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Can you solve these math problems?

Vertimus

Banned
Continuation of the old thread.

These are selected problems I've done in the past few days that I've found interesting. Let's see how the ATOT crowd handles it!

1. Find the sum of all positive integers n for which n^2-19n+99 is a perfect square.

2. In trapezoid ABCD, leg BC is perpendicular to bases AB and CD, and diagonals AC and BD are perpendicular. Given that AB=sqrt(11) and AD=sqrt(1001), find BC^2.

3. The points A, B, and C lie on the surface of a sphere with center O and radius 20. It is given that AB=13, BC=14, CA=15, and that the distance from O to triangle ABC is x. Find x in exact terms.

4. The function f is defined by f(x)=(ax+b)/(cx+d), where a, b, c, and d are nonzero real numbers, has the properties f(19)=19, f(97)=97, and f(f(x))=x for all values of x except -d/c. Find the unique number that is not in the range of f.

5. Consider the polynomials P(x)=x^6-x^5-x^3-x^2-x and Q(x)=x^4-x^3-x^2-1. Given that p, q, r, and s are the roots of Q(x)=0, find P(p)+P(q)+P(r)+P(s).

6 A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects on eof the two verticies where it is not currently located, and crawls along a side of the tirangle to that vetex. Given that the probability that the bug moves to its starting vertex on its tenth move is x, find x.

7. Given that z is a complex number such that z+1/z=2 cos(3), find the least integer that is greater than z^2000+1/z^2000. (Angles are in degrees in this problem)

8. Let v and w be distinct, randomly chosen roots of the equation z^1997-1=0. Let x be the probability that sqrt(2+sqrt(3)) is lesser or equal to the absolute value of (v+w). Find x.

NOTE: Please include solutions with your answers. Tell me how you arrived at your answer.

I solved all of these problems without calculators or computers so you should too. They become too easy and not fun when you use calculators, and are designed without them.

Good Luck! :beer:
 
I must say these problems require some awfully difficult mathematical theory.

What do you think we are? Actuaries?
 
Originally posted by: ActuaryTm
I must say these problems require some awfully difficult mathematical theory.

What do you think we are? Actuaries?

???

I'm sixteen years old, and if I can do all of them, there's defintatly other people that can do them.

And these problems don't require anything involved with calculus +up.
 
Originally posted by: Vertimus
I'm sixteen years old, and if I can do all of them, there's defintatly other people that can do them.

And these problems don't require anything involved with calculus +up.
Please return your sarcasm meter to the place of purchase as it is obviously defective.
 
Originally posted by: ActuaryTm
Originally posted by: Vertimus
I'm sixteen years old, and if I can do all of them, there's defintatly other people that can do them.

And these problems don't require anything involved with calculus +up.
Please return your sarcasm meter to the place of purchase as it is obviously defective.

So what was the point of your original post?
 
If you're truly seeking a challenge, given that x, y, z, and n are integers, prove:
  • x^n + y^n = z^n
Has no nonzero solutions for n>2.
 
Originally posted by: episodic
What he did not tell you was that this was his homework assignment.

That's completely irrevelent, as I already stated in OP that I have already done these problems.

Why would I finish my homework, and then make ATOT to do it again?
 
3) The area of triangle ABC is 84, if you like to remember that kind of stuff. The distance from O to triangle ABC intersects at the circumcenter. Therefore, the circumradius R=13*14*15/(4*84) = 65/8. Therefore, the distance is just sqrt(20^2-(65/8)^2).
 
6) Let P(n) be the probability that the bug is on the starting vertex on the nth move. Therfore P(n)=1/2*(1-P(n-1)). That is, the probability that the bug is on the starting vertex on the nth move is equal to the probability that the bug is NOT on the starting vertex on the (n-1)st move multiplied by 1/2 (since the bug moves to the starting vertex with probability 1/2). A simple iteration follows.

P(0) = 1 (bug starts on the starting vertex, trivially)
P(1) = 1/2*(1-P(0)) = 0
P(2) = 1/2
P(3) = 1/4
P(4) = 3/8
P(5) = 5/16
P(6) = 11/32
P(7) = 21/64
P(8) = 43/128
P(9) = 85/256
P(10) = 171/512
 
8) Don't really want to complete this one, but the sine of the sum of the angles corresponding to the two roots have to be greater than or equal to sqrt(3)/2.
 
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