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Find all configurations of four points in the plane, such that the pairwise distances between the points take at most two different values.
All 6 configurations are shown below: a square, a rhombus with 60°-120°-60°-120°, an equilateral triangle with its center, an isosceles triangle with 75°-75°-30° and its center, a quadrilateral with 75°-150°-75°-150°, and a trapezoid with base angles of 72°.
You are given 3 boxes – one labeled “Apples”, one labeled “Bananas”, and one labeled “Apples and Bananas”. You are told that the labels on the boxes have been completely mismatched, i.e. none of the three labels is put on its correct box. How can you open just one box and pick a random fruit from it, so that after seeing the fruit, you can guess correctly the contents of every box out of the three?
Open the box labeled “Apples and Bananas”. If you pick a banana from it, then the box labeled “Bananas” will contain apples, and therefore the box labeled “Apples” will contain apples and bananas. Similarly, if you pick an apple from it, then the box labeled “Apples” will contain bananas, and therefore the box labeled “Bananas” will contain apples and bananas”.
A cowboy walks into a bar and asks the barman for a glass of water. The barman pulls out a gun instead and points it at the man. The man genuinely says “Thank you” and walks out.
What happened?
The cowboy had hiccups and needed water. The barman shocked him with his gun instead and that cured the hiccups.
Two identical bolts are placed together so their grooves intermesh. If you move the bolts around each other as you would twiddle your thumbs, holding each bolt firmly by the head so it does not rotate and twiddling them in the direction shown below, will the heads:
(a) move inward
(b) move outward, or
(c) remain the same distance from each other?
One of the bolts will be screwing itself, and the other one will be unscrewing itself. This will happen at the same pace and the bolts will remain the same distance from each other. Thus the answer is (c).
Three people check into a hotel room and each of them gets charged $10 – a total of $30. Later the clerk realizes that the bill is just $25, so he sends the bellboy to return $5 to the guests. On his way to the room, the bellboy decides to cheat and pockets $2 of the money, and gives the three men just one dollar each. Now the three men have spent $9 each, for a total of $27. Additionally, the bellboy took $2 for himself, which adds up to $27 + $2 = $29. Since the guests originally handed over $30, the question is what happened to the remaining $1?
The calculation is made the wrong way. The three men originally gave $30, but later $5 of them were sent back, which makes it $30 – 5 = $25 left at the clerk. Each of the men spent $9, so they gave $27 in total, $2 of which ended up in the bellboy’s pocket. $27 – $2 = $25, so no “missing dollar” here.
3 x $9 – $2 = $30 – $5.
Find a move for White, which does not result in a mate.
The move is Rc6+. Then Black can play Rxh7.
What is correct to say – “the yolk of the egg is white” or “the yolk of the egg are white”?
Neither – the yolk of the egg is yellow.
Where is the driver sitting in this car?
Using the positioning of the mirrors, you can conclude that the driver is sitting on the right.
All integer numbers between 1 and 121 are written in the cells of a square grid with size 11 by 11. Then the product of the numbers in every row and the product of the numbers in every column are calculated. Is it possible that the set of all 11 column products coincides with the set of all 11 row-products?
No, it is not possible. There are 13 prime number between 61 and 121. Since there are only 11 rows, two of them, X and Y, appear in the same row. Now that row is divisible by XY, but clearly, no column is divisible by that number.
A scientist has 9 bottles, exactly one of which contains poison. The poison kills any creature which drinks it within 24 hours. If the scientist has 2 lab mice at his disposal, how can he find which is the poisonous bottle within 2 days only?
Label the bottles B1, B2, B3, … , B9.
The first day he lets the first mouse drink B1, B2, B3, and let the second mouse drink B1, B4, and B5. If after 24 hours both mice die, then the poisonous bottle is B1. If only one mouse dies, say the first one, then he lets the second mouse drink B2. If it dies, then the poisonous bottle is B2, otherwise, it is B3. Finally, if neither mouse dies, then he lets the first mouse drink B6 and B7, and lets the second mouse drink B6 and B8. If both mice die after 24 hours, then the poisonous bottle is B6. If only one mouse dies, say the first one, then the poisonous bottle is B7. If neither mouse dies, then the poisonous bottle is B9.
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