10402  Triangle Covering
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Re: mistaken
It was jst a mistake. Thanx for informing this to us. want to replace the whole problem?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.
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 Guru
 Posts: 1080
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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 rightangled 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.
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 rightangled 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.
Re: ATTN : PROBLEMSETTER 10402 (TRIANGLE COVERING)
For t3 there is analytic solution: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.....).
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) );