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Concurrency, What? - Problem 7

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.

Note by Cody Johnson
3 years, 9 months ago

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Without 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 $\prod_{cyc}\frac{\cos(\theta_A+2\theta_B+\theta_C)}{\cos(2\theta_A+\theta_B+\theta_C)}=1.$

- 3 years, 9 months ago