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Ask The Wizard #415

three-squares-question

In the image above, the numbers represent the areas of the four squares. What is the area of the red square?

anonymous

225

 

First, the yellow square doesn't help us at all. It is a red herring, so let's ignore it.

Next, consider the triangle to the left of the orange square, as shown in the following image.

three-squares-solution

BC = sqrt(16) = 4.

AB = sqrt(49) - sqrt(16) = 7-4 = 3.

By the Pythagorean formula, AC2 = 42 + 32 = 25.

AC = sqrt(25) = 5.

The height of all three squares on the right is sqrt(49) + sqrt(16) + sqrt(1) = 12.

The ratio of the side length of the red square to the height of the height of the three squares of the right is going to be the same as the ratio of AC to BC = 5/4.

So, the side length of the red square is (5/4)*12 = 15.

Thus, the area of the red square is 152 = 225.

A similar puzzle is asked and discussed in my forum at Wizard of Vegas.

The Mind Your Decisions YouTube channel also has a similar puzzle.

What is the smallest possible rectangle such that the area equals the perimeter?

anonymous

The rectangle dimensions are 4x4 for an area and perimeter of 16.

 

Let the dimensions of the rectangle be x and y.

We're given: xy = 2x + 2y.
2y - xy = 2x
y(2-x) = 2x
y=2x/(2-x)

Let f(x) = The area of the rectangle = x*y =
x*2x/(2-x) = 2x2/(2-x)

To find the minimum area, take the derivative using the Quotient Rule:

f'(x) = 4x(2-x) + 2x2 / (2-x)2 = 0

4x(2-x) + 2x2 = 0

8x = 2x2

x=4

If x=4, y = 2*4/(4-2) = 8/2 = 4.

Let's solve for y for other values of x close to 4.

If x=3, y=6 for an area of 18

If x=5, y = 10/3 for an area of 16+(2/3).

It's easy to see that the solution at x=4 and y=4 results in a minimum. Thus, the smallest possible rectangle is 4x4 = 16.

In a casino drawing there are the following number of tickets in the drum by the holder of the tickets:

  • Player 1 holds 6 tickets.
  • Player 2 holds 2 tickets.
  • Player 3 holds 1 tickets.
  • 21 other players hold 21 tickets.

The casino will draw five tickets for five equal prizes. Each player may only win once. If a ticket is drawn by a player who already won that ticket will be thrown away and a new ticket drawn.

Player's 1, 2 and 3 agree to split winnings according to their equity in the drawing. What is a fair split?

Mukke

I'm going to simplify the problem by assuming that if a ticket is drawn by a player who already won, that prize will be voided. Otherwise, the math gets much too messy, pretty much necessitating a random simulation.

The probability a player holding n tickets does NOT win a prize is combin(30-n,5)/combin(30,n).

Thus the probability player A does in a prize is 1-combin(24,5)/combin(30,5) = 0.701739.

Thus the probability player B does in a prize is 1-combin(28,5)/combin(30,5) = 0.310345.

Thus the probability player C does in a prize is 1-combin(29,5)/combin(30,5) = 0.166667.

The sum of these probabilities is 1.178750. That is how many wins the group can expect to get.

In my view, each player should get a share equal to his probability of winning a prize divided by the expected total wins of the group.

A gets a 0.701739/1.178750 = 0.595324 share.

B gets a 0.310345/1.178750 = 0.263283 share.

C gets a 0.166667/1.178750 = 0.141393 share.

This question is asked and discussed in my forum at Wizard of Vegas.