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In the attempt to answer this question, I was brought to another one, connected to a certain family of circles.

Consider the semicircle made of the upper half of the unit circle and the horizontal diameter with endpoints $A(-1,0)$ and $B(1,0)$.

Let us now consider the family $(F)$ of circles inscribed into this semicircle as can be seen on Fig. 1.

enter image description here

Fig. 1: Certain circles of family $(F).$

They can be described in a parametric way as circles

$$\text{with center} \ (x_c=-\cos a, y_c=\frac12 \sin^2a) \ \text{and radius} \ y_c, \ a \in (0,\pi) \tag{1}$$

or equivalently as circles $(C_T)$

$$\text{with center} \ \left(x_c=\frac{T^2-1}{(1+T)^2}, y_c=\frac{2T}{(1+T)^2}\right) \ \text{and radius} \ y_c, \ T \in (0,+\infty) \tag{2}$$

(Connection between the two parameterizations: $T=t^2$ where $t:=\tan \tfrac{a}{2}$).

The locus of these centers is parabola $(P)$ with cartesian equation:

$$y=\tfrac12(1-x^2),\tag{3}$$

represented on Fig. 2 as the light blue curve.

Each circle $(C_T) \in F$ is associated with a unique circle $(C_{T'}) \in F$ which is tangent to $(C)$ as depicted on Fig. 2.

enter image description here

Fig. 2: Two circles of family $(F)$ are mutually tangent if their parameters $T$ and $T'$ comply with relationship (4).

Let $I$ be the point of tangency of $(C_T)$ and $(C_{T'})$.

Question: how can we describe the locus of $I$?

(sketched as the magenta curve on Fig. 2.)

My work: I have obtained through tedious calculations an unexpected linear relationship between parameters $T$ and $T'$ which is:

$$T=(3-2 \sqrt{2})T'\tag{4}$$

Using (4), I have tried to obtain the locus of $I$ as the envelope of chords defined by the line segment joining the centers of circles $C_T,C_{T'}$ using homographies but to no avail.


Edit: Nice solution using inversion (a little different from the solution given in a comment by Blue).

'Refer to Fig. 3). Consider inversion with pole $S(0,-1)$ and power $2$ meaning that $M$ is exchanged with $M'$ iff $S,M,M'$ are aligned and $SM.SM'=2$. Its circle of inversion is the black circle (center $S$, radius $\sqrt{2}$). It is easy to see that any circle of family (F) is globally invariant. This is in particular the case for the two blue circles. As the image by inversion of two tangent curves in a certain point $I$ is two tangent curves with contact point $I'$ where $M'$ is the image of $M$, point $I$ is invariant, establishing that $I$ belongs to the circle of inversion.$\square$

As a consequence, tangency points $T_1$ and $U_1$, $T_2$ and $U_2$, are exchanged, meaning that $S,T_1,U_1$ and $S,T_2,U_2$ are aligned.

enter image description here

Fig. 3.

Remark: Two properties of circles of family (F) deserving to be known:

http://www.gogeometry.com/problem/p311_circle_inscribed_semicircle_chord.htm

http://www.gogeometry.com/problem/p310_semicircle_circle_tangent_angle.htm

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  • $\begingroup$ Does $(y_2(x_1,y_1)+y_1(x_2,y_2))/(y_1+y_2)$ give a nice answer? $\endgroup$ – Empy2 Mar 12 at 23:22
  • $\begingroup$ @Empy3 See my comment to the answer of GReyes. $\endgroup$ – Jean Marie Mar 12 at 23:30
  • $\begingroup$ It is equivalent, and not connected to formula $\sin 2A = 2\tan A /(1+\tan^{\color{red}{2}} A)$ (no squared tangent). $\endgroup$ – Jean Marie Mar 13 at 9:58
  • $\begingroup$ @JeanMarie I understood. Thanks, great method! $\endgroup$ – cosmo5 Mar 13 at 15:08
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There's something more fundamental going on here.


Consider any circle $\bigcirc P$ tangent to the semicircle $ACB$ at $T$ and the diameter at $T'$. (My figure is rotated.) Let $\overline{CC'}$ be the perpendicular diameter with $C$ the midpoint of the semicircle.

enter image description here

Thales' Theorem guarantees that right angle $\angle CTC'$ subtends parallel diameters of $\bigcirc P$ and $\bigcirc O$; thus, $C'$, $T$, $T'$ are collinear, and $\triangle CTC'\sim\triangle T'OC'$, so that $$|C'T||C'T'|=|C'C||C'O|=2r^2$$ That is, the power of $C'$ with respect to $\bigcirc P$ is independent of $P$. Consequently,

  • the tangent segments from $C'$ to any circle is $\sqrt{2r^2}=r\sqrt{2}$; and
  • $C'$ lies on the radical axis of any pair of circles.

In the specific case of a pair of inscribed circles that are tangent to each other at a point $I$, the mutual tangent line through $I$ is their radical axis, necessarily meeting $C'$. Moreover, $\overline{C'I}$, being a tangent segment to each circle, has length $r \sqrt{2}$. We conclude that $I$ lies on the circle of radius $r\sqrt{2}$ about $C'$. $\square$

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    $\begingroup$ @JeanMarie: I try. :) ... Incidentally, if you carry the construction beyond the bounds of the semicircle ---ie, taking the centers anywhere along the parabola, and maintaining tangency with the diameter-line--- then the inscribed circles of the semicircle become exscribed circles of the other semicircle, and the locus of $I$ becomes the (almost-) entire circle about $C'$. $\endgroup$ – Blue Mar 13 at 10:09
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    $\begingroup$ @JeanMarie: "I think that a proof using inversion should exist as well." ... Indeed: One can argue without specificity that there's some circle through $A$ & $B$ such that inversion exchanges $\overline{AB}$ w/ $\stackrel{\frown}{ACB}$. The inversion preserves tangencies and thus also the family of inscribed circles, and we conclude that the center of inversion ---wherever it may be--- has the requisite radical axis and tangent segment properties for the circle to be the locus of $I$. ... In fact, this argument works even for non-semicircle arcs and non-diameter chords ... and even two arcs! $\endgroup$ – Blue Mar 13 at 12:30
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    $\begingroup$ A big thank for your comment about the extension to the whole circle and for your inversive proof. $\endgroup$ – Jean Marie Mar 13 at 13:10
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    $\begingroup$ I just saw two interesting properties (1) and (2) of circles of family $(F)$: $\endgroup$ – Jean Marie Mar 14 at 19:43
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    $\begingroup$ @JeanMarie No one usually remembers but angle bisector of a right triangle, $x$, divides hypotenuse into two segments $a,b$ such that $2/x^2=1/a^2+1/b^2$. Similarly (1) should be obvious but didn't strike to me immediately. Nice find. :) $\endgroup$ – cosmo5 Mar 14 at 21:45
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$$(x_1,y_1)=\left(\frac{T-1}{T+1},\frac{2T}{(T+1)^2}\right)\\ (x_2,y_2)=\left(\frac{U-1}{U+1},\frac{2U}{(U+1)^2}\right)\\ (x_1-x_2)^2+(y_1-y_2)^2=(y_1+y_2)^2\\ \frac{4(T-U)^2}{(T+1)^2(U+1)^2}=4y_1y_2=\frac{16TU}{(T+1)^2(U+1)^2}\\ T^2-6TU+U^2=0$$ The tangent point is $(x,y)=$
$$(y_2(x_1,y_1)+y_1(x_2,y_2))/(y_1+y_2)\\ =\left(\frac{U(T^2-1)+T(U^2-1)}{U(T+1)^2+T(U+1)^2},\frac{4TU}{U(T+1)^2+T(U+1)^2}\right)\\ \frac{4x}y=T-\frac1T+U-\frac1U\\ \frac4y=T+2+\frac1T+U+2+\frac1U\\ \left(\frac4y-4\right)^2=\left(\frac{4x}y\right)^2+4(T+U)(\frac1T+\frac1U)\\ (1-y)^2=x^2+2y^2\\x^2+(y+1)^2=2$$

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  • $\begingroup$ Therefore an arc of circle centered in $(0,-1)$ with radius $\sqrt{2}$ ! A (piece of) conic curve... $\endgroup$ – Jean Marie Mar 13 at 9:54
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Observe that the point of tangency divides the segment between the centers in the ratio of the radii, that is, in the ratio $y(T')/y(T)$. Therefore, if you call $(U,V)$ the point of tangency, $$ U(T)=x(T)+\frac{y(T')}{y(T)}(x(T')-x(T));\quad V(T)=y(T)+\frac{y(T')}{y(T)}(y(T')-y(T)) $$ where $T'$ is expressed through $T$ using the relation you found (assuming it is correct).

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  • $\begingroup$ I thought to it (not exactly with this form) but it is not practicaly tractable (for example for solving the initial question). $\endgroup$ – Jean Marie Mar 12 at 23:28
  • $\begingroup$ @Jean Marie I see..Does your parameter $a$ have a nice geometric meaning? $\endgroup$ – GReyes Mar 12 at 23:31
  • $\begingroup$ Yes, for the meaning ("nice", why not ?): If $\Omega$ is the center of the circle with parameter $a$, you draw a vertical line from $\Omega$. Let $J$ be the intersection of this line with the unit circle. Then $a$ is the angle between the $x$ axis and $\vec{OJ}$. $\endgroup$ – Jean Marie Mar 12 at 23:36

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