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Geometry Difficulty 6.7 National Olympiad Prove it Romania

Given a triangle ABCABC, a circle centered at some point OO meets the segments BC,CA,ABBC, CA, AB in the pairs of points XX and XX', YY and YY', ZZ and ZZ', respectively, labeled in circular order: X,X,Y,Y,Z,ZX, X', Y, Y', Z, Z'. Let MM be the Miquel point of the triangle XYZXYZ (i.e., the point of concurrence of the circles AYZ,BZX,CXYAYZ, BZX, CXY), and let MM' be that of the triangle XYZX'Y'Z'. Prove that the segments OMOM and OMOM' have equal lengths.
Figure 1

Solution

We begin by reviewing some basic facts on conics. For an ellipse Σ\Sigma with center NN, foci MM and MM', semiaxes aa and bb, it is known that the orthogonal projections PP and PP' of MM and MM' on any line tt tangent to Σ\Sigma lie on the major auxiliary circle of Σ\Sigma, so that NP=a=NPNP = a = NP'. Application to triangle MNPMNP (respectively, MNPM'NP') of a rotation θ\theta (respectively, θ-\theta) about MM (respectively, MM') and a homothety of ratio secθ\sec\theta with center MM (respectively, MM') yields triangle MOXMOX (respectively, MOXM'OX'), where MO=MOMO = M'O, NO=12MMtanθNO = \frac{1}{2} \cdot MM' \cdot \tan\theta, OX=asecθ=OXOX = a \sec\theta = OX', and X,XX, X' both lie on tt. If tt varies and θ\theta is constant, the locus of XX and XX' is then a circle Γ\Gamma centered at OO.

By Cartesian geometry it is readily checked that Γ\Gamma and Σ\Sigma are bitangent, and the line \ell supporting their common chord is also the radical axis of the circles Γ\Gamma and OMMOMM', with this real geometrical significance even if the bitangency is not real. Since \ell and the circle OMMOMM' are mutually inverse in Γ\Gamma, the inverse points of MM and MM' in Γ\Gamma both lie on \ell. Finally, the distance dd between the parallel lines \ell and MMMM' is given by dMM=2b2tanθd \cdot MM' = 2b^2 \tan \theta. Similar considerations hold for a hyperbola Σ\Sigma.

Consider now an isopair M,MM, M' (two isogonally conjugate in the triangle ABCABC, the foci of a conic Σ\Sigma touching its sides), and take points X,Y,ZX, Y, Z (respectively, X,Y,ZX', Y', Z') on lines BC,CA,ABBC, CA, AB, respectively, so that the lines MX,MY,MZMX, MY, MZ (respectively, MX,MY,MZM'X', M'Y', M'Z') make the same directed angle θ\theta (respectively, θ-\theta) with the perpendiculars to BC,CA,ABBC, CA, AB, respectively; then the isopedal triangles XYZ,XYZXYZ, X'Y'Z' of angles θ,θ\theta, -\theta for the isopair M,MM, M' have their Miquel points at M,MM, M' and are inscribed in a common isopedal circle Γ\Gamma bitangent to Σ\Sigma, centered at a point OO on the perpendicular bisector of the segment MMMM'.
Conversely, for any pair of triangles inscribed in a triangle ABCABC and in a circle Γ\Gamma (as in the statement of the problem), the Miquel points M,MM, M' are an isopair and Γ\Gamma is an isopedal circle of M,MM, M'. (If M,MM, M' are the Brocard points, Σ\Sigma is the Brocard ellipse and Γ\Gamma is a Tucker circle.)

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.