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anonymous
 3 years ago
Circle problem
anonymous
 3 years ago
Circle problem

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Please refer to the attachment.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Intuitively, I think I is correct. But I don't know how to ensure the other two are correct or not..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think you are right about I. The sum of the angles is always 1080 degrees. II is also right, because of this: imagine O is at height h above line CD. If you draw a horizontal line through O, it intersect both AC and BD, lets say in A' and B'. Then it is easy to see that triangles AA'O and BB'O are congruent. This means AA'=BB'=a. SO AC = ha and BD = h+a, total length is 2h, so constant. U can use this also to reason whether ACxBC is constant or not!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Why is ACxBD NOT constant? (so III is false).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1357564433476:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So, C and D are not fixed point, i.e. they'll change when the diameter rotates. AC // BD, so theta + phi must be 180 degrees (int. angles, AC//BD) The two triangles are congruent (ASA), so both sides have the same (magnitude of) change. And AC+BD would remain constant as 2h. And the last point: AC x BD = (ha)(h+a) = h^2  a^2, which is not a constant. So, it's the answer. Thanks everyone!!!
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