\angle AEI=180-(\angle AIE+\angle AEI)=180-((90-2a)+(90-a))=3a,\]](http://latex.artofproblemsolving.com/9/1/c/91c404fb9350c922711219f4c097cb46c95dd9c7.png)
which in turn gives us that 
.
and 
,![\[\frac{IK}{KC}=\frac{EK*\frac{\sin 45}{\sin 2a}}{EK*\frac{\sin (135-3a)}{\sin a}}=\frac{\sin 45\sin a}{\sin 2a \sin (135-3a)}.\]](http://latex.artofproblemsolving.com/2/8/9/289399134c8ec73ba7dd835895632dcdfa8882c8.png)
However, by Angle Bisector Theorem and Law of Sines on ![\[\frac{IK}{KC}=\frac{ID}{DC}=\frac{\sin a}{\cos a},\]](http://latex.artofproblemsolving.com/2/9/c/29ce6875ad7ab2ae1d0e9e7e3025868ddc6564c6.png)
and setting the two equations equal, we get that![\[\frac{\sin 45\sin a}{\sin 2a \sin (135-3a)}=\frac{\sin a}{\cos a} \iff \sin 45\cos a=\sin 2a\sin (135-3a).\]](http://latex.artofproblemsolving.com/7/9/1/791ed788bacbe38c2f0fe4dbb78327e058e5bac1.png)
By Product-to-Sum, this gives us that![\[\frac{1}{2}(\sin (a+45)+\sin(45-a))=\frac{1}{2}(\cos(135-5a)-\cos(135-a)),\]](http://latex.artofproblemsolving.com/8/7/0/870cc322f819ccf24dde22866d711b5da87a2fa9.png)
however, since 
and 
,![\[\sin(a+45)+\sin(45-5a)=0,\]](http://latex.artofproblemsolving.com/9/c/8/9c8b50a92e681e98ba7c99443cd8849f9976ba87.png)
which is![\[2\sin(45-2a)\cos(3a)=0,\]](http://latex.artofproblemsolving.com/1/9/d/19d56d305fcfbf2f8ec78a455d422194ce0957ee.png)
by Sum-to-Product. Since 
,
or 
,
are 
of 
.
,
is an altitude, forcing 
.
tells us 
is cyclic, which induces![\[\angle IKE' = \angle IKE = 90 \implies EKE' \text{ collinear} \implies \angle IKA = \angle EKA = 45\]](http://latex.artofproblemsolving.com/4/1/1/41188d115dbdc640a95a487ebcef91d200188056.png)
![\[\implies \triangle EKA \cong \triangle IKA \implies \angle AIE = \angle AEI,\]](http://latex.artofproblemsolving.com/7/c/6/7c6375814187afcace8122825332428835f19e44.png)
,
.
,
is cyclic.
is a key point.
in 
.
Can we use Ceva's theorem on 
.
.
.
,
or 
.
.
.
so quadrilateral 
is cyclic. We employ barycentric coordinates wrt 
,
,
,
.
.
,
,![\[u=\frac{b^3-2a^2b-ab^2}{2a+b}.\]](http://latex.artofproblemsolving.com/e/9/3/e939a7380d62e47852ebbb507101721016e02461.png)
Plug in 
.![\[(c-a)(c^2+2ac-a^2)=b(c-a)(c+a).\]](http://latex.artofproblemsolving.com/6/a/b/6ab3f4a1a4b501ea79981fe8c6488f40b6ee394a.png)
So 
or we can square both sides and cancel stuff out to get 
,
.
,
.
.
,
.
.
.
,
.
,
or 
;
and since 
,
or 
.
works fine. From hereon, assume 
.
be the reflection of 
and we have 
.
(this is crucial), we must have![\[CD' \cdot CE = CK \cdot CI \implies \tfrac{CD}{CK} = \tfrac{CI}{CE},\]](http://latex.artofproblemsolving.com/2/f/5/2f56d8524f67059a1291ffb4da6a57f9cbd44da3.png)
so by SAS similarity we have 
.
,
.
