[/quote]
,
is written on the board. Anna, Boris, and Charlie can do the following actions: Anna can replace the number with its floor. Boris can replace any integer number with its factorial. Charlie can replace any nonnegative number with its square root. Prove that the three can work together to make any positive integer in finitely many steps.
Prove that
Where 
and 
Prove that
and 
be different points on a circle 
such that 
is not a diameter. Let 
be the tangent line to 
is such that 
.
is chosen on the shorter arc 
of triangle 
intersects 
be the common point of 
meets 
is tangent to 
satisfying
Prove
be a right angled triangle with 
.
And 
be an arbitrary point on 
.
over 
and 
,
and 
lies on 
.
be a trapezoid with parallel sides 
.
and 
lie on the line segments 
and 
,
.
and 
on the line segment 
satisfying ![\[\angle{APB} = \angle{BCD}\qquad\text{and}\qquad \angle{CQD} = \angle{ABC}.\]](http://latex.artofproblemsolving.com/b/1/a/b1ae208955104081bafe221832509d9dd20eeb83.png)
Prove that the points 
and 
are concyclic.
be a positive integer. A board with a format 
is divided in 
(or 
)
positive integers 
such that 
for any 
,
such that 
?
![$A \in \mathbb Q[x]$](http://latex.artofproblemsolving.com/d/1/5/d158f344cf30f8beb5f2199da53272bd8e304704.png)
with ![$\exists Q \in \mathbb Q[x], Q(A(x))= x^{2025!+2}+x^2+x+1$](http://latex.artofproblemsolving.com/1/1/9/119a7b617e09ba16bb266b44ab3c85acc529f88f.png)
and 
.
.
,
is the smallest number different from 
which doesn't divide 
.
.
and 
meet the sides 
and ![\[ AB+AC\geq4DE. \]](http://latex.artofproblemsolving.com/3/a/4/3a49c5f8bd2930adca7bde2601dd045d4bf05ff6.png)
.
and 
lie on the circles 
and 
,
and 
.
and 
are concyclic.
and the recursive relation
for all natural number 
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 
are 
,
are perpendicular bisector of 
,
are concurrent by Pascal on 
.
,
.
.
denote the 
, 
and 
e
, 
and 
the midpoints of minor arcs 
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.
as intersection of 
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 
