The terms (I believe) refer to multiview orthographic projection, the illustration technique used in architectural drawings: The front elevation is the view looking squarely at the “front” of the object, and the plan view looks down from above. What is the side view?
No explanation is given. I had imagined something like a curved disk, but that wouldn’t give a circular plan view.
The nearest similar puzzle I know is the “architect’s puzzle,” in which you’re asked to imagine a single object that can cast three different shadows: a circle, a square, and a triangle. That’s possible, but it’s a distinctly different task, and it’s hard to see how to adapt that object into the one we’re seeking here.
09/27/2023 UPDATE: I’ve received a ton of mail about this puzzle — a million thanks to everyone who’s contributed; I’m constantly impressed by your intelligence, imagination, and resourcefulness. The consensus is that the solution was published upside down! And the puzzle as presented is a bit ambiguous. Reader Catalin Voinescu explains:
Note there are no additional lines to the semicircle that is the front elevation: that’s all there is to see from that direction. The plan view could be filled or not, and we can’t tell. So the figure must be the intersection of the bottom half of a horizontal cylinder surface (like a half-round gutter) and either a vertical cylinder surface, like a gutter downpipe, or a solid vertical cylinder. If you think of what a T joint of two cylinders of the same diameter looks like from the side, the joints appear as straight lines that bisect the angle between the cylinders. (They must be straight lines bisecting the angle for reasons of symmetry.)
So the given answer is one of two possibilities.
When the plan view is hollow, the figure is a closed curve in three dimensions: draw an ellipse with a long diameter times the short diameter, and bend the paper at 90 degrees along the short diameter. The top view is a circle, the side view a semicircle, the other side view is the angle made by the paper.
However, if the plan view was solid, the figure would be a curved surface: what you need to cut out of the half-round gutter to get the vertical cylinder to go through it. Viewed from the side, it would look like a triangle — like the given solution, but with a line at the bottom.
My thanks to Drake Thomas and to Catalin for their work on all this, and thanks again to everyone else who wrote in about it. I’ve actually gone through the subsequent issues of the journal to see whether an erratum notice was ever published, and I don’t find one. I wonder how much mail this generated in 1958!
A problem from the National Bank of New Zealand Competition 2000, via Crux Mathematicorum, November 2006:
Humanity is visited by three alien races, the Kweens, the Ozdaks, and the Merkuns. Kweens always speak the truth, and Ozdaks always lie. In any group of aliens, a Merkun never speaks first; when it does speak, it tells the truth if the previous statement was a lie and lies if the previous statement was truthful. The three alien races can tell one another apart, but to humans they all look the same. A delegation of three aliens visits Earth. At least one of them is a Kween. When they arrive they make the following statements, in order:
Merkuns never speak first, so the first alien is either a Kween or an Ozdak. Suppose it’s a Kween. That means the first statement is true and the second alien is a Merkun, which means that the second statement is false and thus the third alien is a Merkun. But this would require the third statement to be true (since we’re supposing that the third alien is a Merkun and the second statement is false), and that can’t be the case, as we know that a Merkun won’t speak first in a group. So the first alien is an Ozdak, and thus the second is either a Kween or an Ozdak. If it’s a Kween then the third is an Ozdak (the third can’t be a Kween), and if it’s an Ozdak then the third is a Merkun. The latter can’t be the case, because a Merkun would speak the truth after a lie is uttered, and we know that the third statement can’t be true. So the first and third aliens are Ozdaks and the second is a Kween.
Three photographers and three cannibals come to a river. The boat can carry only two people at a time. The cannibals will eat any group of photographers that they outnumber (on either side of the river). How can all six people safely cross the river?
Suppose a 5 × 9 rectangle is partitioned into a set of 10 rectangles with integer dimensions. How can we prove that some two of these smaller rectangles are congruent?
These have areas, respectively, of {1, 2, 3, 4, 4, 5, 6, 6, 7, 8, 8, …}. If no two of the 10 smaller rectangles are congruent, then their minimum total area would be 1 + 2 + 3 + 4 + 4 + 5 + 6 + 6 + 7 + 8 = 46, which is greater than that of the larger rectangle in which they must fit (5 × 9 = 45). So some duplication is unavoidable.
From Quantum, via Ross Honsberger’s Mathematical Delights, 2019.
Pyrrho the philosopher being one day in a boat in a very great tempest, shewed to those he saw the most affrighted about him, and encouraged them, by the example of a hog that was there, nothing at all concerned at the storm. Shall we then dare to say that this advantage of reason, of which we so much boast, and upon the account of which we think ourselves masters and emperors over the rest of all creation, was given us for a torment? To what end serves the knowledge of things if it renders us more unmanly? if we thereby lose the tranquillity and repose we should enjoy without it? and if it put us into a worse condition than Pyrrho’s hog? Shall we employ the understanding that was conferred upon us for our greatest good to our own ruin; setting ourselves against the design of nature and the universal order of things, which intend that every one should make use of the faculties, members, and means he has to his own best advantage?
— Montaigne, “That the Relish for Good and Evil Depends in Great Measure Upon the Opinion We Have of Them,” 1580
There are 12 people in a room. Some always tell the truth, and the rest always lie.
#1 says, “None of us is honest.”
#2 says, “There is not more than 1 honest person here.”
#3 says, “There are not more than 2 honest people here.”
#4 says, “There are not more than 3 honest people here.”
#5 says, “There are not more than 4 honest people here.”
#6 says, “There are not more than 5 honest people here.”
#7 says, “There are not more than 6 honest people here.”
#8 says, “There are not more than 7 honest people here.”
#9 says, “There are not more than 8 honest people here.”
#10 says, “There are not more than 9 honest people here.”
#11 says, “There are not more than 10 honest people here.”
#12 says, “There are not more than 11 honest people here.”
If there are k honest people, then the first k statements are false and the last 12 – k statements are true. True statements are made by honest people, so 12 – k = k; there are k = 6 honest people in the room.
‘There’s a mouse in one of these barrels,’ said the dog.
‘Which barrel?’ asked the cat.
‘Why, the five-hundredth barrel.’
‘What do you mean by the five-hundredth? There are only five barrels in all.’
‘It’s the five-hundredth if you count backwards and forwards in this way.’
And the dog explained that you count like this:
1 2 3 4 5
9 8 7 6
10 11 12 13
So that the seventh barrel would be the one marked 3 and the twelfth barrel the one numbered 4.
‘That will take some time,’ said the cat, and she began a laborious count. Several times she made a slip, and had to begin again.
‘Rats!’ exclaimed the dog. ‘Hurry up or you will be too late!’
‘Confound you! You’ve put me out again, and I must make a fresh start.’
Meanwhile the mouse, overhearing the conversation, was working madly at enlarging a hole, and just succeeded in escaping as the cat leapt into the correct barrel.
‘I knew you would lose it,’ said the dog. ‘Your education has been sadly neglected. A certain amount of arithmetic is necessary to every cat, as it is to every dog. Bless me! Even some snakes are adders!’
Now, which was the five-hundredth barrel? Can you find a quick way of arriving at the answer without making the actual count?
You have simply to divide the given number by 8. If there be no remainder, then it is the second barrel. If the remainder be 1, 2 , 3, 4, or 5, then that remainder indicates the number of the barrel. If you get a remainder greater than 5, just deduct it from 10 and you have the required barrel. Now 500 divided by
8 leaves the remainder 4, so that the barrel marked 4 was the one that contained the mouse.
times the short diameter, and bend the paper at 90 degrees along the short diameter. The top view is a circle, the side view a semicircle, the other side view is the angle made by the paper.
