10545 - Maximal Quadrilateral

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Revenger
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10545 - Maximal Quadrilateral

Post by Revenger »

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 3-ary search) to solve it and get AC. But I want to know if there is another (easier) way to solve it.

little joey
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Post by little joey »

Yes there is. Some time on google will tell you that (or you can try to derive a formula yourself, but it's not easy).

Publishing it would be a spoiler, so look in your PM :)

shahriar_manzoor
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hmm

Post by shahriar_manzoor »

I also don't know the formula :)

Can you pm it to me as well.

--A lazy problemsetter :)

windows2k
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Re: hmm

Post by windows2k »

shahriar_manzoor wrote:I also don't know the formula :)

Can you pm it to me as well.

--A lazy problemsetter :)
I also want to know the formula :)
Can you pm it to me ?

turuthok
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Post by turuthok »

Looks like it's also called "tangential quadrilateral", ... http://www.mathworld.com has some interesting properties (and formula) about it.

-turuthok-
The fear of the LORD is the beginning of knowledge (Proverbs 1:7).

shahriar_manzoor
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hehe

Post by shahriar_manzoor »

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.

sharklu2000
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Post by sharklu2000 »

I tried many times and always get wrong answer. :x I think my algorithm is ok.

Code: Select all

cutted
Can anyone give me some critical input? thanks
Last edited by sharklu2000 on Sat Aug 30, 2003 10:43 am, edited 1 time in total.

little joey
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Post by little joey »

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 :D :D

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 Manzoor-twist, like negative side-lengths, and imaginary radii, when I first read the problem :) , but that wouldn't fit the description, alas :lol: ).

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_manzoor
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Thanks

Post by shahriar_manzoor »

Thanks little joey! :) Although I know you r not so little :D . 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.

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yahoo
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Post by yahoo »

Thanks Shahriar for giving such a good geometrical problem. Though it can be solved by using a formula. During contest i had no time solve it by proving the formula. So i directly use it and got accepted. But i later saw that it is a good one to proof. :lol: :lol: :lol:

subbu
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Post by subbu »


little joey:
It should be the maximal radius, because there are either zero or one quadrlaterals satisfying the given A, B and P.
Isn't a rhombus an exception to the above statement.????!!

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??

Dreamer#1
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Post by Dreamer#1 »

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. :D
Not all of our dreams can be made to come true. But still we live with the hope to turn them into reality someday.

shanto86
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10545 solving formula

Post by shanto86 »

Hi,
A very [b]straight formula [/b]can be used to solve this problem. :P 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!

Taman
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Re: 10545 - Maximal Quadrilateral

Post by Taman »

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. :)

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