.
and 
are chosen on segments 
and 
such that 
.
is the midpoint of the arc 
and 
is the midpoint of the arc 
,
.
that, for all 
and 
,![\[\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j}\geq C \ln n\left(\prod_{i=1}^n x_i\right)^{\frac{1}{n}}\]](http://latex.artofproblemsolving.com/4/8/b/48b9fc4af5a20de7fd9dfc45a74d5aa27e615e46.png)
squares with two game pieces. At the beginning, Alice’s piece is on the first square while Bob’s piece is on the last square. The figure shows the starting position for a strip of 
squares.
with incenter 
= 
= intersection of 
and 
= intersection of 
and 
concycle
and 
intersect the altitude from 
to 
is the circumcenter of triangle 
and 
is the reflection of 
is the second intersection of circumcircles of triangles 
and 
.
are collinear.
be all positive integers smaller than 
such that 
for all 
is the largest positive integer that divides both 
and 
.
.
such that 
is a perfect square.
and 
,
and 
.
.
be real numbers such that 
.
and 
.
,
be 
,
intersects the circumcircle of 
.
and 
intersects 
at 
respectively, prove that 
are concurrent.
Y by farhad.fritl, Frd_19_Hsnzde, Rounak_iitr, cubres, radian_51, dangerousliri, ItsBesi
Let 
be an acute triangle with incentre 
and 
. Let lines 
and 
intersect the circumcircle of at 
and 
, respectively. Consider points 
and 
such that 
and 
are parallelograms (with 
, and 
). Let 
be the point of intersection of lines 
and 
. Prove that points 
, and are concyclic.
be an acute triangle with incentre 
and 
. Let lines 
and 
intersect the circumcircle of at 
and 
, respectively. Consider points 
and 
such that 
and 
are parallelograms (with 
, and 
). Let 
be the point of intersection of lines 
and 
. Prove that points 
, and are concyclic.
and
So 
,
and 
.
as desired. Thus 
cyclic
.
be added to vector 
,
.
.
by symmetry and well-known
and because 
we are done because we have spiral sim.
.
and 
hence![\[\frac{-xy-yz-zx-y^2}{-xy-yz-zx-z^2}=\frac{x+y}{x+z}=\frac{-yz-zx+x^2-y^2}{-yz-yx+x^2-z^2}\]](http://latex.artofproblemsolving.com/0/4/d/04d1adc793381c8c1c195de7ac8ed0e874864a7c.png)
Thus, 
as desired
.
to 
.
Note that![\[
\angle IBR = 360^\circ - \angle ABR - \angle ABI
= 360^\circ - (180^\circ - \angle QAB) - \frac{\angle ABC}{2}
= 180^\circ + \frac{\angle ACB}{2} - \frac{\angle ABC}{2}
\]](http://latex.artofproblemsolving.com/6/e/7/6e71616c2db3b748ce0bdea955bd9f3d61e68464.png)
![\[
\angle ICS = \angle ICA + \angle ACS
= \frac{\angle ACB}{2} + 180^\circ - \angle CAP
= 180^\circ + \frac{\angle ACB}{2} - \frac{\angle ABC}{2}
\]](http://latex.artofproblemsolving.com/9/d/7/9d7f8b5941a57d3f990b0039e80bd29e66f2a11c.png)
![\[
\Rightarrow \angle IBR = \angle ICS
\]](http://latex.artofproblemsolving.com/2/5/d/25d8e1c2fafa855ce6dc2683aca5dbc44bf222b0.png)
![\[
\frac{IB}{BR} = \frac{IB}{AQ} = \frac{IB}{IQ} = \frac{IC}{IP} = \frac{IC}{AP} = \frac{IC}{CS}
\]](http://latex.artofproblemsolving.com/2/e/0/2e06432f2910618941dd8413051fcd229a02561a.png)
![\[
\Rightarrow \triangle IBR \sim \triangle ICS \quad \blacksquare
\]](http://latex.artofproblemsolving.com/8/c/7/8c73a6009be05b959e979f56b857cb327a04f3c9.png)
,
are concyclic. 
and 
.
,
.
,
,
,
,
.
.
.
By Incentre-Excentre Lemma, 
and 
.
,
,
.
and 
,
.
It follows from angle chase:
and 
.
and the claim follows.
,
lie on a circle. Hence 
.
,
.
,
are concyclic.
be the unit circle such that 
This is symmetric in 
so triangle 
is directly similar to 
so 
and 
Then 
and 
The incenter is given by 
Then we must verify that 
or 
which follows upon expansion.
to 
have ratio 
.
and satisfies 
,
.
.
.
.
.
.
.
this claim is done. 
.
.
.
.
.
and similarly 
.
.
.(By the way there is unused 
is cyclic info 
).
so 
.![\[
\frac{RB}{BI} = \frac{QB}{BI} = \frac{QI}{BI} = \frac{PI}{CI} = \frac{PC}{CI} = \frac{SC}{CI}
\]](http://latex.artofproblemsolving.com/2/b/1/2b1c9880ca4e13f74afb1d106e43d0cb6c8b26ea.png)
and 
so we are done.
,
,
。
.
.
.
to be concylic, 
.
It suffices to show that 
and 
.
.
Hence 
Which completes our proof by making 
concyclic.
and 
.
.
.
(WLOG, 
,
.
.
.
and we are done.
and 
.
and 
are similar.![\[\measuredangle IBR = \measuredangle IBA + \measuredangle ABR = \measuredangle IBA + \measuredangle BCI\]](http://latex.artofproblemsolving.com/d/e/6/de6172f0686cb53d2dda92132c52f58af1be2b79.png)
and![\[\measuredangle ICS = \measuredangle ICA + \measuredangle ACS = \measuredangle ICA + \measuredangle CBI\]](http://latex.artofproblemsolving.com/a/0/c/a0cf0a36c680ae58f20bb4e47f56262d6cf1805d.png)
which implies that 
.
we have that![\[\frac{IB}{BR} = \frac{IB}{IQ} = \frac{IC}{IP} = \frac{IC}{CS}\]](http://latex.artofproblemsolving.com/e/e/4/ee4d2cf1b96aa4945b7dbd6851f0664bcf389314.png)
which implies that 
and thus,![\[\measuredangle TRI = \measuredangle BRI = \measuredangle CSI = \measuredangle TSI\]](http://latex.artofproblemsolving.com/e/6/3/e63e83b89368f41269875f268855d3104d6a2ead.png)
which shows the result.
![\[ \frac{RB}{BI}=\frac{QA}{BI}=\frac{QI}{BI}=\frac{PI}{CI}=\frac{PA}{CI}=\frac{SC}{CI} \]](http://latex.artofproblemsolving.com/a/b/d/abddba6ddea4633db83ebceb3c02158201b3eaf7.png)
But also we have when 
and the similar holds for the other by an analogous process which means from SAS criteria that 
which is sufficient to finish thus we are done 
.
are concyclic
Indeed, note that 
and 
and that 
and 
from Fact 
)
are concyclic.
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[/asy]](http://latex.artofproblemsolving.com/9/e/8/9e8de5a7c6ca1f36663b6ead9f6ef04e01f7e4c8.png)
are concyclic.
,
.
as desired.
.
,![\[
\frac{RB}{BI}=\frac{AQ}{BI}=\frac{\sin \tfrac{1}{2}\angle C\cdot 2R}{r\cdot \tfrac{1}{\sin \tfrac{1}{2}\angle B}},
\]](http://latex.artofproblemsolving.com/8/7/a/87a4a9543f07fcdf84eb25be024c4909b7958d5b.png)
![\[
\frac{SC}{CI} =\frac{PA}{CI} = \frac{\sin \tfrac{1}{2}\angle B\cdot 2R}{r\cdot \tfrac{1}{\sin \tfrac{1}{2}\angle C}}
\]](http://latex.artofproblemsolving.com/6/1/d/61d6e2dd558ef73d98622b6eba9bf622ddf7c26d.png)
which is the same value, where 
is the radius of circumcircle, incircle of 
so we are done.
when you're working with directed angles... So, @above try to patch your solution by writing your angles completely in terms of directed angles, or just use normal ones throughout the solution...
and 
instead of 
,
is smaller in size, hence it's harder to notice things about it.
and 
.
.
are isogonal is 
of 
,
similarly
Additionally we get that
Similarly
Now using these two relations and the ones from above, we get 
cyclic.
by angle chase, then length chase to get 
and WLOG 
and 
we get that triangles 
are similar 
and 
is an isosceles triangle 
,
from the parallelogram so: 
is an isosceles triangle 
,
from the parallelogram so: 
combining with previous claim we get: 
and 
we get that: Triangles 
are similar 
Points 
and 
it is easy to see that 
.
.![\[\frac{IB}{BR} = \frac{IC}{CS} \iff \frac{IB}{IC} = \frac{AQ}{AP}\]](http://latex.artofproblemsolving.com/6/6/b/66bdc795125139be64efe48fd8fac6987c8d50f8.png)
Which is obvious since 
by angle chasing.![\[ \measuredangle RTS = \measuredangle(RT, ST) = \measuredangle(QA, PA) = \measuredangle QAP. \]](http://latex.artofproblemsolving.com/f/7/6/f7681fcf763b446ab20c6f0126c95eb8af9e2813.png)
Hence, it suffices to calculate 
.
we can conclude that ![\[ \measuredangle(\vec R - \vec I, \vec S - \vec I) = \measuredangle(\vec B + \vec Q - \vec I - \vec A, \vec C + \vec P - \vec I - \vec A) = \measuredangle NKM \]](http://latex.artofproblemsolving.com/5/f/9/5f948b58fbbd24837452a07acca957685ebb01a7.png)
where 
is the midpoint of 
,
and 
.![\[ \measuredangle QAP = \measuredangle NKM. \]](http://latex.artofproblemsolving.com/3/8/2/38219d5b42e0d47e633cdaf437c054e7575af375.png)
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[/asy]](http://latex.artofproblemsolving.com/8/d/e/8de1364f36fa7d851672abf1e4018b39ef823bbf.png)
.
and similarly 
.![\[ \triangle BIC \overset{+}{\sim} \triangle NKM \overset{+}{\sim} \triangle QAP \]](http://latex.artofproblemsolving.com/d/d/a/dda12381dd541e94080ba81bd2e70eca3adacaf6.png)
since each vertex of 
is the midpoint of the two corresponding vertices of 
.
