Spiral Knights - <<RIDDLE ME THIS, RIDDLE ME THAT>> Puzzle Contest! {**PUZZLE 5 being created....**}

<<RIDDLE ME THIS, RIDDLE ME THAT>> Puzzle Contest! {PUZZLE 5 being created....}

75 replies [Last post]

Sun, 02/05/2012 - 15:20

Sorbcell

So, after finally coming to realize there are no legitimate puzzle contests in the Spiral Knights community, I have decided to run a puzzle contest that runs on a monthly basis. This is how it works:

Everyday, I will be releasing a single (1) puzzle on the forums (in this thread) that range from brain teasers, to logic problems, to numeric conundrums, to riddles. The first person to answer each puzzle correctly by posting their answer through this thread will automatically qualify for the GRAND FINALE ENIGMA held on the last Saturday of every month. This GRAND FINALE ENIGMA will be held at a particular location and time, yet to be specified, in Spiral Knights. It will consist of 5 questions/puzzles purely related to Spiral Knights. Whoever answers the most correctly will win the grand prize (**plus an additional 250ce if the winner managed to be the first to get all 5 questions correct**). Prizes will be given to 2nd and 3rd place as well.

To add another element to the contest, the first five people who just "show up" for the event, will gain automatic entry into the GRAND FINALE ENIGMA.

There will also be a draw prize given out for the best looking costume as well, judged by the knights of Benevolent Knights and THE MR. HIPSTER(and any other sponsors).

The Grand, 2nd, 3rd, and draw prizes have yet to be specified.

I am looking for sponsors to make this event a more enjoyable occasion. The more sponsors I get to help out with prizes, the more valuable the prizes will be.

PLEASE make sure you leave your IGN listed with your answer to any of the puzzles in this thread.

I am open to any suggestions or comments.

Cheers,
Sorb

Current Puzzle (Correctly answer by Lord):

I can circumnavigate the world
When I am Googled, I can't be found
I know no limits or bounds
I can organize governments and revolutions
Even though I can't be seen
Many say I am the puppet master of sorts

What am I?

Past Puzzle #3 (Correctly answered by Fortune):

Lord works for the CIA :O. He was presented with the following anagrams that are ALL types of sports except for one...
Picking out the one that is not a sport will give him a pay raise at the CIA :D.
Which of the following anagrams is not a type of sport:

1)TIN NAB MOD
2)CANT REED
3)ILL PECK LAB
4)CAR SOLES
5)RAT WAR

**Please name off the 4 types of sports as well in order to be correct with your answer. Also, if you can spell the non-sport word you get a cookie :D**

Past Puzzle #2 (Correctly answered by Lord):

Xairathan was playing a game of pool one night, and questioned, "hmmmmm, I wonder what the area of the triangular frame for the pool balls is?"
He knows that each ball is 2 inches in diameter and that all 15 coloured pool balls fit exactly into the triangular frame.
What is the area of this particular pool frame?
**Note: Remember that the edges of a pool triangular frame are rounded at the corners**

List of Qualifiers for the GRAND FINAL ENIGMA:

-Lord (2x)
-Fortune
-Xairathan

Sun, 02/05/2012 - 15:30

Shadownox

This is awesome! Except that

This is awesome!

Except that i'm very confused...

I'm not sure what you are talkiing about. Are the questions hidden around the forums? do u have to answer ingame?

Anyway, my IGN is Shadownox.

Sun, 02/05/2012 - 15:44

Sorbcell

Woops

Yea, for the bi-weekly questions I will post them through this thread.
For the GRAND FINALE ENIGMA, it will be ingame at a location I haven't said yet.

Sun, 02/05/2012 - 16:15

Windsickle

IGN: Windsickle

Windsickle hungers for mental stimulation.

Sun, 02/05/2012 - 16:53

Sorbcell

And so it begins...

First official puzzle released! (*see bottom of original post)

Sun, 02/05/2012 - 17:13

Isisdelltion

IGN: Votabbot

4328, because it is the only repeating number (it appears twice).

Sun, 02/05/2012 - 17:55

Xairathan

IGN: Xairathan

5449 is the odd one out. The outer numbers multiplied = the inside numbers. However, 5x9 =/= 44

Sun, 02/05/2012 - 19:39

Shadownox

dang, i think xairathan got

dang, i think xairathan got it... plus, i cant think of anything else XP

so, i guess i wont answer. but does that mean u dont get that 50CE prize anymore for each question, but i just qualifies u for the enigma thingy?

Sun, 02/05/2012 - 22:26

Frostrex

...

Sun, 02/05/2012 - 22:47

Windsickle

5449

I'm going to guess 5449 also, but for a different reason than Xairathan. Though I haven't run it through any spectacular series of tests, I am led to believe it's the only prime number of the group.

Mon, 02/06/2012 - 01:14

#10

Skold-The-Drac

Same name

A particular hint would be nice. However 5449 and 4328 stand out as the most suspect numbers for two alternate reasons.
5*9 =/= 44 (indeed)
4328 appears twice in the series.

Now the 2nd place tie goes to the two odd numbers on the board... (due to the possible hint of "odd" being within the question) which 5449 again would appear to follow.

With all that on top of 5449 being prime whilst the others are not...

5449 stands out the most as the odd number out.

Mon, 02/06/2012 - 14:17

#11

Grittle

9273 is the odd one, its the

9273 is the odd one, its the only with the 1st number 9

Mon, 02/06/2012 - 16:15

#12

Fallenhope

IGN Fallenhope

5408 is the odd one out. its the only one that has a 0.

Mon, 02/06/2012 - 16:14

#13

Sorbcell

Changing of the Guard

Ohhhh what the heck! Looks like the SK community is really enjoying the puzzles, so now I will be releasing one everyday.
Xairathan was correct with his answer to the puzzle. He qualifies for the GRAND FINAL ENIGMA at the end of the month.

@Windsickle: Your theory proves correct, but Xai answered before you :D
@Isis: Skold summed up why the solution you chose isn't the correct answer. But nice try :)

As for Shadow's question, ce won't be released to people who answer the puzzles in the forums, because of the strain it would put on my funds. I'm trying to rally up some people that would be willing to donate some prizes, along with myself, for the grand finale at the end of the month.

I will be leaving the past 3 puzzles in the original post just in case someone missed a day and wanted to take a crack at the puzzle for themselves.

Happy Solving,
Sorb

Mon, 02/06/2012 - 16:44

#14

Xairathan

Guess...

My guess is that the rounded things don't affect the area; therefore diameter 10 x height 10 = 100, divided by 2 = 50 square units.

I might change this answer later :/ I'm not too sure of myself

Mon, 02/06/2012 - 17:49

#15

Shadownox

Guess too... XD

okay.. so each one has a diameter of 2, and the area of a circle is pieRsquared, so 3.14 times 1 is 3.14, and 3.14 times 3.14 is 9.8596. fifteen ball total,

so ties 15, and we get... 147.894 sq.units. :/ i tink i got that wrong... XD but the u have to include the hiehgt, so times two, and my answer is...

295.788 sq.units. XP

IGN, Shadownox

Mon, 02/06/2012 - 18:28

#16

Fallenhope

maths..... great..... -_-

Mon, 02/06/2012 - 21:40

#17

Mindspark

Well, I'm not sure how to

Well, I'm not sure how to account for the rounded corners, but if it were all straight edges, I'm not sure, I got 43.30127019. I have a feeling that it is incredibly easy to account for the corners.

ING = MindSpark

Mon, 02/06/2012 - 20:20

#18

Thimol

-----

I know how to subtract the area off the rounded edges (tee-hee) but I'm not entirely certain as to how to get the dimensions of the whole triangle.

Quite a conundrum here.

Mon, 02/06/2012 - 20:24

#19

Shadownox

guys, guys... its rounded,

guys, guys... its rounded, but so are the balls, so it fits perfectly! DUH!

Mon, 02/06/2012 - 20:42

#20

Thimol

-----

No, this answer was wrong. Lulz.

Mon, 02/06/2012 - 21:29

#21

Windsickle

I believe that Xairathan mentioned that she was female somewhere

1+2+3+4+5+6+7+8+9+10+11+12+13+14+15=120
Finding the actual area is deceptively difficult. Just taking the area of all the balls (in 2 dimensional terms) does not quite make it as there will be space between the balls that will not be included in that calculation but will still count as part of the area of the triangular frame (which we assume that what is being asked for is the area contained within the frame and not the frame's actual area). It seems to me that the best way to go about this would be to calculate it in the way that Xairathan tried, just taking the area of the triangle pool frame. What Xairathan did doesn't work exactly though because the height of the triangle is not given. Assuming that this is a regular 60-60-60 triangle, we still can't be sure of the exact height yet because some of the balls overlap and leave spaces. That is my deduction so far though I find myself no closer to the answer. All I seem to have accomplished is to show that everyone else is wrong. I probably should have waited to show this until I found the answer myself. Oh well.

Mon, 02/06/2012 - 21:46

#22

Xairathan

EDIT

He said area. I'll assume that means cubic meters

Take Length x Width x Depth

Length = 5 balls = 10 units
Width = 5 balls = 10 units
Depth = 1 ball = 2 units

So... if my theory is correct, then the answer is still.. 100 units. :/

EDIT: I meant 'cubic inches herp'
Anyways. Still. 100 cubic inches. :/

Dang it, I'm gonna brb. Gotta work on this.

Mon, 02/06/2012 - 21:35

#23

Mindspark

Well it appears Thimol has

Well it appears Thimol has figured out what I can't figured out... but it appears he can't figure out what I figured out, so we're slightly at a stalemate.

Uh... Xairathan, he said two inches... why would it be cubic meters?

Congratulations Sorbcell. You've stumped us all.

Mon, 02/06/2012 - 22:30

#24

Nechrome

Oh wat? You moved them?

Oh wat? You moved them? Wait... THis is confluzzling. There's one in the GD and on here... o wait, the one in the GN is the glitch one. XP my bad.

BUZZ. Answer=15pi inches squared. 20pi if you want the balls volume times pi instead.

Unless there are spaces between the balls. that's take some time. But I'm good at math. :)

Edit: after massive amounts of calculations, I now know the answer. it is 24 + 16root3 + pi

although I sort of cheated. I got my mom's help. XP

Mon, 02/06/2012 - 22:04

#25

Shadownox

... my sister just told how

...

my sister just told how to calculate for spheres T_T

but im too lazy to change my answer and do the math, so i'll just wait for tomorows question :D

does the person who gets it closest win if no one gets it?

Mon, 02/06/2012 - 22:40

#26

Skold-The-Drac

Depending on the triangle (wish I had the pool triangle on me)

@shadownox, close no cigar
A rounded triangle with holes is formed when you find the area of the balls, which is 15pi once everything is condensed and reduced I leave any number multiplying pi alone.. since pi is far more than 3.14 simply.

Let's find the area of the outer holes of the rack, holes between the balls and the rack itself. I'll use rectangles in this case that start at the tangent of the two outermost circles and the pool rack without it dropping from the rectangular shape To find the area of the holes, I'll cut out the area of the balls the rectangle will impede upon.

4xball diameter = 8in. = length of filled rectangle
.5xball diameter = 1in = width of filled rectangle
Now, in drawing a diagram of what I'm trying... which is a tad different than the notes, it still shows that the area of 4 halves of the ball can be taken out of the filled rectangle (in order to make the original area found 15pi sq. in. fit) the found area has to be multiplied by three. For the three sides.

3((8 -2(pi*1/2))= area of filled holes(1) = 14.575 sq. in. After that point.. I can breakdown the rest of the holes with an equilateral with 4xBd for it's side lengths (stay with me).

AoFH(1)+AoFH(2)+15pi sq. inch. = A of Pool rack.

AoFH(2)
Equilateral is 27.7128 sq. in. subtracting the area of 5 total balls due to how it cuts them up. (love geometrics here)
27.7128-5pi = AoFH(2) = 3.1416 sq. in. approx. (or exactly if you follow scientific rule)

SO! now to make everything fit together. And get the area of this thing.

14.575+3.1416+15pi= 64.840 sq. in. rounded to the thousandth. Guess what?

The pool rack's area is 64.840 sq. in. Enjoy geometry.

I'll put all this on paper and scan if you guys don't exactly understand how I got to my answer.

Mon, 02/06/2012 - 22:46

#27

Shadownox

Mon, 02/06/2012 - 22:46

#28

Xairathan

Answer

After lots of calculatoins I have a different answer

16(root 3) + 24 + pi
= 54.85

Mon, 02/06/2012 - 23:18

#29

Skold-The-Drac

@Xairathan

Care to show your process? As best you could? If not it's fine, just asking as to how you performed the math.

Mon, 02/06/2012 - 23:24

#30

Xairathan

@Skold

Well I kinda like... put circles around certian areas and... it's kinda hard D:

like... kinda like I split the triangle into four, and assigned values and whatnot per different area, and.... long story short, I got that >.>;

Mon, 02/06/2012 - 23:29

#31

Skold-The-Drac

'Tis fine. I did a lot of that in High School. Form something up in my brain and somehow it fits perfectly... just don't ask how exactly XD
Still do it a lot... I just had to talk myself through this problem since my problems usually consisted of acceleration and other concepts.

Mon, 02/06/2012 - 23:30

#32

Xairathan

Hmm

Guess we have to wait for tomorrow to find out who was right XD

Tue, 02/07/2012 - 16:39

#33

Sorbcell

I admit this was a bit of a tricky one...

Well after seeing all your answers, Lordofnecromancers (I'm presuming this is your IGN) had the first correct answer (even if it was from his mom :D).
He now qualifies for the GRAND FINALE ENIGMA at the end of the month!

Nice try Skold, but you duplicated a couple of areas within the area of the pool triangle :D

Here's the official work:

So the very first step is to make a trianlge (we will call this triangle, TRI 1) within the pool triangle. The sides of TRI 1 will go through all of the centres of the outtermost balls.
Since the radius of each ball is 1 inch, then the length of each side of TRI 1 will be 8 inches (there are 8 raddis on each side of this newly created triangle). Cut TRI 1 in half and use Pythagorean theorem to calculate the height of this triangle. The height of the newly acquired triangle is 4root3, so the height of TRI 1 is also 4root3. To calculate area of TRI 1, multiply the base (8 inches) by the height (4root3 inches), then divide by two. From this, the area of TRI 1 is equal to 16root3 square inches.

Now by using the sides of TRI 1, create 3 rectangles that (one on each side of the pool triangle) that borders the side of the pool trianlge and TRI 1. Since the sides of TRI 1 are 8 inches long and the distance from the side of TRI 1 to the side of the pool triangle is 1 inch (because the sides of TR1 pass through the centres of the outtermost pool balls and the radius of the balls is 1 inch), then by using the area of the rectangle forumla (lw=A), each rectangle has an area of 8 square inches. So the total area of the 3 rectangles combined is 24 square inches. Now are we done yet? Nope...

We still have to deal with the curved sections of the pool triangle. The only sections left to solve for the total area are parts of the 3 pool balls that are at the corners of the pool rack. So, the area of a circle can be represented by (pi)r^2. But these 3 pool balls aren't complete circles...hmmmmmm. Well to approach this, we need to know how much of each circle we have left. To do this we have to go back to the triangle we created earlier (TRI 1). Now, TRI 1 passed through the centres of all the outtermost balls in the pool triangle so it passed through the 3 corner pool balls as well. Since TRI 1 was an equilateral triangle, each vertex of TRI 1 has a measure of 60 degrees. And since the vertices of TRI 1 were in the centre of each corner ball, we can take away 60 degrees from the total degrees a circle has (360 degrees) to get 300 degrees of a circle left for each corner ball. YES, WE FINALLY KNOW HOW MUCH OF THE 3 CORNER CIRCLES WE HAVE LEFT!! or do we....

Well we haven't accounted for the rectangles we created after making TRI 1. Since a rectangle has a measure of 90 degrees at each vertex, we must now take out the 90 degrees that chewed into each of the 3 corner balls. But wait...it's not only an additional 90 for each ball, it's an additional 180 degrees because rectangles were created on two sides of each of the corner pool balls. So subtract the 180 degrees from the 300 degrees to get 120 degrees.

So we now have 120 degrees left of a circle with a radius of 1 inch. To find the total area of this part circle, we know that 120 degrees is 1/3rd of 360 degrees, so the total area of the part circle is 1/3rd of (pi)1^2. Which is (pi)/3 square inches. Since we have 3 corner balls, then the total combined area of the three is just pi square inches.

So now to get the total area of the pool triangle, we add 16root3 square inches + 24 square inches + pi square inches=54.85 square inches

Alright well time for the next puzzle! This one will be for more of you wordy folk out there...

EDIT: Remember to include proof for how you got to your answer for future questions! :)

Happy Solving,
Sorb

Tue, 02/07/2012 - 17:14

#34

Shadownox

Rail rein-airlinerNod

Rail rein-airliner
Nod goal-gondola
Aim loo tube-automobile
Carve forth-hovercraft

I think i got this one... XD

oldpear is the non-type of transport.

IGN: Shadownox

EDIT: i just searched it up on google, and i found the website with that puzzle :O

Tue, 02/07/2012 - 17:17

#35

Thimol

-----

So now to get the total area of the pool triangle, we add 16root3 square inches + 24 square inches + pi square inches=54.85 square inches

CRAP
That was close to my initial answer (52.something). I discarded that because I thought that the height of the triangle was the same as the length of the sides of the triangle.

Anyway, I'd used some trigonometry to figure out the length of the sides, and used a triangle circumscribed about a circle to figure out how to account for the rounded corners in calculating the area of the rack.

AWWGH

Tue, 02/07/2012 - 18:32

#36

Sorbcell

LOL @Shadow

Thanks for your honesty Shadow. I'm sending out a new puzzle as we speak. Sorry Shadow but this won't qualify you for the GRAND FINALE at the end of the month. I thought I google proofed this question...guess not. :D

Tue, 02/07/2012 - 23:14

#37

Sorbcell

Change of Heart

Since you pointed out that it was on a website Shadow I will give you the qualification. It was my fault that I didn't make sure it couldn't be googled.

Wed, 02/08/2012 - 16:12

#38

Fortuneangel

IGN: Fortuneangel

Umm.... I don't know the fourth sport but I'll give it a try
1. Badminton
2.Dent Race (Is this even a real sport? Probably not but whatever)
3. Pickleball
4. Lacrosse
5. Art War ( LOL this and Dent race are the ones that I am unsure of)

I know I'm wrong but I think I got 3 right.

Wed, 02/08/2012 - 14:54

#39

Windsickle

I got the same as

I got the same as Fortuneangel. I was able to come up with a variety of words for #2 but hardly anything for #5 so I can't yet be sure.

Wed, 02/08/2012 - 15:57

#40

Sorbcell

I take it you guys are stumped

The fourth sport that you two are unsure about originates in the Thai-Malasyia area. It is relatively unknown to North Americans. I'll give you guys a hint: It's a mix of two games that both use a ball.

Wed, 02/08/2012 - 16:06

#41

Windsickle

Sepak takraw? I'm stumped.

Wed, 02/08/2012 - 16:29

#42

Shadownox

For the RAT WAR, i got

For the RAT WAR, i got trawar... apparently, its somehting for poker. i think RAT WAR isnt the sport. i didnt get number two either though...

OMG thanks sorb! i was doing the recreated one last night, but then my mom wanted to use the comp, and then i had to go the sleep, so i never finshed my answer.... XD

Wed, 02/08/2012 - 17:09

#43

Fortuneangel

Is it just takraw?

Takraw is a mixture of volleyball and soccer and uses a rattan ball

Wed, 02/08/2012 - 18:51

#44

Sorbcell

No one has pointed the odd one yet

I still haven't seen the odd one yet mentioned by anybody LOL. The puzzle is still up for grabs :D

Wed, 02/08/2012 - 19:15

#45

Fortuneangel

So we need to say the odd one

So we need to say the odd one too? I thought it was only for a cookie? (:

Wed, 02/08/2012 - 19:20

#46

Sorbcell

Yea it is

But you have to list which anagram is not the sport. You don't have to spel it out.
.

Wed, 02/08/2012 - 19:30

#47

Fortuneangel

Is it cantered?

Is it cantered? (CANT REED?)

Wed, 02/08/2012 - 19:53

#48

Xairathan

Guess

looking at everyone's, I'd have to guess 'CANT REED' is not a sport.

Wed, 02/08/2012 - 20:04

#49

Sorbcell

Congrats

Yep CANT REED is the anagram that isn't a sport. The word it spells out is a decanter.
Congrats Fortune! You have qualified for the GRAND FINAL ENIGMA at the end of the month.

Alright the next puzzle is going to be a riddle.
Good Luck!

-Sorb

Wed, 02/08/2012 - 21:13

#50

Windsickle

So what was the fourth sport?

So what was the fourth sport? I must know.

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<<RIDDLE ME THIS, RIDDLE ME THAT>> Puzzle Contest! {**PUZZLE 5 being created....**}

<<RIDDLE ME THIS, RIDDLE ME THAT>> Puzzle Contest! {PUZZLE 5 being created....}