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Isodynamic Points and Apollonius Circles of a Triangle.
Peter Y. Woo, Biola University, 8/22/99
Theorem 1, on Apollonius Circles. The locus of a variable point whose distances from 2 fixed points are at a constant ratio e, is a circle for e ¹ 1 and the perpendicular bisector of the two points for e = 1. The family of such loci for all real values of e form a coaxial family of circles with the two fixed points as limit circles.
Proof. Let L be the line through the two fixed points A, B. Let P be a variable point such that PA/PB = e is a constant. WLOG [without loss of generality] assume e > 1. Let P1 and P2 be two points on L, P1 between A and B, and P2 outside, such that PiA/PiB = e for i = 1,2.
Then since PA/PB = P1A/P1B = area(DPAP1) / area(DPAP2), P1 is equidistant from PA and PB, so that PP1 bisects ÐAPB internally. Similarly, P2 is equidistant from lines PA and PB, hence PP2 bisects ÐAPB externally. \PP1 ^ PP2. \P lies on the circle Ze on P1P2 as diameter.
Let M be the midpoint of AB. \MP1/MB = (½)(AP1-BP1) / (½)(AP1 + BP1) = (e-1)/(e+1) = (½)(AP2-BP2) / (½)(AP2 + BP2) = MB/MP2.
\MB2 = MP1 × MP2. \The tangent from M to Ze has length MB, or AB/2, which is independent of e. So for each real value of e, the different circles Ze all have L as diameter and have tangents from M having AB/2 as length, thus forming a radical family of circles with the perpendicular bisector L' of AB as radical axis. The points A and B corresponds to the cases when e = 0 and ¥ and are null circles. For e = 1, the locus is the perpendicular bisector L' of AB. QED
Definition. For any triangle, the Apollonius circle
ZV through a vertex V is the locus of points whose ratio of
distances from the other two vertices is the same as that of V.
Theorem 2, on isodynamic points. Given any
non-equilateral triangle ABC,
the three Apollonius circles intersect one another at exactly two points,
called the 1st and 2nd isodynamic points, the first one, I', inside the
largest vertex angle and the other, I", outside.
Then (a) its distances of I' from the 3 sides of the triangle have ratios
sin (A+p/3) : sin (B+p/3) : sin (C+p/3), and those of I" have ratios
sin |A-p/3| : sin |B-p/3| : sin |C-p/3|;
(b) I'A : I'B : I'C
= 1/a : 1/b : 1/c; (c) the pedal triangle formed by the three feet of
perpendiculars from I' or I" to the three sides of ABC is equilateral; (d) if
none of the angles of ABC exceeds 120° then I' is inside the triangle.
Proof. WLOG assume A being the greatest angle.
Let I = incenter, and B'=BIÇAC, C'=CIÇAB.
Within ÐA, since A is inside the circles ZB and ZC, and
that BC'B'C is a convex quadrilateral, the arc BB' of the first circle
and the arc CC' of the second must intersect at some point I'. Hence
the two circles must intersect in 2 points, one inside ÐA, called
the 1st isodynamic point I', and the other outside ÐA, called the
2nd isodynamic point I".
Let I' be one of these isodynamic points. Let A1, B1, C1 be
the projections of I' on BC, CA, AB.
Since I' lies on the Apollonius Circle through B,
I'A/I'C = BA/BC = c/a.
Similarly, I'A/I'B = CA/CB = b/a. Dividing them, we get
I'B/I'C = c/b = BA/CA, proving that I' also lies on ZA.
This also proved that ZA, ZB, ZC intersect one another
at exactly 2 points, I' and I", and so their centers A3, B3, C3
lie on the perpendicular
bisector of I'I". [It is called the Brocard Line.] Also it follows
that I'A : I'B : I'C = 1/a : 1/b : 1/c. This proves (b).
Next, I'B1AC1 is a concyclic quadrilateral, with I'A
as diameter. Hence I'A = B1C1 / sin A. Similarly I'B = C1A1 / sin B,
and I'C = A1B1 / sin C.
\ I'A : I'B : I'C = B1C1/sin A : C1A1/sin B : A1B1/sin C
= B1C1/a : C1A1/b : A1B1/c = 1/a : 1/b : 1/c from (b).
Hence B1C1 = C1A1 = A1B1, and (surprise!)
A1B1C1 is an equilateral triangle. This proves (c).
So far (b) and (c) are true for both isodynamic points.
Now let I' be the isodynamic point inside ÐA.
From (b) we have I'A = k/a, I'B = k/b, I'/C = k/c for some
constant k.
Then I'C1 = 2 × area(DI'AB)/AB = (k/a) (k/b) sin
ÐAI'B / c = (k2/a b c) sin ÐAI'B .
Now ÐAI'B = ÐAI'C1 + ÐBI'C1 = ÐAB1C1 + ÐBA1C1
[due to cyclic quadrilaterals]
= p-A-ÐAC1B1 + p-B-ÐBC1A1
= (p-A-B) + (p-ÐAC1B1-ÐBC1A1) = C + p/3.
\I'C1 = (k2/a b c) sin (C+p/3).
Similarly I'B1 = (k2/a b c) sin(B + p/3), I'A1 = (k2/a b c)
sin(A + p/3), so that I'A1 : I'B1 : I'C1 =
sin (A+p/3) :sin (B+p/3) :sin (C+p/3).
Similarly, for I" outside ÐA, let its projections on
AB, BC, CA be C2, A2, B2.
Then ÐAI"B = ÐBI"C2-ÐAI"C2 = ÐBA2C2-ÐAB2C2
[due to cyclic quadrilaterals]
= B±ÐBC2A2-p+A+ÐB2C2A
= -(p-A-B) +(ÐB2C2A±ÐBC2A2)
= p/3 -C, and
similarly for ÐBI"C and ÐCI"A. So we also get
I"A1 : I"B2 : I"C2 = sin |p/3-A| : sin |p/3-B| : sin |p/3-C|.
QED
Theorem 3. Let I' be one of the isodynamic points of a
nonequilateral triangle ABC. Then any circle inversion with I' as center
will map A, B, C into an equilateral triangle.
Proof. Let A''', B''', C''' be the image of A, B, C under
the inversion, center I'.
Since I'A : I'B : I'C = 1/a : 1/b : 1/c,
so I'A''': I'B''': I'C''' = a : b : c, or
I'A''' = K a, I'B''' = K b, I'C''' = K c for some real number K.
\DI'A'''B''' is
similar to DI'BA, so that
A'''B''' = BA (I'A'''/I'B) = c (K a / (k / b)) =(K / k) a b c, which
is symmetric in a, b, c.
\ A'''B''' = B'''C''' = C'''A'''. QED
Exercise 1. Find a quick proof that I' lies on BC iff
ÐA = 120°, and I" lies on BC iff ÐA = 60°. Minimise
the use of trigonometry.
Exercise 2. Discuss the case when ABC is isosceles.
