Pizza and Problems/Pizza and Problems 3 (4/18/08)

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Use this page to post solutions to our seventh Pizza and Problems session, held 4/18/08. Examine the original problem set.

A great time was had by all at our meeting of Pizza and Problems. We had some very cool problems, good pizza, and enthusiam and chuckles were high. Finally, we had some very good student presentations of solutions. Some students worked together on problems, arguing their case and developing ideas! A very cool strategy.

1 Student Solutions to the Problems in Pizza and Problems Set #3

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Student Solutions to the Problems in Pizza and Problems Set #3

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Problem 1: Solution by:

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Problem 2: Solution by:

Jonathan Gent stated that three points determine a circle and labeled the points as follows:

He then plugged each of the points A, D, and B into the equation of the circle, $x 2 + y 2 = r 2$ . This gave three equations in the unknowns a, b, and r, which he claimed was easily solvable. I checked. He's right and the answer is $r=\sqrt{65}/2$ , making the diameter $\sqrt{65}\,$ . Very cool solution.

Maybe some brave soul will provide the details?

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Problem 3: Solution by: Bruce Wagner

The correct answer is 4/ $π 2$ .

Start by dividing the unit circle in small pieces $P i$ , each of length $Δθ$ . Note that the probability of $p$ landing in any particular $P i$ is exactly $(Δθ) / (2π)$ .

Now given a choice of $p = (a, b)$ , let Pr( $p$ ) be the condition probability that the entire rectangle determined by $p$ and $q$ lies inside the circle. The rectangle will lie inside the circle precisely when $q$ lies in the inscribed rectangle $R p$ determined by $p$ , namely $R_p=[-|a|,|a|]\times[-|b|,|b|]$ . Therefore, Pr( $p$ ) equals the ratio

$\frac{\mbox{area of }R_p}{\mbox{area of the circle}}.$

Since $p = (cos(θ),sin(θ))$ for some $θ$ , the area of $R p$ is $(2 | cos(θ) | )(2 | sin(θ) | )$ , and therefore,

$\mbox{Pr}(p)=\frac{4|\cos(\theta)\sin(\theta)|}{\pi}$

The last step is to choose points $p i = (cos(θ i),sin(θ i))$ in each subinterval and add up the conditional probabilities. The final probability is then approximately equal to the Riemann sum

$\sum_{i=1}^{n} \frac{\Delta\theta}{2\pi}\cdot\frac{4|\cos(\theta_i)\sin(\theta_i)|}{\pi} =\sum_{i=1}^{n} \frac{2|\cos(\theta_i)\sin(\theta_i)|}{\pi^2}\Delta\theta$

Taking the limit as $\Delta\theta\to 0$ , we find that the final probability is

$\int_0^{2\pi}\frac{2|\cos(\theta)\sin(\theta)|}{\pi^2}d\theta$

Since the integrand has period $π / 2$ , this is equal to

$4\int_0^{\pi/2}\frac{2\cos(\theta)\sin(\theta)}{\pi^2}d\theta,$

and a final calculation of this integral yields the final answer $4 / π 2$ .

As there was some disagreement about this problem, I thought that I should check my answer using a simulation. This lead to a rather interesting programming problem. For the details, please see [1].

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Problem 4: Solution by:

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Problem 5: Solution by: David Arnold

In the following argument, we assume all numbers are integers. We define $a \equiv b \pmod{n}$ if and only if $n | a - b$ , which means "n divides a - b".

Now, suppose that we know two things:

$a\equiv b\pmod n$ , and
$c\equiv d\pmod n$ .

These lead to the following two facts.

$n | a - b$ , and
$n | c - d$ .

Because n divides each of a - b and c - d, we have

$a - b = k 1 n$ , and
$c - d = k 2 n$ ,

where $k 1$ and $k 2$ are integers. Multiply the first of these equations by c, the second by b to obtain the following results.

$a c - b c = k 1 c n$ , and
$b c - b d = k 2 b n$ .

Adding these two equations produces the following result.

a c - b d = (k 1 c + k 2 b) n

Consequently, $n | a c - b d$ and

$ac\equiv bd\pmod n.$

Hence, we have an important result. If $a\equiv b\pmod n$ and $c\equiv d\pmod n$ , then we can multiply left- and right-hand sides of these congruences together to produce the result $ac\equiv bd\pmod n$ .