## Tuesday, February 5, 2013

"Proof" without words? There must be a dozen other ways...

No takers yet? My free offer will expire by Fri 2-8-13!

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Jacob Antony said...

For the given problem, center of circle has to be at the center of the chord. And it won't satisfy for any other points inside circle as the length of the chord won't decrease linearly with its distance from center.

Let R be the radius and the chord of 2x length is r units away from the center.
R^2 = r^2 + x^2 ( 1 )
Also given that the chord is R-x units away from the center
R-x = r ( 2 )
Substituting x = R-r in equation 1
R^2 = r^2 + (R-r)^2
= r^2 + R^2 + r^2 - 2rR
therefore,
2r^2 = 2rR
R = r or r = 0
x = 0 or x = R

The chord should pass through the center or it should of be of zero length !

Dave Marain said...

Jacob? Antony?
Nice job! I like how you first gave a conceptual overview then went into an Alg derivation. I'll choose the top 3 explanations by Fri, then I will send out the prize. Of course, yours may be the only submission! I am looking for originality as well as mathematical elegance...

Couple of questions...
Are you a teacher? Student? (same thing of course!)
Professor? Math enthusiast?

Also, did my visual proof make sense? Did it mesh with your way of thinking?

Dave Marain said...

Remember folks, I'm looking for the 3 most original and elegant solutions. Submissions will be accepted until 12 noon EST on Fri 2-8-13.

I will be going back into hibernation soon. I hope you're enjoying MathNotations again for now...

Mr H said...

Angle ACD = angle BCD = 45° as triangles ACD and triangle BCD are right-isosceles.
So angle ACB = 90°
Angles in a semi-circle are 90° therefore chord AB is a diameter.

Dave Marain said...

Thank you, Mr. H. A straightforward argument but a nice one. Would you expect your 'B' students to arrive at your solution? Your 'A' students? I'm still looking for other approaches, particularly those which rest on first principles.

You're definitely a candidate for the prize.