# Construct a circle tangent to sides $BC$ and $CD$ and s.t. its meetings with the diagonal $BD$ are tangent points from tangents draw from point $A$

Given square $$ABCD$$

I want to construct (with ruler and compass) the circle in the interior of the square such that it is tangent to sides $$BC$$ and $$CD$$ and such that it's meetings with the diagonal $$BD$$ are tangent points from tangents draw from point $$A$$:

It is clear that the center of the circle must lie in $$AC$$. I tried finding some cyclic quad somewhere and I failed miserably. I then thought about puting $$K$$ in hyperbola with focii $$A$$ and the center $$O$$ of the square. Then again $$K$$ lies outside the segment $$A O .$$

This problem is hard because we would think of looking at the locus of the centers of circles such that the meetings of the circle with line $$BD$$ are the tangents from $$A$$.

But that this locus is exactly the same as the locus of the centers of the circles tangent to $$CD$$ and $$BC$$: line $$AC$$.

The proof is simple: as the tangents from $$A$$ must have the same lenght the meetings $$M$$ and $$N$$ of $$BD$$ with the circles must be reflections of each other with respect to the center $$O$$ of the square $$ABCD$$ thus the center of the circle must lie in line $$AO$$ which is line $$AC$$.

The real geometric constrain is between the distance of the centers (all of which lie on line $$AC$$) to point $$A$$ and the radius of the circles.

Let $$P$$ be in line segment $$OC$$.

$$PA = x$$

$$r$$ the radius of the circle centered at $$P$$.

$$a=AB$$

we have that $$r^2 = x^2 - x \frac{a\sqrt2}2$$ and $$x = a\frac{\sqrt2}4 + \sqrt{r^2+\frac{a^2}8}$$

and these weird relations are the "locus" that I desire to work with.

• What have you tried, and where did you get stuck? This information can help answerers tailor their responses to best serve you, without wasting time (theirs or yours) explaining things you already know, duplicating your effort, or using techniques beyond your skill level. ... BTW, if the center of the circle is $K$, then $|AK|/|KC|$ is an interesting value.
– Blue
Commented May 21, 2020 at 13:53
• i tried finding some cyclic quad somewhere and I failed miserably. I then thought about puting $K$ in a hyperbola with focii $A$ and the center $O$ of the square. Then again $K$ lies outside the segment $AO$. Commented May 21, 2020 at 13:56
• Related issue asked yesterday here Commented May 21, 2020 at 18:26
• @JeanMarie indeed, thanks for showing me! Commented May 21, 2020 at 19:29
• What inspired this? I love constructions like this. Commented May 22, 2020 at 23:29

## 2 Answers

Let center $$O$$ of the circle lie on diagonal $$\overline{AB}$$ with midpoint $$M$$, and define $$a:=|OA|$$, $$b:=|OB|$$. Let the circle meet the other diagonal at $$R$$, and define $$r:=|OR|$$; note that $$r=b/\sqrt{2}$$.

\begin{align} \underbrace{\frac{|OR|}{|OA|}=\frac{|OM|}{|OR|}}_{\triangle ORA\sim\triangle OMR} &\quad\to\quad \frac{r}{a}=\frac{a-\frac12(a+b)}{r} =\frac{a-b}{2r}\tag{1} \\ &\quad\to\quad a(a-b)=2r^2=b^2 \tag{2} \\[8pt] &\quad\to\quad \frac{a}{b}=\frac{b}{a-b}=\phi \tag{3} \end{align} (ignoring a negative solution) where $$\phi := \frac12(1+\sqrt{5})$$ is the Golden Ratio.

Consequently, the construction reduces to dividing diagonal $$\overline{AB}$$ in the ratio $$\phi:1$$. A simple method for doing so is described under "Dividing a line segment by interior division" in the Wikipedia entry. $$\square$$

• thanks, you did find a solution. If nobody comes up with a less algebraic solution I'll accept yours Commented May 21, 2020 at 15:11
• hey blue which is the software used for drawing pictures? Commented May 21, 2020 at 16:16
• @endgameendgame: I use the desktop version of GeoGebra "Classic" to draw my figures.
– Blue
Commented May 21, 2020 at 16:40
• oo the pictures are awesome and catchy Commented May 21, 2020 at 17:09
• @Blue are you aware of the similar problem asked yesterday here ? Commented May 21, 2020 at 18:11

Just a hint: If $$K$$ is the center of the circle, $$O$$ is the center of the square and $$L$$ is the tangency point lying on $$BD$$ then triangles $$KOL$$ and $$KLA$$ are similar. This gives $$KO\cdot KA=KL^2$$. This allows you to calculate the radius of the circle.

• yes $O$ and $A$ are inverses with respect to the circle Commented May 21, 2020 at 14:19
• can you explain further Commented May 21, 2020 at 14:45
• @endgameendgame it is just basic circle inversion: $BD$ is the polar line of point $A$ with respect to the circle Commented May 21, 2020 at 14:48
• the radius is $\frac{3 - \sqrt5}2$ times the side of the square Commented May 21, 2020 at 14:57
• @hellofriends I got exactly the same expression for the radius. The construction of the circle should be straightforward now. Commented May 21, 2020 at 15:04