Please don't spend too much time on this ;)
The first hint will likely save you a lot of time. Without it, I would never have guessed the direction of the solution suggested by the authors of this problem.
I'll try posting it tomorrow after i wake up.
Here a figure that should help you figure out the method. Or at least, point you in the direction of a possible solution. Points are equally spaced on the circle, and the black shape is a convex quadrilateral.
A first hint would be to look up Ptolemy’s inequality.
I'll wait a bit before giving out the next hints ;)
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Angle units are .
Angle sums in triangles must be 7 (=180°), in quadrilateral 14 (=360°).
Note two pairs of similar triangles. One pair is isosceles.
Ptolemy says
Hence
Equals
Equals
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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 is involved. At least I did for too long. Thank you!
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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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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 will cancel out. The remaining sine is the one that will remain in the final expression.
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stackers have outlawed this. turn on wild west mode in your /settings to see outlawed content.