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CTK Exchange
darij
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Sep-04-04, 08:57 AM (EST) |
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1. "RE: Seemingly simple problem"
In response to message #0
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This is a problem from an Indian mathematical olympiad of the year 2002. It's actually one of those op02 geometry problems I didn't solve. Anyway, I'll give it another try. It actually does NOT look simple to me, as usually such problems with implicite conditions aren't too simple. Darij |
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grobber
Member since Jul-27-04
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Sep-09-04, 03:35 PM (EST) |
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4. "RE: Seemingly simple problem"
In response to message #3
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You might want to try to prove this first: Given a cyclic quadrilateral ABCD, let M, N be the midpoints of AC, BD respectively. Show that AC bisects angle BMD iff BD bisects angle ANC. This can be used to show that the quadrilateral we're concerned with is cyclic (some sort of "converse", although not exactly a converse). This, I think, gives further insight. Moreover, the cyclic quadrilateral satisfying this is a special kind: a harmonic quadrilateral, i.e. one for which the tangents to the circumcircle in A, C concur on BD (and those in B, D concur on AC). |
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