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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_3Hence
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)}reply
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.
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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!reply
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
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Next hint: law of sines...
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As an alternative to the solution in #743121, With the law of sines, one can get to the same conclusions. The triangles to be built are each using an edge of the quadrilateral, an edge going through the center of the circle, and the remaining edge connecting the two endpoints of the first two edges. This construction always gives a right angle with a sine equal to one, and the
R^2 will cancel out. The remaining sine is the one that will remain in the final expression.reply
P_iare equally spaced on the circle, and the black shape is a convex quadrilateral.