Oompa, Loompa, doompa-dee-do
I have a perfect puzzle for you
remember The Disastrous “Willy Wonka Experience” In Glasgow a few weeks ago? Parents paid £35 per ticket and were lured by an AI-generated advert depicting a lush candy paradise, but were greeted by an almost empty warehouse with a few small decorations. No one believes Willy Wonka anymore. Maybe we should never do this. After all, this is the man who invited five children to his factory and exposed them to a horrific fate.
This week Wonka will reveal you for the impostor you are. Wait a moment. Hit that one. Twist it.
Did you miss last week’s puzzle?check it out here, and find its solution at the bottom of today’s article. If you haven’t solved last week’s question yet, be careful not to read too far into it!
Puzzle #34: Fool’s Golden Ticket
Willy Wonka is selling new chocolate bars. They are rectangular bars consisting of a 3×7 array of individually filled chocolate squares. Some cubes are filled with sparkling drinks, while others are filled with noseberry filling. The flavor arrangement at each bar is randomly assigned.
Note that in the column above, the four squares labeled 1 form a rectangle whose corners are all snozzberries, while the square labeled 2 forms a rectangle whose corners are all sparkling drinks (two by two and three by three are still is a rectangle)).Wonka promises anyone who buys the bar No four squares of the same type can form a rectangle Will win a tour of his factory. Your Uncle Joe starts shelling out your life savings to buy chocolates, but you sense a scam. How do you convince Uncle Joe that Wonka’s winning gold bar doesn’t exist?
I’ll be back next Monday with answers and new puzzles. Do you know a cool puzzle that you think should be featured here?Leave me a message on X @JackPMurtagh or email me gizmodopuzzle@gmail.com
Solution to Puzzle #33: Pi Day
Did you run in circles? last week’s Puzzle?speak out Riding a Rabbit 111 In order to solve these two problems.
A rope is wrapped tightly around the Earth’s equator. You can splice extra rope to add enough slack so that you can (in principle) put the new longer rope a foot off the ground all over the world. How much rope was added? How much rope would need to be added to the rope wrapped around the basketball to raise it one foot?
In both cases, you will need to add 2π or approximately 6.283 feet of rope.
There are two things about this solution that I find surprising. One is that 6 feet of rope is tiny compared to the circumference of the Earth, and I’m surprised it would cause so much slack to be distributed around the globe. Another is that the answer does not depend on the size of the sphere at all. Marbles, basketballs, and globes all require the same adjustments.
To solve this problem, remember that the circumference of a circle with radius r is 2πr. The central question of this puzzle is: How much longer does the circumference become when the radius increases by one foot? The circumference of the longer rope is 2π(r+1). The length difference between the longer string and the original string is 2π(r+1) – 2πr = 2π.
The second puzzle asks whether the yellow, blue, or red area in the image below is the largest:
In fact, all three areas are the same! You could solve this problem by comparing the radius of the circle to the side length of the square in each case, but there is a view I prefer.
Whenever you inscribe a circle within a square, the area of the circle is always exactly π/4 or 78.5% of the area of the square. To see this, assume that the radius of the circle is r, and note that the side length of the square is 2r and the area is 4r². Divide the area of the circle (πr²) by the area of the square to get π/4. Again, the radius cancels and we’re left with a number that has nothing to do with the size of the shape.
We can imagine the blue square divided into four smaller squares, each smaller square having an inscribed circle, as follows.
The circles occupy approximately 78.5% of the area of each small square, and therefore 78.5% of the area of the large square. The same argument applies to all three colors. Since the large squares are all the same size, all three color regions have the same area.
