Two Apollonius circles whose radical axis goes through the circumcenter

1hpClutch·Mar 2, 2025(edited)

The ratios 21:20:19 look numerical, but the conclusion feels structural. Could this be an Apollonius-circle argument rather than a long coordinate expansion? Problem: Let

A

B

C

P

and

Q

be five pairwise different points in the plane. Suppose that

A

B

and

C

are not collinear and that

\frac{AP}{BP} = \frac{AQ}{BQ} = \frac{21}{20},

\frac{BP}{CP} = \frac{BQ}{CQ} = \frac{20}{19}.

Prove that the line

PQ

contains the circumcentre of the triangle

ABC

. Source: Baltic Way 2020 · MNET-01H2

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4 replies

parry_this·Mar 3, 2025(edited)

Put the circumcenter O at the origin and write the common circumradius as R. The locus

|X-A|/|X-B|=\lambda

has equation

|X|^2-2U\cdot X+R^2=0

after dividing by

1-\lambda^2

, with

U=(A-\lambda^2B)/(1-\lambda^2)

1hpClutch·Mar 3, 2025(edited)

Then its power at O is R². The B:C ratio produces another circle with the same power at O, so O lies on their radical axis.

parry_this·Mar 3, 2025

Right, and P,Q are two distinct common points, so their line is that radical axis. The two circles must be distinct: equal centers would make a nontrivial affine relation expressing A as a combination of B and C, contradicting noncollinearity.

midnaMain·Mar 5, 2025

The ratios being different from 1 ensure we have actual Apollonius circles rather than perpendicular-bisector lines. The specific numbers are less important than they first appear.

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