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if longest side of triangle is the longest diameter of oval, the rest point is located on the side of oval, you can vary it to change the triangle

and except the longest side, set one of side of triangle be variable t

how to the find another side in terms of t? if needed, you can add other variables

beside longest side of triangle is the longest diameter of oval (ellipse),

there is another case is that longest side of triangle is the length of between two focus of oval (ellipse),

if possible, use least number of variables to find another side, great one is another side's equation is in terms of only t. if impossible, then please show other variables in a diagram

Remark: this remark question is not the required one, if can not show in one variables, will it mean that the two sides that can have different equations if in terms of two or more variables? what are they?

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Oval? do you mean something like $-b \le x \le 0$ part of the elliptic curve $y^2 = x(x+b)(x-a)$ or something else? If yes, let $s$ be the length of the other side of the triangle, then $s^2 - (t^2 + b^2)$ is linear in $x$ and you need to solve $x$ as a root of a cubic equation. –  achille hui Aug 1 '13 at 7:24
    
yes, it is like elliptic curve in mysite.du.edu/~jcalvert/math/ellipse.htm, what is x and y ? if possible, could you show in a diagram? –  bolo Aug 1 '13 at 7:37
    
@bolo: The term "elliptic curve" means something else (see Wikipedia). –  Zev Chonoles Aug 1 '13 at 7:42
    
sorry, it should be Ellipse –  bolo Aug 1 '13 at 7:43
    
remove the tag "elliptic curve". It seems now you are asking 2 or 3 questions but I really don't understand what you are asking. BTW, what is $f(t)$, what is $t\$2$ ???? –  achille hui Aug 1 '13 at 8:42

1 Answer 1

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Ellipse

Let $a > b$ be the semi-axes of the ellipse and $c = \sqrt{a^2 - b^2}$ be half of the distance between the two foci. With a suitable choose of origin and coordinate axes, one can make the ellipse satisfies:` $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad\iff\quad - \frac{c^2}{a^2b^2} x^2 + \frac{1}{b^2}(x^2 + y^2) = 1\tag{*1} $$

Let $s$ be the length of the other side of your triangle, one have:

$$\begin{cases} t^2 = (x-a)^2 + y^2\\ s^2 = (x + a)^2 + y^2 \end{cases} \implies \begin{cases} s^2 - t^2 = 4ax\\ s^2 + t^2 = 2(x^2 + y^2 + a^2) \end{cases} $$

Substitute this back into $(*1)$, one obtain a quadratic equation in $s^2$:

$$-\frac{c^2}{a^2b^2}\left(\frac{s^2 - t^2}{4a}\right)^2 + \frac{1}{b^2}\left(\frac{s^2 + t^2}{2} - a^2\right) = 1$$

Solving this give us: $$s = \sqrt{\frac{4 a^2 \sqrt{c^2 t^2+b^4}+ c^2 t^2 +4a^4}{c^2}}$$

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ode := diff(u(t),t)-f(t)=0 ,diff(f(t),t)+t*f(t)=0,diff(f(t),t$2)+((s)^2)*f(t)=0; can be used to create distribution function, after i put a = 0.5, b = 0.3, this system can not be solved, it seems my discovery only suitable for right angled triangle, i guess need to dig deeper –  bolo Aug 1 '13 at 14:07

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