Pi Day 2021 – the one special occasion every Math aficionado should have celebrated without fail, on 14th March. We absolutely did, trying to involve the student community in the celebration. Publishing four intriguing Math puzzles and asking the students to submit the answers for the same, we also went about doing what we do best – come up with the most intuitive and elaborate solution for each of them.

### Solution #1 – Two squares/tiles

Firstly, the question.

*Two square tiles of same size are stacked upon each other such that their centres coincide, but the top tile does not completely cover the bottom tile. If the length of the sides of the tiles is 1 unit. Show that the area that is common between the two tiles is always greater than 3/4.*

Let us randomly pick an orientation of the two squares where we fix the bottom square and rotate the top square.

Observe that the common area between the two squares now = (The entire area of the bottom square) – ( A + B + C + D)

Also observe that for any orientation of the two squares, A, B, C, and D will always be equal, thanks to symmetry.

The common area between the two tiles will be as low as possible when the area of A + B + C + D is as large as possible. Since A = B = C = D, this happens when A is as large as possible.

Focusing on A alone…

As we start rotating the top tile, area of A keeps on increasing from zero, reaches a maximum at the middle and starts decreasing, to finally reach zero again.

The case for maximum area of A occurs when the figure of overlapping squares is symmetric, or technically, when you rotate it by 45^{o}.

Therefore, the minimum possible value of the common area between the two squares is,

Even in the worst case scenario, the common region between the two squares is greater than 3/4. Therefore, we can say that the common area between the two squares is always greater than 3/4.

But why is are we doing all this on π day? There should be some thing special about this problem!

##### Perception it is.

Mark a point on the centre of the top side of the squares when they are perfectly overlapping. Trace the point as you turn the upper square.

*Won’t the point trace a circle?*

Now it gets beautiful. Whatever be the orientation of the top square, this circular region (which has a radius of 1/2) is common to both the squares.

Therefore, it is guaranteed that the common area between the two squares will **at least** be,

Now since π is greater than 3, the common area is further guaranteed to be greater than 3/4.

π and circle – the simplest and most symmetric of all figures – come to our rescue again!

Watch an animation of this problem here. See how the square goes around the circle!

### Solution #2 – The Chalice problem

*Two semicircles are placed in a circle as shown in the figure below. Find the area of the shaded region w.r.t the area of the circle.*

At first it feels like this problem is just undoable. There is no relation between the ratios of the semicircles or the circle given. It is just a figure. Almost feels like the question has inadequate information. But this problem goes on to reveal a beautiful secret about circles and semicircles.

If there should at all be some relation between the area of the shaded region and the area of the circle, since we lack the knowledge of their exact sizes, shouldn’t such relation be globally true for any given size of the semicircles, as far as they are contained in the circle touching each other tangentially?

That is, the area of the coloured region in all these following cases should be the same!

More importantly, As we push the upper semi circle to be smaller and smaller, we get the following:

**Bang! The area of the shaded region was half the area of the circle all the while!**

Though we got to the answer logically, let us also prove it mathematically.

Let r_{2} be the radius of the smaller semi circle, r_{1} be the radius of the larger semi circle and R be the radius of the circle.

From the figure,

R^{2} = (r_{2} + k)^{2} + r_{2}^{2}

R^{2} = (r_{1} – k)^{2} + r_{1}^{2}

(r_{2} + k)^{2} + r_{2}^{2} = (r_{1} – k)^{2} + r_{1}^{2}

(r_{2} + k)^{2} – (r_{1} – k)^{2} = r_{1}^{2} – r_{2}^{2}

(r_{2} + r_{1})(r_{2} – r_{1} + 2k) = (r_{1} + r_{2}) (r_{1} – r_{2})

r_{2} – r_{1} + 2k = r_{1} – r_{2}

k = r_{1} – r_{2}

We know that, R^{2} = (r_{2} + k)^{2} + r_{2}^{2}

Putting k = r_{1} – r_{2} in R^{2} = (r_{2} + k)^{2} + r_{2}^{2}

R^{2} = r_{1}^{2} + r_{2}^{2}

Therefore, Sum of areas of the semi circles = Half the area of the Circle.

Watch an animation of this problem, here.

### Solution #3 – SMO Circles

*Two semi-circles and a circle are arranged in a rectangle as shown in the figure below. If the length of the rectangle is 10 units, find the area of the shaded region.*

We are told that the length (longest side) of the rectangle is 10 units.

The vertical measure of the rectangle should be less than the diameter of the semicircle since there is overlapping.

Hence, the longest side = Diameter = 10.

Radius of the semicircles = 10/2 = 5

Let the radius of the inner circle be r.

Since, triangle AOB, is an isosceles right angled triangle, angle OAB = 45^{o}

(Area of sector CAB) – (Area of triangle AOB) will leave us with one-fourth of the required intersection region.

Since angle CAB = 45^{o} = 180/4

Watch an animation of this problem, here.

### Solution #4 – Pythagoras Pie Puzzle

*Pythagoras wanted to celebrate Pi day by sharing Pie with 100 of his friends. He serves 1% of the Pie to the first guest, 2% of the remaining Pie to the second guest, and 3% of the remaining to the third guest, and so on. Which guest gets the largest share of the Pie?*

Though this question seems very random at the outset, the beauty lies in unravelling it layer by layer.

Let’s tabulate for the first few guests, what they take and what remains thereof of the pie.

Guest | Takes | Remains |

1 | 1% of 100% | 99% of 100% |

2 | 2% of 99% of 100% | 98% of 99% of 100% |

3 | 3% of 98% of 99% of 100% | 97% of 98% of 99% of 100% |

4 | 4% of 97% of 98% of 99% of 100% | 96% of 97% of 98% of 99% of 100% |

… | … | … |

… | … | … |

… | … | … |

It is so satisfying to see a pattern emerging out here!

If we compare what two consecutive guests take, we can determine who takes more.

Suppose we compare what the 3^{rd} and the 4^{th} guests take:

3^{rd} Guest takes 3% of 98% of 99% of 100%

4^{th} Guest takes 4% of 97% of 98% of 99% of 100%

To compare these two quantities, we need not calculate the entire chunk; it is enough to compare **3%** with **4% of 97%**.

Suppose to compare what the 4^{th} and the 5^{th} guests take:

4^{th} Guest takes 4% of 97% of 98% of 99% of 100%.

5^{th} Guest takes 5% of 96% of 97% of 98% of 99% of 100%.

Again, to compare these two quantities, it is enough to compare **4%** with **5% of 96%**.

Or, in general, to compare what the (n-1)^{th} guest takes to what the n^{th} guest takes, we just need to compare **(n-1)%** with **n% of (101-n)%**.

As long as **n% of (101-n)% > (n-1)%**, the n^{th} guest gets a larger share of the pie than his predecessor.

So, all we have to do is to know the largest value of n, which satisfies the inequality, **n% of (101-n)% > (n-1)%.**

n% of (101-n)% > (n-1)%

n% of (101-n) > (n-1)

101n – n^{2} > 100n – 100

100 + n – n^{2} > 0

The max value of n will be 10

So the 10^{th } guest gets the maximum share of the pie.

##### The beauty of this question doesn’t end there.

If we look closely at the shares of the pie that each guest takes, we can generalize and write an equation for the percentage of the pie that the n^{th }guest takes.

*Note: We can also get to the answer of who gets the maximum share by putting the condition as*

More interestingly, by putting n = 10 in P_{n}, the 10^{th} guest who gets the maximum share of the pie gets 6.281565095552947%, which is a close approximation of 2π

We start with **a** “pie” and the maximum share is 2π !

Take a look at this graph. Observe how the differences in the shares of consecutive guests keep decreasing and become negative after 10. Also take a look at the negligible share size of the last few guests.

Well, it surely seems to have been a great Pi(e) Day for Pythagoras and his guests!

Hope you all had a great Pi Day 2021! Keep your Math hat on every now and then to solve these kinds of interesting, non-routine problems 🙂 CAT or not, these are good for your intuition.

The video solutions of our problems celebrating Pi Day 2021.

**Stay safe and best wishes.**

Oh, and yes, the best responses from our students will be published as well! Stay tuned.

*Rajesh Balasubramanian** takes the CAT every year and is a 4-time CAT 100 percentiler. He likes few things more than teaching Math and insists to this day that he is a better teacher than exam-taker.*

*Raghavendra Potluri loves Math, Math and Math. When not engrossed in nerdy Math questions, he loves to code and watch tele-series.*

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