Math Problems and Teasers

[Cipher X]

This whole section is kinda getting boring so i thought hey why not have some math problems that are really freaky and somewhat hard.

1) If x > ½, find the simplest radical form expression for (1+sqrt(2x-1)) / (sqrt( x + sqrt (2x-1)))

IT might be easier to write it out on paper seeing that i couldnt put in the squareroot symbol oh btw sqrt = squareroot

Good Luck

[Cipher X]

The Length of each side of a certain right triangle is teh reciprocal of a different integer. What is the least possible sum of these three integers?

This one has got me stumped so Good Luck!

[Cipher X]

Together Al, Barb, Cal, Di, and Ed earned a total of $150, but in unequal amounts. In order to equalize their earnings exactly, Barb gave half (1/2) her earnings to Al. Next Cal gave 1/3 of his earnings to Barb. Then Di gave 1/4 of her earnings to Cal. Finally Ed gave 1/6 of his earnings to Di. How many dollars did Al earn before equalization?

This one just a little bit of thinking and you shoudl be able to get it

Have Fun

Cipher X

[rbayer]

Originally posted by Cipher X
The Length of each side of a certain right triangle is teh reciprocal of a different integer. What is the least possible sum of these three integers?

This one has got me stumped so Good Luck!

15, 20, and 12 will give you 47(). I'm not sure how you would do this with normal math stuff, but it's a great computational problem. If anyone has some kind of cool discrete math trick for stuff like this, I'd definately be interested in seeing it.


The Al, Barb, Cal, Di, Ed one isn't too hard:
Al had $11.


The first one is a little too much thinking for me at this time of night, and I assume the point would be that symbolic calculators are strictly forbidden. Maybe tomorrow during math I'll crunch through it...

[Richard Wallace]

sqrt(2)

[Gadget470]

Ok, first off. I'm stupid. It's late and I've done too much math and physics today. I was about to refute "Richard" .. as I typed it I realised he was right and that sqrt() IS radical. [sigh].. I've been sitting here working out a RATIONAL.. [/sigh]

I'm not going to bother reasoning out the other two tonight. I think I'm gonna sleep tonight.

w00t for sl33p. b00 for th1n|<

[Caleb Fulton]

I'm tired/lazy, so I wrote a program on my calculator and I'll check it in the morning

[Cipher X]

Originally posted by Richard
sqrt(2)

Yeah the answer to that one is Sqrt(2). the only reason i know is because i put into my Ti-83 and had an epiphany that all the answers were sqrt(2). But if anybody knows how to solve it using algebra or some wierd rule in math that nobody knows about. I wanna know how to solve it !! lol



And as far as i can see the Right triangle problem. the only equation that i can come up with that might be worthwhile is (1/a)^2 + (1/b)^2 = (1/c)^2 where a,b and c are integers. Since it says that the numbers have to be reciprocal of an integer

Cipher X

[Richard Wallace]

square the original expression

this gives 1 + 2*sqrt(2x-1) + 2x - 1 in the numerator

and x + sqrt(2x-1) in the denominator

(note the ones cancel in the numerator)

so the square is 2

so the original expression is sqrt(2)

qed

[Cipher X]

Hey thanks richard,
i can finally rest. omg i feel like an idiot i squared it but i didnt see that teh ones cancel and yeah......
but anyways here is another one

What are all the ordered pairs of real numbers (x,y) for which

y^(x^2 - 7x + 12) = 1 and x + y = 6


pretty easy

[Yan Wang]

Take a weird shaped field 100'x26'. At one end is a goal 10' wide centered in the field's width (8' on each side). If you were only allowed to shoot at the goal from the sideline, you would want an optimal angle... hence, what would be that distance down the sideline and what would be the corresponding pheta degree value to 2 decimal places?

It's a challenge to do this without the first derivative of arctan but with a geometric method.

Note that the field is not to scale.... and that a geometric answer does not mean going into CAD and constraining a field of the following dimensions.

[AJ Quick]

Hey! Why not do my math homework!?!

I know I could do it, but it would really cut into my sitting around time.

[Richard Wallace]

Monsieurcoffee, thanks for a very engaging problem.

I changed ISPs this weekend, so my connection was down until now. This gave me some time to think of several solutions.

The obvious one is to express theta in terms of x using the transcendental arctan function, differentiate and solve for the maximum theta.

A little less obvious is to use the pythagorean theorem and the law of sines to express sin(theta) as an algebraic function of x, differentiate and solve for the maximum sin(theta).

More elegant is the following:

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.

Now express alpha and beta as:

alpha = artan(x/18), beta = arctan(8/x)

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

The solution is x = 12.

The corresponding value of theta is 22.62 degrees.

[Richard Wallace]

OK, it is my turn to pose a problem --

A regular polygon having N sides is circumscribed on a circle having unit radius. [Note: all sides of a regular polygon are equal, and all of its angles are equal. 'Circumscribed' means that the midpoints of each of the polygon's sides are tangent to the circle.]

As a function of N, find an expression for the fraction of the polygon's area that lies outside the circle. In other words, what fraction of the polygon would have to be removed to leave the circle?

Added challenge (ala Monsieurcoffee's previous problem): show a derivation of your expression that does not make use of transcendental functions such as sine, cosine, tangent, etc.

[Yan Wang]

Originally posted by Richard
More elegant is the following:

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.

Now express alpha and beta as:

alpha = artan(x/18), beta = arctan(8/x)

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

The solution is x = 12.

The corresponding value of theta is 22.62 degrees.

That is one solution... however, if you want elegant, check this out:

Draw a circle radius 13 within the rectangular field.
Translate it to the right so that the goalposts lie on the circle.
The goal = 10, the section below = 8
The missing part = x

From a theorum about secants/tangents, the part*(part+whole)=other part*(other part + whole)

so... 8(8+18)=x*x or 8(18)=x^2 which simplifies to 12.