Real Mathematics – Geometry #15

Drawing a square

I am dealing with geometry and I imagine that I am in ancient Greece again. Aegean sea is in front of me and I am sitting on a marble between two huge white columns while holding an unmarked ruler and a compass.

First I draw a circle that has center at A and has radius r:

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Then I draw the same circle but taking its center at B this time:

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I connect the points A and B with a straight line. Then I draw two perpendiculars from the endpoints of the line AB:

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I connect the point E to the point F and end up with the ABEF square which has side lengths r:

Biggest circle that can be drawn into ABEF will have diameter r and touch the square at exactly four points:

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Area

In order to find the area of a square one can take the square of one side that gives r2.

To find a circle’s area one should multiply the square of the radius with π. In our inscribed circle we calculate the area as πr2/4.

Ratio of these areas would give π/4.

Weight

Now let’s make an experiment. For that all you need is some kind of cardboard cut as a square and a precision scale. Using the scale find the weight of the square-shaped cardboard.

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Then draw the biggest possible circle inside this square. Cut that circle out and find its weight with the scale.

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Since we are using the same material ratio of the weights should be equal to the ratio of the areas. From here one can easily find an approximation for the number π:

0,76/0,97 = π/4

3,1340… = π

One of the main reasons why we only found an approximation is that the cardboard might not be homogeneous. In other words the cardboard might not have equal amount of material on every point of itself.

Another reason for finding an approximation is that I didn’t cut the square and the circle perfectly.

One wonders…

Draw a circle and then draw the biggest-possible square inside that circle. Find their areas and measure their weights. See if you found an approximation.

M. Serkan Kalaycıoğlu

Real Mathematics – Life vs. Maths #2

Matches

For thousands of years people tried find a precise value for the number π (3,1415192…). At first this special number was thought to be seen only when there is a circle around. Within time π started to appear in places where scientist didn’t expect it to be. One of them was an 18th century scientist Georges Buffon.

Buffon came up with a probability problem named “Buffon’s needle problem” in 1777 when he came across with the number π. As I didn’t possess that many needles, I modified the problem as “Serkan’s matches problem”.

Buffon’s Needle Problem: Take a piece of paper and draw perpendicular lines on it with specific amount of space between them. Buffon wondered if one can calculate the probability of a needle that will land on one of the lines.

To start Serkan’s matches problem you need at least 100 matches, a piece of empty paper, a ruler, pen/pencil and a calculator.

First of all, draw perpendicular lines with 2 matches-length spaces between them.

Then just throw the matches on the paper randomly.

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Start collecting the matches which land on a line. At last you should use your calculator to divide the total number of matches to the number of matches landed on a line.

In my experiment out of 100 matches, 32 of them landed on a line. That gave me 3,125 which is close to the magical number π.

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In fact, 100 matches are not enough for this experiment. In my second try 34 matches landed on one of the lines which gave 100/34=2,9411… Obviously this is not close to π. More matches we use, closer we will get to π.

In an experiment back in 1980 2000 needles were used to analyze Buffon’s needle problem. Result was 3,1430… which is seriously close to the number π.

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You could go to https://mste.illinois.edu/activity/buffon/ and use this simulator which uses 1000 needle. In my first try I got 3,1496… You should try and see the result yourself.

In the future I will be talking about why a needle (or a match) is connected to the number π.

One wonders…

Try to do your own experiment and repeat Buffon’s needle problem for five times. Take the arithmetic average of your solutions and see how close you are to π?

M. Serkan Kalaycıoğlu