How to construct figure without constructible segments

Question

I can't seem to find a way to draw this circle which is tangent to the 2 sides of the square and the semicircle. The radius is $4-2\sqrt{3}$ if I recall correctly. I would want a syntethic solution, without trying to replicate the $\sqrt{3}$ or any other similar.

Answer 1

Omitting straightforward construction of midpoints $M$ and $A$ as shown ...

• Use an arc about $M$ to transfer $A$ to $A'$ on the side of the square.

• Use an arc about $A'$ to transfer vertex $B$ to $B'$ on the side of the square.

• Use an arc about $B'$ to transfer vertex $B$ to $B''$ on the diagonal of the square.

• The target circle has center $B''$ and passes through $B'$.

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FYI, if the first step transfers $A$ to the "other" $A'$ on the side-line of the square, then the rest of the construction gives the circle tangent to the side-lines of the square and the "lower" semicircle.

Answer 2

If the side of the square is $2$, then the radius of the small circle is indeed $4-2\sqrt3$. So how do we create such segment? Note that $$\sqrt3=\sqrt{2^2-1^2}$$or$$2\sqrt3=2\sqrt{2^2-1^2}=\sqrt{4^2-2^2}$$ Then do the following construction:

1. Stating from your square, extend $AB$ and $CD$, such that $GB=AB$ and $HD=CD$
2. Draw a circle, with center in $A$, with radius $AG$
3. The intersection of the circle with $DH$ is $I$. Then $IC=\sqrt{IA^2-AC^2}=\sqrt{4^2-2^2}=2\sqrt 3$
4. Length of $HI$ is $HI=CH-CI=4-2\sqrt 3$, which is exactly the radius of our circle.
5. Draw a circle with center $H$ and radius $HI$. The intersection with $GH$ is $J$
6. Draw a parallel to $HC$ through $J$. The intersection with $BC$ is the center of your circle.

Answer 3

Let's draw a line $GT$ parallel to $CD$ and at a distance $AM$ from it. The centre $E$ of the circle is then equidistant from that line and $M$: it is thus the intersection between line $BC$ and a parabola having $GT$ line as directrix and $M$ as focus.

To find that intersection we can follow the method explained here:

• construct $M'$ symmetric of $M$ about $BC$ (and midpoint of $BD$);
• construct the intersections $H$ and $Q$ of line $MM'$ with $BC$ and $GT$;
• construct two half-circles having $FF'$ and $HQ$ as diameters;
• let $R$ be the intersection of those half-circles;
• construct the circle of centre $Q$ passing through $R$;
• let $T$ be an intersection of the circle with line $GT$;
• the perpendicular from $T$ to $GT$ intersects $BC$ at $E$.

Answer 4

A construction with GeoGebra, using conic sections:

Locus of center of circles touching both the given circle and the upper side of the square (or better: the line containing this side) is a parabola.
This parabola is also locus of point equidistant to $E,$ which is center of the given circle, and a line $p$ parallel to the lower and upper sides of the circle.

Therefore, $E$ is focus and $p$ is directrix of the parabola.