[/quote]
,
,
to the sides 
are 
,
is the orthocenter. 
is the midpoint of segment 
.
and 
intersect at 
.
from 
and 
intersect at 
are colinear
be an integer and let 
be positive integers. In a group of 
people, some games of chess are played. Two people can play each other at most once. Prove that it is possible for the following two conditions to hold at the same time:
.
with 
,
games of chess.
.
is chosen from the minor arc 
such that 
.
be a point on 
is perpendicular to 
,
be a point on line 
is perpendicular to 
and 
,
.
satisfying the following equations: 
such that: 
.
be positive real numbers such that 
.
.
be real numbers such that 
Prove that
Let 
Prove that
Let 
Prove that
.
be a real numbers such that 
P
Where 
P
Where 
I
be positive real numbers, show that:
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.
This is symmetric in 
and 
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.
.
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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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.
