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.

#Proofathon

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Comments

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 $\prod_{cyc}\frac{\cos(\theta_A+2\theta_B+\theta_C)}{\cos(2\theta_A+\theta_B+\theta_C)}=1.$

Jack D'Aurizio - 7 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$