Was thinking about hyperbolic geometry, the Poincare Disk Model and Sweikarts constant and combined them all in a construction puzzle that I was unable to solve.

My construction puzzle:


  • A circle $Circle_0$ with centre $Centre_0$ and radius $r$
  • On $Circle_0$ we have 2 points $I_1$ and $I_2$
  • Trough point $I_1$ orthogonal (perpendicular) to $Circle_0$ is circle $Circle_1$
  • Trough point $I_2$ orthogonal (perpendicular) to $Circle_0$ is circle $Circle_2$
  • $Circle_1$ and $Circle_2$ have the same radius
  • $Circle_1$ and $Circle_2$ are orthogonal to eachother.
  • point Q is the point inside $Circle_0$ where $Circle_1$ and $Circle_2$ cut eachother.

Wanted: construct point Q

the only limits I could find are:

  • Q is on the line perpendiculer to $ I_1I_2$ going to the midpoint of $ I_1I_2$
  • Q is on the same site as side of $Centre_0$ as $ I_1$ and $I_2$
  • $ \angle I_1Centre_0I_2$ is smaller than a right angle

I did manage the opposite:

Given point Q (different from $Centre_0$ ) construct the points $I_1$ and $I_2$

so if it helps somebody:

  • Draw ray $r$ from $Centre_0$ trough Q
  • Draw line l trough Q perpendicular to ray r
  • Point $ I_c$ where line l cuts $Circle_0$ (any of the two)
  • Draw segment $Circle_0$ to Point $ I_c$
  • Draw line $j$ trough $ I_c$ perpendicular to the segment$Circle_0$ $ I_c$
  • Point $ I_Q$ where line $j$ cuts ray $r$
  • Point $ I_m$ is the midpoint of the segment $Q$ $I_Q$
  • Line $m$ trough $ I_m$ perpendicular to ray $r$
  • Draw $Circle_m$ centre $ I_m$ trough Q
  • Point $ Centre_1$ where line $m$ cuts $ Circle_m$ (one of the two)
  • Point $ Centre_2$ where line $m$ cuts $ Circle_m$ (the other one)
  • Draw $Circle_1$ centre $ Centre_1$ and trough Q
  • Draw $Circle_2$ centre $ Centre_2$ and trough Q

  • Point $I_1$ is where $Circle_1$ cuts $Circle_0$ nearest to Q

  • Point $I_2$ is where $Circle_2$ cuts $Circle_0$ nearest to Q

But now from $ Circle_0 $ , $I_1$ and $I_1$ how can I construct $Point Q$ ?

  • $\begingroup$ What do you mean by circles being orthogonal to each other? $\endgroup$ – sid Mar 26 '14 at 21:56
  • $\begingroup$ @Sid That the tangent of one of the circles at the intersection goes trough the centre of the other circle. Or that the tangents at the intersection cut eachother at right angles. Or that the angle centre of one circle - intersection - centre of the other circle is a right angle. All these decriptions are equivalent,so just take the one that you find easiest. $\endgroup$ – Willemien Mar 26 '14 at 22:37

Not an answer, but too long for a comment.

Let your $Circle_0$ be the unit circle centered at the origin, and let $I_1$ and $I_2$ be mutual reflections in the $x$-axis. Take $J_i$ to be the center of a circle through $I_i$ orthogonal to the unit circle; then $J_i$ lives on the line tangent to the circle at $I_i$. The intersections, say, $P$ and $Q$, of $\bigcirc J_1$ and $\bigcirc J_2$ (each of sufficiently-large radius, $r$) live on the $x$-axis. If the circles are to be orthogonal at these points, then their tangents at $P$ and $Q$ make $45^\circ$ angles with the $x$ axis; thus, $\overline{PQ}$ is necessarily one side of an (the) axis-aligned square inscribed in $\bigcirc J_1$ (and, likewise, of the corresponding square in $\bigcirc J_2$). This says that each $J_i$ must be at distance $r/\sqrt{2}$ from the $x$-axis.

We may assume the coordinates of $I_1$ are $(\cos\theta, \sin\theta)$ for some $0 \leq \theta \leq \pi/2$; and the coordinates of $J_1$ become $(\cos\theta \pm r\sin\theta, \sin\theta \mp r \cos\theta)$, where the $\pm$ and $\mp$ signs depend upon which direction $\overrightarrow{I_1J_1}$ points along the tangent line at $I_1$. The distance from $J_1$ is therefore $\sin\theta \mp r \cos\theta$. By the above, we must have $$\sin\theta \mp r \cos\theta = r/\sqrt{2} \qquad \to \qquad r = \frac{\sqrt{2}\sin\theta}{1\pm\sqrt{2}\cos\theta}$$

To construct $Q$ (and $P$), then, all you need to do is construct a radius of length $r$. Radius $r$ is certainly constructible, but I don't (yet?) have a construction that fits nicely with the rest of the figure.


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