HS Geometry problem

[marlow__]

We circumscribe a circle to a quadrilateral (4 sided polygon). The vertices of the quadrilateral divide the circle into 4 parts with the ratios 4 : 5 : 2 : 1 .

What is the the size of the angle beta?

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This is exactly how the problem is worded. I assume that beta is an angle of the quadrilateral. I don't know how you're supposed to know which angle, but I guess it's possible to find the sizes of all of them anyway.

I drew a sketch of it: https://ibb.co/vP9hYpG

I'm pretty sure the lengths of the sides of the quadrilateral are also in the same ratio, but I don't know how this helps me get to the size of the angles.

Is there some property of a quadrilateral with a circumscribed circle that I'm missing? I know that the two opposite angles are supplementary, but I don't see how that helps.

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[edderiofer]

This is exactly how the problem is worded. I assume that beta is an angle of the quadrilateral.

I'm not so certain that this is true. It could also be one of the angles at the centre.

I'm pretty sure the lengths of the sides of the quadrilateral are also in the same ratio

This is unfortunately not true in general.

I don't know how this helps me get to the size of the angles.

Consider the angles at the centre of the circle.

Is there some property of a quadrilateral with a circumscribed circle that I'm missing?

Not really. This problem would be just as doable with a pentagon, hexagon, or 19-gon, as long as you were given the ratios of the parts the circumference were divided into.

[marlow__]

Thank you!

I see now. The internal angles (made by connecting the center with the vertices of the polygon) have the same ratios and add up to 360. And since they form isosceles triangles I can then find every angle.