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Topic: Integral Square (Read 1110 times) |
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Sir Col
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Integral Square
« on: Aug 29th, 2007, 9:55am » |
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A right triangle has legs with integer lengths x and y. A square is placed inside the triangle with one vertex at the right angle and the opposite vertex on the hypotenuse. Determine the conditions for the square to have an integral side length.
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ThudnBlunder
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Re: Integral Square
« Reply #1 on: Aug 29th, 2007, 10:08am » |
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x+y divides xy
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ecoist
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Re: Integral Square
« Reply #2 on: Aug 31st, 2007, 3:24pm » |
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The existence alone of a square so located inside a right triangle restricts x and y.
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Sameer
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Re: Integral Square
« Reply #3 on: Aug 31st, 2007, 4:56pm » |
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I get If a is length of square, then a2 = (x-a)*(y-a) . I can't think of what to express it in words yet...
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"Obvious" is the most dangerous word in mathematics. --Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself. Sir Arthur Eddington, quoted in Bridges to Infinity
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ecoist
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Re: Integral Square
« Reply #4 on: Sep 1st, 2007, 6:27pm » |
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What integers satisfy ThundanBlunder's condition? If a and b are arbitrary positive integers, then x=a(a+b) and y=b(a+b) satisfies his condition, with a side of the inscribed square of length ab. But, alas, x=18 and y=9 also satisfies ThudanBlunder's condition, with square of side length 6, yet the above formula doesn't cover this case.
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Eigenray
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Re: Integral Square
« Reply #5 on: Sep 1st, 2007, 7:13pm » |
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An integer parametrization is x = ka(a+b), y = kb(a+b), where gcd(a,b)=1.
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Sameer
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Re: Integral Square
« Reply #6 on: Sep 1st, 2007, 7:21pm » |
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on Sep 1st, 2007, 6:27pm, ecoist wrote:yet the above formula doesn't cover this case. |
| I see. It's weird. I started with this: Square of length a x = a+b y = a+c So sqrt(a2 + c2) + sqrt(a2 + b2) = sqrt( (a+c)2 + (a+b)2) Simplifying which gives me a2 = bc I wonder why doesn't the case of x=18 and y = 9 get covered!!
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« Last Edit: Sep 1st, 2007, 7:22pm by Sameer » |
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"Obvious" is the most dangerous word in mathematics. --Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself. Sir Arthur Eddington, quoted in Bridges to Infinity
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Grimbal
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Re: Integral Square
« Reply #7 on: Sep 2nd, 2007, 2:53pm » |
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on Sep 1st, 2007, 7:21pm, Sameer wrote:I wonder why doesn't the case of x=18 and y = 9 get covered!! |
| Doesn't it? a = 6, b = 3, c = 12 a2 = b·c = 36 x = a+b = 9, y = a+c = 18
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« Last Edit: Sep 2nd, 2007, 3:15pm by Grimbal » |
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Sameer
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Re: Integral Square
« Reply #8 on: Sep 2nd, 2007, 3:00pm » |
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on Sep 2nd, 2007, 2:53pm, Grimbal wrote: Isn't it? a = 6, b = 3, c = 12 a2 = b·c = 36 x = a+b = 9, y = a+c = 18 |
| Duh!?! Thanks!! Seems like I have been overworking again!! However my original question still holds. How do you geometrically or in words express a2 = bc!!
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"Obvious" is the most dangerous word in mathematics. --Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself. Sir Arthur Eddington, quoted in Bridges to Infinity
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Grimbal
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Re: Integral Square
« Reply #9 on: Sep 2nd, 2007, 3:01pm » |
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Inserting the square creates 2 smaller triangles, one with sides b and a, the other with sides a and c. The fact that they are similar implies b/a = a/c, which directly gives a2 = bc
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« Last Edit: Sep 2nd, 2007, 3:02pm by Grimbal » |
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Sir Col
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Re: Integral Square
« Reply #10 on: Sep 2nd, 2007, 3:07pm » |
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And using similarity, (x-a)/a = a/(y-b), which gives a2 = (x-a)(y-a) = xy-a(x+y)+a2, leading to T&B's result: a = xy/(x+y).
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Grimbal
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Re: Integral Square
« Reply #11 on: Sep 2nd, 2007, 3:09pm » |
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The fact that the 2 smaller trianges are similar implies (x-a)/a = a/(y-a) <=> (x-a)(y-a) = a2 <=> xy-ax-ay-a2 = a2 <=> xy-a(x+y) = 0 <=> a = xy/(x+y) So, for a to be an integer, xy must be a multiple of (x+y), which conveniently removes the unknown a from the equation. And it proves what ThudanBlunder said straight away and Sir Col minutes ago.
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« Last Edit: Sep 2nd, 2007, 3:12pm by Grimbal » |
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Sameer
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Re: Integral Square
« Reply #12 on: Sep 2nd, 2007, 3:11pm » |
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Thanks for the great simple explanation!! I unnecessarily seem to make things complicated, don't I?
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"Obvious" is the most dangerous word in mathematics. --Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself. Sir Arthur Eddington, quoted in Bridges to Infinity
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ThudnBlunder
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Re: Integral Square
« Reply #14 on: Sep 3rd, 2007, 7:59pm » |
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on Sep 2nd, 2007, 3:12pm, Sir Col wrote: At least you proved it. I think that T&B's was just a lucky guess. |
| And the great thing is the more I study, the luckier I get.
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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