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Author Topic: Check my Answer  (Read 2745 times)
Roby
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« on: April 11, 2009, 10:45:48 AM »

My high school teacher was right.  There is a "real world" use to all of that geometry and algebra crap they force fed us, and I really did need to know it later in life. 

Here is a real life math problem that presented itself to me, and also why I have not been on the forum much lately.  I was trying to guess-timate the answer for a week, and after cutting multiple pieces for my next puzzlebox that just would not work, I finally decided to try the math approach. 

This really is all of the information that I had to go on, and it took me about 4 hours to figure out.  (I've been out of school for a while now!!)  See if you guys can confirm my answer - and perhaps beat my time.  I would offer a prize for the winner, but you guys already have all of my spare puzzles, so it would just be for bragging rights.

The entire illustration is symetrical and all triangles are "perfect - 90 degrees".   

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« Last Edit: April 11, 2009, 10:50:15 AM by Roby » Logged

rolly_wood
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« Reply #1 on: April 11, 2009, 10:55:42 AM »

What is S? does it mean that colour are equal?
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Roby
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« Reply #2 on: April 11, 2009, 10:57:49 AM »

"S" is just a variable for "Side" and shows that they are the same length.
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Canuck
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« Reply #3 on: April 11, 2009, 11:02:21 AM »

 Huh?
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http://www.puzzleparadise.ca/


"May you find hidden treasures in every pothole, real or imagined, and may your childhood never really end"  Stewart T. Coffin
Roby
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« Reply #4 on: April 11, 2009, 11:08:06 AM »

Come on John, you're not even trying.
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rolly_wood
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« Reply #5 on: April 11, 2009, 11:10:44 AM »

Are you sure there is ONE solution? it seems that S can be chosen freely within certain limits... and the same for X..
the measurement of the "diagonal" does not add info if S is free.... maybe I am wrong this is my first impression
the "checkerboard pattern" is also free? Must green triangles be equal?

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« Last Edit: April 11, 2009, 11:14:56 AM by rolly_wood » Logged

Roby
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« Reply #6 on: April 11, 2009, 11:16:07 AM »

No, I don't think there can be more than one answer - as the lenght of the entire project is set.
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Roby
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« Reply #7 on: April 11, 2009, 11:19:11 AM »

Green triangles are all identical, but I think that with a set length that would automatically happen.  You have a .525 section on the left and right sides followed by .25 sections inside of that, and that leaves a set distance for the triangles to intersect.  This in turn sets an exact distance for the intersection of the triangles on the width.

All triangles are the same size, but I suppose that could be proven because it is symetrical.
« Last Edit: April 11, 2009, 11:25:07 AM by Roby » Logged

Chinnomotto
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« Reply #8 on: April 11, 2009, 11:24:41 AM »

Master Master, pick me, pick me ( sorry I can't help myself  Grin )

Simple unitary method, I just measure off the screen

Right side if 66mm=4 inch, then x= 10mm = 10 x 4 divided by 66=.606 inch, give or take 1 inch

But then, what do I know, I pull teeth,,,,,,,,and crack eggs

No No, not the whip, again

Your turn Roly,,,,,
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I Have Such Sights To Show You

Chinny from  Sydney, Japan
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« Reply #9 on: April 11, 2009, 11:28:39 AM »

But Chinimotto I am surfing currently on my iPod touch so my measurements are much smaller Grin
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Roby
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« Reply #10 on: April 11, 2009, 11:29:41 AM »

Ahh, I tried your way Stephen for a week.  Just ended up with a bunch of useless puzzle parts.  You are within a half inch, but remeber that our projects require variances of less than 1/1000.  

Also, to keep you cheaters from cheating, the distances are not to scale on purpose.  The solution has to be done mathamatically based only on the number and the fact that it is symetrical.
« Last Edit: April 11, 2009, 11:30:37 AM by Roby » Logged

rolly_wood
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« Reply #11 on: April 11, 2009, 11:32:37 AM »

Roby if you do not specify anything about triangles, what would forbid a situation like this?

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Roby
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« Reply #12 on: April 11, 2009, 11:40:07 AM »

OK, I had to think about that for a bit.  Yeah, you are right.  I do need to specify that all green parts are equal.
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rolly_wood
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« Reply #13 on: April 11, 2009, 11:41:50 AM »

and...even assuming traingles as equal, what forbids a situation like this (being both X and S free)?

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Canuck
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« Reply #14 on: April 11, 2009, 11:42:53 AM »

Roby, stop messin' with my head! Angry Grin
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http://www.puzzleparadise.ca/


"May you find hidden treasures in every pothole, real or imagined, and may your childhood never really end"  Stewart T. Coffin
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