Apollonian_Circles

Apollonian circles

Given a segment AB and a positive real number k, the Apollonian circle of the oriented segment AB, for the ratio k, is defined to be the geometric locus of points P, such that |PA|/|PB| = k.

[1] If PC and PD are the bissectors of the angle APB, it is a well-known elementary theorem that |CA|/|CB| = |DA|/|DB| = k, hence C, D, which are well defined points on the line AB, belong to the locus. But the angle CPD is a right one. Thus, every point P of the locus looks the segment CD under a right angle. Thus, it belongs to the circle with diameter CD.
[2] Well, the angles OPA and PBO are equal. To see this, add to OPA the angle APC and get OPC. Add to PBO the angle CPB = APC (because PC bisector) to get OCP = OPC (isosceles).
[3] These angles beeing equal implies that triangles OPA and OPB are similar. This implies that |OP||OP| = |OA||OB|. This means: (i) that OP is tangent to the circumcircle c of ABP, (ii) that that the Appolonian circle a is orthogonal to the circumcircle c ABP.
[4] Well, we move slowly to deeper waters: The enclosing circles of all triangles APB (P beeing on the Apollonian) build a "circle-bundle" X of circles passing through A, B. The Apollonian circle belongs to the orthogonal bundle Y of X, whose members are characterized by beeing orthogonal to every member of X. Thus, the Apollonian circle is characterized by beeing orthogonal to both, the enclosing circle of PAB and the segment AB and passing through the point P.
[5] Let us consider now the Apollonian circles of the (oriented) sides of a triangle ABC. By this we mean the following three circles:
c₁: the locus of points P, such that |PB|/|PC| = |AB|/|AC|, (passes through A),
c₂: the locus of points P, such that |PC|/|PA| = |BC|/|BA|, (passes through B),
c₃: the locus of points P, such that |PA|/|PB| = |CA|/|CB|, (passes through C).
Denote further the circumcircle of ABC with d.

[6] The three circles pass through two common points K and L. In fact, assume that K is one common point of c₁ and c₃. Then |KB|/|KC| = |AB|/|AC| and |KA|/|KB| = |CA|/|CB|. Multiplying side by side: |KA|/|KC| = |AB|/|CB|. Thus K belongs to c₂.
[7] It follows that the centers U, V, W of the three circles are on a line. This is called the Lemoine line of the triangle.
[8] The centers U, V, W of c₁, c₂, c₃ are respectively on the lines BC, CA, AB. In addition lines UA, VB, WC are tangents to d, at A, B, C respectively and c₁, c₂, c₃ are all orthogonal to d (see [3] above).
[9] d being orthogonal to the bundle of intersecting type containing the three circles c₁, c₂, c₃ implies that K, L and O (center of d) are collinear. The line containing K, L and O is called Brocard axis of the triangle.
[10] It follows that, K and L are inverse with respect to the circumcircle d.
[11] Consider now the inversion F₁ with respect to c₁:
(i) F₁ permutes the members of the bundle of c1,c2,c3, which are all orthogonal to d.
(ii) F₁(B) = C (shown at the beginning). Hence F₁(c₃) = c₂.
(iii) It follows that the inversions F₁, F₂, F₃ w.r. to the three circles, applied to these circles permute them. Because inversion preserves angles, all angles between the circles c₁,c₂ and c₃ at K (and L) are equal to each-other and equal to 60 degrees.
[12] Besides, each center of the c1, c2, c3 is center of similarity for the other two circles.
[13] The tangency property of UA (see [8]) implies that it is the harmonic conjugate of the symmedian through A w.r. to the sides AB, AC. This in turn implies that the Lemoine axis is the trilinear polar of the symmedian point (not drawn) of the triangle. See the references below for some more discussions on that.
A lot of facts. And this is just the beginning. The points K, L are called isodynamic points of the triangle ABC.
EucliDraw has a special tool to construct the Apollonian circles of ABC:
1) Select the tool from the menu-item [ Shape-Tools \ Circle Tools \ Apollonian Circle ].
2) Click three times 1st: on AB, 2nd: on AC, 3rd: on BC. This defines circle c₁.
3) Click three times 1st: on BC, 2nd: on BA, 3rd: on CA. This defines circle c₂.
4) Click three times 1st: on CA, 2nd: on CB, 3rd: on AB. This defines circle c₃.
Switch to the selection-tool (CTRL+1). Catch and modify the triangle. Watch how the shape changes.

Apollonian circles

See Also