Workshop garrotte

[Alan Haisley]

...
One of the most famous equations in mathematics is:

e^(i*pi) = -1
...
Prove that this is true.

To darn much stuff online. It is fairly easy to get to Euler's Identity, from there to Euler's Formula, and from there to several proofs.
The first one is more interesting in that I need to actually work it out (somehow), rather than just look up a solution and so it's the one I'm thinking about.
I remember sitting in class while the professor put a proof of Euler's Identity on the board, but I certainly don't remember it. Still, in terms of mathematical elegance I would say that it certainly beats out anything else.
I have to thank Vi Hart for one of her recreational mathematics videos in the Kahn Academy for reminding me where to search. I try to review her exposition on pi and tau every pi day; have to admit though that I haven't celebrated tau day yet.
Alan

[Tennessee Whiskey]

I hear bongos in the desert. You really do respect the mans work, cheers my friend.

E

[Alan Haisley]

Marv,
This is what I came up with:
Hard as this was, it was also fun
Alan

[Maryak]

Sine me on Cos I'm infinitely variable even to the point of cycling backwards. Others just think I'm minus one. My Dad threatened to nail my other foot to the floor to stop me going round in circles.

Best Regards
Bob

[mklotz]

Marv,
This is what I came up with:
Hard as this was, it was also fun
Alan

Congratulations, Alan, you got the correct answer.  Your method was ingenious but way too complicated.

The secret to getting an answer simply is to realize that the angle (theta in the attached diagram) made with the string must be equal on both sides.  [If they were not there would be a component of gravity operating on the weight to pull it to the position where the angles are equal.]

Armed with that insight, one can compute the angle and from there on it's just some simple trig grinding to get the answer.

[Alan Haisley]

I suspected that I had found one of the hardest ways to solve it.    But, not only did my earlier comments pretty much force me to post some kind of answer, I was sure that until I did you wouldn't post yours 
Alan

[mklotz]

I suspected that I had found one of the hardest ways to solve it.    But, not only did my earlier comments pretty much force me to post some kind of answer, I was sure that until I did you wouldn't post yours 
Alan

Hey, Alan, be proud of yourself.  Not only did you solve it and produce a verification of my equal angle assertion in the process, but you were the only one to attempt a solution.  That fact flatters not only your mathematical abilities but your character as well.

[Tennessee Whiskey]

You guys make me realize that the three years I spent in the 10th grade were all wasted. Well, maybe not, the cheerleaders were fun.

Eric

[Lew Hartswick]

For the string problem, I got 3 equations in 3 unknowns but havent gotten around to solving them yet.
   ...lew...

[Mosey]

You guys make me realize that the three years I spent in the 10th grade were all wasted. Well, maybe not, the cheerleaders were fun.

Eric

Weren't all of your years in school wasted?
Your Yankee pal,
Mosey

[mklotz]

Want to really wind your math neurons tight as a harp string?  If so, try this problem but, I warn you, this is not for the faint of heart.

A 6 unit cubical box is pushed tight against a vertical wall.  A 25 unit long ladder is placed against the wall such that it just touches the edge of the box.  What is the height of the point at which the ladder touches the wall?

Remember, the numerical answer is not the object here.  Any fool with a CAD program can find that.  What is required is the mathematical reasoning that leads to the answer.  The numerical answer is then just a quick way to check the validity of the reasoning.

[Mosey]

Please clarify what is meant by 6 unit cubical box. 6 units per edge, side?
Mosey

[mklotz]

Please clarify what is meant by 6 unit cubical box. 6 units per edge, side?
Mosey

Each edge is 6 units long.

[Tennessee Whiskey]

Mosey, none of them were. It's where I learned about sex, chemistry, and Rock-n-Roll. Hey, it was the 70's. At my senior prom I had the band announce that I would be buying drinks for all faculty members in the lounge. My bar bill was over $200 bucks, and that was in '77. I took an English teacher from another high school to that one. My, how times have changed. She would be on the news at 6:00 for that now.

Whiskey

[Captain Jerry]

Mosey, none of them were. It's where I learned about sex, chemistry, and Rock-n-Roll. Hey, it was the 70's. At my senior prom I had the band announce that I would be buying drinks for all faculty members in the lounge. My bar bill was over $200 bucks, and that was in '77. I took an English teacher from another high school to that one. My, how times have changed. She would be on the news at 6:00 for that now.

Whiskey

I don't see why she would be in trouble.  You were 21 then, weren't you?

Jerry