reply on: [Daily puzzle] Proving a curious equality \ stacker news ~science

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Angle units are

\frac{1}{7}\pi

Angle sums in triangles must be 7 (=180°), in quadrilateral 14 (=360°).

Note two pairs of similar triangles. One pair is isosceles.

Ptolemy says

 $c_1*c_2 + c_1*c_3 = c_2*c_3$

Hence

 $sin(\frac{1}{7}\pi)*sin(\frac{2}{7}\pi) + sin(\frac{1}{7}\pi)*sin(\frac{3}{7}\pi) = sin(\frac{2}{7}\pi)*sin(\frac{3}{7}\pi)$

Equals

 $sin(\frac{1}{7}\pi) = \frac{sin(\frac{2}{7}\pi)*sin(\frac{3}{7}\pi) }{sin(\frac{2}{7}\pi)+sin(\frac{3}{7}\pi)}$

Equals

 $\frac{1}{sin(\frac{1}{7}\pi)} = \frac{1}{sin(\frac{2}{7}\pi)} + \frac{1}{sin(\frac{3}{7}\pi)}$

I guess the Ptolemy hint did the trick. Or maybe with just the figure, you'd have figured it out too...

There are some beautiful other problems in Math Horizons, I'll have to take some time to curate the ones that are easy to post here.

What makes this so tricky is that there is no right triangle involved, yet you look for one because

sin()

is involved. At least I did for too long. Thank you!

Yeah, with the law of sines, one can get such right angles, but this is basically by constructing a new set of triangles. See #743146