such that 
,
,
are pairwise tangent to each other. Find the radius of the circle that is externally tangent to all three of these circles.
,
,
,
b
is increasing and 
i
have the same horizontal asymptote at 
,
?(Or can this number not be specified?)
independent and uniformly random letters from A to Z. Bob naturally tries to see if any subsequence of the said string of random letters spells "ILOVEYOU", and we call him delulu if he manages to find at least one such subsequence. What is the least value of 
?
,
where the curve 
is the boundary of the portion of plane 
in the first octant, traversed in the counterclockwise direction as viewed from above, using the definition of line integral and the Stokes's theorem.
![\[
A =
\begin{bmatrix}
1 & 2 & 3 & \cdots & n-1 & n \\
2 & 3 & 4 & \cdots & n & 1 \\
3 & 4 & 5 & \cdots & 1 & 2 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
n-1 & n & 1 & \cdots & n-3 & n-2 \\
n & 1 & 2 & \cdots & n-2 & n-1
\end{bmatrix}
\]](http://latex.artofproblemsolving.com/7/e/5/7e564602cc0cc14e1501f6654b00a4dfeac50f70.png)
satisfy the functional equation![\[[f(x)]^{2}=1+x f(x+1)\]](http://latex.artofproblemsolving.com/3/4/d/34de57b7e476314c43e0dc4cbae1e329511cbdb8.png)
![\[\frac{x+1}{2}\leq f(x) \leq 2(x+1)\]](http://latex.artofproblemsolving.com/5/5/8/55858a3de3307af5aac073230338d67840c1a45b.png)
.
.
.
where 
and 
is twice continuously differentiable on 
,![\[
\lim_{x \to +\infty} x f(x) = 0 \quad \text{and} \quad \lim_{x \to +\infty} x f''(x) = 0.
\]](http://latex.artofproblemsolving.com/3/e/5/3e536d8b5357b75e68d24c12b5cb0202edd52c8c.png)
Prove that:![\[
\lim_{x \to +\infty} x f'(x) = 0.
\]](http://latex.artofproblemsolving.com/4/d/3/4d3487c0be2069b8e1a44e7fe88b8d3b4456557d.png)
such that
.
is invertible
Y by doxuanlong15052000, Awesome_guy, ImSh95, Adventure10, Mango247, ehuseyinyigit
Let ABC be a scalene triangle. Let 
, 
and 
be the respective intersections with BC of the internal angle bisector, external angle bisector, and the median from A. The circumcircle of 
intersects 
a second time at point 
different from A. Define 
and 
analogously. Prove that the circumcenter of 
lies on the Euler line of ABC.
(The Euler line of ABC is the line passing through the circumcenter, centroid, and orthocenter of ABC.)
Proposed by Ivan Borsenco
, 
and 
be the respective intersections with BC of the internal angle bisector, external angle bisector, and the median from A. The circumcircle of 
intersects 
a second time at point 
different from A. Define 
and 
analogously. Prove that the circumcenter of 
lies on the Euler line of ABC.(The Euler line of ABC is the line passing through the circumcenter, centroid, and orthocenter of ABC.)
Proposed by Ivan Borsenco
This post has been edited 2 times. Last edited by v_Enhance, Aug 23, 2016, 12:45 AM
Reason: Add author
Reason: Add author
cut 
at 
.
is cyclical, then from the link we get 
i.e. 
i.e. 
hence 
ortocenter and 
feet of 
onto 
let 
be the projection of 
are collinear.
then 
is polar of 
WRT 
so 
so the lemma proved.
is the 
-
)
let 
then 
.
is cycli
so 
hence 
is cyclic.
and 
we have 
(
(
we have 
in triangle 
:
so 
so 
thus 
similarly 
projections of 
and let 
.
then 
and 
so in order to prove the problem we have to show that 
are collinear.let 
midpoint of 
line 
is perpendicular bisector of 
(
projections of 
on 
so 
.
is cyclic we have 
thus 
similarly 
so 
(ie that the circumcenter is the midpoint of 
from the givens and then observe this lies on the circle with diameter 
.
is pretty easy to make.
,
is tangent to 
and 
.
.
be the antipode of 
.
is a parallelogram 
are collinear.
is the midpoint of 
be the orthocenter, centroid, respectively, of 
From the Internal and External Angle Bisector Theorems, we have 
Thus, the division 
is harmonic. Hence, it is 
Therefore, 
,
Combining this relation with 
,
Similarly, we have 
By using these similar triangles, we find ![\[\angle BX_aC = \angle BX_aM_a + \angle CX_aM_a = \angle ABM_a + \angle ACM_a = 180^{\circ} - \angle BAC = \angle BHC.\]](http://latex.artofproblemsolving.com/d/9/0/d90914d42a4930222424b6a296b0342ecadd4f2a.png)
It follows that 
are concyclic. Therefore, 
Meanwhile, from 
,
Therefore, 
Therefore 
Similarly, 
and 
,
is just the circle of diameter 
Consequently, its center is the midpoint of 
,
,
and 
.
has equation 
;
.
and plug in 
for 
;
whence 
in Conway's triangle notation. Scaling to a more friendly version of the coordinates, we see that 
.
,
and 
also lie on the circle. Again, plugging in 
which is identical to our previous equation for 
root corresponds to 
be the antipode of 
Let 
for a second time at 
Then by Power of a Point and a ![\[M_aA \cdot M_aX_a = M_aK_a \cdot M_aL_a = -M_aB \cdot M_aC = -M_aA \cdot M_aX.\]](http://latex.artofproblemsolving.com/9/c/b/9cbe4cbb2f4773e89a492040f1e2cd75c5dab7d9.png)
Therefore, 
is the midpoint of 
But since 
(well-known), it follows that 
is a parallelogram. Hence, 
Then if 
lies on the circle 
of diameter 
,
is just the midpoint of 
,
and 
;
,
is the intersection of the line through 
similarly. 
is the midpoint of minor arc 
is the midpoint of major arc 
.
,
are concurrent; we claim that the concurrency point is the intersection of the tangents to 
at 
.
be the intersection of the tangents from 
so that 
to prove that the intersection of the tangents from 
denote 
for almost every point 
the orthic triangle of 
,
.
and 
.
is harmonic, which means 
is the 
and 
and 
,
and 
.
intersect 
again in 
.
.
and 
are both harmonic, so 
,
,
.
be the inversion that swaps 
,
,
,
lie on the radical axis of 
maps to the circumcircle of 
.
is centered at the midpoint of 
or equivalently that 
is the 
be the 
However, letting 
we know that
where the first line follows from the Three Tangents lemma. Thus in a similar fashion we can show 
and 
lie on the circle with diameter 
.
![\[\angle HX_aA = \angle HX_aG = 90,\]](http://latex.artofproblemsolving.com/6/6/4/6645c3e0e76c1ef6f6796cc36f11c96dd150acad.png)
.
is the 
,
and 
be the feet from 
,
respectively. Now it is well-known that 
,
and 
.
where 
is the midpoint of segment 
is the 
,
,
,
concur at 
lies on 
is the 
,
,
lie on 
be the 
denote the circumcenter and 9 point center respectively then 
.
where 
are the orthocenter and centroid respectively.We also know that 
.
and 
lies on the euler line.
from the above claim so the circumcenter is on 
where 
.
so from P1 we have 
.
.
are tangent to 
is the 
are 
is perpendicular to 
,
lies on the circle with diameter 
.
is cyclic and its diameter is 
center of 
so we know 
By PoaP, 
which implies 
say 
and 
we notice 
so 
the orthocenter. By Brocard, 
However we know 
also passes through the inverse of 
which is 
and 
so 
Hence 
all lie on circle with diameter 
and we are done.
where 
and 
because the humpty point is defined as the perpendicular from 
![\[\frac{XB}{XC} = \frac{AB}{AC},\]](http://latex.artofproblemsolving.com/3/c/9/3c9a7ff20b1c21e62b219799da631b448ebdfac7.png)
.
.
,
(where 
therefore the circumcenter of 
.
,
.
has circumcenter 
.
,we have that 
where 
.
similarly
lie on the circle with diameter 
lie on 
forms a harmonic quadrilateral. Using well-known properties, 
bisects angle 
Clearly, 
from right angles and bisector on point 
Thus, 
It is well known 
Thus, the Humpty point lies on 
This finishes the desired claim.
