10545  Maximal Quadrilateral
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10545  Maximal Quadrilateral
I thoght much about the problem. It is not so hard. I try to find out the formula that solved this problem but fail I used some kind ob binary search (actually, it was 3ary search) to solve it and get AC. But I want to know if there is another (easier) way to solve it.

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hmm
I also don't know the formula
Can you pm it to me as well.
A lazy problemsetter
Can you pm it to me as well.
A lazy problemsetter

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Looks like it's also called "tangential quadrilateral", ... http://www.mathworld.com has some interesting properties (and formula) about it.
turuthok
turuthok
The fear of the LORD is the beginning of knowledge (Proverbs 1:7).

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hehe
OH! I know these formulas. I thought from one of the previous posts that there is a direct formula for for maximal radius.
This is a problem for online contest, geometric knowledge cannot be judged. Everything is available in the Internet.
This is a problem for online contest, geometric knowledge cannot be judged. Everything is available in the Internet.

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I tried many times and always get wrong answer. I think my algorithm is ok. Can anyone give me some critical input? thanks
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Last edited by sharklu2000 on Sat Aug 30, 2003 10:43 am, edited 1 time in total.

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Yep, that's the one. It's as direct as it gets. I dindn't want to publish it, 'cause I think it's a spoiler. But now it's in the open, I don't have to PM it to all you guys
Shahriar: or did you expect anything else? It should be the maximal radius, because there are either zero or one quadrlaterals satisfying the given A, B and P. ( Well, to be honest, I expected a little Manzoortwist, like negative sidelengths, and imaginary radii, when I first read the problem , but that wouldn't fit the description, alas ).
I personaly like Revenger's approach better than using the direct formula. It takes much more run time, but makes it a programming challenge, and not a math contest.
sharklu2000: giving critical inputs would spoil the only fun left in the problem, now you published the formula...
But think of the problemsetter, and his great talent for spotting the trouble zones in simple looking problems. I realy admire him for that. Solving his problems made me a much better programmer. Although it hurts, at first
Shahriar: or did you expect anything else? It should be the maximal radius, because there are either zero or one quadrlaterals satisfying the given A, B and P. ( Well, to be honest, I expected a little Manzoortwist, like negative sidelengths, and imaginary radii, when I first read the problem , but that wouldn't fit the description, alas ).
I personaly like Revenger's approach better than using the direct formula. It takes much more run time, but makes it a programming challenge, and not a math contest.
sharklu2000: giving critical inputs would spoil the only fun left in the problem, now you published the formula...
But think of the problemsetter, and his great talent for spotting the trouble zones in simple looking problems. I realy admire him for that. Solving his problems made me a much better programmer. Although it hurts, at first

 System administrator & Problemsetter
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 Joined: Sat Jan 12, 2002 2:00 am
Thanks
Thanks little joey! Although I know you r not so little . Words like these really will help me keep going. Although I also appreciate good critisisms. but I don't have to hear much critisism these days as Derek Kisman does it successfully while writing alternate solutions.
I always feel that someday I will meet all the prominent members of this site, although I know this is not to be.
I always feel that someday I will meet all the prominent members of this site, although I know this is not to be.
Isn't a rhombus an exception to the above statement.????!!
little joey:
It should be the maximal radius, because there are either zero or one quadrlaterals satisfying the given A, B and P.
infact there are infinitely many rhombuses with the same sides,
containing an inscribed circle.
the maximal radius in this case would be for the square.
I still am unable to figure out how to do this problem...
Can nebody suggest me more ideas or ways of implementing ideas??
The length of two adjacent sides of the quadrilateral are given as A & B. The total perimeter, i.e. the sum of A, B & the rest two sides is also given as P. So the length of the other two sides not specified is what you can vary & in the process of varying those lengths keep track of the maximum radius of the circle that you can inscribe within the quadrilateral. About the shape, I think the quadrilateral with the largest inscribed circle will never be a rhombus because if the length of the two given sides are equal & if the perimeter is double of their length's sum then obviously a square would be what can have the largest inscribed circle.
If you find it too complicated PM me for the mathematical relation. But I guess this problem wasn't meant to be solved using a formula.
If you find it too complicated PM me for the mathematical relation. But I guess this problem wasn't meant to be solved using a formula.
Not all of our dreams can be made to come true. But still we live with the hope to turn them into reality someday.
10545 solving formula
Hi,
A very [b]straight formula [/b]can be used to solve this problem. Thanks to turuthok for his mentioning mathworld site. I just searched there as tangential quadrilateral and found a topic from there I went to Bretschneider
A very [b]straight formula [/b]can be used to solve this problem. Thanks to turuthok for his mentioning mathworld site. I just searched there as tangential quadrilateral and found a topic from there I went to Bretschneider
Self judging is the best judging!
Re: 10545  Maximal Quadrilateral
It can be solved without any hint of the mathworld. I found it to be one of the easiest geometrical problem set by Shahriar Manzoor sir.
I agree with what little joey said. There exist either one or zero quadrilateral that satisfies the conditions.
I agree with what little joey said. There exist either one or zero quadrilateral that satisfies the conditions.