2

Given two distinct points $A$, $B$ and a random point $T$ in between them:

  • $c_0$ is the semicircle of diameter $AB$, center $C$.

  • $c_A = \odot(A,AT)$

  • $c_B = \odot (B,BT)$

  • $I_A = c_A \cap c_0$ and $I_B = c_B \cap c_0$

  • $O = I_AI_B \cap AB$

  • $H_A$ is the center of in-similitude of $c_0$ and $c_A$

  • $H_B$ is the center of in-similitude of $c_0$ and $c_B$

  • $c_1$ is the circle internally tangent to $c_0$ and externally tangent to both $c_A$ and $c_B$.

  • $P = c_0 \cap c_1$.

I need an answer on why the magic circle $c_2 = \odot(O,OT)$ passes through $P$. I made this construction kinda by accident, comparing it with the construction of the soddy circle of a random triangle. The more I look at it, the more it doesn't make sense.

I've already made a question related to this configuration, these three circles share a common tangent line. enter image description here

hellofriends
  • 2,048
  • 1
    If my calculations are correct, then you are right to be skeptical. Writing $a$ and $b$ for the radii of $\bigcirc A$ and $\bigcirc B$, if the perpendicular bisector of $TP$ meets line $AB$ at $O$ (ie, the center of the circle through $T$ and $P$), and line $I_A I_B$ meets line $AB$ at $O'$, then $$|OC|=\dfrac{a^2+ab+b^2}{2(a-b)}\qquad |O'C|=\frac{\left(a \sqrt{a(a+2b)} + b\sqrt{b(b+2a)};\right)^2}{2(a+b)^2(a-b)}$$ which is to say: $O$ and $O'$ are not the same point. They can deceptively close, however. Eg, if $a=3$ and $b=5$, the distances are about $12.25$ and $12.17$, respectively. – Blue Jun 26 '21 at 17:04
  • but I walk point T and the circles still look tangent – hellofriends Jun 26 '21 at 17:18
  • To be clear: I'm starting with $\bigcirc A$, $\bigcirc B$, $\bigcirc C$, and I've defined $P$ as the point where the little circle tangent to all three touches $\bigcirc C$. (I've ignored $H_A$ and $H_B$. Also, I should've switched my $O$ and $O'$ defns so that my $O$ matched yours, but be that as it may.) ... The upshot is that, given the tangent circle configuration, the line $I_AI_B$ does not meet the line $AB$ at the center of a circle through $T$ & $P$; for $a=3, b=5$, the calculated distances from that pt to $T$ & $P$ are $11.1696$ & $11.1740$(!). I'll check my work and post an answer. – Blue Jun 26 '21 at 17:46

1 Answers1

1

Let (semi-)circles $\bigcirc A$, $\bigcirc B$, $\bigcirc C$ have radii $a$, $b$, $c:=\frac12(a+b)$, and suppose $\bigcirc R$ of radius $r$ is externally tangent to $\bigcirc A$ and $\bigcirc B$, and internally tangent to $\bigcirc C$ at $P$. (It shouldn't matter, but in case of sign error, I'm assuming $b>a$, as in the figure.)

enter image description here

(I'll note that the lines from $P$ through the other points of tangency necessarily meet $\overline{AB}$ at OP's centers of in-similitude $H_A$ and $H_B$. This is a property of the configuration, not a constraint, so we'll ignore it.)

Now, as $\overline{CR}$ is a median of $\triangle ARB$, Apollonius gives us this relation: $$\begin{align} |AR|^2+|BR|^2=2(|CR|^2+|AC|^2) &\quad\to\quad (a+r)^2+(b+r)^2=2((c-r)^2+c^2 ) \\[4pt] &\quad\to\quad r = \frac{a b}{2(a+b)} \tag1 \end{align}$$

If we coordinatize, with $C=(0,0)$, $A=(-c,0)$, $B=(c,0)$, we can write $P= (c\cos\theta, c\sin\theta)$, where $\theta := \angle BCP$. The Law of Cosines in $\triangle CBR$ gives us $$\cos\theta = \frac{-|BR|^2+|CB|^2+|CR|^2}{2|CB||CR|}=\frac{-(b+r)^2+c^2+(c-r)^2}{2c(c-r)}=\frac{(a-b)((a+b)^2+ab)}{(a+b)((a+b)^2-ab)} \tag2$$

Let the perpendicular bisector of $\overline{PT}$ meet $\overleftrightarrow{AB}$ at $Q'$ (which is therefore the center of a circle through $P$ and $T$). Using coordinates, we obtain $$|Q'C|=\frac{a^2 + a b + b^2}{2\,|a-b|} \tag{3}$$

Shifting focus to the points $I_A$ and $I_B$ where $\bigcirc A$ and $\bigcirc B$ meet $\bigcirc C$, we can calculate $$I_A = \left(\frac{a^2-2ab-b^2}{2(a+b)},\frac{av}{a+b}\right) \qquad I_B=\left(\frac{a^2+2ab-b^2}{2(a+b)},\frac{bu}{a+b}\right) \tag{4}$$ where $$u := \sqrt{a(a+2b)}\qquad v := \sqrt{b(b+2a)}$$ Let $\overleftrightarrow{I_AI_B}$ meet $\overleftrightarrow{AB}$ at $Q$. (This corresponds to the definition of OP's point $O$.) We find that $$|QC|=\left|\frac{a v (a^2 + 2 a b - b^2) - b u (a^2 - 2 a b - b^2)}{2 (a + b) (a v - b u)}\right|=\frac{(a u + b v)^2}{2 (a + b)^2 |a - b|} \tag{5}$$ Since $(5)$ and $(3)$ do not match (unless $a=b$, a case we can ignore), we conclude that $Q$ and $Q'$ are distinct points.

enter image description here

This says that OP's point $O$ is not the center of a circle through $T$ and $P$; the "magic" is an illusion. $\square$

It's a pretty good illusion, though. One can calculate the discrepancy relative to the diameter of $\bigcirc C$ as $$\frac{|QQ'|}{|AB|} = \frac{a b (u - v)^2}{2 (a + b)^3 |a - b|} \tag{6}$$ For, say, $(a,b)=(3,5)$, this gives $|QQ'|/|AB|\approx 0.0100$. Further, one can show $$|QT|=\frac{a b (a^2 + a b + b^2 + u v)}{ (a + b)^2|a - b|} \qquad |QP|=\frac{u v (a u + b v) \sqrt{a b}}{ (a + b)^2|a - b| \sqrt{a^2 + a b + b^2}} \tag{7}$$ For $(a,b)=(3,5)$, the relative discrepancy in these values is $$\frac{\left|\;|QP|-|QT|\;\right|}{|AB|}= 0.000550\ldots$$ which is easy to miss, visually. I was almost convinced of the "magic" myself, but moving my GeoGebra sketch to make $a$ and $b$ dramatically different revealed how things didn't quite match up.

Blue
  • 83,939