600
   

NeoPets Riddles (Lenny Conundrums) and Answers Here

 
 
amusicluffer
 
  1  
Reply Wed 18 Nov, 2009 03:12 pm
I will be MIA for about 2 hours this afternoon... I hope I can still get on and submit in time! :O
Cross your fingers for me! Smile
0 Replies
 
willzi8
 
  1  
Reply Wed 18 Nov, 2009 03:48 pm

Congratulations! You have guessed correctly in the Lenny Conundrum game (round 333). You have won 691 NP!

Because you were in the first 250 to guess correctly, you also have been awarded a Capn Threelegs Training Sword, and receive a trophy and the Lenny Conundrum avatar!

I got gold Razz
lennyfan
 
  1  
Reply Wed 18 Nov, 2009 04:06 pm
@willzi8,
Nice work - you must have caught it just as it was posted.
willzi8
 
  1  
Reply Wed 18 Nov, 2009 04:09 pm
@lennyfan,
Yeah, I got it within the first 10 minutes Smile
0 Replies
 
willzi8
 
  1  
Reply Wed 18 Nov, 2009 04:10 pm
NEW LC OUT 30 seconds ago!

A Chomby was walking down a path one day when Jhudora appeared in front of it. "Solve this puzzle, and you may pass," she said.

Suppose you have two identical circles that overlap such that the center point of one circle is on the edge of the other circle. If the overlapping area is exactly 20000 square centimetres, what is the radius of the circles? Please round to the nearest whole centimetre, and please submit only a number as your answer
Waunted
 
  1  
Reply Wed 18 Nov, 2009 04:13 pm
@willzi8,
Ew, this one's hard. Good thing I got the avatar today. Smile
0 Replies
 
willzi8
 
  1  
Reply Wed 18 Nov, 2009 04:16 pm
@willzi8,
A hint, think of the circle segments Wink
Kutusita
 
  1  
Reply Wed 18 Nov, 2009 04:16 pm
I tried to do it, but the math is beyond me.

Good luck, everyone!
0 Replies
 
Waunted
 
  1  
Reply Wed 18 Nov, 2009 04:18 pm
@willzi8,
I think I'll just guess. Razz
willzi8
 
  1  
Reply Wed 18 Nov, 2009 04:20 pm
@Waunted,
Note it is a sector (the angle being in the middle) minus a triangle * 2
For the nearly elliptical shape.
Waunted
 
  1  
Reply Wed 18 Nov, 2009 04:21 pm
@willzi8,
Yeah, well the combination of algebra and geometry doesn't work for me. Razz

Thanks for trying to help, though, and congrats on your avatar! I got mine today, too! Last week's was SO easy!
willzi8
 
  1  
Reply Wed 18 Nov, 2009 04:22 pm
@willzi8,
Also look for equilateral triangles, as they will help you calculate the central angle
0 Replies
 
willzi8
 
  1  
Reply Wed 18 Nov, 2009 04:23 pm
@Waunted,
I already had the avatar. I got it in my third week =P
willzi8
 
  1  
Reply Wed 18 Nov, 2009 04:24 pm
@willzi8,
You also need to use the law of cosines
Waunted
 
  1  
Reply Wed 18 Nov, 2009 04:25 pm
@willzi8,
Well, there's only room for one math wiz here, so you can take that role. Razz

Congrats again, and talk to you next week when I try again to get a gold trophy. Very Happy
0 Replies
 
Sniffy
 
  1  
Reply Wed 18 Nov, 2009 04:27 pm
@willzi8,
law of cosines? I have to google, lol. This one is going to take me all week.
willzi8
 
  1  
Reply Wed 18 Nov, 2009 04:34 pm
@Sniffy,
It's really quite easy to calculate the intersection of the two equations with a graphing calculator

I got a whole # am I supposed to add .0?
willzi8
 
  0  
Reply Wed 18 Nov, 2009 04:40 pm
@willzi8,
NVM didn't get a whole number, I set my calculator to round for me xD
0 Replies
 
Kutusita
 
  0  
Reply Wed 18 Nov, 2009 05:00 pm
I was able to find enough bits and pieces for this type of puzzle on the Internet, but I am almost certainly wrong. I got an answer where I rounded up.
0 Replies
 
medy
 
  0  
Reply Wed 18 Nov, 2009 05:09 pm
Solution is a bit hard to describe... the two overlapping circles will create an equilateral triangle between them (all three sides of the triangle are of the length 'radius').
As all angles in an equilateral triangle are 60degrees, you know the sections of the circles that overlap are both 120degrees (or 1/3rd of circle surface).

We can now express both the surface area of these triangles in the circle and the area of the section of the circle in their radius. If we were to calculate twice the area of such a circle-section (once for each circle) we would have added in the surface of the two triangles within twice, so we'll have to subtract these again.

Surface area of the overlap in r is then: 2 * (1/3 * pi * r²) - 2 * ( root(0.75)*0.5 *r²). This will have to equal to 20000, working back we can conclude that r = 127.4, rounded down as the LC requests makes for 127!
 

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