NeoPets Riddles (Lenny Conundrums) and Answers Here

Hmm, by very, very large, I meant bigger than 7 digits :/

My answer was 7 digits that added up to 27, for the record.

I thought it was pretty easy to find proper equations. They're all on wikipedia. As long as you remember your properties of regular polygons it's not bad at all. Took me less than 20 minutes (Although we'll find out next week if it was as easy as I thought it was)

Thank you... I have revised my math and have also gotten an answer with 7 digits, adding up to 27.

Agreed, 7 digits adding up to 27.

For those still working on it, a triacontakaihenagon is a 31-sided polygon. We are given the length of a side of 1 meter, but this is the outside edge of the wall. If we find the distance from the center of the polygon to the center of one of the sides, the radius of an inscribed circle, we can then subtract 30 cm from that to get the radius of an inscribed circle for the inside of the tower. From that, you can find a formula to calculate the area of the inner polygon, multiply it by the height and use that fact that a cubic meter contains 1000 liters.

Don't forget to subtract 30cm from the height as well to take into account the thickness of the roof.

Uhhh.. i read that, and I'm like: WHHAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAT??????

A TRIAHENTAHECKWHATAGON?

HAHA! Apparently I did it the hard way. I didn't even think about inscribed circles. I used trapezoids to figure out the length of the inside walls and then worked from there. At least we probably got the same answer anyway. *sheepish grin*

Your approach works fine. You can find the angle of the wider part of the trapezoid (PI/62 radians) and then use the tangent to find out how much shorter the interior wall is than the outside wall. As you said, either way should give you the same answer.

I would like to point out that if you don't assume that a 31 sided polygon approximates a circle, and you actually take the time and posses the knowledge to do so, apply calculus, specifically integration in polar coordinates, one could find a more precise answer.

Personally I ran the calculations in three different ways. The nieve methods of estimating with a circular base .3 m less than the polygon measures at the circumradius gave an answer about .7% above the actual value. The other circular method, .3 m less than the apothem length, gave a value about .3% less than actual.

For those curious enough to get my better answer, I'll leave you with the hint, r=.5cot(pi/31)/cos(x) - .3; for 0<=x<=pi/31.

what about any thickness for a floor?

I considered that, but it doesn't mention anything about a floor at all. Specifically just the walls and roof.

He didn't approximate a circle. He used an INSCRIBED circle to find the apothem (distance from the center of the polygon to the edge) and used that to calculate the area. It's a much easier way to do it than polar calculus. It's super-basic geometry applied on a 31-sided figure.

The only reference I used to figure out this problem was
http://en.wikipedia.org/wiki/Apothem
It has all of the formulas to find the apothem and to use it to find the area of any regular polygon (which, fortunately, this tower is). The hardest thing I had to do was figure out the inside length of the walls, which I was able to do because I know the interior angle (b/c of the regular polygon) and could make each wall segment into a trapezoid (long side=one meter, height=.3 meters).
No calculus necessary!

And as for your "better" answer, what did the digits add up to? Did you get an answer with digits adding up to 27 like the rest of us? Our answer should be the same, since it isn't an approximation at all.

I jumped to conclusions before I had read all of the posts here. Yes, my answer will be marked wrong. However, the thickness of the walls using a smaller 31 sides figure on the interior would be inconsistent. Sure, the thickness at the apothem is .3 meters, but what about at the circumradius? I'm sure your digits adding to 27 is more likely to be the desired answer, however depending on interpretation of thickness, it is not the only "correct" answer.

Yes, the thickness measured between a vertex of the inner polygon and the corresponding vertex of the outer polygon will be more then .3 meters. I based my calculation of the thickness at the apothem based on two things:
1. Previous experience with LC's that have wall thicknesses, and
2. An assumption that when measuring the thickness of any wall, the proper way to do it is along a line that is perpendicular to the planes of the inner and outer wall surfaces. This could be at the apothem or at any other location where the inner and outer surfaces are parallel. Knowing that the apothem was at such a point gave me a convenient way to calculate the apothem of the inner polygon and then calculate its area.

A particularly speedy Meerca decided he was going to run around the entire planet of Neopia.

He started completely stationary, and began accelerating. His acceleration increased at a rate of 0.1 metres per second per second per minute.

Assuming the planet has a diameter of 1600km and is a perfect sphere, and disregarding any pesky obstacles like water or mountains or Snowagers, how many seconds will it take the Meerca to circumnavigate the planet? Round UP to the nearest second, and submit only a number with no other information.

Prize is a Grey Doughnut.

With a quick calculation, I get 4 digits, sum of 16.

I can confirm 4 digits with a sum of 16 after way more work than it should have been. It was almost 10 minutes that I realized I should be working from Jerk rather than acceleration (it's CHANGE in acceleration). And then I thought I remembered you said a sum of 14, and so I checked my formulas twice before I looked and realized I had been right all along.

And yay I got in the first 250 two weeks in a row. Awesome!

You may even hit a gold trophy this week. People seem to be having terrible trouble understanding that the rate given is jerk instead of acceleration. I haven't seen our answer posted elsewhere yet and it is 40 minutes after you posted.

This is a particularly fun one to help people because I can get snarky and say "jerk" and they'll think I'm insulting them even though I'm "helping" them.

Is it too much to ask where you would suggest looking for the appropriate equations? I've never used jerk before, and after almost an hour of searching, I haven't been able to come up with much more than a definition. I doubt they would post a LC that requires a graduate physics degree, so I must be doing the wrong google searches.