Problem 1

A coin rolls in the interior of a square along its sides without slipping. When the coin has returned to its starting position, it has made 1 whole revolution. If the coin’s radius is 1, what is the side length of the square?

Problem 2

Two circles intersect a rectangle, dividing the top and bottom sides into 3 segments. What is the length of the segment marked with a question mark?

Problem 3

In the diagram below, there are 4 squares and 3 areas are known. What is the radius of the circle?

As usual, watch the video for a solution.

3 Problems About Circles And Rectangles

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“All will be well if you use your mind for your decisions, and mind only your decisions.” Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon.

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Answer To 3 Problems About Circles And Rectangles

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Problem 1

In each corner of the square, we can construct a square with side length equal to the coin’s radius r. As the coin makes 1 revolution after rolling along 4 sides, the coin makes 1/4 of a turn on each side, so 1/4 of its circumference (2πr/4) is the distance between the small squares.

Thus the side length is:

r + πr/2 + r
= r(2 + π/2)

With r = 1 the side length is 2 + π/2.

Problem 2

Construct a blue rectangle with horizontal side length equal to the smaller chord. Then by symmetry the two rectangles on its sides are congruent. Do this for both circles, as shown.

The first calculated length is 12 – 8 = 4 for the left green rectangle. Shifting to the right side will remove 4 from 24, leaving 20. This means the next rectangle’s length is 26 – 20 = 6. Shifting to the right side gives 6 + x = 22. The missing length is x = 16.

Problem 3

Calculate the sides of the squares whose 3 areas are known. These lengths will be 4, 10, and 5. Let x be the remaining distance of the horizontal chord.

By the chord-chord power theorem, 5x = 4(10) = 40, so x = 8. Construct a chord 4 units from the bottom of the circle, so the distance between the two horizontal chords is 10 – 4 = 6. We can complete a rectangle with sides 8 + 5 = 13 and 6. The rectangle’s right angle is an inscribed angle of the circle, so the chord it subtends is a diameter with length 2r. So half the rectangle is a right triangle with hypotenuse 2r, and legs 6 and 13.

Thus we have:

6² + 13² = (2r)²
205 = 4r²
50.25 = r²
50.25π = πr²

The circle’s area is 50.25π.

References

Problem 1 Australian Math Competition (sent by email)

Problem 2 Math Kangaroo 2022, Q30, Level Student, Grades 11-13

Problem 3 power chords
https://twitter.com/cshearer41/status/1030405543255592960?lang=en
solution
https://twitter.com/jldavilaa01/status/1030409264911863808
other solutions
https://math.stackexchange.com/questions/3601104/what-is-area-of-circle-given-areas-of-4-squares
https://mathematrec.wordpress.com/2018/10/13/power-chords/