𝛑 day is upon us. The 14^{th} of March represented as 3.14 is called 𝛑 day. (I would have preferred the 22^{nd} of July). It is a mostly forgotten day because E-commerce has not yet co-opted it. But, we at 2IIM are definitely fascinated by it.

As a part of 𝛑 day celebrations, we posted a blog article in which we gave you 4 interesting geometry questions from different topics all related to 𝛑 to keep you engrossed. Here are the solutions to the questions.

#### Qn 1: Five points are placed on the surface of the sphere. Prove that some four of them must be contained in some closed hemisphere.

This is a delightful question based on a fabulous idea called “Pigeonhole Principle”. Go on Google it! Once you read about it on Google, you will realize that what the mathematicians call as “Pigeonhole principle” is nothing but common sense :). But the beauty in Pigeonhole principle is that it helps you redefine a number of challenging questions and give elegant proofs to them. It takes about 20 minutes to understand “Pigeonhole Principle”, I can assure you that you will have fun.

Let us jump into the question. Draw a sphere with 5 points, A, B, C, D, and E:

We can cut through the sphere in many ways such that some four of them are in one hemisphere. For example:

or another one:

**Key point to remember**: This is a sphere and not a circle. So, there is a 3-dimensional angle to it.

Now, let us go into the proof for the question. Pick some two points such that you draw a circle through these two points. Not any circle, draw a “Great Circle” between these two points.

What is a Great Circle? A Great Circle is a circle that passes through two points on the surface of the sphere and whose center is also the center of the sphere. Now, draw a circle with the center of the sphere and passing through 2 points. There are three points remaining.

We now have a hemisphere and three points that are yet to be distributed. We can distribute the 3 points in the following ways:

a) 3 points in hemisphere 1 and 0 points in hemisphere 2

b) 2 points in hemisphere 1 and 1 point in hemisphere 2

c) 0 points in hemisphere 1 and 3 points in hemisphere 2

d) 1 point in hemisphere 1 and 2 points in hemisphere 2

Example: 0 points in hemisphere 1 and 3 points in hemisphere 2

From the scenarios, we can see that there will be at least one hemisphere where there will be 2 points or including the points that are on the Great circle, there will be ** at least 4 points** in one hemisphere.

In our example, there are 5 points in hemisphere 2 as all the 5 points, A, B, C, D and E are in the same hemisphere.

**Hence proved**.

The Pigeonhole angle to this proof: We have 3 points to be put in 2 hemispheres. There will be **at least one hemisphere** that will have **at least 2 points (4 points overall)** as after distributing one in each hemisphere, the remaining point has to go in one of the two hemispheres :).

Beautiful question to solve. Rearranging everything and imagining the sphere in such a manner gives a unique perspective.

#### Qn 2: 100 people standing in a circle in an order 1 to 100. No. 1 has a sword. He kills the next person (i.e. No. 2) and gives the sword to the next (i.e. No. 3). All people do the same until only 1 survives. Which number survives at the last?

Let us draw a circle and arrange the people in order from 1 to 100.

Person 1 has a sword, he kills the next person (No. 2) and gives the sword to No.3. No.3 kills No.4 and gives the sword to No.5 and so on. This process happens until only 1 person is remaining. How does this happen?

As they are in a circle, after 100, we come back to person 1 and the process continues. You can go step by step and then get to the answer or follow the elegant solution to this question. Let us dive in.

Step 1 is easy: 1 will kill 2, 3 will kill 4, 5 will kill 6 and so on. All the persons in even positions will be killed. 50 people will be killed and 50 people will be remaining.

Number of people remaining = 50 or the set {1, 3, 5, 7, 9, 11, 13, 17,……97, 99} survives.

Step 2: Now, 1 will kill 3, 5 will kill 7, 9 will kill 9 and so on. We can clearly spot a pattern. We will now focus on the pattern rather than on the diagram.

{1, 5, 9, 13, 17, 21, ….. 97} will survive. Multiples of 4 + 1 are the survivors. At the end, 97 will kill 99 and now, we have 25 people remaining.

Now, 1 gets the sword. 1 kills 5, 9 will kill 13, 17 will kill 21 and so on. {1, 9, 17, 25, ….. 97} or multiples of 8 + 1 will survive.

No. of people remaining is 13 or 12 people will get killed. 97 is the last guy and now has the sword.

Step 3: 97 will kill 1. This gets tricky. If 1 survives, we are OK. Now, this opens a lot of questions.

This is where the question gets interesting!! When 100 people were there, 1 survives, When 50 people were there, he survives, When 25 people were there, he survives as he starts. But in the next round, when 13 people are there, 1 gets killed ** or** when there are an odd no. of people, 1 gets killed. This is obvious.

Let us take a smaller example. If there were 25 people from 1 to 25, in the first cycle, all the even numbers will be killed. At the end of the first cycle, 25 has the sword and then 25 kills 1.

So, the takeaway is when there are odd number of people remaining, 1 will die.

Now, let us flip this question. Let us rearrange this question in such a way that 1 wins. Now, when will 1 win? Or in other words, when will 1 be the survivor.

1 will survive whenever there are even number of people remaining after each round. or after every time you kill, if there are an even number of people remaining or in other words whenever we are halving the number and we still get an even number, 1 will survive.

Or we are looking at powers of 2 -- 2, 4, 8, 16, 32, 64, 128 and so on. If there are 128 people remaining, 1 will survive.

In our case, we do not have 128 people remaining. But, we have 100 people.

What do we do now? If the number of people remaining is 64, we know the answer. So, from 100 keep killing until the number of people remaining is 64. From hereon, we know the answer.

For 64 people to be alive, we have to kill 36 people out of the 100. The moment we kill 36 people, 64 people are remaining and we know the answer. So, how do we kill 36 people?

1 will kill 2, 3 will kill 4, 4 will kill 6 and so on or the list of people getting killed is {2, 4, 6, 8, 10… }. In this list, the 36th number is 72. After 72 is killed, **73** will now have the sword. Now, consider this as the starting point. If this is the starting point and if person 73 starts the process of killing, 64 people will be remaining in the circle. Who will win? **73**!!

The winner or the last person standing will be 73.

**Key takeaway**: Think about the highest power of 2 less than 100: 64. If 64 people are remaining, 36 should get killed. When the 36th person is killed, whoever is handed the sword, that person will be the guy who has the sword when 64 people are remaining or in other words will be the winner.

73 is the winner. 73 is a delightful number. It gets special attention in an episode of “Big Bang Theory” as well. Now, what is so special about it? **P.s**. think about prime numbers.

#### Qn 3: The radii of the smaller semi circles are 2 and 1 cm respectively. Find the radius of the red circle.

Let us draw the diagram with the radii.

We can see that the diameter of the large circle is 6 cm. Therefore, the radius of the large circle is 3 cm. We need to find the radius of the red circle.

Points to remember:

1. When two circles touch each other internally, the difference between their centers is “R -- r”.

2. When two circles touch each other externally, the difference between their centers is “R + r”.

Let us proceed with this info and further mark out some territory.

We are yet to use one more idea -- The big circle touches the red circle internally. Let us use this idea as well.

We need to find “r”. ∆ABC is carved out into ∆ABO and ∆OAC. We will need to use this somehow. How do we do this?

We have been given the three sides of both ABO and OAC. If we know the three sides, we can find the area of the triangle.

Area of ∆ABC is BC * height.

Area of ∆OAC is OC * height and

Area of ∆ABO is OB * height. We know, OC = 2 * OB

or Area of ∆OAC = 2 * Area of ∆ABO

=> 2 -- r = 8 -- 8r => 7r = 6 or

#### Qn 4: Data Sufficiency: Use the following answer choices for the question below.

A. Statement 1 alone is sufficient but statement 2 alone is not sufficient to answer the question asked.

B. Statement 2 alone is sufficient but statement 1 alone is not sufficient to answer the question asked.

C. Both statements 1 and 2 together are sufficient to answer the question but neither statement is sufficient alone.

D. Each statement alone is sufficient to answer the question.

E. Statements 1 and 2 are not sufficient to answer the question asked and additional data is needed to answer the statements.

Is the area of a certain circle greater than the area of a certain equilateral triangle?

1) The perimeter of the equilateral triangle is less than the circumference of the circle.

2) The circumference of the circle is less than the perimeter of the equilateral triangle.

A bit of intro about Data Sufficiency: A question is given followed by two statements. Guidelines on how to choose an answer option based on the statements are given above.

Let us dive into the question. Is the area of a certain circle greater than the area of a certain equilateral triangle?

According to **Statement 1**, the perimeter of the equilateral triangle is less than the circumference of the circle. Let us flip the question and look at a slightly different scenario. Consider that perimeters are equal for an equilateral triangle and circle.

In this case, definitely, Area of circle > Area of equilateral triangle.

How so? Imagine you have a string and you spread it out, you can imagine the area increasing for the same perimeter.**Rule of thumb**: More the no. of sides, higher the area for a given perimeter.**Rule of thumb**: More the symmetry of a shape, higher the area for a given perimeter.

Given a perimeter, when will be the area of a triangle be the largest?

When the triangle is an equilateral triangle.

Given a perimeter, which quadrilateral will have the largest area?

Square will have the largest area.

Now, for a given perimeter, which will have the larger area between a Square and an Equilateral triangle?

Since Square has more sides, the square will have the larger area.

Given perimeter, when will the area be maximum?

A circle has the maximum sides (infinite) and most symmetry. Hence, the circle will have the largest area.

Now, according to **Statement 1**, the perimeter of the equilateral triangle is less than the circumference of the circle.

We know that if perimeters of triangle and circle are equal, the circle will have the greater area.

If perimeter of triangle < perimeter of circle, definitely, the circle will have more area.

So, Statement 1 is **sufficient** to answer the question.

According to **Statement 2**, the circumference of the circle is less than the perimeter of the equilateral triangle.

Can we now say if the area of a certain circle greater than the area of a certain equilateral triangle?

It need not be. Let us take an example.

Assume perimeter of equilateral triangle is 18 units and perimeter of circle is 18 units. In this case, Area of circle > Area of equilateral triangle.

Let us take another example.

Assume perimeter of equilateral triangle is 18 units and perimeter of circle is 17.98 units. Even in this case, probably, Area of circle > Area of equilateral triangle.

However, if perimeter of equilateral triangle is 18 units and perimeter of circle is 1 units. Then, Area of circle < Area of equilateral triangle.

So, with Statement 2, we have two opposite scenarios. Hence, Statement 2 is **not** **sufficient** to answer the question.

Choice **A**. Statement 1 alone is sufficient but statement 2 alone is not sufficient to answer the question asked.

That’s all folks! Hope you had as much fun as I had solving these wonderful questions. Remember CAT preparation can be made fun if the concepts are learnt the right way :).

*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.*

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