117 replies to this topic

[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.

[phil]

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 a "proof" 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 a puzzle 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 by phil, 06 September 2011 - 03:15 AM.

[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 ...

Spoiler for

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.

Spoiler for Since no one else has posted a solution, here's what I got

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).

[phil]

Spoiler for Since no one else has posted a solution, here's what I got

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.