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22 sats \ 6 replies \ @south_korea_ln OP 28 Oct \ parent \ on: [Daily puzzle] Proving a curious equality science
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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500 sats \ 3 replies \ @Scroogey 28 Oct
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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11 sats \ 2 replies \ @south_korea_ln OP 28 Oct
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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0 sats \ 1 reply \ @Scroogey 28 Oct
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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1 sat \ 0 replies \ @south_korea_ln OP 28 Oct
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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22 sats \ 1 reply \ @south_korea_ln OP 28 Oct
Next hint: law of sines...
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1 sat \ 0 replies \ @south_korea_ln OP 28 Oct
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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