2 Olympiad Geometry problems I like

[sirio11]

Tom, in my solution I used n mini circles with r_k = d_k / 2 and the big circle with R = (3/2)r.
So, A_k = pi (d_k^2/4) and using the areas:

pi (d_1^2/4)+pi (d_2^2/4)+......+pi (d_n^2/4) <= pi(9/4)r^2

Therefore d_1^2+d_2^2+......+d_n^2 <= 9r^2

And yes, my 1st thought was that the 9r^2 was way too high. It would be interesting to try to solve the problem for 7r^2, as soon as I have some time I'll try.

[sirio11]

[ QUOTE ]
Good catch. If ACE is obtuse, then it's tougher to cover ACF and/or BCF, where C/F are the obtuse angles, than it is to cover ACE, and the shortest distance across the hexagon is CF, not AC/CE/AE. Then the problem reduces to the parallelogram ACEF, and by similar analysis as for ACE in the other post, the CF distance is maximized to sqrt(2) as EFA reduces to 90 degrees from above.

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If ac,ce or ea is <= sqrt2 we're done; so wlog we assume ea >= ac>= ce> sqrt2
d(c,ae) <= 1 therefore angle ace must be > 90

Now, wlog let angle fce > 45, d(f,ce) = fc*sin(fce) >= sqrt2*sin45 > 1, similarly we obtain d(e,fc) > 1; therefore {a,c,e} can not be covered by a strip of width 1 contradicting the hypothesis.

[TomCowley]

Your solution for half-radius circles is the same as mine, but it doesn't explain why quarter-radius circles (or anything else) wouldn't be better.

[sirio11]

[ QUOTE ]
Your solution for half-radius circles is the same as mine, but it doesn't explain why quarter-radius circles (or anything else) wouldn't be better.

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You wrote:

[ QUOTE ]
Assuming that points are evenly spaced (tedious but not hard to show that non-evenly spaced points can never be optimal)

[/ QUOTE ]

My solution does not need this (and the tedious work to show it)

Explaining why quarter-radius (or anything else) wouldn't be better is quite trivial.

Edited to add: And probably you already meant that, but my solution is not about half-radius circles, using the d_k/2 as radius is quite different from R/2 and avoids several of the problems involved with using R/n. Not to mention the algebra is nicer, you don't need any limits .... [img]/images/graemlins/smile.gif[/img]

[TomCowley]

Ahh, I see it now. I wan't reading your notation right. Nice.

[jay_shark]

I think we can do better for the first problem .
I conjecture that the lhs is <=7*r^2 .

Equality occurs when we have a regular hexagon inscribed in a circle .

[TomCowley]

I made the same observation. No clue how to prove it though.