Math 3118 Exam 2 Solutions

(15 points) 1. Find the area of the "slug-shaped" region below.

Solution: The right triangle has angles 45,45, 90, so both legs are equal, call them x. So x^2+x^2=2*x^2=4^2, or x=\sqrt8). So the area of the triangle is 1/2*\sqrt(8)*\sqrt(8)=4, the area of the semicircle is 1/2*\pi*2^2=2\pi, and the area of the rectangle is 7*\sqrt(8)=14*\sqrt(2). The total area is 4+2\pi+14*\sqrt(2), about 30.08.

(20 points) 2. State, in complete sentences, the Pythagorean theorem. Explain in complete sentences and in your own words a proof of this theorem. You may refer to a drawing.

Solution: See the text.

(25 points) 3. A regular octagon which is inscribed in a circle of radius 1 has a
side length of \sqrt{2-\sqrt{2}}, and an area of 2\sqrt{2}.
A regular octagon which is circumscribed in a circle of radius 1 has a
side length of 2\sqrt{2-\sqrt{2}}/\sqrt{2+\sqrt{2}}, and an area of 8\sqrt{2-\sqrt{2}}/\sqrt{2+\sqrt{2}}.

Use these facts to give inequalities for \pi. Be sure to explain the reasons for your inequalities.

Solution: The inscribed polygon has perimeter 8*\sqrt(2-\sqrt(2)). Since it connects points on a circle of radius 1 by straight line segments, its perimeter must be less than than circumference of a circle of radius one, which is 2*\pi. Thus

\pi is greater than 4*\sqrt(2-\sqrt(2)) which is about 3.06.

The area of the inscribed polygon is 2*\sqrt(2), must be less than the area of the circle which is \pi. So

\pi is greater than 2*\sqrt(2) which is about 2.83.

For the circumscribed octagon, the area must be greater than \pi,

\pi is less than 8\sqrt{2-\sqrt{2}}/\sqrt{2+\sqrt{2}}, which is about 3.314.

It is not automatically true that the perimeter of the circumscribed octagon is larger than the perimeter. But here is does work you get exactly the same inequality for \pi that the area gives.

(20 points) 4. Find all primitive Pythagorean triples (a,b,c) such that c=b+1. Be sure to explain why you have them all. (Possible hint: Try to factor c-b.) If you cannot find them all, then list as many such triples (a,b,c) as you can. Perhaps you can notice a pattern from your list.

Solution: CLAIM; All such triples are given by

(2r+1, 2*r*(r+1), 2*r*(r+1)+1), where r is a postive integer.

So the first few examples are

(3,4,5), (5,12,13), (7,24,25), (9,40,41),

Another way to state this is to start with any odd integer at least 3 for a. Then a^2/2 is not an integer, and take b and c to be the two integers closest to a^/2. For example if a=9, 9^2/2=40.5, so b=40, c=41.

To see this, let b=2*r*s, and c=r^2+s^2, Then c-b=r^2-2*r*s+s^2=(s-r)^2=1, How could a square of a positive integer equal 1? Only if that integer is 1, so s-r=1, or s=r+1. So

a=s^2-r^2=(r+1)^2-r^2=2r+1,
b=2*r*s=2*r*(r+1)
c=s^2+r^2=(r+1)^2+r^2=2r^2+2r+1.