Curve in a Box
Is it possible to design a simple closed curve inside a box, such that its projections on all 6 walls of the box are trees, i.e. curves without loops?
SOLUTION
Yes, it is possible, as shown on the image below.
Category: Puzzles
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Donuts and Candies
Huey has 3 donuts, Dewey has 5 donuts. Louie comes along and the three of them split the donuts equally. In exchange, Louie offers 8 candies to Huey and Dewey. What is the fair way to split the candies?
SOLUTION
Huey must take 1 chocolate, and Dewey must take 7. This is because each of them ate 8/3 donuts, and therefore Huey gave away 1/3 of a donut, whereas Dewey gave away 7/3 of a donut.
Inside Outside
You throw away the outside and cook the inside. Then you eat the outside and throw away the inside. What do you eat?
SOLUTION
The answer is CORN.
Puzzle Sum 16
What two animals do these sums spell?
SOLUTION
PANTS + TRAP – STRAP + HERD + APPLE – DAPPLE = PANTHER
HIP + POP + EAR – PEAR + POTATO + EMU – TOE + SCOT – COT = HIPPOPOTAMUS
False Statements
None of these statements is correct.
At most 1 of these statements is correct.
At most 2 of these statements are correct.
…
At most 98 of these statements are correct.
At most 99 of these statements are correct.
How many of these statements are correct?
SOLUTION
If the number of true statements is X, then statements 1, 2, … , X are wrong, and the rest are correct. Therefore X = 100 – X and X = 50. Thus, there are 50 correct statements.
Mastermind
You must find a 3-digit number. You know that:
- 682 shares one digit with the number, and it is well placed;
- 614 shares one digit with the number, but it is wrongly placed;
- 296 shares two digits with the number, but they are wrongly placed.
What is the number?
SOLUTION
The first and second clues imply that the 6 is not a part of the number. The third clue then implies that 2 and 9 are parts of the number. Now the first clue implies that 2 is the last digit of the number, and the third clue implies that 9 is the first digit of the number. Finally, the second clue implies that 4 is the second digit of the number. Therefore, the number is 942.
The Bank Robber
Harry Ape robbed a bank yesterday and has been hiding in the bitterly cold forest. Max Mouse found Harry’s clothes near a hole in the pond ice. Why does Slylock Fox believe that Harry is nearby?
SOLUTION
Slylock thinks Harry fell through the ice, and after climbing out of the water, took off his wet clothes due to the frigid temperatures. The sleuth suspects the ape is nearby because the wet clothes are still unfrozen.
Islands and Bridges
You need to cross a river, from the north shore to the south shore, via a series of 13 bridges and six islands, which you can see in the diagram below. However, as you approach the water, a hurricane passes and destroys some (possibly none/all) of the bridges. If the probability that each bridge gets destroyed is 50%, independently of the others, what is the chance that you will be able to cross the river after all?
SOLUTION
Imagine there is a captain on a ship, who wants to sail through the river from West to East. You can see that he will be able to do this if and only if you are not able to cross the river. However, if you rotate the diagram by 90 degrees, you can also see that the probability that you cross North-South is equal to the probability that he sails West-East, and therefore both probabilities are equal to 50%.
The Lane Cafe
Get from START to FINISH in this landscape maze, created by Matthew Haussler.
SOLUTION
The solution is shown below.
Every Acute Triangle
Consider an arbitrary acute triangle ABC. Let E be the intersection of the bisector at vertex C and the bisection of the side AB. Let F and G be the projections of E on AC and BC respectively.
Since E belongs to the bisection of AB, we must have AE = BE. Also, since E belongs to the bisector of C, we must have EF = EG. However, this would imply that triangles AEF and BGF are identical, and then AF = BF. We also have that CF = CG, which implies that AC = BC. The arbitrarily chosen triangle ABC is isosceles!
Can you find where the logic fails?
SOLUTION
The bisector of C and the bisection of AB always intersect outside the triangle, on the circumcircle. One of the points F and G always lies on the segment AC or BC, and the other one does not.