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Here's about what I think the curve looks like: enter image description here

In working through a theory of mine, I have come across a curve I cannot identify. Along with $x$ and $y$, this curve needs an additional input to complete the curve. Things I know about this curve:

The curve is identical when reflected around the line $y=x$.

The curve always contains the points $(0,10), (10,0), (5,a),$ and $(a,5)$.

The area under the curve from $0$ to $10$ is always $10a$.

At $a=5$, the curve matches perfectly with the line $10-x$.

I am nearly certain that the curve is hyperbolical in nature.

Knowing this, what is the equation of this curve?

PinkyWay
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FurryTheFury
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    Can you provide a picture? – Blue May 01 '21 at 20:13
  • At a=8 it does not look like a hyperbola. – Narasimham May 01 '21 at 20:52
  • In case it helps: if one makes the change of variables $u=\frac{x-y}2$ and $v=\frac{x+y}2$, then the resulting function $v=g_a(u)$ is an even function of $u$, and satisfies $g_a(\pm10)=10$, $g_a(\pm(a-5))=a+5$, $g_5(u)=10$ identically, and $\int_{-10}^{10} g_a(u),du = 20a+100$. – Greg Martin May 01 '21 at 21:35
  • If you are willing to consider the area requirement approximately, then the Mobius transform $f(x,a)=5a\frac{10-x}{5a-(a-5)x}$ meets the other criteria exactly. It has the correct area for $a\to 0$, $a \to 10^{-}$ and at $a=5$ – Sal May 01 '21 at 22:09
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    Something doesn't seem right here ... Naming the points $P=(0,10)$, $Q=(a,5)$, $R=(5,a)$, $S=(10,0)$, with $a\leq 5$, the area under the polygonal path $PQRS$ is $$10a + \frac12(25-a^2)$$ In fact, the area under segments $PQ$ and $RS$ alone accounts for the $10a$, which is the total area allotted. For small-ish $a$ (say, $a<2$), it's clear that a "shallow" curve joining $P$ to $Q$, and its mirror image from $R$ to $S$, won't carve-out enough area to reasonably fill the gap between $Q$ and $R$. Are you sure about your area condition? – Blue May 01 '21 at 23:03
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    @Blue Once more, we are on the same tracks... See my answer... – Jean Marie May 01 '21 at 23:08
  • As for the hyperbola conjecture ... The four given points, and an arbitrary fifth point on the line $y=x$, determine a conic, with hyperbolas appearing (for $a<5$) when that fifth point is at some threshold distance from the origin. But the area under such hyperbolas is generally way too large. The smallest area occurs with the degenerate "crossed lines" hyperbola (determined by lines through the four given points), but that area is $100a/(5+a)$, which is rarely equal to $10a$. – Blue May 01 '21 at 23:20
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    @Blue The question is tagged "functions" but the OP asks about "curves". I guess a clarification would be in order here. A "curve" could technically zig-zag any way it wants between the 4 points as to satisfy the area requirement. – dxiv May 01 '21 at 23:21
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    @dxiv: I was just going to comment to @ JeanMarie that maybe we shouldn't assume function-ness. :) That said, OP's drawings (and the hyperbola conjecture) suggest that zig-zags aren't the intention here. ... Without clarification from OP, I guess there's no point in speculating about how to make this work. – Blue May 01 '21 at 23:24
  • Hey there. Thanks for trying to solve this one. It looks like it's not possible in the way I was hoping. That explains why I couldn't solve it myself. I did actually mean a function here. – FurryTheFury May 01 '21 at 23:26

1 Answers1

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In fact, the constraints are such that you will not obtain a smooth curve in general. Let us consider, for the sake of simplicity the case $a=1$ for which the objective is to have a total area equal to $10$.

enter image description here

If we consider the figure below, you see that $90\%$ of the area is already taken by the rectangular shapes. It remains a tiny one-unit area to be "distributed" under the 3 arcs of the curve. It means that each arc must stick to the borders, generating uneasthetic spikes, rather far from the shape(s) you desire.

As $a$ increases, the spikes will be attenuated till $a=5$. A symmetrical issue will take place beyond $a=5$.

Jean Marie
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    +1. ... Why does it feel like I'm rewarding myself? ;) – Blue May 01 '21 at 23:31
  • @Sal You shouldn't have removed you answer! it was an astute answer with respect to the curves given by the OP. The extraneous condition about the area was bringing too much constraint, that's all. If you undelete it, let me know it. I will be happy to upvote it. – Jean Marie May 03 '21 at 05:42