http://s205.photobucket.com/albums/bb192/falzoon/L…
Equation to figure is : (x^2 + y^2)^2 = x^2 – y^2
Additionally, of the 3 remaining areas in the loop, another remarkable
property to find, is the ratio of the largest area to one of the smaller areas.
The lemniscate looks like an infinity sign, so I thought a few
relevant quotes may be of interest as you tackle this problem.
1. “It is good a philosopher should remind himself, now and then,
that he is a particle pontificating on infinity.”
– Ariel Durant (writer, 1898 – 1981)
2. “I cannot help it; – in spite of myself, infinity torments me.”
– Alfred de Musset (dramatist, poet, novelist, 1810 – 1857)
3. “Yet it is in this loneliness that the deepest activities begin. It is here that
you discover act without motion, labor that is profound repose, vision in
obscurity, and, beyond all desire, a fulfillment whose limits extend to infinity.”
– Thomas Merton (Trappist Monk, writer, poet, 1915 – 1968)
Update:
Scythian, you are as close as I was. Where’s everyone else? Sometimes I feel as if I
create these problems just for you. And it doesn’t seem to matter what the time of day
or night it is. You do sleep, don’t you? Anyway, good answer, being no doubt, an easy
one for you, but the results were surprising enough to me, that I just had to share.
Update 2:
Superbly done, mathsmanretired. You certainly made it easier by changing to a
polar equation. Scythian, I thought you were being facetious when you said you
were guessing, but because of your edit, I now don’t feel so bad at not giving you
BA. You are correct in your gracious deference.
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✅ Answers
? Favorite Answer
From the shape of the curve, I think it reasonable to assume that the triangle of maximum area will have one corner at the origin and the others on the same x coordinate each side of the x axis.
If this is correct and corners are (0, 0), (x, y), (x, -y) then area of triangle = xy.
Atr^2 = (x^2)(y^2)
The polar equation for the curve is r^2 = cos(2T) with the right loop from pi/4 to -pi/4.
Area loop = INT [-pi/4, pi/4] (1/2)r^2 dT = (1/4)*sin(2T) [-pi/4, pi/4] = 1/2
Atr^2 =(x^2)(y^2) = (r^2*(cosT)^2)(r^2*(sinT)^2) = (1/4)(r^4)*(sin2T)^2
= (1/4)(cos2T)^2*(sin2T)^2 = (1/16)(sin4T)^2
2Atr*dA/dt = (1/2)cos4T
For maximum area dA/dt = 0 —-> 4T = pi/2 —-> T = pi/8
Atr^2 = (1/16)*sin(pi/2)^2 —-> Area of triangle = 1/4
Let B be the larger area inside the loop but outside the triangle, i.e. to the right of the triangle in the right hand loop.
Area B = area of “sector” – area triangle
= INT [-pi/8, pi/8] (1/2)r^2 dT – 1/4
= (1/4)*sin4T [-pi/8, pi/8] – 1/4
= (1/4)*[sin(pi/2) – sin(-pi/2) – 1]
= (1/4)*[sqrt(2) – 1]
Let C be each smaller area inside the loop but outside the triangle, i.e. above and below the triangle.
2C = Area of loop – Area of triangle – Area B
2C = 1/2 – 1/4 – (1/4)(sqrt(2) – 1)
2C = 1/2 – (1/4)*sqrt(2)
C = 1/4 – (1/8)sqrt(2)
Area B/Area C = [(1/4)(sqrt(2) – 1)] / [1/4 – (1/8)*sqrt(2)] = sqrt(2)
by rationalising denominator.
It is questions like this that make it worthwhile trawling through questions such as “what’s wrong with this proof that 2 = 1” or “can you solve this linear equation”.
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Well, immediately there’s {3, 5, 16}. Any others are going to be hard to find, so I doubt if it’ll be in OEIS if there isn’t an infinite number of them. I’ll keep looking. Edit: Rita has found others as I have, 5, 7, 10, 12, 15, 16, but seems to be missing the most obvious one, 3. What I’m not sure about, is this sequence infinite? Just because the variables can be put into an elliptic equation doesn’t seem to mean that there’s an infinity of solutions, because of the requirement that both the area and one side be integers and the same. Edit 2: Dragan K, but an infinity of DIFFERENT integers? Oftentimes I come up with something that while the generated sequence is infinite, that does’t mean that there’s that many distinct solutions. But I will have a look into this. Edit 3: If that’s so, that’s interesting. Since Rita has a head start on this, I’ll follow the conversation here. Unless I get some brilliant insight into this, but I doubt it. Edit 4: I found some more values, matching a few of what Rita has, and including 72 and 135, seemingly with no end in sight. However, I have no idea of how to prove that there’s an infinity of them, but certainly if some of the solutions are in parametric form, there is. I am as curious about this as Rita is too. Edit 5: This problem is equivalent to finding all integer sums of : (a² – b²) / ab + (c² – d²) / cd where a, b, c, d are integers, both positive and negative. I agree now that the sequence is infinitely long, you’ve proved it with your parametrization. OEIS doens’t have this series, but I think it should. Maybe you can pitch it to OEIS? I think it’s a mathematically interesting and distinctive series. Edit 6: Oh, that’s interesting, Rita is saying the same thing. Finding all integer sums of: (S – 1/S) + (T – 1/T) where S and T are rational numbers. But proof isn’t hard. Consider all triangles with an altitude of 1 with rational hypotenuses. Then ask what’s the base? The base has to be the sum of 2 more rational numbers, it cannot be of the form p + r, q – r, where p, q are rational but r is not, no such solution exists for which this is true. Note that I said “altitude of 1”, when 2 might be expected, but don’t worry, work it out. Also, note that (a² – b²) / 2ab = 2mn / (m² – n²) using a simple linear substitution, so we don’t have to consider two separate cases, it’s all the same series. Edit 7: Does anyone have an impossibility proof for c = 4? I think I have one, but it’s very long and very shaky. I just wonder that the conditions for such impossibilities might lead to the dual series, c = 1, 2, 4, 6, 8, 9, 11, 13, 14, 17, 20, 21, 22, etc. which might be easier. But I’m out on a limb here. Edit 8: This is the same as finding all the integer sums of the following, for integers a, b, c: (a²b² – c²) / abc + (a² – b²c²) / abc That is to say, S = ab / c, T = a / cb I don’t have the room to give the proof, nor I’m 1% sure of it. But it certainly makes for a faster search or confirmation of the numbers in the series. This is useful because then we know that both of the following have to be integers, their product being the sum of the above: (1 + b²) / ac (a² – c²) / b so that if the sum is p, a prime number, then one of them = p, and the other = 1. We can use this to rule out some particular sums, such as 4. Anyway, I’m out of space now, Y!A is going to start truncating me. Edit 9: Correction. Above analysis is a bit off, but it does lead to a superfast way of finding more terms. If for any given x, a²x² + 4a^4 + 8a² + 4 is a square for some a, then x is part of the sequence. It bugs me that it fails SOME of the time, i.e., can’t find a to make it a square. Edit 10: Eh, maybe this just needs an additional adjustment, hold on to see if I can improve on this. Edit 11: Okay, final edit, out of room now. If for some positive integers a, b, c, d, all relatively prime with each other, (a² – b²) / cd = p, an integer (c² + d²) / ab = q, an integer then pq is in the sequence, and conversely as well. This is one way to find out if a prime number is part of the sequence, for example, as in such cases, either p or q has to be 1. In general, S would be ac / bd while T would be ad / bc. As an example, S = 65*7 / 2*9 and T = 65*9 / 2*7 yields 67, which is a prime number, as 7² + 9² = 130 = 65*2, so that q here is 1, and p is 67. Edit 12: I’m going nuts with this. Let N = xyz be a candidate for the sequence, where x, y, z are any integer factors, including 1. Then if there exists a positive integer m such that the following is rational: √( (xy² – 4m²) / (xz²m² + 4) ) then N is part of the sequence. Note that given x, y, z, the search based on m is finite, so it can make a determination whether or not any number N is part of the sequence. Edit 13: This needs just a tiny bit more work.
The total area of the loop is 1/2, and the are of the maximum triangle is 1/4. The ratio of the larger remaining area to one of the smaller ones is √2.
I’m guessing all of this, of course. Am I close?
Edit: Let it be known that guessing doesn’t count. If others are able to show that the area of the loop is 1/2, and that the said ratio of remaining areas is √2, they obviously deserve BA. But I do like these kinds of neat “suprising results” myself.
Lagrange Multipliers.
Source(s): http://en.wikipedia.org/wiki/Lagrange_multipliers