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 $Point Q$ ?