10402 - Triangle Covering

All about problems in Volume 104. If there is a thread about your problem, please use it. If not, create one with its number in the subject.

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dwyak
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Post by dwyak » Sat Oct 18, 2003 12:23 pm

ooh... that's quite amazing.

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shamim
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Post by shamim » Sun Oct 19, 2003 4:42 pm

Just for the record, what is actually the trouble with this problem. :lol:

Andrey Mokhov
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Post by Andrey Mokhov » Mon Oct 20, 2003 7:07 am

Hello everyone!

Mr. Andrew Stankevich, you are right !!! I'm quite sure that the problem is in the "sin" instead of "tan" as you said :D :P !

Thank's a lot!
Andrey.

helpless
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Post by helpless » Mon Oct 27, 2003 4:14 am

there is ambiguity in the picture of T4 and may be judge's solution wrong for T6

Sajid
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Re: mistaken

Post by Sajid » Fri Oct 31, 2003 12:28 pm

shahriar_manzoor wrote:Triangle Covering 239 ( 0.8 %)

I will have to set a new problem to replace it so the replacement is taking time.
It was jst a mistake. Thanx for informing this to us. want to replace the whole problem?

*** I am one of your fan and had a chance to see you in my Uni. 8) :wink:
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little joey
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Post by little joey » Sun Mar 21, 2004 8:08 pm

Is there any news about this problem yet?
I see quite some people 'solved' it by now, but I haven't heard from Shahriar Manzoor that the problem was OK. Or did I miss something?
Why is the problem still submittable if it's incorrect?

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little joey
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Post by little joey » Thu Sep 29, 2005 2:16 pm

Well, still no news, so we'll have to accept the fact that this problem will remain unfixed forever.
Since people almost got killed over this problem, let me try to summarize what you have to do to get Accepted. Thanks to Ivan Golubev and Andrew Stankewich who addressed the problems in other threads.

T2 and T3 are OK.

T4:
To get a full coverage, let the bottom two squares intersect at the base of the triangle (P3 in Ivan's picture). Now the intersections of the top two squares (P1 and P2 in Ivan's picture) will be just outside the triangle, thereby violating assumption b) of the problem statement. But to get Accepted you'll have to ignore this.

T6:
The problemsetter used a wrong formula to calculate the ratio.
If you look at the picture, you see that the middle square at the bottom of the triangle is tilted slightly to the left. Call the angle of tilt alpha (it's just over 9 degrees).
Now the part of this square that's below the base of the triangle forms a right-angled triangle with sides s, s/cos(alpha) and s*tan(alpha) (s is the side of the square, as in the problem statement).
So the length of the smallest side is s*tan(alpha), but the problemsetter erroniously used s*sin(alpha) in his calculations of the ratio. To get Accepted you'll have to make the same mistake, which adds 0.000309018459573 to the ratio, as Andrew reported.

Hope it helps some people who spent fruitless hours solving this problem, only then to discover on the forum that the problem is actually broken. Now you can at least get Accepted as a reward for your efforts, albeit for the wrong reasons...

You learn a lot from your own mistakes, but you learn almost as much from other people's mistakes, and it feels a lot better.

feodorv
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Re: ATTN : PROBLEMSETTER 10402 (TRIANGLE COVERING)

Post by feodorv » Fri Jan 06, 2017 9:16 pm

Mahbub wrote:So the prob is for t3 and t6..where he stated that t3 = 2.108+ and t6 =
3.168+..which do not hav analytic solutions. (He didnt wrote the equation
either...:(..).
For t3 there is analytic solution:

Code: Select all

double q3 = sqrt( 3 );
double t = (2 + q3 - sqrt( 4 * q3 + 1 )) / 3;
double a = q3 * sqrt( 1 + (1 - t) * (1 - t) );
And judge's solution for t6 is really nasty :evil:

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