Wikipedia:Reference desk archive/Mathematics/2006 August 26

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[edit] Integral lengths

Stimulated by a recent newspaper puzzle, I got to thinking - can a point on the circumcircle of an equilateral triangle be an integral distance from all 3 vertices? Obviously it can if the point is at, or directly opposite, a vertex, but I'm not sure otherwise.

For triangle ABC with centre O, circumdiameter d and P a point on the circle st 0 < angle AOP < 60deg (i.e. A is the nearest vertex), the distances from P to the vertices are d*sin(alpha), d*sin(60-alpha) and d*sin(60+alpha), where alpha is half angle AOP.

By suitable choice of d and alpha, can all 3 be integral?

Semiable 10:22, 26 August 2006 (UTC)

I think you got the expressions for the distances wrong. In any case, the easiest solution is to set α = 0 (If I understood you correctly that makes P coincide with A). Then, from the perspective of P, A is on zero distance no matter what d we choose. Both B and C are on the same distance and d can be chosen so that this distance is integral. —Bromskloss 14:11, 26 August 2006 (UTC)

I see nothing wrong with my expressions, and anyway I considerd this trivial case, and another one, before defining my diagram above.

Semiable 18:46, 26 August 2006 (UTC)

OK, I probably misunderstood your question, then. —Bromskloss 15:34, 27 August 2006 (UTC)
Absolutely. For distances to the vertices a, b, c resp., set
\begin{matrix} \alpha & = & \arctan{\frac{\sqrt{3}a}{b+c}} \\ d & = & a \csc \alpha & = & \sqrt{a^2 + \frac{(b+c)^2}{3}} \end{matrix}
EdC 14:31, 26 August 2006 (UTC)
The trick is to expand the angle-sum formulae, giving dsinα and \frac{\sqrt{3}}{2}d\cos\alpha \pm \frac{1}{2}d\sin\alpha. The first distance can be made integral by setting d as above, while a suitable α will set cosα and sinα in a ratio commensurate to \sqrt{3}, acheived by making α the arctangent of that ratio. EdC 14:38, 26 August 2006 (UTC)

Thanks, EdC. I note too that c = a + b, i.e. for any point on the circumcircle of an equilateral triangle, the distance to the farthest vertex is the sum of the distances to the other two. This may be a standard result, but I wasn't aware of it.

Semiable 18:46, 26 August 2006 (UTC)

Indeed it is. Possibly the easiest way to see this is by Ptolemy's theorem; since APBC is a cyclic quadrilateral, letting the triangle have side x, we have (product of diagonals) cx = ax + bx (sum of products of opposite sides). EdC 02:46, 28 August 2006 (UTC)
This also allows us to derive d in an elegant manner. Letting K be the point where AB intersects CP, note that by angles subtending chords, ∠ACP = ∠ABP and ∠CPA = ∠CBA = ∠CAB = ∠CPB, so △KCA ∼ △ACP ∼ △KBP. So AC/CK = PC/AC, giving x² = c·CK; also CP/PA = BP/PK, giving c·KP = ab; since CK + KP = c, x² = c² - ab; since d² = 4/3x², d = \sqrt{\frac{4}{3}(c^2 - ab)} = \sqrt{a^2 + \frac{(b+c)^2}{3}}. EdC 04:21, 28 August 2006 (UTC)

[edit] Decimals to Fractions.

Okay... i find this a bit embarrassing to ask especially depending on my age... but I recently found out i do not know how to convert a decimal to a fraction without using a calculator or without having prior knowledge (i memorized .33333333 = 1/3 .666667 = 2/3 .16667 = 1/6 etc.) Thanks for any input --Agester 21:45, 26 August 2006 (UTC)

If you know that the decimal terminates, like 0.625, then it is equal to 625/1000 and then cancel it down. If you know that it recurs, like 0.212121..., then you can do this: x=0.212121..., so 100x=21.212121..., and then 99x=21, so x=21/99. If it neither recurs nor terminates, then it is an irrational number, so there is no fraction. Madmath789 21:58, 26 August 2006 (UTC)

So .234234... would be 1000x=234.234234.... 999x=234 x=234/999 ?? (i think thats how it was explained somewhere else) but i would like to confirm it. In addition how about .16666... (1/6) 100x= 16.66666... 99x=16 x=16/99???? --Agester 22:12, 26 August 2006 (UTC)

Yep - quite right about .234234..., but as for .1666..., you could do something like this: x=.1666..., so 10x=1.666..., and 100x=16.666..., then look at the difference between the last 2 equations: 90x=15, so x=15/90=1/6. Madmath789 22:18, 26 August 2006 (UTC)
By the way, there's a pretty thorough account at Recurring decimal. Melchoir 23:17, 26 August 2006 (UTC)

Madmath you are a genius!!!!!!!!! thanks a lot for your help! you saved me a lot of time! i spent all day googling this stuff and i got no where. Thank you very very much! That article on Recurring decimal is a great reference article too! Thank you all again! --Agester 02:05, 27 August 2006 (UTC)

If there is no such cycle (i.e. the number is irrational), or if it's inconveiently long, you can use the method of continued fractions to find a sequence of rational approximations as accurate as you like. —Tamfang 09:59, 27 August 2006 (UTC)
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