ZA Furries

Rakuen is quite correct: the answer is 562. Thank you for participating.

I am beginning to think I ought to have this thread renamed to "Galahad's Puzzles & Riddles". I do not mind it, but I think the era in which those who answered also submitted puzzles and riddles of their own might be over.

In any case, here is another for our resident brainiacs! I was inspired to make this one after I had to wash, iron and pack away my laundry (solving a puzzle based on it is much more fun than actually doing it, trust me):

It is laundry day, but it has happened again: your favourite pair of briefs - because we all know boxers are the inferior choice of underwear - has been misplaced.
In your bedroom, you have five small closets in a row next to one another. From left to right, these closets are numbered one to five. You know your beloved briefs are in one of these, but you are not sure which. As such, you ask your friend, who did the laundry:

"Oh... I cannot remember, either. But I do remember some facts that might help you locate them immediately!

a) I remember putting your jersey either in closet 1, 4 or 5. Your briefs are not with your jersey.
b) They are definitely not in a closet that is next to the closet that has your hat. (Your hat, jersey and briefs are each in a different closet.)
c) The remaining two closets are empty. I also remember that there are two closets between the empty closets.
d) The closets containing your jersey and your hat are not next to each other.

Well, genius, in which closet are your briefs?"

You smile. "Pfft, I'll tell you more than that. I'll tell you exactly what is in each closet."

What do you say?

Sent you a PM of the answer.

I like solving the puzzles. Don't know many myself though and don't have too much inclination to make my own.

My goodness, I have yet to reveal the solution to this one! My apologies for the delay!

Rakuen and CyntheWightRabbit both submitted answers to me, and both were correct. Well done, gentlemen!

The solution for the previous puzzle is:

Closet 1: Hat
Closet 2: Empty
Closet 3: Briefs
Closet 4: Jersey
Closet 5: Empty

Thank you for your participation.

Now, how about another puzzle? This one, however, will be somewhat different - a little more technical. I am taking a slight risk here, as I strive to ensure puzzles do not require extensive background knowledge to solve. The following puzzle is an analytical geometry problem, but all you need to solve it is to know the Pythagorean theorem, the area of a square and the area of a circle: so, basic high school maths. However, I shall provide all three in the problem so you do not need to go searching.

If you did maths in high school, then you know enough to solve this.

Have a look at the diagram below:

Geometry Puzzle.png (4.52 KiB) Viewed 5042 times

It shows a quarter of a circle enclosed in a square, with a whole circle on its shoulder. Assume all the shapes are perfect (even if on the diagram they may be just slightly deformed) - for example, the sides of the square are all equal, and the circle is perfectly round. The length of the side of the big square is 1. (Units of measurement don't matter.)

What is the area of the RED region in the corner?

You can use a calculator to give an exact answer, or you can give it in the form of a sum you worked out.

Formulas you will need:
Area of a square = (Length of side) × (Length of side)
Area of a circle = (π) × (radius) × (radius)
Pythagorean theorem: a × a = (b × b) + (c × c) [Where a is the diagonal side of a right-angled triangle, and b and c are the two other sides]

If enough people are able to solve this, I might make more complicated geometric problems in the future. This is just a test to see how keen my solvers are on this sort of problem.

the answer is not big enough... assuming it's a square pizza (I didn't do too well at maths... but I did great at cooking...)

Well, I certainly know whose place to visit for lunch should I ever go to the U.K.

Hate geometry questions!

Solved it. But I don't want to make a riddle...

Spoiler

I named the sides:
D C
A B

And cut the smaller circle in half. Called them E (Between A and D) and F(Between D and C)

So. The area of the smaller area EDF is the area of ADC - Area of Trapezium ACFE - 1/4 area of the smaller circle.

Radius of the bigger circle is equal to BC or AB.
Using Pythagoras: 1.414

It's touching the smaller circle. And thus. The radius of the smaller circle is 0.414
Area of that is thus 0.414(pi)
1/4 of that is 0.065

We ennd the diagonal length of the trapezium to calculate it's area. So we use sin(theta) = A/h
A = 0.6748

Ao the area of the Trapezium is given by [(a+b/2)h]
= 0.48

Lastly. The big triangle ADE is simple substitution.
=0.71

THUS:
0.71 -0.48 -0.065

=0.165unit^2

Can't you draw that out?

Rakuen Growlithe wrote:Can't you draw that out?

Ummm... I guess I can show my calculations. Though my sketch ability is really bad.

But it's really hard to translate those back on to the figure.

Unfortunately, no-one got this one right. Funnily enough, the first answer I drafted for the puzzle memo was incorrect as well, so the puzzle even confounded the puzzler for a time. However, I think the time has come for the solution to be revealed:

The answer is about 0.006. To get it:

1) Find the length of the diagonal yellow line above - drawn from the top left corner to the edge of the bigger circle. To get it, simply subtract the radius of the big circle (1 unit) from the diagonal length of the big square (√2, calculated by means of the Pythagorean theorem).

2) Think about what this equals. If we call the unknown radius of the smaller circle r, then that yellow line is equal to the diameter of the smaller circle (2 × r) plus the small portion of the yellow line that crosses the shaded area (between the top left corner and the edge of the smaller circle). Let's call that portion of the line that crosses through the shaded region x.

3) The length of x is equal to r × (√2 - 1). This is because if you were to imagine an even tinier square that encloses the top left quarter of the smaller square, what that tiny square contains would be like a microscopic version of the biggest square and circle - it has thus been scaled down according to a proportion or ratio. So, the proportion by which the radius has been scaled down, will be the same as the proportion by which that diagonal line has been scaled down, and we know that the bigger diagonal line is equal to (√2 - 1) when r = 1.

4) You can now work out r. Once you have that, the area of the shaded region is simply the area of the smaller square ((2r)^2) minus the area of the smaller circle (π(r)^2) divided by four. (Divided by four because, before it is divided by four, it is the area of the four corners around the circle.)

I think it is time to attempt to revive this thread once more.

While playing Skyrim, I came across a puzzle that inspired me. I created one based on it, that was similar but just a little harder than the one that appeared in-game, albeit still moderately easy.

You are an explorer/adventurer/Indiana Jones cosplayer, and, while exploring the ruined tomb of an ancient empire, came across three triangular stone obelisks, with an image etched on each of the three sides of each of the three obelisks.

Going counter-clockwise around the sides, the images are:
An owl; a raven; a crocodile.

You find that you can turn each pillar clockwise (and only clockwise), so that the side facing you changes by one turn. So, for example, you can turn an obelisk with a raven once, and now the owl will show.

From a carved inscription on the ceiling of the chamber, you can see that, from the first to the third, the images on each obelisk are supposed to be in this final order: Crocodile, Owl, Raven. You are expected to turn each in such a way as to create this order.

However, there are three catches!
1) When an obelisk is turned by one turn, it also turns all the obelisks behind it by one turn. So, if you turn the third obelisk once, it also turns the first two once. If you turn the first obelisk once, it just turns once by itself, because there are no obelisks behind it.
2) You have to begin with the first obelisk, and work to the third - in one direction, only.
3) You can only manipulate each obelisk once; in other words, once you are done turning an obelisk (however many times you wish), you cannot come back to it.

In their initial order, the obelisks from first to third, are: Owl, Owl, Crocodile.

How many times will you turn each obelisk?

It's a bit confusing with the third one having the others behind it because shouldn't it be first... You could turn them as many times as you like to get the final pattern.

Also if the pattern is going counter clockwise then turning a pillar showing a raven will show the crocodile, not the owl! Let's assume a clock face. The pattern is counter-clockwise, so owl at 9 o'clock, raven at 6 o'clock and crocodile at 3 o'clock. If you are facing the raven (6 o'clock) and turn it clockwise, then the owl will be at 12 o'clock, the raven will be at 9 o'clock and the crocodile will be at 3 o'clock.

To make it easier, a circular pattern can be visualised as a repeating linear pattern. The starting position will be this.
1 Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile
2 Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile
3 Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile

I can then turn the third position twice to Raven, which also turns the other two as well.

1 Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile
2 Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile
3 Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile

Then I turn the second one once to owl which also moves the first one but doesn't change the third.

1 Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile
2 Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile
3 Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile

And finally I turn the first one twice to crocodile which will not affect anything else.

1 Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile
2 Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile
3 Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile, Owl, Raven, Crocodile

Rakuen Growlithe wrote:Also if the pattern is going counter clockwise then turning a pillar showing a raven will show the crocodile, not the owl!

Um.

Galahad wrote:You find that you can turn each pillar clockwise (and only clockwise)

And also...

Galahad wrote:2) You have to begin with the first obelisk, and work to the third - in one direction, only.

Perhaps my wording was a little vague. I thought I had made it clear enough, but perhaps I should clarify:

I described the image sequence in a counter-clockwise direction, but you can only turn each obelisk clockwise. So one has to do more thinking.
And as for the sequence you work in: simply imagine a row of obelisks stretching out in front of you. The "first" is the one closest to you, with the two others in front of it - it is an owl. The "second" is in the middle - it is also an owl. The "third" is the last one, with the two others behind it - it is a crocodile. You begin with the first, and work forward.

Whoops. Missed point two. ^^;

But I'm still right about turning them. Try drawing it. If you describe the pictures going counter-clockwise but can only turn it clockwise then they will come back to you. Like going counter clockwise the numbers on a clock are 6, 5, 4, 3, 2, 1... But if you turn the clock clockwise then they will pass also pass you 6, 5, 4, 3, 2, 1...

Your pillar has three images and the order going counter clockwise (blue arrow) is owl, raven, crocodile.
Then if the red eye is the viewer and they are looking at the raven, then rotating the pillar clockwise (green arrow) will move the crocodile into their view and not the owl.

Fair enough. I should have phrased it differently (I was thinking along the lines of the clock hand turning, rather than the clock turning), but it doesn't matter. I understand what you're saying.
Just to clarify it once and for all for those who also want to try it: Let's go with turning referring to the pillar, as I initially said and as Rakuen interpreted it to be.

In that case, I will turn the first one twice, the second one once and the third one twice.

Alright. We have one answer thus far. I'll reveal the solution on Saturday, in case anyone else wants to try it.

You never told me I won.

Cat - 7kg
Rabbit - 3kg
Dog - 17kg

Total: 27kg

So did you get that by struggling through simultaneous equations or really simple logic?

Simple logic. Compare 20kg (dog + rabbit) versus 24kg (dog + cat). Cat must weigh 4kg more than the rabbit.
By similar logic, the dog must weigh 10kg more than the cat. You can then subtract this difference from a total and divide it by two to get the weight of the lighter animal, and substitute from there.

What logic! The picture is highly misleading. The Dog would have to be a puppy or medium size breed to be 17kg. And your Cat and Rabbit should go on a diet.

Ah. I guess that's a third way. My way was even simpler. If you add all three known weights together you know that 54 kg is the weight of 2 dogs, 2 rabbits and 2 cats. Then just halve that to get one of each.

Cape_F0X wrote: ↑Tue Jan 02, 2018 8:41 pm What logic! The picture is highly misleading. The Dog would have to be a puppy or medium size breed to be 17kg. And your Cat and Rabbit should go on a diet.

Eh... average growlithe is 19 kg and the cat could be a Norwegian Forest.

A fine and effective method.

... Until they ask you what each animal weighs.

That part is easy too. Cause then all 3 are 27. So difference between any other group tells you the weight of the missing animal.
e.g. 27 for all 3, 20 without cat, therefore cat is 7. Cat and rabbit are 10 so rabbit is 3. Dog must be 17.

Rakuen Growlithe wrote: ↑Tue Jan 02, 2018 8:42 pm Eh... average growlithe is 19 kg and the cat could be a Norwegian Forest.

But it's not a growlithe! It's clearly a Labrador, our neighbors lab tips the scale at 40kg.
Children should not be misinformed about the weight of healthy pets!

Excuse me, I have a Siberian forest cat and a healthy adult neutered male of that breed can weigh up to 11kg.