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This is a "fun" question and I have already a solution. I asked this question so that I may see a different approach or an elegant solution.

Let $P$ be a plane with equation $x+y+z=1$. Find an ellipse on $x,y$ plane so that its "shadow" on $P$ is a circle.

Edit: By "shadow", I mean $(x,y)\to(x,y,z)$ where $(x,y,z)\in P$

Mark McClure
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mesel
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  • Does "shadow" mean it's orthogonal projection to $P$? Or any projection? – fgp May 21 '14 at 09:31
  • $(x,y)\to (x,y,z)$ where $(x,y,z)\in P$ – mesel May 21 '14 at 09:34
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    Have you tried to reverse engineer this? Take a circle on P, and let the ellipse be its "pre-shadow" on the $xy$-plane. For example the circle via the points $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$. That circle lies on the cylinder $$\Vert(x,y,z)-\frac{x+y+z}3(1,1,1)\Vert^2=\frac23.$$ – Jyrki Lahtonen May 21 '14 at 09:38
  • @mesel Yes, but what are the constraints on this transform $F$? You can, for example, find a rotation+translation that maps the $xy$-plane into $P$, and the problem is trivial then, since rotations and translations map circles to circles. But that's quite certainly not what you want. The minimal requirement you probably want to impose is probably that $F$ is idempotent, meaning that applying it twice is the same as applying it once. That makes $F$ a projection. But there are still many such projections (because you can pick the direction the light comes from)... – fgp May 21 '14 at 09:46
  • @mesel Ah, OK, I think I see what you want. You want a projection onto $P$ in the $z$-direction, i.e. You map $(x,y,0)$ to $(x,y,1-x-y)$, right? – fgp May 21 '14 at 09:49
  • @fgp:we set $F(x,y)=(x,y,z)$ and $(x,y,z)\in P$ which means $z=1-x-y$ so $F(x,y)=(x,y,1-x-y)$ which uniqly determine the $F$ ?. – mesel May 21 '14 at 09:50
  • @fgp: now, it is okey. – mesel May 21 '14 at 09:51

3 Answers3

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The circle with radius 1 and center $(0,0,1)$ in the plane $x+y+z=1$ can be parametrized by \begin{align} \vec{p}(t) &= \langle 0,0,1 \rangle + \cos(t)\langle 1,-1,0 \rangle/\sqrt{2} + \sin(t) \langle 1,1,-2 \rangle/\sqrt{6} \\ &= \left\langle \frac{\sin (t)}{\sqrt{6}}+\frac{\cos (t)}{\sqrt{2}},\frac{\sin (t)}{\sqrt{6}}-\frac{\cos (t)}{\sqrt{2}},1-\sqrt{\frac{2}{3}} \sin (t) \right\rangle. \end{align} It's shadow can be parametrized by just dropping the $z$ coordinate. $$\vec{s}(t) = \left\langle \frac{\sin (t)}{\sqrt{6}}+\frac{\cos (t)}{\sqrt{2}},\frac{\sin (t)}{\sqrt{6}}-\frac{\cos (t)}{\sqrt{2}} \right\rangle.$$

This is enough to visualize.

enter image description here

Once we have the parametrization, we can generate points on the ellipse by simply plugging in $t$ values. Here are five points on the ellipse in the plane, for example.

$$ \begin{array}{l|l} t & (x,y) \\ \hline 0 & \left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right) \\ \hline \frac{\pi }{3} & \left(\frac{1}{\sqrt{2}},0\right) \\ \hline \frac{\pi }{2} & \left(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}}\right) \ \\ \hline \pi & \left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right) \\ \hline \frac{3 \pi }{2} & \left(-\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}}\right) \end{array} $$

Since five points determine an ellipse, we can find the Cartesian formula easily enough. My favorite technique is to set the following determinant equal to zero.

$$ \left| \begin{array}{cccccc} x^2 & x y & y^2 & x & y & 1 \\ \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 1 \\ \frac{1}{2} & 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 1 \\ \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & 1 \\ \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 1 \\ \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & -\frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{6}} & 1 \\ \end{array} \right| $$

After some simplification, this yields $$ 2 x^2 + 2 xy + 2 y^2 = 1. $$

Mark McClure
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Without loss of generality, I'll use the plane $x + y + z = 0$ instead of $x + y + z = 1$.

Suppose the circle on the plane $x + y + z = 0$ is a unit circle centered at $(0, 0, 0)$. It intersects the $xy$-plane at $\frac{1}{\sqrt 2}(-1, 1, 0)$ and $\frac{1}{\sqrt 2}(1, -1, 0)$. It intersects the plane $x = y$ at $\frac 1{\sqrt 6}(1, 1, -2)$ and $\frac 1{\sqrt 6}(-1, -1, 2)$. These four points, projected onto the $xy$-plane, define the major and minor axes of the ellipse on the $xy$-plane. (In particular, the major and minor axes have length $2$ and $\frac 2{\sqrt{3}}$, and they make $\pi/4$ angle with the standard $xy$-axis.)

Tunococ
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Let $C$ be the unit circle on the plane $P$ centered at $R=(0,0,1)\in P$ and $(x_0,y_0)$ be projection of any point of $C$ to $x,y$ plane.

Then $T=(x_0,y_0,1-x_0-y_0)\in C$ and by constraction $d(R,T)=1.$ So, $$|(x_0,y_0,-x_0-y_0)|=1$$ $$x_0^2+y_0^2+x_0^2+2x_0y_0+y_0^2=1$$ $$2x_0^2+2y_0^2+2x_0y_0=1$$ Hence, The equation of ellipse is $2x^2+2y^2+2xy=1$

mesel
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