[/quote]
,
be a triangle with incentre 
and circumcircle 
.
be the midpoints of the arcs 
of 
respectively. Let 
be the point of 
.
and the midpoint 
of 
lie on a line.
on a plane with centers 
,
is feet of the perpendicular from 
to 
and 
and 
are intersections of line 
with circle 
.
and 
can move around the circle 
and 
,
.
be the circumcircle of 
and let 
be A-excircle .Let the two common tangents of 
cut 
in 
.
.
,
and 
.
can be partitioned into several 
the following inequality holds:
prove that
is a prime number. We have a number of cards, each of which has a number written on it such that each of the numbers 
appears at most once and 
and 
,
If card 
,
If card 
,
must not win over card 
.
Y by mjkkra, CoordinateBash13, Adventure10, Mango247, and 1 other user
In a small pond there are eleven lily pads in a row labeled 
through 
. A frog is sitting on pad 
. When the frog is on pad 
, 
, it will jump to pad 
with probability 
and to pad 
with probability 
. Each jump is independent of the previous jumps. If the frog reaches pad it will be eaten by a patiently waiting snake. If the frog reaches pad it will exit the pond, never to return. What is the probability that the frog will escape being eaten by the snake?
through 
. A frog is sitting on pad 
. When the frog is on pad 
, 
, it will jump to pad 
with probability 
and to pad 
with probability 
. Each jump is independent of the previous jumps. If the frog reaches pad it will be eaten by a patiently waiting snake. If the frog reaches pad it will exit the pond, never to return. What is the probability that the frog will escape being eaten by the snake?
This post has been edited 2 times. Last edited by LauraZed, Dec 27, 2018, 4:41 PM
Reason: fixed capitalization
Reason: fixed capitalization
.
and we use our intuition about symmetry to find that 
.
?
be the probability from lily pad 
(so 
,
)
Adding them all, we get![\[ 1 = p_{10} = \left(
\frac{1}{\binom90} + \frac{1}{\binom91}
+ \frac{1}{\binom92} + \dots + \frac{1}{\binom99}
\right) p_1. \]](http://latex.artofproblemsolving.com/2/2/6/2268bed3c36a298b8959e4a42661246ae870cc70.png)
Equivalently, one can also add up to just 
to derive the equivalent![\[ \frac12 = p_{5} = \left(
\frac{1}{\binom90} + \frac{1}{\binom91}
+ \frac{1}{\binom92} + \frac{1}{\binom93} + \frac{1}{\binom94}
\right) p_1. \]](http://latex.artofproblemsolving.com/e/1/c/e1c234216c2dc9c5988d35e86d3b4100eb849016.png)
In any case, we have![\[
p_1 =
\frac{1}{2\sum_{k=0}^4 \binom9k ^{-1} } = \frac{63}{146}.
\]](http://latex.artofproblemsolving.com/e/0/a/e0a6ad11d4ffc7e2e85f89b117aa7fe34ef956e9.png)
and similar equations to what you have above?
.
?
denote the probability that starting from the 
t
Using our recursion on 
and 
again we obtain 
from which we obtain 
.
