Sunday, June 16th, 2019

Harmonic subdivision and the circle of Apollonius

[ Thursday, 21 October, 2010 ]

Đây là một bài thi giỏi toán Hàn Quốc chưa có bài giải trên mạng. This is a problem that has no solution posted in the web.

Problem 2 of the Korean Mathematical Olympiad 2007

ABCD is a convex quadrilateral, and AB ≠ CD. Show that there exists a point M such that AB/CD = MA/MD = MB/MC.

Solution by Vo Duc Dien

Extend BA a segment of AE and AB a segment of BF such that AE = BF = CD. Link EC and DF. From A draw a line to parallel EC and meet BC at Q. From B draw a line to parallel DF and meet AD at P.

We have

AB/CD = AB/EA = BQ/QC = AB/BF = AP/PD (i)

Construct a harmonic subdivision for segment AD by drawing the two circles with incenters D and P and with their radii being DP and PA, respectively.

Draw an arbitrary line from circumcenter D to cut its circle at I, and from A draw AJ (J on the circle with center A) so that AJ || DI. Link and extend IJ to meet the extension of DA at K. The four points D, P, A and K are said to be in harmonic order, and we have

AP/PD = AK/DK.

Similarly, constructing the harmonic subdivision for segment BC, we get the point N such that BQ/QC = BN/CN.

Draw the circles of Apollonius with diameters KP and NQ. These two new circles intercept at two points (in our graph) M whereas, by Ray’s theorem (see proof below), we have
∠AMP = ∠DMP, or MA/MD = AP/PD = AB/CD (according to (i)), and

∠BMQ = ∠CMQ, or MB/MC = BQ/QC = AB/CD.

And, at long last, AB/CD = MA/MD = MB/MC.

Proof

First construct a harmonic subdivision for segment AB using the method above. Connect and extend IJ to meet the extension of AB at D. Next, draw a so-called Apollonius circle with diameter DC. The locus of the points on the plane, for which the ratio of the distances to two fixed points A and B is a constant, in this case equals to AC/BC, is the Apollonius circle.

Indeed, pick an arbitrary point P on the Apollonius circle. Draw a line through point C that is also parallel to PD; this line meets AP at E. Link and extend PB to meet the extension of EC at F. By Ray’s theorem, EC/PD = AC/AD and CF/PD = BC/BD. We have r/R (the ratio of the two radii) = BC/AC = IB/IA = BD/AD (since IA || IB), or AC/AD = BC/BD, and thus EC=CF. Also because EF || PD, and ∠CPD = 90°, ∠ECP = 90°, and ΔPCE is congruent to ΔPCF, implying that PC is bisecting ∠EPF.

Therefore, PA/PB = AC/CB.

This problem is only for people who are interested in it.