Pizza and Problems/Pizza and Problems 1 (3/7/08)

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Use this page to post solutions to our fifth Pizza and Problems session, held 3/7/08. 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 #1

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

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Problem 1: Solution by:

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Problem 2: Solution by:

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Problem 3: Solution by Ian Snyder:

Problem: Given that $i 2 = - 1$ , for how many integers n is $(n + i) 4$ an integer?

In this problem I deceided to expand the binomial, $(n + i) 4$ , as follows:

$(n + i)^4\,$

$= n^4 + 4n^3i + 6n^2i^2 + 4ni^3 + i^4\,$

$= n^4 + 4n^3i - 6n^2 - 4ni + 1\,$

Now, in order for all terms of the polynomial involving i to cancel, $4 n 3 i$ must equal $4 n i$ .

$4n^3i = 4ni\,$

$n^3 = n\,$

$n(n^2 - 1) = 0\,$

$n = \{ -1,0,1 \}\,$

(sorry I can't figure out how to get set bars)

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Problem 7: Solution by: David Arnold

OK. No one else is posting a solution to this problem, so I will give it a go.

Triangle $\triangle ADC$ is inscribed in a semicircle, so it is a 'right' triangle with right angle at D. That is, $D = π / 2$ .

The area of sector DCBAD is given by the formula $(1 / 2)θ r 2$ , where it is easily shown (see right triangle $\triangle DEC$ that $r=DC=\sqrt{2}$ . Hence, the area of sector DCBAD is

$\mbox{Area DCBAD} = \frac{1}{2}\cdot\frac{\pi}{2}\cdot (\sqrt{2})^2= \frac{\pi}{2}.$

To find the area of the sector AOCBA, we must subtract the area of triangle $\triangle ADC$ from the area of sector DCBAD. Because the area of triangle $\triangle ADC=(1/2)(2)(1)=1$ , we have