2 math problems

[Safari]

Hi!
We'll let’s start

1.

Show that no matter how you're placing the numbers (1 - 12) in a clock you'll still get >= 20 when adding three numbers that are after each other.

2.

The first circle has the circumference P1 = 2{pi}r. Decide Pn

[Alvaro]

1. Divide the clock in four disjoint sets of three consecutive numbers each. If each one had a sum of at most 19, the total sum would be at most 76, but 1+2+3+4+5+6+7+8+9+10+11+12 = 78

2. Each iteration divides the radius by sqrt(2). You need little more than Pythagoras's theorem to prove it.

[Safari]

Alvaro: Where did you get sqrt(2) from?

[Jetru]

Saf:Pythagoras.

I didn't get the second question. 1+2+3 <20

[Alvaro]

Let's look at the first and second circles. The radius of the first is 1. The radius of the second is the distance between the origin and any of its points. One of its points is (.5,.5), where the circle is tangent to the line that goes from (1,0) to (0,1). The distance from the origin to (.5,.5) is 1/sqrt(2).

[Alvaro]

I didn't get the second question. 1+2+3 <20

That was the first question. What he meant is that no matter how you arrange the numbers from 1 to 12 on a clock, there is always going to be a group of three consecutive numbers whose sum is at least 20. Actually, the result can be improved to 21.

[Safari]

Okey, thank you. I understand the first one and I'll try to understand the second one

[Jetru]

Sorry, first question,yes. It's asking that there's going to ATLEAST ONE group of three numbers whose sum is >=20?

[Alvaro]

Sorry, first question,yes. It's asking that there's going to ATLEAST ONE group of three numbers whose sum is >=20?

Yes, that's what I said. And you can actually prove that there is at least one group of three consecutvie numbers whose sum is >=21.