[/quote]
,
be nonnegative positive integers.
triangle with 
.
,
.
,
lines intersect at point 
,
be a cyclic quadrilateral with 
and let 
and 
to the lines 
and 
,
bisects 
.
be a function which is differentiable on 
and satisfies![\[
\lim_{x\to+\infty}\bigl(f'(x)/e^x\bigr)=0.
\]](http://latex.artofproblemsolving.com/a/f/b/afbd725dc4a4cbb1ecfdc1d4923f9a7edcb53397.png)
Prove that![\[
\lim_{x\to+\infty}\bigl(f(x)/e^x\bigr)0.
\]](http://latex.artofproblemsolving.com/8/c/a/8caa7f6cc28759036af097a634f000c6daf6bb79.png)
![$P\in \mathbb{R}[x] $](http://latex.artofproblemsolving.com/1/1/9/1194a1382b5ae7a9945cf8d9a6281954cf8cdb3d.png)
with even-degree
,
.
.
meets 
at 
.
and 
the circumcircles of 
and 
,
meets 
and 
,
meets 
.
and 
,
and 
of 
.
on segment 
is tangent to 
.
meets 
.
be a circle thaat passes through 
be a circle that passes through 
and tangents to 
is tangent to circles 
and 
= 2∠AMP.
I
be positive real numbers, show that:
such that 
and 
are also primes.
be pairwise distinct integers. Show that the equation 
has at most one integral solution.
be a positive integer and let 
be odd primes. Prove that the positive integers 
can be colored in 
colors, such that for any 
of the same color, 
is not divisible by 
.
be the Gergonne point of the triangle 
is parallel with line 
.
be a polynomial with real coefficients such that 
for all 
.
for some real constant 
and 
.
Y by Rounak_iitr
Given a 
with incenter 
. Line 
and 
intersects 
and 
at 
and 
, respectively. Let 
and 
be the midpoints of and . Line 
meets the external bisector of angle 
at 
and line 
meets the external bisector of angle 
at 
. Prove that 
are concylic.
with incenter 
. Line 
and 
intersects 
and 
at 
and 
, respectively. Let 
and 
be the midpoints of and . Line 
meets the external bisector of angle 
at 
and line 
meets the external bisector of angle 
at 
. Prove that 
are concylic.This post has been edited 1 time. Last edited by crocodilepradita, Aug 22, 2024, 5:51 AM
at 
at 
by 
and the midpoint of 
are 
,
are perpendicular bisector of 
,
are concurrent by Pascal on 
.
.
,
.
.
denote the 
, 
and 
e
, 
and 
the midpoints of minor arcs 
and 
and 
implying that 
is the perpendicular bisector of segment 
passes through 
and 
are parallel.![\[\frac{KI_c}{M_cM} = \frac{FI_c}{FM_c} \text{ and } \frac{LI_b}{M_bN} = \frac{EI_b}{EM_b}\]](http://latex.artofproblemsolving.com/8/4/d/84dff54cfc5ce43e372238385bb8afebb7f47656.png)
from which we have![\[\frac{KI_c}{LI_b} = \frac{FI_c}{FM_c} \cdot \frac{EM_b}{EI_b} \cdot \frac{MM_c}{NM_b}\]](http://latex.artofproblemsolving.com/7/8/4/784ffdb7907931ed5a02cb28ade142763b94911f.png)
Now, since 
is cyclic we have that![\[FI \cdot FI_c = FA \cdot FB = FM_c \cdot FC\]](http://latex.artofproblemsolving.com/e/5/9/e59854f7b9bfce258bbc13cbb196a8cfe6b7614e.png)
from whence we have![\[\frac{FI_c}{FM_c} = \frac{FC}{FI}\]](http://latex.artofproblemsolving.com/7/3/c/73c450147e973dc0135b0594a265731774ab8629.png)
and similarly,![\[\frac{EM_b}{EI_c} = \frac{EI}{EB}\]](http://latex.artofproblemsolving.com/2/d/c/2dc98677edcb57213e8a563f9a355138473c3ad7.png)
Thus,![\[\frac{KI_c}{LI_b} = \frac{FC}{FI} \cdot \frac{EI}{EB} \cdot \frac{MM_c}{NM_b} \]](http://latex.artofproblemsolving.com/b/9/c/b9c2c18a89a5288b75fe2af6093a04d43e147f27.png)
Now, by the Law of Sines we have![\[\frac{FC}{FB} = \frac{\sin B}{ \sin \frac{C}{2}} \text{ and } \frac{FB}{FI} = \frac{\sin (90 - \frac{A}{2})}{\sin \frac{B}{2}} \]](http://latex.artofproblemsolving.com/5/6/f/56f7edc0be53d0aac29d023b64bc2e1e62476ae9.png)
from which it follows that![\[\frac{FC}{FI} = \frac{\sin B \sin (90 - \frac{A}{2})}{ \sin \frac{B}{2} \sin \frac{C}{2}}\]](http://latex.artofproblemsolving.com/b/e/b/bebc0bcd18ae75a6a258b26860f0c95fbf414bb3.png)
and similarly,![\[\frac{EB}{EI} = \frac{\sin C \sin (90 - \frac{A}{2})}{ \sin \frac{C}{2} \sin \frac{B}{2}}\]](http://latex.artofproblemsolving.com/1/4/1/141cbfc91708a781d5faf3efc3b460181390c7c5.png)
Thus,![\[\frac{KI_c}{LI_b} = \frac{\sin B \sin (90 - \frac{A}{2})}{ \sin \frac{B}{2} \sin \frac{C}{2}}\cdot \frac{ \sin \frac{C}{2} \sin \frac{B}{2}}{\sin C \sin (90 - \frac{A}{2})}\cdot \frac{\sin \frac{C}{2}}{ \sin \frac{B}{2}} = \frac{\sin B \sin \frac{C}{2}}{\sin C \sin \frac{B}{2}} = \frac{\cos \frac{B}{2}}{\cos \frac{C}{2}}\]](http://latex.artofproblemsolving.com/5/b/e/5be568c199ee2c05956ed344312819a24554d331.png)
Further, since 
is cyclic,![\[\frac{I_aI_c}{I_aI_b} = \frac{I_aC}{I_aB} = \frac{\sin \angle CBI_a}{ \sin \angle BCI_a} = \frac{\sin (90 - \frac{B}{2})}{\sin (90 - \frac{C}{2})} = \frac{\cos \frac{B}{2}}{\cos \frac{C}{2}}\]](http://latex.artofproblemsolving.com/f/1/5/f152c3a01985d4ab124bfeeb2a6a287b4f05bc59.png)
Thus,![\[\frac{KI_c}{LI_b} = \frac{I_aI_c}{I_aI_b}\]](http://latex.artofproblemsolving.com/e/c/c/ecca8312a7a08718a31068d7a66c115dd73b9492.png)
which implies that 
as desired.
,
as well. Hence,![\[\measuredangle LKI_a = 90 + \measuredangle II_aB = 90 + \measuredangle ICB = \measuredangle BCL\]](http://latex.artofproblemsolving.com/7/f/0/7f022095805329740c3139e49b45ced0e7efe908.png)
which implies that 
is indeed cyclic, as claimed.
to be the ex-central triangle. See that 
and similarly 
because 
,
Similarly 
and done. (200th post on HSO lessgo)
.
.
then 
so they're parallel, so 
.
so it's cyclic.
and 
,
i got stuck for a while and did not understand how to move the points to 
but then i saw that u can project through 
