Goodbye Maze
Escape from START to END in this maze, created by Ben Uelk.
The solution is shown below.
Ben U. is the creator of the website Maze Dojo. Ben, an Indiana native, has been drawing mazes for fun since 2007 and his work has since been featured in print magazines, online publications, and music album artwork. Ben is currently working on a book of mazes of each U.S. State, due to be completed in 2019.
Escape from START to END in this maze, created by Ben Uelk.
The solution is shown below.
A regular hexagon is split into small equilateral triangles and then the triangles are paired arbitrarily into rhombuses. Show that this results into three types of rhombuses based on orientation, with equal number of rhombuses from each type.
Color the rhombuses based on their type and imagine the diagram represents a structure of small cubes arranged in a larger cube. If you look at the large cube from three different angles, you will see exactly the three types of rhombuses on the diagram.
Alternatively, the problem can be proven more rigorously by considering the three sets of non-intersecting broken lines connecting the pairs of opposite sides of the hexagon, as shown on the image below. The type of each rhombus is determined by the types of the broken lines passing through it. Therefore, there are n² rhombuses of each type, where n is the length of the hexagon’s sides.
Is it true that for every closed curve in the plane, you can use a rope to recreate the layout, so that the rope can be untangled?
Said otherwise, you have to determine at each intersection point of the closed curve, which of the two parts goes over and which one goes under, so that there aren’t any knots in the resulting rope.
Start from any point of the curve and keep moving along it, so that at each non-visited intersection you go over, until you get back to where you started from.
You have 100 blue and 100 red points in the plane, no three of which lie on one line. Prove that you can connect all points in pairs of different colors
Connect the points in pairs of different colors so that the total length of all segments is minimal. If any two segments intersect, you can swap the two pairs among these four points and get a smaller total length.
Place 8 queens on a chessboard, so that no two of them attack each other. For an extra challenge, make sure that no three of them lie on a straight line.
The original puzzle has 12 unique solutions, up to rotation and symmetry. With the additional restriction imposed, there is only one solution.
David Justice and Derek Jeter were professional baseball players. In 1997 they had the following conversation:
David: Did you know that in both 1995 and 1996 I had better batting averages than you?
Derek: No way, my batting average over the last two years was definitely higher than yours!
It turned out that both of them were right. How is it possible?
This is the so called Simpson’s paradox. The reason it occurred is that during 1996 both players had high averages and Derek Jeter had many more hits than David Justice. In 1996 both players had low averages and David Justice had many more hits than Derek Jeter. You can see their official statistics below.
Player | 1995 | 1996 | 1995-1996 |
Derek Jeter | 12/48 (.250) | 183/582 (.314) | 195/630 (.310) |
David Justice | 104/411 (.253) | 45/140 (.321) | 149/551 (.270) |
Can you figure out which word is depicted by this rebus?
Each of the images in the first row depicts HAIR. Each of the images in the second row (except the second cell) depicts a BUG. Each of the images in the third row depicts ER.
The I from HAIR on the first row gets moved right after the B from BUG on the second row. Also, the U from BUG gets flipped upside down. Therefore, we get HARBINGER.
Escape from the black hole to the upper right corner in this maze, created by Ben Uelk.
The solution is shown below.
A man leaves home and makes three left turns, only to return home facing two men wearing masks. Who are those two men?
The person is a baseball player. The men are a catcher and an umpire.
Split the grid into four identical regions, such that each region contains a lion and a crown.
The solution is shown below.
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