I came across the picture below through random means.

ellipse magic

What is being demonstrated? All I could think of is maybe the center of the triangle is moving back and forth between the focii of the ellipse, but even if that's true (which it may or may not be - it's purely conjecture on my part), there's clearly more going on here.

  • 1
    $\begingroup$ It seems that the side lengths of the triangle are fixed, and two of the vertices are each constrained to be on one of the two lines. The resulting locus of the third vertex is an ellipse. I can't think of a deeper explanation right now. $\endgroup$
    – JimmyK4542
    Jun 27, 2015 at 4:59
  • 1
    $\begingroup$ The origin of this appears to be a blog post about the "Van Schooten Ellipse", which has some discussion and links to other ellipse-drawing mechanisms. $\endgroup$
    – Blue
    Jun 27, 2015 at 6:14
  • $\begingroup$ I foresee a bunch of calculus/analytic geometry students doing battle with this some time in the future. $\endgroup$ Jun 27, 2015 at 12:07
  • $\begingroup$ The points moving along the diameters are the feet of perpendiculars to those diameters from a point moving along the circumference. $\endgroup$
    – Blue
    Jun 27, 2015 at 13:37

2 Answers 2


We need to prove the following: when two vertices of a fixed triangle slide along two arms of a fixed angle, the locus described by the third vertex is an ellipse.

In the figure all the letters excepting $t$ and $P(x,y)$ are data of the problem; the angle $t$ determines the position of the vertex $P(x,y)$ so we choose $t$ as a parameter to be eliminated in order to find out the searched locus.

enter image description here

We have $$OB=\frac {c\sin (\theta+t)}{\sin\theta}$$ then $$x= \frac {c\sin (\theta+t)}{\sin\theta} \,–\, a\cos(\beta+t)$$ $$y=a\sin(\beta+t)$$ i.e. $$x=A\cos t-B\sin t$$

$$y=C\cos t+D\sin t$$

Where $$A=c-a\cos\beta$$$$B=c\cot \theta +a\sin\beta$$ $$C=a\sin\beta$$ $$D=a\cos\beta$$

Solving for $\sin t$ and $\cos t$ and because of $$\sin^2 t+\cos^2 t=1$$ it follows

$\det^2\begin{pmatrix}x&-B\\y&D\\ \end{pmatrix}$+$\,\det^2\begin{pmatrix}A&x\\C&y\\ \end{pmatrix}=$$\,\det^2\begin{pmatrix}A&-B\\C&D\\ \end{pmatrix}$

Thus $$(Dx+By)^2+(Ay-Cx)^2=(AD+BC)^2$$i.e.$$(C^2+D^2)x^2+(A^2+B^2)y^2+2(BD-AC)xy=(AD+BC)^2$$ This is a quadratic form of two real variables in which coefficients of $x^2$ and $y^2$ are both positive and distinct. This shows that the locus is an ellipse.

  • $\begingroup$ Not necessarily distinct. If $\theta = \frac\pi 2$ (the two lines are perpendicular) and $\cos\beta=\frac c{2a}$ (the triangle is isosceles) then $A^2+B^2=C^2+D^2$ and the ellipse may become a circle (but the triangle would probably have to have a zero height for that). $\endgroup$
    – CiaPan
    Jun 28, 2015 at 23:29
  • $\begingroup$ @CiaPan:When $\theta = \frac\pi 2$ the triangle $\Delta ABC$ have no restriction to $\alpha=\beta$; you can choose it as you want (inside the reasonable of course and, if I am not wrong, this case of $\theta = \frac\pi 2$ was the only one studied by Van Schooten in the 17th century) $\endgroup$
    – Piquito
    Jun 29, 2015 at 0:00
  • $\begingroup$ Of course we're not limited to isosceles triangles. However, if you choose $\alpha\approx 0$ and $\beta \gg 0$, that is a $C$ vertex close to $B$, then you get an ellipse close to the horizontal axis (degenerated to a line segment if $C=B$). I know the resulting curve is a circle when $C$ is a middle of the $AB$ segment—and I suspect, it's only then. $\endgroup$
    – CiaPan
    Jun 29, 2015 at 0:14
  • $\begingroup$ @CiaPan: I wanted to answer this way but the resulting equation was difficult to simplify: $$x=\sqrt {A^2+B^2}(\sin\theta_1 cost-\cos\theta_1 sint)=\sin(\theta_1-t)$$ Similarly $$y=\sqrt{C^2+D^2}\sin(\theta_2+t)$$ i.e.$$\theta_1-t=arc\sin \frac{x}{\sqrt{A^2+B^2}}$$ $$\theta_2+t= arc\sin \frac{y}{\sqrt{C^2+D^2}}$$ then $$\theta_1+\theta_2=arc\sin \frac{x}{\sqrt{A^2+B^2}}+arc\sin \frac{y}{\sqrt{C^2+D^2}}$$ $\endgroup$
    – Piquito
    Jun 29, 2015 at 1:19

You are given a circle with two diameters that are not perpendicular to each other. Two points go back and forth along the diameters. A third point creates a triangle. As the two points move along the diameters, the third point of the triangle traces an ellipse.

  • $\begingroup$ Will that always be true? That is given any two non-perpendicular diameters and any third point, will it always create an ellipse? $\endgroup$
    – Ben Thul
    Jun 27, 2015 at 12:37
  • $\begingroup$ Sure. Taking perpendicular lines will give ellipse, too (possibly a special case of ellipse, namely a circle, if a triangle was isosceles). $\endgroup$
    – CiaPan
    Jun 28, 2015 at 23:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .