Math Problems and Teasers

[Richard Wallace]

Monsieurcoffee, I like your big circle solution. Can you provide a reference to the 'theorem about secants and tangents' on which it is based?

Revising my previous post to remove all reference to transcendental functions yields a solution that is more purely geometric:

Let the near corner of the field be called O, the desired point on the sideline A, the near goalpost B, and the far goalpost C.

Let the angle ACB be called alpha, and the angle BAO be called beta. Note that (in degrees) the desired angle theta = 90 - alpha - beta.

Further note that as x increases, alpha increases and beta decreases.

Conclusion: theta is largest when alpha = beta.

So theta is largest when triangles ACO and BAO are similar.

So theta is largest when x/18 = 8/x.

The solution is x = 12.

[Yan Wang]

Um... looking around. It's in the Course III math textbook here at Ithaca High School in the circles section. "McDougal, Littell Integrated Mathematics Course 3". It's Canadian.

Rather than using that theorum, after drawing that circle, you could have drawn radii of 13 to each goal post, forming a triangle, 13,10,13... half of which is 13,5,y... y=12. You can justify that y=x because x is greatest at the point of tangency of the circle and also that x and y are parallel and some other things which make them equal. I like your proof very much because it justifies things clearly and easily.

[Yan Wang]

Um, decided to draw it up from the book:

[jon virgi]

wow, great problems, TOO BAD THEY ARE STRAIGHT FROM THE PA MATH LEAGUE. wow, at least give them credit.

[Yan Wang]

I don't know what the PA Math League is (though now I guess I do since the title's pretty self explanatory)...

Anyway, I do the Mandelbrot math competition (national thing) and the NY Math League (NYML). That problem I had was from extra credit problems in math...

Anyway, I'll find some more

[Kevin Watson]

This is one of my favorite enigmas:

Three guys walk into a hotel. They pay ten dollars apiece for their room, which costs thirty dollars a night. After the three guys get to their room, the manager realizes that he over charged them five dollars, so he sends the bellhop up with the extra five dollars to repay them. On his way, the bellhop decides that he will give a dollar to each of the men and keep two dollars as a tip because he can't figure out how to divide five dollars between the three guys.

Here's the real problem: Since each guy got one dollar back, each only paid nine dollars apiece for the room or twenty-seven dollars altogether. When you add the two dollars the bellhop kept, you get twenty-nine dollars. Where'd the last dollar go?

-Kevin

[srjjs]

It didn't go anywhere. The hotel has 25 dollars, the bellhop has 2, and each of the 3 friends has 1. Your phrasing is just a little off.

[Kevin Watson]

Originally posted by srjjs
It didn't go anywhere. The hotel has 25 dollars, the bellhop has 2, and each of the 3 friends has 1. Your phrasing is just a little off.

Not much wrong with the phrasing (grammer, yes ). It's designed to be deceptive. It's one of those puzzles that you either get immediately (wondering what the fuss is all about), or spend the next couple of hours wondering if you need non-Euclidean geometry to solve it.

This problem and many others can be found on a great web site devoted to "hardcore tech-interview style riddles":

http://www.ocf.berkeley.edu/~wwu/riddles/intro.shtml

-Kevin

[JJG13]

Here's my all time favorite math problem. It's really easy if you know the trick.

What is the sum of the first two numbers that are perfect squares, cubes, and fourth powers.

[Greg Ross]

Originally posted by JJG13
What is the sum of the first two numbers that are perfect squares, cubes, and fourth powers.

1?

[JJG13]

Originally posted by gwross
1?

That's the first perfect square, cube, and fourth, but whats the second?

[Richard Wallace]

4097 = 2^12 + 1

[JJG13]

Correct.

[Greg Ross]

Originally posted by JJG13
That's the first perfect square, cube, and fourth, but whats the second?

No! 1 is the second, and 0 is the first. So 1 is also the sum!

[tatsak42]

I didn't think 0 counted.

Here's a problem from a competition i just did:
abcd is a 4 digit number between 1000 and 9999. When you multiply a 3 digit number (efg) and a 2 digit number (hi) and a 1 digit number (j) you get abcd. a,b,c,d,e,f,g,h,i, and j use the numbers 0-9 once and only once. What are the two combinations that work?
abcd=efg*hi*j