Happy 𝜋 Day!

Happy 𝛑 Day! What better day to celebrate our favorite, familiar, fantastic irrational number? This 𝛑 Day we are bringing you a 𝛑-centric approximation activity, a detailed explanation of why this activity approximates 𝛑, and a special bonus math problem and solution (in case you want to spend your 𝛑 Day doing math, math, and more math :) ).

First, the activity:

Select your favorite straight thin object you don’t mind tossing at random. For example: birthday candles, crayons, toothpicks, Q-tips, etc. The more the merrier.* Whatever you pick, you want them all to be about the same length. We will use birthday candles for our example.

Make parallel lines on a flat surface. Make the distance between the lines twice as far as your throwing object is long. So, for birthday candles, your lines should be two candle lengths apart. To make lines you can use masking tape or Sharpie on butcher paper.

Now toss all your throwing objects onto the flat, lined surface. Carefully count how many objects cross a line. Now divide the number of objects you tossed by how many crossed one of the lines. Does this ratio look familiar?

*If you don’t have 100 or more of your tossing objects you may wish to do the experiment multiple times using fewer objects. Just keep a tally of the total number of objects you have tossed and the total number that have crossed a line. And use your tossing object tally and crossing tally to find the ratio at the end.

Why in the world does this work? 

First, let's think about our set-up. The lines are two tossing objects apart so from here on our we will just use the length of our tossing object as our units. In our case we were tossing birthday candles, so for our calculations and graphs 1 unit = 1 candle length. Okay, now let's get into calculations. To simplify the problem we’ll start by thinking about just one of our candles, and see if we can work out the probability that it should cross a line when tossed randomly. 

Zooming in on just one of our candles we have the following situation. (We included the two candles along the edge just as a reminder that the lines are 2 units apart).

Examining this picture we see that there are two variables, the angle at which the candle falls, θ, and the distance from the center of the candle to the closest line, D. Theta can vary from 0 to 𝛑 degrees and is measured against a line parallel to the lines in the experiment. The distance from the center of the candle to the closest line can never be more than half the distance between the lines, which means D ≤ 1 unit. 

The candle in the picture misses the line. We know that candle will hit the line if the closest distance to a line (D) is less than or equal to the length of the dotted line, L. Thanks to trigonometry, we can find the length of this dotted line in terms of θ, and it is ½ (sin(θ)), so L is a function of θ, and we can write L(θ) = ½ sin(θ). If you aren’t familiar with trig functions yet, hang in there, you can still follow most of this! You just have to take our word on the graph of L(θ).)

In the graph below, we plot D along the y-axis and L(θ) along the x-axis. The values on or below the curve represent a hit, namely D ≤ L. The chances that we will hit the line are given by taking the area below the curve (that is the area of values of D and L that yield a hit) and dividing it by the area of the rectangle that represents all possible values for D and L (this rectangle is shown with the dotted green lines). 

The area of the rectangle is of course easy to find and it is just 𝛑. The area of the shaded region does require some calculus to find and it is given by the definite integral of L(θ) evaluated from 0 to 𝛑, which comes out to just 1. Which means that the chances a value of (D,θ) randomly chosen fall within the shaded area are 1/𝛑. Which is the same as saying that the chances we hit the line on any given throw are 1/𝛑. That means that if we threw as many candles as we could randomly, about 1/𝛑 of them should hit the line. So:

Which means:

So there we have it, our famous and familiar ratio, 𝛑, showing up in an unusual and unexpected place. This explanation was largely based on this handy overview and explanation of Buffon’s needle problems: https://mste.illinois.edu/activity/buffon/. It also includes a simulation, which is the less messy option for trying this experiment if you don’t have a looooot of tossing objects at home.

And if you’d like to celebrate  𝛑 Day with a little extra math, try out the challenge problem below. (The answer appears after the problem.) 

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Following BEAM Discovery (our summer program for rising 7th graders), students are sent monthly Challenge Sets, fun math puzzles that students can do independently (and win prizes!). This is a math problem from the first Challenge Set of this school year.  (It comes from the Math Kangaroo Contest.)

Myriam chooses a 5-digit whole number, and deletes one of the digits to make a 4-digit number. When she adds the 5-digit and 4-digit numbers together, she gets 52713. 

What is the original 5-digit number?

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Solution:

Let’s use A, B, C, D, and E to represent the digits of our 5-digit number, where A, B, C, D, E are all whole numbers from 0 to 9. So our 5-digit number looks like ABCDE. If this is the case, then we have 5 options for our 4-digit number: ABCD, ABCE, ABDE, ACDE, and BCDE. You might notice that four of these options end in the same digit as our original 5-digit number, namely E.

Well, what happens when we add two numbers that end in the same digit? In the ones place of our new number we will get the ones digit of E + E = 2E. We know that 2E must be an even number, so its ones digit is 0, 2, 4, 6, or 8. But we are looking for a ones digit of 3, because our 5-digit number and 4-digit number need to sum to 52713. That means we know that we don't want our 4-digit number to end in E. So, in fact, ABCE, ABDE, ACDE, and BCDE aren’t options.

Note: We interrupt our original post to provide an alternate (and really lovely) solution suggested by a BEAM supporter:

Now that we know the last digit is the one that must be dropped, we have an approximate equation X + X/10 ~ 52713. Solving this equation gives X ~ 47920.9090..., from which we can immediately deduce that A = 4, B = 7, C = 9, and D = 2, and very likely E = 1. Testing shows that this is indeed the solution, making short work of this problem!

This clever alternate solution path was submitted by Marc-Paul Lee, a long-time BEAM supporter who, unlike the theoretical mathematician who suggested this problem, knows how to approximate! :-D

And now, back to our original solution:

Now we know that Myriam must have deleted E, the digit in the ones place. So, our 4-digit number must look like ABCD. We also know that ABCDE + ABCD = 52713. We might think at this point that D + E = 3, but remember that just the ones place of D + E needs to be 3, so D + E could equal 3 or 13. We know that 0 <= D + E <= 18 (because D and E are 1-digit numbers so neither of them can be greater than 9) so D + E can’t equal 23, or anything else ending in 3 except 3 or 13.

Moving on to the tens column, we might think that D + C = 1 or 11, but is that right? If D + E > 9, then we will have to carry a 1, which means that D + C might actually be 0 or 10 and we would still get the 1 we need in our tens place. So D + C = 0 or 1 or 10 or 11. In a similar way C + B = 6 or 7 or 16 or 17, and B + A = 1 or 2 or 11 or 12. Finally, we know that A = 4 or A = 5.

Since A > 3, B + A (in the thousands place) can’t equal 1 or 2 (two of our options above). So, B + A = 12 or 11. Which means we carried a 1 to the ten-thousands place and A = 4 not 5.

So B = 8 or B = 7. If B + A = 12, then we didn't carry a 1 when we added C and B, so C + B = 6 or 7, but that’s not right because if B + A = 12, then B = 8. So B + A = 11, which means B = 7.

Now we know A = 4 and B = 7, and B + A = 11, which means we did carry a 1 from adding C and B, so C + B = 16 or 17. Thus, C = 10 or 9, but it can't be 10 because it is a 1-digit number. So C = 9.

Now we know that A = 4, B = 7, and C = 9, and C + B  = 16. Because C + B = 16, we know that to get the desired 7 in our hundreds place, we must have carried a 1 from our addition of D and C. So D + C = 11 or 10. But because we know that C = 9, we know that D = 1 or 2, which means that E + D < 12. So E + D = 3, and we know that we didn’t carry 1 from our addition there, which means that D + C is not equal to 10,  leaving us with D + C = 11. Plugging in C = 9, we can now find that D = 2 and E = 1. 

Putting all the pieces together we get that Myriam’s original number, ABCDE, is 47921. And just to check, let’s do one last step: 47921 + 4792 = 52713.