be a cyclic quadrilateral with 
and 
.
and 
are selected on segment 
such that 
.
and 
are selected on segment 
such that 
.
is a cyclic quadrilateral.
![$$\int_0^{\pi/2} \Bigl[
\log\bigl(\sqrt{5}-\sin\theta+1\bigr)
+\log\bigl(\sqrt{5}+\sin\theta-1\bigr)
-\log\bigl(\sqrt{5}-\sin\theta-1\bigr)
-\log\bigl(\sqrt{5}+\sin\theta+1\bigr)
\Bigr]\,d\theta.
$$](http://latex.artofproblemsolving.com/a/f/f/affd7cb3d94d4fbcab6f9912e71ebcffeb86d960.png)
, s.t. ecuation: 
have 2 real solutions .
,
.
be a ring with unity such that for every 
there exist 
such that 
.
then 
then the result from a) may no longer hold.
and 
be fixed positive integers. Show that the set of primes that divide at least one of the terms of the sequence 
is infinite.
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 
.
