Hungry Lion

A hungry lion runs inside a circus arena which is a circle of radius 10 meters. Running in broken lines (i.e. along a piecewise linear trajectory), the lion covers 30 kilometers. Prove that the sum of all turning angles is at least 2998 radians.

Imagine the lion is static, facing North, and instead, the center of the arena moves around. Then, each time the lion runs X meters in some direction, this translates into the center moving X meters South. Each time the lion makes a turn of Y radians, this translates into the center moving along an arc of Y radians.

Thus, the problem translates to a point inside the arena alternating between traveling straight South and then moving along arcs around the center of the arena. Since the total distance traveled straight South by the point is 30KM and the distance between the starting and the ending points is at most 20M, the total distance traveled North must be at least 30KM – 20M = 29980M. Therefore, the total length of the arcs traversed by the point is at least 29980M, and since the radius of each arc is at most 10M, the total angle of the arcs must be at least 2998 radians. The sum of all turning angles of the lion is the same, so this concludes the proof.

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10 Dots, 10 Coins

If you have 10 dots on the ground, can you always cover them with 10 pennies without the coins overlapping?

Assume the dots lie in a plane and the radius of a penny is 1. Make an infinite grid of circles with radii 1, as shown on the picture, and place it randomly in the plane.

If we choose any point in the plane, the probability that it will end up inside some circle of the grid is equal to S(C)/S(H), where S(C) is the area of a coin and S(H) is the area of a regular hexagon circumscribed around it. A simple calculation shows that this ratio is larger than 90%. Therefore, the probability that some chosen point in the plane will not end up inside any circle is less than 10%. If we have 10 points, the probability that neither of them will end up inside a circle is less than 100%. Therefore, we can place the grid in the plane in such a way that every dot ends up in some circle. Now, just place the given coins where these circles are.

Married Couples

In a small village, there are 100 married couples living. Everyone in the village lives by the following two rules:

  1. If a husband cheats on his wife and she figures it out, the husband gets immediately killed.
  2. The wives gossip about all the infidelities in town, with the only exception that no woman is told whether her husband has cheated on her.

One day a traveler comes to the village and finds out that every man has cheated at least once on his wife. When he leaves, without being specific, he announces in front of everybody that at least one infidelity has occurred. What will happen in the next 100 days in the village?

Let us first see what will happen if there are N married couples in the village and K husbands have cheated, where K=1 or 2.

If K = 1, then on the first day the cheating husband would get killed and nobody else will die. If K = 2, then on the first day nobody will get killed. During the second day, however, both women would think like this: “If my husband didn’t cheat on me, then the other woman would have immediately realized that she is being cheated on and would have killed her husband on the first day. This did not happen and therefore my husband has cheated on me.”. Then both men will get killed on the second day.

Now assume that if there are N couples on the island and K husbands have cheated, then all K cheaters will get killed on day K. Let us examine what will happen if there are N + 1 couples on the island and L husbands have cheated.

Every woman would think like this: “If I assume that my husband didn’t cheat on me, then the behavior of the remaining N couples will not be influenced by my family’s presence on the island.”. Therefore she has to wait and see when and how many men will get killed in the village. After L days pass however and nobody gets killed, every woman who has been cheated on will realize that her assumption is wrong and will kill her husband on the next day. Therefore if there are N + 1 couples on the island, again all L cheating husbands will get killed on day L.

Applying this inductive logic consecutively for 3 couples, 4 couples, 5 couples, etc., we see that when there are 100 married couples on the island, all men will get killed on day 100.

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