Lateral thinking puzzles

[Chronos]

The plumb-bob method won’t work for the cake: Picture a “cake” shaped like a dumbbell, with almost all of the mass in two lumps at the ends, with a rigid but lightweight rod connecting them. Now suppose that the two lumps at the end have different masses, say 2 kg and 3 kg. The center of gravity will be about 40% of the way from the heavy lump to the light lump, but if you cut there, you’ll get one 2 kg piece and one 3 kg piece.

[typoink]

re: the bear question. My own personal answer when I first encountered that was white, because that’s the only bear with a habitat specific enough that a goofy question like this might make sense.

The “real” answer makes sense, but I find mine more elegant.

[BrotherCadfael]

@Khadaji

But of course, I made it clear that because a bear is involved, it is probably at the north pole.

In Ernest Thompson Seaton’s Two Little Savages, several farm boys spend part of a summer (1917) living as nearly as they could like Canadian Indians, whom they admired. One of the boys, who had some formal education, demonstrates the use of geometry to measure heights and distances:

This White-man’s Woodcraft was easy for him, and he volunteered to teach the other Indians, but they thought it looked “too much like school.”

“Say, Beaver,” said Woodpecker sadly, harking back, “if a Dog’s front foot is 3-1/2 inches long and 3 inches wide, what colour is the end of his tail?”

“White,” was the prompt reply; “'cause a Dog with feet that size and shape is most likely to be a yaller Dog, and a yaller Dog always has some white hairs in the end of his tail.”

“Well, this 'un hadn’t, 'cause his tail was cut off in the days of his youth!”

[Boozahol_Squid_P.I]

Take a tic-tac-to board and draw four straight lines without lifting your pen from the paper, but going through the central point of each and every square.

[SleepyDuck]

Quiz for Executive Ability

Too outside the box? Not boxy enough? Eh, I found it mildly amusing.

[Damuri_Ajashi]

Most of these have been answered but for the ones that haven’t.

@Hampshire

This one was in the latest Professor Layton game. I got it fairly quick but still thought it was clever:

A teacher has 10 students. She randomly places a red or blue hat on each student (they can not see their own hat). She then tells each student that if they can see 5 red hats or more they can have a red balloon. If they can not they get a blue balloon. In the end she handed out some red balloons and some blue balloons. How many students had blue hats on?

5 of each.

[quote=“Robot_Arm, post:47, topic:513008”]

[spoiler]Find the centroid of the uncut cake. (The centroid of a geometric object is the point at which any straight line will divide the shape into two parts of equal area. In this case, the center of the rectangle.) Next, find the centroid of the shape that has been cut out of the rectangle. On the line connecting those two points, make a straight cut all the way across the cake.

Any cut through the centroid of the cake will divide it into equal pieces. If the cut also passes through the centroid of the missing piece, each half will be missing exactly half of the empty void.

Can anyone spot the flaw with this answer?[/spoiler]

You would get two peices of cake with equal amounts of cake but they would not look the same. For example if you took off a corner you would have different corners of the cut peices missing a corner.

@Max_Torque

One I swiped from a Piers Anthony book:

You have a balance scale, and twelve coins. The twelve coins are, to all outward appearances, identical. However, one of the coins is definitely a counterfeit. The only way it differs from the other coins is by weight. The counterfeit may be lighter than the other coins, or it may be heavier; you only know that the weight is different. You cannot determine the weight by touch.

In no more than three weighings on the balance scale, find the counterfeit. Note: there is no “trick”, only a method.

In this box is the method I know:

[spoiler]First weighing: 1 2 3 4 against 5 6 7 8. Note which side is heavier and which is lighter.

Second weighing: 1 2 3 5 against 4 9 10 11. Again, note which side is heavier and lighter. This is where the gimmick lies: swapping one coin from each side of the scale to the other.

Now, let’s talk about the possibilities, of which there are only three. If the same side is heavier or lighter in the second weighing, then you know that the counterfeit must be 1, 2, or 3, and you know whether the counterfeit is heavy or light. Just weigh 1 against 2 for your final weighing to find the counterfeit.

If the opposite side is heavy or light for the second weighing, you know that the counterfeit must be either 4 or 5, because those coins switched sides, but you don’t know yet if the counterfeit is heavy or light. Weigh 4 against a known good coin to find the counterfeit. If they unbalance, it’s 4; if they balance, it’s 5.

And if the second weighing results in a balance, you know that the counterfeit is one of the coins you removed (6, 7, or 8), and you know whether the counterfeit is heavy or light. For the third weighing, weigh 6 against 7.

And of course there was an additional possibility at the first weighing that it would result in a balance. If weighing number one balances, there’s no need to swap coins; just make the second weighing 1 2 3 against 9 10 11, and go from there.

Note that it’s possible to find the counterfeit without determining whether it’s heavier or lighter in certain situations. However, that doesn’t matter; the only problem the puzzle poses is “find the counterfeit”, not determining whether it’s a heavy or light counterfeit. Simply weighing the coins as groups of four will not find the counterfeit in all situations, but the method of switching places with two coins for the second weighing works no matter where the counterfeit is.[/spoiler]

I have heard the same puzzle with a similar answer with 6 orbs and 3 weighing (but you have to tell me if the odd ball is heavier or lighter) and I have heard the same puzzle with 8 balls in 2 weighings but you know the odd ball is heavier.

@Boozahol_Squid_P.I

Take a tic-tac-to board and draw four straight lines without lifting your pen from the paper, but going through the central point of each and every square.

Isn’t this the same as the original “think outside the box problem”?

[Damuri_Ajashi]

@Malleus_Incus_Stapes

Actually, it IS for a game.

Not that I’m admitting that I play, mind you mumble mumble mumble event horizon of geekery mumble mumble my brothers made me mumble…

…But my brothers want me to finish the layout for their new Dungeons and Dragons adventure, and they want it yesterday, and I had this lovely riddle prepared as the centerpiece. Even sign up and the OOTS forum to ask for some advice. Unfortunately, my brother also went to the OOTS forum to browse, and now I’m out a riddle. :smack: I need something that’s moderately difficult and references a physical object.

You could use the princess and the tiger. Two doors two guys standing next to the doors one always lies and one always tells the truth. You get to ask one question and then you have to open a door. pick the right door a princess comes out, pick the wrong one and a tiger comes out.

[Robot_Arm]

@col_10022

You would get two peices of cake with equal amounts of cake but they would not look the same. For example if you took off a corner you would have different corners of the cut peices missing a corner.

The two pieces wouldn’t be the same shape, but they would be equal in area, which is what really counts when you’re dealing with cake.

The flaw I was thinking of is that the missing piece may be concave, such that the centroid is not within the shape. Imagine the missing piece is in the shape of a U, oriented so that one side faces the center of the cake. A line through both centroids will cut across the U, dividing the cake into three pieces. (Two large pieces, and a small half-circle that was the bottom of the interior of the U.) You can still combine two of them to be equal in area to the third, however.

[Chronos]

Just dug up an old favorite of mine:

You are an archaeologist that has just unearthed a long-sought pair of ancient treasure chests. One chest is plated with silver, and the other is plated with gold. According to legend, one of the two chests is filled with great treasure, whereas the other chest houses a man-eating python that can rip your head off. Faced with a dilemma, you then notice that there are inscriptions on the chests:

Silver Chest:
This chest contains the python.

Gold Chest:
One of these two inscriptions is true.

Based on these inscriptions, which chest should you open?

[ZenBeam]

@Chronos

Just dug up an old favorite of mine:

Either one. If you find the bones of a long dead python, you can always open the other one.

[Boozahol_Squid_P.I]

@col_10022

Isn’t this the same as the original “think outside the box problem”?

Damn, it is. Didn’t see the link, and completely didn’t connect the fact that it’s the origin of the phrase.
:smack:

[Chronos]

Clever, ZenBeam, but if this is for a D&D game, it might be a zombified python, or a half-demon one, or an animated clockwork python assembled by an insane gnome artificer, or whatever. Or, of course, you could replace it with some other sort of trap.

[Captain_Awesome]

In its strict sense, the puzzle is insoluble since ‘one of these two inscriptions is true’ doesn’t exclude the possibility that both are true. More loosely, I presume you’re aiming for the silver chest housing the treasure.

[Chronos]

OK, then, change the gold plaque to “exactly one of these two inscriptions is true”.