Maths is broken Mk II

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zed0
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Maths is broken Mk II

Postby zed0 » Fri Oct 12, 2007 12:37 am UTC

Sorry about the diagram, I just did it quickly in GIMP.
Image
As shown in the image above:
ABC is defined as a right angle;
BCD is defines as 90 degrees minus x;
AB and CD are defined to be of equal length;
G is defined as the point where the two perpendicular bisectors (EG and FG) meet;
As ADG is isosceles AG and DG are clearly equal;
As BCG is isosceles BG and CG are clearly equal;
Angle GBC and angle BCG are clearly equal due to being on the inside of an isosceles;
(GBC=BCG)
Triangles ABG and CDG are congruent so angle ABG is equal to angle DCG;
(ABG=DCG)
ABG+GBC=DCG+BCG
90=90-x
???

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Mouffles
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Re: Maths is broken Mk II

Postby Mouffles » Fri Oct 12, 2007 1:16 am UTC

Spoiler:
D is actually on the other side of GC.
In the spirit of taking things too far - the 5x5x5x5x5 Rubik's Cube.

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Strilanc
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Re: Maths is broken Mk II

Postby Strilanc » Fri Oct 12, 2007 6:18 am UTC

Spoiler:
F should be to the right of EG (with x > 0 like in your picture, it's to the left with x < 0). I would bet the error stems from this, but don't really want to go through the motions.
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Token
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Re: Maths is broken Mk II

Postby Token » Fri Oct 12, 2007 8:48 am UTC

Alky wrote:
Spoiler:
F should be to the right of EG (with x > 0 like in your picture, it's to the left with x < 0). I would bet the error stems from this, but don't really want to go through the motions.

If D is left of C, why would F not be left of E?
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Seraph
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Re: Maths is broken Mk II

Postby Seraph » Fri Oct 12, 2007 10:33 am UTC

Spoiler:
Why the "???"?
You've shown that x = 0, which is a correct answer.

This leaves point G can be undefined, but with x = 0 any point that lies on both the perpendicular bisectors will satisfy the other parts of the problem.
Last edited by Seraph on Fri Oct 12, 2007 1:17 pm UTC, edited 1 time in total.

bittyx
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Re: Maths is broken Mk II

Postby bittyx » Fri Oct 12, 2007 12:00 pm UTC

Spoiler:
Yeah, D should be on the other side of GC, so, instead of
BCD = BCG + DCG
we have
BCD = BCG - DCG

Therefore, you cannot write
ABG+GBC=DCG+BCG
90=90-x
since DCG+BCG != 90-x

It's hard to see that the orientation will be different, but actually the main problem is that the picture is wrong - if ABCD is a rectangle, GBA and GCD will be differently oriented; if you tilt CD for an angle x, than limx->0+FG = +∞, and they will have the same orientation.

Yay!

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DrStalker
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Re: Maths is broken Mk II

Postby DrStalker » Fri Oct 12, 2007 12:31 pm UTC

Here's a to-scale version, for some value of X I didn't bother to calculate.

Spoiler:
Image


And with the triangles in red and moved aside for easy viewing:

Spoiler:
Image
There are two types of people in the world: 1) those that can extrapolate from incomplete data.


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