[/quote]
,
be an acute triangle with 
and let 
and 
be its orthocenter and circumcenter, respectively. Let 
.
and the circle of radius 
cross 
and 
and circle 
are concurrent.
of functions from the set of real numbers to itself that satisfy ![\[g(f(x+y)) = f(x) + (2x + y)g(y)\]](http://latex.artofproblemsolving.com/0/a/3/0a3bc25299a0bcbd3f3602a0e6881d68fce750c5.png)
for all real numbers 
and 
.
such that for all 
be given pairwise coprime positive integers where 
.
be positive integers. We call 
to be a grandson of 
if and only if, for all possible piles of stones whose total mass adds up to 
of primes (greater than 5) such that : 
of positive integers such that:
.
and incircle touching 
be the antipode of 
is chosen on the internal angle bisector of 
such that 
.
be the midpoint of arc 
,
be the midpoint of 
.
be distinct natural numbers each of which has exactly 
positive integer factors. For instance, 
has exactly 
.
divides any of the 
'
and 
intersect at points 
.
and 
intersects 
,
.
,
.
be such that![\[u_{1} = 1; \quad u_{n+1} = \dfrac{u_{1} + u_{2} +...+u_{n}}{n}+n-1 \quad \forall n \in \mathbb{N^{*}}\]](http://latex.artofproblemsolving.com/1/5/e/15e5bf0cc14794e7028dbfe4a8754f6803e850ea.png)
and 
.
.
denote the set of positive real numbers. Find all functions 
such that for each 
,
satisfying 
and primes 
such that 
Y by AdithyaBhaskar, MathLuis, Adventure10
In 
, let point 
be the tangency point of incircle 
with side 
. Analogously define 
and 
for sides 
and 
respectively. Let 
be the feet of the perpendicular from to 
. Let 
be the midpoint of arc 
of circumcircle 
of . Prove that the lines 
and 
concur on .
Anant Mudgal, IMOTC-2016, Senior Batch
, let point 
be the tangency point of incircle 
with side 
. Analogously define 
and 
for sides 
and 
respectively. Let 
be the feet of the perpendicular from to 
. Let 
be the midpoint of arc 
of circumcircle 
of . Prove that the lines 
and 
concur on .Anant Mudgal, IMOTC-2016, Senior Batch
This post has been edited 4 times. Last edited by Ankoganit, May 19, 2016, 2:48 PM
Reason: Typo corrected thanks Prof65
Reason: Typo corrected thanks Prof65
denote the angles of 
Then , note that, making extensive use of sine rule gives![\[ \frac{EP}{PF} = \frac{\tan \frac{C}{2}}{\tan \frac{B}{2}} \Longrightarrow \frac{\sin \angle CAP}{\sin \angle PAB} = \frac{FP}{PE} = \frac{\tan \frac{B}{2}}{\tan \frac{C}{2}} \Longrightarrow \frac{\sin \angle BCQ }{\sin \angle QBC} = \frac{\tan \frac{C}{2}}{\tan \frac{B}{2}} \Longrightarrow \frac{BD}{DC} = \frac{\tan \frac{C}{2}}{\tan \frac{B}{2}} \cdot \frac{\sin \angle BQD}{\sin \angle DQC} \]](http://latex.artofproblemsolving.com/9/b/3/9b377c7fb93af1ebee17ba586a497b68db66b39f.png)
But we have 
Thus 
which implies the result.
be a point on the circumcircle such that 
.
,
antipode of 
,
meet 
respectively, 
.
are collinear, and from 
collinear. Both 
and 
are diameters 
.
.
that touches 
is the tangency point of 
similarly, then 
concur at 
,
respectively. Let 
of circumcircle 
. it's clear 
but 
then 
is 
b
and 
so 
besides 
thus 
therefore 
pass through 
and 
,
and 
.
and 
,
and 
are similar. Then 
which implies that 
is an angle bisector of 
,
intersect 
and 
intersect 
is cyclic as 
.
and 
.
and 
,
.
. Here'a Short Proof (The Solution looks a bit lengthy because of adding so many details. ) which I, 
and let 
.
swaps 
and as 
we get that 
bisects 
and also 
.
is a harmonic bundle. Now let 
and let 
.
.
.
where 
swaps 
.
But clearly 
is a harmonic bundle as 
.
concurs on 
and 
and 
as desired.
and 
and 
and 
send to 
sent to 
colinear. Also note that 
thus we have 
colinear. (spiral sim). Now by projecting:
Thus 
and we are done 
and 
.
are collinear.
is harmonic,so:-
And
proving the required.
