The NSA sponsors a great math competition that ISN'T your basic sit down do math in x minutes type contest. You get a whole month to brood over 5 excellent problems, 4 times a year. I suggest you just go there are read all 345 posted problems. http://www.nsa.gov/programs/mepp/usamts.html

My own problem: What is the area of land that can be seen from a watch tower if it is 50 feet high? Assume a spherical Earth.
Answers that do not use calculus:yikes:deserve a pat on the back. Yes, the answers, not the person.

math
 

there is a cylinder that is a gas tank in a car that lays on its side (so if it wasn't held down it would roll back and forth). The meter is broken so you can't tell how much is in there but you do know if you stick a stick in through a hole in the top where it is usually filled and the tank is half way full the stick will be coated a quarter of the way.

How high is the stick coated when you only have one fourth of a tank of gas?

Here's one, what are the four magic cubes? Figure this out without using a computer program. Hint: If you don't know what a magic cube is, 153 is one of them because 1^3=1, 5^3=125, and 3^3=27. So 1+125+27=153. Well, there's one for you, find the other three!

Nice problem, but not too bad. The maximum distance a person can see from the watch tower is a line tangent to the earth, which means this line is PI/2 rads in relation to the center of the earth. This gives a right triangle with a hypotenuse of R+50 (R= earth's radius), and one side of length R .

To get the angle between these to lines take the arc-cosine of (50+R)/R = arccos(1 + 50/R)= theta.

Now, the length of an arc equals theta times radius, so the length of the arc visible by the person in the watch tower is R*arccos(1+50/R).

Of course, this isn't the area, as requested. to get that, we imagine this is a straight line and use the area of a circle = PI* r^2. Therefore the total area is PI*R^2*arccos(1 + 50/R) ^2!!

I'm feeling lazy right now. can someone please plug in the value of R and see what that comes out to? Thanks!

Ummm.... I don't think that works. First, I believe you actually mean theta=acos[R/(R+50)], since cos(theta)=a/h.

Second, I'm not sure where that area formula came from, but I don't believe that pretending like it's a straight line is valid without the use of some kind of fudge factor (my professor's words, not mine) like a Jacobian.

Anyways, using a bit of Calc (sorry, I don't know the formula for area of a partial sphere of the top of my head :D), I came up with:

A = 2*pi*r^2*(1-cos(phi)), where phi is the angle in spherical coordinates that defines where the line of sight and the sphere intersect.

Plugging in r=R and phi=acos[R/(R+50)], I get:

A=2*pi*R^2*[50/(R+50)]

I know you said a calc-free answer was best, but what really defines calc-free? After all, our good old friend A=4*pi*r^2 comes from calc, as does just about every other area/volume formula. If anyone does come up with a solution that doesn't somehow use any form of calc, they (and the solution!) definately deserve a pat on the back.

--Rob

Here's one our team has a lot of fun with.

A, B and C are digits. Find the solution to the following

A+B^C = ABC

:)

This works

1+1^2 = 2

I'm thinking A, B, C, are different digits, and ABC is a 3 digit number...

And you want 7+3^6=736

Thanks rob! you are right, it should be arcos[R/(R+50)]. Also, looks like the "straighten it out" part doesnt work either. sorry about that everyone!!! I have got to stop doing math at 3am........

tatsak42 you are correct! Well done.
How about the form (A+B)^C=ABC.

:)

for (x,y)
(2,4) and (3,3)

simply factor (x^2 - 7x + 12) = 1 so that x must = 2 or 3 to make the expression always 0 so what ever y = the whole thing becomes 1. Fill in 3 and 4 for y.

Right idea, but i think you may want to double-check your factoring. (x^2 - 7x + 12) = 0 will factor to (x-3)(x-4)=0, giving x=3 or x=4 as solutions. Also, don't forget that if y=1, then the whole thing is 1 regardless of the exponent. Therefore, the solutions I come up with are:

(3, 3), (4, 2), (5, 1)

--Rob

Not sure if this is a good problem or not...

1, 2, 6, 42... what is the nex number in the pattern?

1806

3263442

After that my calculator can't keep track of the digits.

Re: Math Problems and Teasers
 

(3+4)^3 = 343