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I'm solving some of my countries olympiad problems, and after manipulating all of the data I was able to get the title of the question. I have an isosceles trapezium $ABCD$ which has all of its vertices inscribed in a circumference, and I need to find the circle's radius. The side lengths are $AB=4$, $AD=BC=2$, and $CD=6$. The angles are $\angle ADC = \angle DCB = 60$ and $\angle BAD = \angle CBA = 120$.

I know there is a ridiculous formula which uses diagonals and semi perimeters, but I don't think that is the intended answer. In fact, I suspect the answer is half 3, half of $CD$, but I haven't been able to prove it.

I'm not sure if this matters anymore, as I used it to get to this part of the problem, but $AD$ and $BC$ intersect at $E$, and form an angle of $60$ (all numbers on degrees).

Can't provide a sketch, but it should be easy to draw.

Dollar Store Richard
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  • Are you sure you have the side lengths correct? AB,AC,AD,CD don't form sides of a trapezium. Anyway, the answer cannot be half 3 (i.e. 3/2), or half CD (i.e. 3). – user10354138 Oct 19 '21 at 11:51
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    Hello, why can't you provide a sketch? Also every time you post a question on the site, you should share your attempt. – Math Lover Oct 19 '21 at 11:52
  • Also I think you should have written $AD = BC = 2$? – Math Lover Oct 19 '21 at 12:05
  • Sorry, just read all of this. I did indeed right stuff wrong, will correct. Cant provide a sketch cause I frankly don't know how to. And I didn't include my attempt, as this isn't the full problem. It involved using the fact that $\angle AED$ was 60 to find that $DC$ was 6. Afterwards I got stuck, and tried to, unsuccessfully, prove arc $CD=180$ – Dollar Store Richard Oct 19 '21 at 13:38
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    Software for drawing geometry diagrams, Actually we welcome even a diagram drawn by hand. The only purpose is to understand your question better. – ACB Oct 19 '21 at 13:56
  • @DollarStoreRichard it still does not justify you posting a question without showing your efforts. Here is a hint - the center of the circle $O$ is on perpendicular bisector of $AB$ and $CD$ (parallel sides of trapezium) and as $\angle D = 60^\circ$, what angle does diagonal $AC$ subtend at center $O$? So if you find $AC$, you are done. – Math Lover Oct 19 '21 at 14:26
  • I'm too bad at this to conclude anything. Sorry. I have no idea the angle $AC$ subtends at $O$ (or what that even means). – Dollar Store Richard Oct 21 '21 at 22:42

1 Answers1

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Let O is center of circumference, E is middle of AB, F is middle of CD, OA=$R$, OE=$x$. Then OF=$|\sqrt{3}-x|$.

Then $R^2=OA^2=AE^2+OE^2=4+x^2$, $R^2=OD^2=DF^2+OF^2=9+3-2\sqrt{3}x+x^2$.

Then $4+x^2=9+3-2\sqrt{3}x+x^2 \Rightarrow$ $2\sqrt{3}x=8 \Rightarrow$ $x=\frac{4}{\sqrt{3}} \Rightarrow$ $R^2=4+\frac{16}{3}=\frac{28}{3} \Rightarrow$ $R=2\sqrt{\frac{7}{3}}$.

Ivan Kaznacheyeu
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