You know you're a math nerd when...

[Guest]

When you spend your time in Geometry class reading a Precalc book.

When you go to a math team meet rather than your friends birthday party,

When you have all your volume settings on exponents of 2. (2,4,8,18,32,64)

When I wish I could get the phone number 2416256 (2,4,16,256 or 2, 2^₂,2^₂2 ,2^₂2²)

When you teach your friends how to program the graphing calculator.

When you draw probability curves for your test grades--during the test.

-when you get aquestion wrong and everybody in your class is shocked

I know right? (Actually, this normally arises because I don't listen to a single word the teacher says)

Edited August 30, 2011 by phil

[MissKitten]

@phil Exactly!

When you read somewhere that nerds are gonna take over the world...and totally agree. Mainly because you know that if that ever happens, you'll be on the winning side [a.k.a. the Nerds].

[plasmid]

You might be a math nerd when in geometry class you hear about laws of congruency for triangles with the Side-Side-Side (SSS) rule, the Side-Angle-Side (SAS) rule, and the Angle-Side-Angle (ASA) rules, and are very disappointed that there is not an Angle-Side-Side rule.

You're definitely a math nerd when this fact disappoints you so much that you come up with of the Angle-Side-Side rule that stumps the rest of your class.

You might be a math nerd if you try to solve math/logic puzzles on BrainDen.

You're most probably a math nerd if you post your own math/logic puzzles on BrainDen.

You're definitely a math nerd if you post that no one has tried to answer for nearly a week now.

[MissKitten]

When you actually read plasmid's proof, and try to find the flaw [and then fail ].

When in math, you are more concerned with trying to steal your classmate's stuffed cow than you are with actually reviewing for your big test next week.

[Guest]

When you ask your physics teacher if the graders will still count it if you use calculus on the trig-based AP test.

When by the end of the first month of school you're considering taking both the trig and calc-based tests.

When you spend your free time proving that the 3 kinematics equations are all related using calculus.

(true stories about a friend of mine, not me)

[Aaryan]

When in math, you are more concerned with trying to steal your classmate's stuffed cow than you are with actually reviewing for your big test next week.

This can be taken two ways.

[MissKitten]

I meant the math-nerd-related way. As in, I already know the stuff so well that I don't need to worry about doing the optional review, so I busy myself with stealing his stuffed cow so I don't bore myself to death.

[Guest]

You might be a math nerd when in geometry class you hear about laws of congruency for triangles with the Side-Side-Side (SSS) rule, the Side-Angle-Side (SAS) rule, and the Angle-Side-Angle (ASA) rules, and are very disappointed that there is not an Angle-Side-Side rule.

You're definitely a math nerd when this fact disappoints you so much that you come up with of the Angle-Side-Side rule that stumps the rest of your class.

You might be a math nerd if you try to solve math/logic puzzles on BrainDen.

You're most probably a math nerd if you post your own math/logic puzzles on BrainDen.

You're definitely a math nerd if you post that no one has tried to answer for nearly a week now.

I don't have anything against you, but...

1. Your proof is flawed. The example triangle you have looks like a right triangle. That would mean you are only proving the Hypotenuse-Leg Theorem (didn't they teach you that?). There is a definite counterexample for the SSA theorem right here.

2. The riddle was not well explained. There are too many possible variables for a trapeze act.

Edited September 6, 2011 by phil

[plasmid]

Yep, my proof is flawed all right! The reason it's interesting as a riddle is that the flaw is rather subtle and can easily be missed. It's kind of like a "proof" that I once saw that I can now only vaguely remember, but went something along the lines of ...

4 = (2 + 2)

define a = b = 2, so

4 = (a + b)

4 = (a + b) x 1

4 = (a + b) x (1 / 1)

4 = (a + b) x [ (c - d) / (c - d) ]

note that this would hold regardless of the value of c and d, but to make things simple I'll also make c = d = 2

4 = (a + b) x (c - d) / (c - d)

4 = [ a(c - d) + b(c - d) ] / (c - d)

4 = [ a(c - d) + b(c - d) + 0 ] / (c - d)

4 = [ a(c - d) + b(c - d) + (c - c)] / (c - d)

since c = d we can substitute the last c in the numerator with a d

4 = [ a(c - d) + b(c - d) + (c - d)] / (c - d)

4 = [ (a + b + 1) x (c - d) ] / (c - d)

4 = (a + b + 1) x [ (c - d) / (c - d) ]

4 = (a + b + 1) x [ 1 / 1 ]

4 = a + b + 1

4 = 2 + 2 + 1

4 = 5

That was, if nothing else, rather entertaining the first time I saw it.

As for the trapeze, well, I had enough information to come up with a solution . Admittedly, there's a fair deal of slogging through equations that's easy to get bogged down in, and that can make you lose sight of the bigger picture and concepts behind how to attack it in such a way that you can reach the solution. That probably made it seem like less of a "riddle" and more of a "problem" for anyone who might have tried to take it on. I'll have to remember to avoid that if possible in future riddles.

To simplify the deduction, I'll set the height of the platforms and the trapeze fulcrums to be equal to the length of the trapeze ropes. I'll call this height "H". (It should be fairly obvious you can move the entire setup up or down en bloc without affecting the answer.)

I'll start by solving for the angle at which the acrobat should release from the first trapeze (denoted "A" and measured as the angle away from vertical). The acrobat would release at height H*(1-cos A), at a horizontal distance H*sin A from the axis of the trapeze.

The acrobat starts his swing from the platform at height H with potential energy U(initial) = mass * HG, where "G" is gravitational acceleration (9.8 m/s^2). After swinging on the trapeze and releasing at height H(1-cos A), the acrobat will have potential energy U(release) = mass * [H(1-cos A)]G, and therefore will have a kinetic energy of E = U(initial) - U(release) = mass * (H-[H(1-cos A)])G = mass * GH cos A. Since we now know the acrobat's kinetic energy, we can use it to solve for his velocity with the euquation E = mass * velocity^2 / 2, so

velocity = sqrt(2E / mass)

= sqrt(2[mass * GH cos A] / mass)

= sqrt(2GH cos A)

Because of the symmetry of the problem, it should be obvious that wherever the acrobat reaches the peak of his jump between the two trapezes will be half of the total distance between the platforms.

First, we can split the acrobat's initial velocity (at release from the trapeze) into x-velocity and y-velocity based on his release angle: vx(0) = velocity cos A = sqrt(2GH cos A) cos A; vy(0) = velocity sin A = sqrt(2GH cos A) sin A.

After time (t), the y-velocity would be vy(t) = vy(0) - Gt = sqrt(2GH cos A) sin A - Gt.

At the peak of the jump, the y-velocity would be zero, so we can solve for the time after release at which the jump peaks (T) by solving for vy(T) = 0

sqrt(2GH cos A) sin A - GT = 0

T = sqrt(2GH cos A) (sin A) / G

The x position (measured from the axis of the first trapeze) at time T is equal to the initial x position at release plus the x velocity multiplied by time T.

x(T) = x(0) + vx(0) * T

= H sin A + sqrt(2GH cos A) (cos A) * sqrt(2GH cos A) (sin A) / G

= H sin A + 2GH (cos A) (cos A) (sin A) / G

= H sin A + 2H (cos A)^2 (sin A)

= H (sin A) [1 + 2 (cos A)^2]

To find the maximal x position at the peak of the jump, find out at what angle A the derivative of this expression is zero.

d(x(peak)) / dA = d/dA ( H (sin A) [1 + 2 (cos A)^2] )

0 = [1 + 2 (cos A)^2] d/dA (H sin A) + H (sin A) d/dA (1 + 2 (cos A)^2 )

0 = [1 + 2 (cos A)^2] (H cos A) + H (sin A) * 2 * d/dA ((cos A)^2)

0 = [1 + 2 (cos A)^2] (H cos A) + H (sin A) * 2 * (2 cos A * -sin A)

0 = [1 + 2 (cos A)^2] (H cos A) - 4H (sin A)^2 (cos A)

0 = [1 + 2 (cos A)^2] - 4 (sin A)^2

0 = 1 + 2 (cos A)^2 - 4 (sin A)^2

4 (sin A)^2 - 2 (cos A)^2 = 1

Making use of the fact that (sin X)^2 + (cos X)^2 = 1,

4 (sin A)^2 - 2 (1-(sin A)^2) = 1

6 (sin A)^2 - 2 = 1

(sin A)^2 = 1/2

sin A = sqrt (1/2) = sqrt(2) / 2, so A = 45 degrees (is anyone else surprised?)

Now going back and solving for x(T)

x(T) = H (sin A) [1 + 2 (cos A)^2] (as shown previously)

= H sqrt(1/2) [1+ 2 (1/2)]

= 2H sqrt(1/2)

Remember that x(T) is the distance from the axis of a trapeze to the peak of the jump, so the total distance between the two platforms is 2H + 2x(T) = 2H + 4H sqrt(1/2).

[Guest]

To simplify the deduction, I'll set the height of the platforms and the trapeze fulcrums to be equal to the length of the trapeze ropes. I'll call this height "H". (It should be fairly obvious you can move the entire setup up or down en bloc without affecting the answer.)

I'll start by solving for the angle at which the acrobat should release from the first trapeze (denoted "A" and measured as the angle away from vertical). The acrobat would release at height H*(1-cos A), at a horizontal distance H*sin A from the axis of the trapeze.

The acrobat starts his swing from the platform at height H with potential energy U(initial) = mass * HG, where "G" is gravitational acceleration (9.8 m/s^2). After swinging on the trapeze and releasing at height H(1-cos A), the acrobat will have potential energy U(release) = mass * [H(1-cos A)]G, and therefore will have a kinetic energy of E = U(initial) - U(release) = mass * (H-[H(1-cos A)])G = mass * GH cos A. Since we now know the acrobat's kinetic energy, we can use it to solve for his velocity with the euquation E = mass * velocity^2 / 2, so

velocity = sqrt(2E / mass)

= sqrt(2[mass * GH cos A] / mass)

= sqrt(2GH cos A)

Because of the symmetry of the problem, it should be obvious that wherever the acrobat reaches the peak of his jump between the two trapezes will be half of the total distance between the platforms.

First, we can split the acrobat's initial velocity (at release from the trapeze) into x-velocity and y-velocity based on his release angle: vx(0) = velocity cos A = sqrt(2GH cos A) cos A; vy(0) = velocity sin A = sqrt(2GH cos A) sin A.

After time (t), the y-velocity would be vy(t) = vy(0) - Gt = sqrt(2GH cos A) sin A - Gt.

At the peak of the jump, the y-velocity would be zero, so we can solve for the time after release at which the jump peaks (T) by solving for vy(T) = 0

sqrt(2GH cos A) sin A - GT = 0

T = sqrt(2GH cos A) (sin A) / G

The x position (measured from the axis of the first trapeze) at time T is equal to the initial x position at release plus the x velocity multiplied by time T.

x(T) = x(0) + vx(0) * T

= H sin A + sqrt(2GH cos A) (cos A) * sqrt(2GH cos A) (sin A) / G

= H sin A + 2GH (cos A) (cos A) (sin A) / G

= H sin A + 2H (cos A)^2 (sin A)

= H (sin A) [1 + 2 (cos A)^2]

To find the maximal x position at the peak of the jump, find out at what angle A the derivative of this expression is zero.

d(x(peak)) / dA = d/dA ( H (sin A) [1 + 2 (cos A)^2] )

0 = [1 + 2 (cos A)^2] d/dA (H sin A) + H (sin A) d/dA (1 + 2 (cos A)^2 )

0 = [1 + 2 (cos A)^2] (H cos A) + H (sin A) * 2 * d/dA ((cos A)^2)

0 = [1 + 2 (cos A)^2] (H cos A) + H (sin A) * 2 * (2 cos A * -sin A)

0 = [1 + 2 (cos A)^2] (H cos A) - 4H (sin A)^2 (cos A)

0 = [1 + 2 (cos A)^2] - 4 (sin A)^2

0 = 1 + 2 (cos A)^2 - 4 (sin A)^2

4 (sin A)^2 - 2 (cos A)^2 = 1

Making use of the fact that (sin X)^2 + (cos X)^2 = 1,

4 (sin A)^2 - 2 (1-(sin A)^2) = 1

6 (sin A)^2 - 2 = 1

(sin A)^2 = 1/2

sin A = sqrt (1/2) = sqrt(2) / 2, so A = 45 degrees (is anyone else surprised?)

Now going back and solving for x(T)

x(T) = H (sin A) [1 + 2 (cos A)^2] (as shown previously)

= H sqrt(1/2) [1+ 2 (1/2)]

= 2H sqrt(1/2)

Remember that x(T) is the distance from the axis of a trapeze to the peak of the jump, so the total distance between the two platforms is 2H + 2x(T) = 2H + 4H sqrt(1/2).

Yep, you're definitely a math nerd!

[MissKitten]

Wow.....I used to think that one of the kids I know was the biggest math nerd in the world, but congrats, Plasmid, the incredibly long proof has placed you in the highest position of honor in my mind! :congrats:

[peace*out]

when youve finished the worksheet by the time the teacher tells you to start

[MissKitten]

when youve finished the worksheet by the time the teacher tells you to start

Haha, I do that all the time!

When you sit in class, drawing and coloring pictures of characters from different mangas/animes, and still manage to learn what the teacher is teaching and correctly answer at least two or three of her questions.

When you do the above and still manage to make it look like you're actually paying attention.

[peace*out]

When you understand the nerdy math jokes on the posters in your friends room.

[MissKitten]

When you are usually the cause of the nerdy jokes.

When your MathCounts t-shirt usually has some kind of joke that you get, but nobody else does.

When you have your own brand of math-related humor that makes people look at you funny, and when those looks don't really affect you anymore.

[Molly Mae]

When you prove that women are evil. Observe.

The two things required to get a woman are time and money.

Woman = Time*Money

Time is money.

Time = Money

Hence

Women = Money²

Money is the root of evil.

Money = sqrt(Evil)

So Evil = Money²

Women = Money² = Evil

Women = Evil

Edited September 16, 2011 by Molly Mae

[Thalia]

When you prove that women are evil. Observe.

The two things required to get a woman are time and money.

Woman = Time*Money

Time is money.

Time = Money

Hence

Women = Money²

Money is the root of evil.

Money = sqrt(Evil)

So Evil = Money²

Women = Money² = Evil

Women = Evil

I believe that requires addition, not multiplication.

[Molly Mae]

I believe that requires addition, not multiplication.

Noooooooooooooooo!

[Guest]

I believe that requires addition, not multiplication.

Never had a woman, huh Thalia

But it may actually be time ^ money

[MissKitten]

Ha! I've seen that before, and I think I actually put it up as a choice for our MathCounts t-shirt, as a braind of math-related humor. Funny thing is, even though it's so absurdly simple, my little sis, who is also in MathCounts, doesn't get it. It makes me worry for her MathCounts career.

[OmegaScales]

How about proving that 42 is the answer to everything?

Can anyone do that?

[MissKitten]

Nope, no matter what the Hitchhiker's Guide to the Galaxy says, the answer to everything is not 42. It's 64. Either that or duct tape, we haven't been able to establish which one works the best yet.

[OmegaScales]

Actually, it's 42 64 or goriila tape. Gorilla tape is much stronger than duct tape. However, the universally accepted answer is "Chuck Norris". Here's proof Chuck is the answer to everything:

Everything=God

God=Religion

God+Religion=Power

Power=Fear

Fear=Pain+Suffering

Pain+Suffering=Chuck Norris

Everything=Chuck Norris

Might need a little tweeking, but you get the idea.

Now can you prove 64 is also the answer to everything?

[MissKitten]

Yes:

64=everything

And in return, can you prove that gorilla tape is the answer to everything?

[flamebirde]

When you wish that you can write out Graham's Number.