Mastering Buffon's Coin Algebra with Simple Algebraic Problem Solving Techniques

[musicgold]

Homework Statement

This is not a homework problem. I am stuck

Relevant Equations

The attached picture shows the figure and the equation I am trying to solve.

I seem to have stuck with this simple algebra problem.$$ \frac {(L - 2.r)^2 } { L^2} = 1/2 $$
$$ 2( L^2 + 4r^2 - 4rL) = L^2 $$
$$ L^2 + 8 r^2 - 8r L = 0 $$
$$ (L - 2 \sqrt 2 r )^2 = 8rL - 4\sqrt 2. rL $$
$$ (L - 2 \sqrt 2 r )^2 = 4rL. (2 -\sqrt 2) $$

I don't know how to proceed from here to get
$$ L = ( 4 + 2\sqrt 2 ) . r $$

 

[PeroK]

$$ L^2 + 8 r^2 - 8r L = 0 $$

Isn't that a quadratic equation in $L$?

 

[musicgold]

Isn't that a quadratic equation in $L$?

Ah! I missed the forest for the trees. Thanks.

BTW, how should one interpret the second root of the eqaution?
$$ L = (4 - 2\sqrt 2 ). r$$

 

[PeroK]

Ah! I missed the forest for the trees. Thanks.

BTW, how should one interpret the second root of the eqaution?
$$ L = (4 - 2\sqrt 2 ). r$$

Well, $L$ must be greater than $2r$.

There was an easier way to do this. You have:
$$(L-2r)^2 = \frac 1 2 L^2 \ \Rightarrow \ L - 2r = \frac L {\sqrt 2}$$

 

[musicgold]

$$ \ L - 2r = \frac L {\sqrt 2}$$

Not sure how to derive a conclusion from this statement.
It is saying as long as L is greater than 2r, $\frac {L}{ \sqrt2}$ is a postive number, right?

 

[Office_Shredder]

I think it's just the same thing basically. You know that L > 0 and L > 2r, so that's the only version of the square roots that line up (or you can take the negative of both sides).

 

[PeroK]

Not sure how to derive a conclusion from this statement.
It is saying as long as L is greater than 2r, $\frac {L}{ \sqrt2}$ is a postive number, right?

1) We start with a physical problem where some things are positive numbers. In this case $L$ and $L - 2r$.

2) We set up a quadratic equation involving these quantities.

3) We take the solution that meets the physical constraint of positivity.

Sometimes the additional solutions mean something, but not in this case. There is no sense in a negative $L$.