Help on math see picture

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Help on math see picture

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Google says: " A hexagon is inscribed in a circle radius 1. Alternate sides have length 1. Show that the midpoints of the other three sides form an equilateral triangle" -http://mks.mff.cuni.cz/kalva/putnam/putn67.html "Each side length 1 forms an equilateral triangle with the centre. Let the other three sides subtend angles 2A, 2B, 2C at the centre. The midpoints of the sides subtending 2A and 2B are distances cos A and cos B from the centre and the line joining them subtends an angle A+B+60 at the centre. So, using the cosine formula, the square of the distance between them is cos2A + cos2B - 2 cos A cos B cos(A+B+60) (*). But A + B + C = 90, so cos(A+B+60) = cos(150-C) = -√3/2 cos C + 1/2 sin C. Hence we may write (*) as (cos2A + cos2B + cos2C) + √3 cos A cos B cos C - cos A cos B sin C - cos2C. But we can write cos C as sin(A + B) = sin A cos B + cos A sin B and hence cos2C as sin A cos B cos C + cos A sin B cos C. Thus (*) has the symmetrical form (cos2A + cos2B + cos2C) + √3 cos A cos B cos C - (cos A cos B sin C + cos A sin B cos C + sin A cos B cos C), which establishes that the triangle is equilateral." - http://mks.mff.cuni.cz/kalva/putnam/psoln/psol677.html
I didn't read it myself, but I hope this helps you understand the proof!

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

The only thing that seems different is that this link has the radius be 1, while the radius in your problem has the radius be r.
|dw:1351743906628:dw|
I don't follow the proof :P Good luck!
ok thanks alot i read over this and try to work it out myself
Cool; Good luck!

Not the answer you are looking for?

Search for more explanations.

Ask your own question