1. Originally Posted by beero1000

An irregular octagon is inscribed in a circle. The edge lengths of the octagon are 4,4,4,4,2,2,2,2. Find the area of the octagon. Bonus points: Find the radius of the circle.
What did I do wrong?

4b + 4c = 2pi, where b is the arc of the 2 unit long line segment, and c is the other.

c = pi/2 - b

r = 2/sin(c)

r = 1/sin(b)

2sin(b) = sin(c)

2sin(b) = sin(pi/2 - b) = cos(b)

b = tan^(-1)(1/2)

r = 1/sin(b) = 2.236...

2. Originally Posted by beero1000
2. If n is a positive even number, is 23n-1 + 5(3n) divisible by 11?
Yes.

Change variables and write the formula mod 11. Let n = 2m+2, m a non-negative integer. So

23n-1 + 5(3n) = 26m+5 + 5(32m+2)
= 32(26)m + 5(9)(32)m
= -1(9m) + 1(9m) (mod 11)
= 0 (mod 11)

3. Originally Posted by Bomb#20
Originally Posted by beero1000
2. If n is a positive even number, is 23n-1 + 5(3n) divisible by 11?
Yes.

Change variables and write the formula mod 11. Let n = 2m+2, m a non-negative integer. So

23n-1 + 5(3n) = 26m+5 + 5(32m+2)
= 32(26)m + 5(9)(32)m
= -1(9m) + 1(9m) (mod 11)
= 0 (mod 11)
I used induction.

n = 2m, where m is any natural number.

Assume, 26m - 1 + 5(32m) = 11q, where q is another natural number.

m = 1: 77 = 11q

m + 1: 26m + 5 + 5(32m + 2) = 26m - 1 + 5(32m + 2) + 11

4. Originally Posted by ryan
I used induction.

n = 2m, where m is any natural number.

Assume, 26m - 1 + 5(32m) = 11q, where q is another natural number.

m = 1: 77 = 11q

m + 1: 26m + 5 + 5(32m + 2) = 26m - 1 + 5(32m + 2) + 11
You must have dropped a term -- for m=1, that equation works out to 2453 = 448.

(Which isn't to say that induction couldn't work.)

Originally Posted by ryan
What did I do wrong? ... sin(c) ...
Tried to use trig!

(Which isn't to say that trig couldn't work. But I got away with using the Pythagorean Theorem and a couple of pages of algebra.)

5. Originally Posted by Bomb#20

You must have dropped a term -- for m=1, that equation works out to 2453 = 448.

(Which isn't to say that induction couldn't work.)
thanks

Let/assume 26m - 1 + 5(32m) = 11q.

m + 1: 26(26m - 1) + 5(32m + 2

= 26(26m - 1) + 32*5(32m)

= 26(26m - 1) + 9(11q - 26m - 1)

= 26(26m - 1) + 9(11q) - 9(26m - 1)

= 26m - 1(26 - 9) + 9(11q)

= 26m - 1(55) + 9(11q)

= 11((26m - 1)(5) + 9q)

Originally Posted by ryan
What did I do wrong? ... sin(c) ...
Tried to use trig!

(Which isn't to say that trig couldn't work. But I got away with using the Pythagorean Theorem and a couple of pages of algebra.)
But where did I go wrong? Every step seems correct.

6. Originally Posted by beero1000
1. If x,y,z are positive real numbers with xyz = 1, show that x2+y2+z2 ≤ x3+y3+z3
Assume without loss of generality that x <= y <= z. z = 1/xy. If all three were less than 1 their product would be less than 1; likewise, if all three were greater than 1 their product would be greater than 1. Therefore x <= 1 and z >= 1. There are two cases to consider, y <= 1 and y > 1. Here's a proof for y <= 1.

To show that x2+y2+z2 <= x3+y3+z3 is to show that f = x3+y3+z3 - (x2+y2+z2) >= 0.

Let r = 1 - x, s = 1 - y, t = z - 1. r, s, t >= 0.

x(1 + r/x) = x + r = x + 1 - x = 1. Therefore 1 + r/x = 1/x. Similarly, 1 + s/y = 1/y.

z = 1/xy = (1/x)(1/y) = (1 + r/x)(1 + s/y) = 1 + r/x + s/y + rs/xy

t = r/x + s/y + rs/xy

f = x3+y3+z3 - (x2+y2+z2)

f = (1-r)3+(1-s)3+(1+t)3 - ((1-r)2+(1-s)2+(1+t)2)

f = (3 + 3(t-r-s) + 3(t2+r2+s2) + t3-r3-s3)
- (3 + 2(t-r-s) + t2+r2+s2)

f = t-r-s + 2(t2+r2+s2) + t3-r3-s3

Let g = 2(t2+r2+s2)

f = r/x + s/y + rs/xy - r - s + g + (r/x + s/y + rs/xy)3-r3-s3

(r/x + s/y + rs/xy)3 = (r/x)3 + (s/y)3 + [a series of 25 more products of sets of three terms].

Let h = that series of 25 products I'm not going to painfully write out.

f = r/x + s/y + rs/xy - r - s + g + (r/x)3 + (s/y)3 + h - r3 - s3

f = (r/x - r) + (s/y - s) + rs/xy + g + ((r/x)3 - r3) + ((s/y)3 - s3) + h

f = r(1/x - 1) + s(1/y - 1) + rs/xy + g + r3((1/x)3 - 1) + s3((1/y)3 - 1) + h

f = r(1 + r/x - 1) + s(1 + s/y - 1) + rs/xy + g + r3((1 + r/x)3 - 1) + s3((1 + s/y)3 - 1) + h

f = r(r/x) + s(s/y) + rs/xy + g + r3((1 + r/x)3 - 1) + s3((1 + s/y)3 - 1) + h

By inspection, every individual term in the above sum of terms is non-negative. Therefore f >= 0. Q.E.D.

The proof for y > 1 is analogous and is left as an exercise for the masochist.

7. Originally Posted by Bomb#20

You must have dropped a term -- for m=1, that equation works out to 2453 = 448.

(Which isn't to say that induction couldn't work.)
Induction will work, as long as he is careful with his algebra. As written, there are errors.

Originally Posted by Bomb#20
Originally Posted by ryan
What did I do wrong? ... sin(c) ...
Tried to use trig!

(Which isn't to say that trig couldn't work. But I got away with using the Pythagorean Theorem and a couple of pages of algebra.)
Trig will work, if you're careful, but I agree it's not the best approach.

The "sneaky" approach:

Note that the order of the sides does not matter, so take the octagon side sequence to be alternating 4's and 2's. Take a square of side s and symmetrically chop off the corners at 45 degrees. You are left with an octagon that has two groups of four equal edges, and it is clearly cyclic. We want the small sides to have length 2, so we cut $\sqrt{2}$ off each edge. Then we want $s - 2 \sqrt{2} = 4$, so $s = 4 + 2\sqrt{2}$. Therefore, the area is $(4+2\sqrt{2})^2 - 4 = 20 + 16\sqrt{2}$.

The circle's radius is then $\sqrt{(2+\sqrt{2})^2 + 2^2} = \sqrt{10 + 4\sqrt{2}}$

8. Originally Posted by beero1000

Trig will work, if you're careful, but I agree it's not the best approach.
I am so frustrated because each step seems right. Please - I beg you - take me out of my misery!

9. Originally Posted by ryan
Originally Posted by beero1000

Trig will work, if you're careful, but I agree it's not the best approach.
I am so frustrated because each step seems right. Please - I beg you - take me out of my misery!
r = 2/sin(c/2) not r = 2/sin(c)

10. Originally Posted by beero1000
Originally Posted by ryan

I am so frustrated because each step seems right. Please - I beg you - take me out of my misery!
r = 2/sin(c/2) not r = 2/sin(c)
Oh for the love of ...

Thanks!!!!!!

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