In \(\triangle ABC\), let \(M\), \(N\), and \(O\) be the midpoints of \(BC\), \(CA\), and \(AB\), respectively. \(\triangle ABC\) is reflected over an arbitrary line \(\ell\), forming \(\triangle A'B'C'\). Show that

- the lines parallel to \(B'C'\), \(C'A'\), and \(A'B'\) through \(M\), \(N\), and \(O\), respectively, are concurrent.
- the lines perpendicular to \(B'C'\), \(C'A'\), and \(A'B'\) through \(M\), \(N\), and \(O\), respectively, are concurrent.

Conjecture (proof unnecessary, but interesting if presented) as to how this can be generalized.

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TopNewestWithout affecting any angle, we can assume that the line \(l\) goes through the circumcenter of \(ABC\), hence \(A',B',C'\) lie on the circumcircle \(\Gamma\) of \(ABC\). We can further assume that \(\Gamma\) has a unit radius, then \(A=e^{i\theta_A},A'=e^{-i\theta_A}\) and so on. Now we can prove both concurrencies by invoking the trigonometric form of the Ceva theorem with respect to the triangle \(MNO\), having its sides parallel to the sides of \(ABC\). Hence the first concurrency follows from \[\prod_{cyc}\frac{\sin(\theta_A+2\theta_B+\theta_C)}{\sin(2\theta_A+\theta_B+\theta_C)}=1,\] while the second concurrency follows from \[\prod_{cyc}\frac{\cos(\theta_A+2\theta_B+\theta_C)}{\cos(2\theta_A+\theta_B+\theta_C)}=1.\] – Jack D'Aurizio · 2 years, 9 months ago

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