meet at point 
and 
lie on 
,
respectively in such a way that 
and 
.
and 
passes through the mass-center of 
Prove that
.
and 
be positive integers with 
.
.
be positive integers with 
.
be a polynomial of degree 
such that the polynomial 
divides 
is zero. Prove that 
be positive integers. Prove that there exists a positive integer 
such that for every odd integer 
,
are all greater than 
index this yr (118.5 on amc10+ 9 on AIME) i’m in 7th rn btw first year comp math also. I will grind so hard until like 30 hrs/week. I’m ok at proofs. made mc nats
Y by
Let 
be an equilateral triangle, and 
is a point on its incircle. Let 
be the circle tangent to 
passing through and 
. Similarly, let 
be the circle tangent to 
passing through and 
, and 
be the circle tangent to 
passing through and 
.
Prove that the circles, , has a common tangent line.
Proposed by Ivan Chan Kai Chin
be an equilateral triangle, and 
is a point on its incircle. Let 
be the circle tangent to 
passing through and 
. Similarly, let 
be the circle tangent to 
passing through and 
, and 
be the circle tangent to 
passing through and 
.Prove that the circles
, , has a common tangent line.Proposed by Ivan Chan Kai Chin
This post has been edited 1 time. Last edited by navi_09220114, Saturday at 1:06 PM
be the midpoint of 
be the reflection of 
(so that 
is an isosceles trapezoid), 
be the intersection of 
,
be the intersections of the line tangent to 
)
on 
are tangents we have 
,
by isosceles trapezoid so we have 
.
by symmetry so 
.
and 
we have 
.![\[JP=\sin\angle PTJ\cdot PT=\sin\angle PCT\cdot \frac{PC\cdot PQ}{PM}=\sin\angle CPM\cdot\frac{CP\cdot CM}{PM}.\]](http://latex.artofproblemsolving.com/8/9/2/892468fba13d699b2923c7e8ae100517dbf76356.png)
But ![\[\sin\angle CPM=\frac{\sin\angle PMC\cdot CM}{PC}=\frac{r\cdot PM\cdot CM}{PC},\]](http://latex.artofproblemsolving.com/d/3/3/d331b4ba5319df1d5121e6fd896c4d7335d5ba1f.png)
so substituting we have 
which is a constant.
such that they are perpendicular to 
have equal distance to 
and the point circle 
midpoints of 
respectively, then in fact they make the intouch triangle because of 
being equilateral and thus if we consider point 
such that 
is an isosceles trapezoid then from reflection spam and the angles and pythagorean theorem we have that 
is tangent to 
,
are also isosceles trapezoids (notice here we will WLOG 
)
and define cyclically 
then from the isosceles trapezoids we have that tangents from 
to 
respectively are all parallel to 
and thus perpendicular to 
,
is tangent to 
and thus 
and similarily for 
and this means 
all lie on the radax of 
however their tangents to their respectively circles were all parallel to this line (by being perpendicular to 
.
to be the circle through 
tangent to 
to be the circle through 
and tangent to 
be the center of triangle 
be a point such that 
.
,
be the midpoint of 
be the intersection of 
different from 
)
.
to be the reflection of 
,
hence 
and 
are parallel.
be the perpendicular bisector of 
,
,
