## Monday, August 15, 2011

### The Problem of the Week Day 1: Week of 8/14 - 8/20

Since I just recently came back from CTY, the problems will mainly be a problem we discussed in class. The easy one will be a problem about a chess tournament, and the hard will be about slicing pizzas. However, we are going to start off with some triangles as always!

Easy Problem: In the last two problems, we worked with the famous Pythagorean Theorem, developed to find the missing side of a right triangle when given two. Just to remind you, the shortest side is labeled a, the medium labeled b, and the longest labeled c. The formula states that a^2 + b^2 = c^2.

If you have a right triangle with a equalling 26.25 and with c equalling 43.75, what is the value of b? Remember to round to the nearest tenth!

b = ___

You won't need b's value until Wednesday, so hold onto it after you find it.

Hard Problem: Last month, we used the button on the calculator that turns a sine into its angle. Let me review how it works.

"You might not have it on your average calculator, but if you hit the 2nd button on a scientific calculator or iPhone calculator, you will get a button where the sine function has a little -1 above it, in the place of an exponent. That button takes the sine of an angle, and turns it into the angle. So, you could divide the side opposite to an angle by c and get the sine, and then hit that button to retrieve your angle."

If a right triangle's sides are lengths 27, 29.5, and 40, what is the measurement of the angle opposite to side a? This angle will be referred to later on as z.

z = ___

You will not need this value until Friday.

Additional Challenge: At camp, we also had a puzzle called region revenge. It was pretty difficult, but I'd like to share it with you. I will put up the answer in two months.

The problem is basically to create a vertex and see how many shaded areas are in the circle. So, for one vertex, there is one shaded region. Now, make two vertexes and connect them with a line. That makes two shaded areas. Then, you make three, and connect every vertex to one another, making a triangle. There, there would be four shaded areas. Keep doing this for five to seven circles, and look for differences. Can you find an explicit formula? It is not what you think it is.

Hint: The formula is a quartic equation (equation of degree four).