Mathathon 2021 - Sample Problem

Okay, so to get the most obvious answer out of the way, let $l$ be the diameter of the circle. It follows that the circle has area $A_\circ = \pi l^2$ and the square has area $A_\Box = (2l)^2 =4l^2$ .

Intuitively, we can say that the area of the square encompasses all possible points, whereas the area of the circle is comprised of only the points that we 'want'. Thus, it makes sense to take the ratio of the two and interpret it as the probability that a randomly chosen point lies inside the circle. Let this probability be $p_\text{cartesian}$ . It's easy to see that $p_\text{cartesian} = \frac{A_\circ}{A_\Box} = \frac{\pi l^2 }{4l^2 } = \frac{\pi}{4} .$ NOW, what's always important with probability is to explain your method of selecting values. I will show why this matters. You can, for example, compare the probabilities when using Cartesian and polar coordinates using code and see that you get different results with the probability in polar coordinates $p_\text{polar}\approx 90\%$ .

When thinking more deeply about the problem, the fact that the probability is related to the areas of the circle and square required some explanation to me. Since the probability does not in any way seem to be proportional to the radius/side length of either the circle or the square, I will simplify the following calculations by letting the square be have side length 2 and the circle be a unit circle. If we say that we place the two in a Cartesian coordinate system, both centred at the origin, we get the following description: $\bigcirc = \left\{(x,y)\in\mathbb{R}^2 \;\rvert\; x^2 +y^2 \leq 1\right\},\quad \Box = \left\{(x,y)\in\mathbb{R}^2 \;\rvert\; \mid x\mid \leq 1 \text{ and }\mid y\mid \leq 1\right\} .$ Probability in Cartesian coordinates:

The first method that I came up with to randomly choose a point goes as follows:

• Randomly choose a value $x_p\in\mathbb{R}, \mid x_p\mid \leq 1$ . All $x$ -values inside the interval have the same probability of being chosen.

• Now choose a value $y_p\in\mathbb{R}, \mid y_p\mid \leq 1$ . All $y$ -values inside the interval have the same probability of being chosen.

• If $x_p^2 +y_p^2 \leq 1$ , the point $p=(x_p,y_p)$ is considered inside the circle. If $x_p^2 +y_p^2 >1$ , the point is outside the circle.

To yield the same probability without initially resorting to areas, I will use probability density functions. This class of functions has, among others, the following properties:

• The integral between two points of its domain represents the probability that an object is within the interval between the two points.

• The integral over its entire domain is normalised to equal 1. This makes sense since the object has to be somewhere.

I use this kind of function because for a point in the reals, the actual probability for that point to be selected is 0 because there are infinitely many points around it that could also be selected.

Now comes the juicy part! To calculate the total probability I will make use of the fact that the first two steps of the algorithm above can be assigned a probability density function. Multiplying these functions is like multiplying discrete probabilities, only that we have to integrate over the desired domain (the circle) to get a real probability.

So let $\Psi_x$ be a probability density function on $\mathbb{R}$ with domain

$\{x\in\mathbb{R} \;\rvert\; -1 \leq x \leq 1 \} .$ and $\Psi_y$ be a probability density function also on $\mathbb{R}$ with domain: $\{y\in\mathbb{R}\;\rvert\; 0\leq y\leq 1\} .$ Note that I only include half of the square, which, as a quick calculation using the area method shows, should not change $p_\text{cartesian}$ . The same probability $p_\text{cartesian}$ will be calculated as $p_\text{cartesian} = \iint_{\frac{1}{2}\circ} \Psi_x\cdot \Psi_y\text{d}x\text{d}y,$ where $\Psi_x, \Psi_y$ are the probability densities that a random point gets picked.

To determine $\Psi_x$ and $\Psi_y$ , note that they have to be constant since the probability is constant. This will be useful when normalising: $\begin{aligned} \int_{-1}^{1}\Psi_x \text{d}x &\overset{!}{=} 1 \\ \Psi_x\int_{-1}^{1} \text{d}x = 2\Psi_x \implies \Psi_x &= \frac{1}{2}\\[10pt] \int_{0}^{1} \Psi_y \text{d}y &\overset{!}{=}1\\ \Psi_y \int_{0}^{1} \text{d}y =\Psi_y \implies \Psi_y &= 1 \end{aligned}$ Parametrizing the circle leaves the following integral to be evaluated: $p_\text{cartesian} = \int_{x=-1}^{x=1} \int_{y=0}^{y=\sqrt{1-x^2 }} \Psi_x\cdot \Psi_y\text{d}x \text{d}y .$ And this is where the area comes into play. Because both probabilities are constant, we only have an integral over the semicircle left to be evaluated. This corresponds to its area. $\begin{aligned} \int_{x=-1}^{x=1} \int_{y=0}^{y=\sqrt{1-x^2 }} \Psi_x\cdot \Psi_y\text{d}x \text{d}y &= \frac{1}{2}\int_{x=-1}^{x=1} \sqrt{1-x^2 } \text{d}x\\ \text{let } x &= \cos(\phi), \phi\in[0,\pi]\\ \text{d}x &= -\sin(\phi)\text{d}\phi\\[10pt] \implies p_\text{cartesian} &= -\frac{1}{2} \int_{\phi=\pi}^{\phi=0}\sqrt{1-\cos^2(\phi)}\sin[](\phi) \text{d}\phi\\ &= \frac{1}{2}\int_{\phi=0}^{\phi=\pi} \sin^2(\phi) \text{d}\phi\\ &= \frac{1}{2}\cdot \frac{\pi}{2}\\ p_\text{cartesian} &= \frac{\pi}{4} \end{aligned}$ and, oh wonder I get the same answer!

But this is where the story gets interesting, because I now took a look at polar coordinates.

Probability in polar coordinates

To explain the higher probability when using polar coordinates, I constructed a similar algorithm for choosing a random point. Only note that the maximum length of $r$ is now a function of $\vartheta$ , which i will evaluate only in the first eighth of the circle. To see why this function is $r=\sec(\vartheta)$ , you can draw the square and see that $\cos(\vartheta)=\frac{1}{r}$ for $\vartheta\in[0,\frac{\pi}{4}]$ because the square's side for positive $x$ -values stays fixed at 1. This saves some computation effort and seems reasonable, as all other parts of the circle are symmetrical to its first eighth and the area method gives the same probability for this section of the circle as for the whole thing. So here's the algorithm:

• Randomly choose a value $\vartheta_p\in\mathbb{R}, 0\leq \vartheta_p\leq \frac{\pi}{4}$ . All $\vartheta$ -values inside the interval have the same probability of being chosen.

• Now choose a value $r_p\in\mathbb{R}, 0\leq r_p\leq \sec(\vartheta_p)$ independently of your choice for $\vartheta_p$ . All \r -values inside the interval have the same probability of being chosen.

• If $r_p\leq 1$ , the point $p=(r_p,\vartheta_p)$ is considered inside the circle. If $r_p>1$ , the point is outside the circle.

Once again, for computing the probability $p_\text{polar}$ , I will resort to probability density functions which will give the following integral to evaluate: $p_\text{polar} = \iint_{\frac{1}{8}\circ} \Psi_\theta\cdot \Psi_r \text{d}r\text{d}\vartheta .$

Determining $\Psi_\vartheta$ goes as follows: $\begin{aligned} \int_{\vartheta=0}^{\vartheta=\frac{\pi}{4}}\Psi_\vartheta \text{d}\vartheta &\overset{!}{=} 1\\ \Psi_\vartheta \int_{\vartheta=0}^{\vartheta=\frac{\pi}{4}}\text{d} \vartheta= \frac{\pi}{4}\Psi_\vartheta &= 1\\ \implies \Psi_\vartheta &= \frac{4}{\pi} \end{aligned}$ and although $r$ is now a function of $\vartheta$ , $\Psi_r$ still has to be constant for any $\vartheta$ because every value of $r$ has the same probability of being chosen on the length which is determined by $\vartheta$ . $\begin{aligned} \int_{r=0}^{r=\sec(\vartheta)} \Psi_r \text{d}r &\overset{!}{=}1\\ \Psi_r\int_{r=0}^{r=\sec(\vartheta)} \text{d}r = \frac{\Psi_r}{\cos(\vartheta)} &= 1\\ \implies \Psi_r &= \cos(\vartheta) \end{aligned}$ Plugging this into the integral, we get $\begin{aligned} p_\text{polar} &= \frac{4}{\pi}\int_{\vartheta=0}^{\vartheta=\frac{\pi}{4}} \int_{r=0}^{r=1} \cos(\vartheta) \text{d}r \text{d}\vartheta\\ &= \frac{4}{\pi}\left[\sin(\vartheta)\right]_{\vartheta=0}^{\vartheta=\frac{\pi}{4}}\\ &= \frac{4}{\pi}\cdot \frac{1}{\sqrt{2}}\\ p_\text{polar} &= \frac{2\sqrt{2}}{\pi} \approx 90\% \end{aligned}$ which completely agrees with the observations.

Using probability density functions, I have shown that evaluating probabilities for processes like the one described in the problem depends heavily on the method of selecting values.

Let the length of the square's segments be a. The required probability that a randomly selected point inside the square will also fall within the circle = (Area of circle / Area of Square ) . Now , radius of circle = a/2. Therefore Area of circle = pi*(r^2) = pi * (a / 2) 2 = pi * a^2 / 4 . Area of square = a^2 . Therefore, the required probability = (pi * a^2 / 4) / a^2 = pi / 4

Collection of points that lie inside a polygon $=$ Surface area of the polygon

Points inside the circle $= πr^2$ , where $r$ is the radius of the circle, note that $r$ is half the length of the side of the square which inscribes the circle (use the fact that all four of its sides are tangent to the circle). Thus the side of the square is $2r$ , which gives the area to be $(2r)^2=4r^2$ .

Note that every point that lies inside the circle will by itself be also lying inside the square, for every point inside the circle , but the converse is not necessarily true, there's a collection of points outside the circle, which is still inside the square, but not inside the circle, that make up an area of $4r^2- πr^2$ , which should make sense as $4>π$ .

In conclusion, we can say that the points inside the square make up the set of all sample points that lie inside the square and sometimes might also lie inside the circle, or the set of Cartesian points inside the square, $S$ is the superset of the set of all points inside the circle, $C$ , and since we're projecting all our random choices from $S$ , $S$ is also the Universal set here, so while calculating the possibility of a sample point being in $C$ , we will calculate it out of the total given sample points, $S$ . Thus,

$P(C)= \dfrac{\text{Number of sample points in }C}{\text{Number of sample points in }S}= \dfrac{\text{Composition of sample points in }C}{\text{Composition of sample points in }S}= \dfrac{\text{Area of the Circle}}{\text{Area of the square}} =\dfrac{πr^2}{4r^2}$

$=\boxed{\dfrac{π}{4}}$

A probability definition is the likelihood of an event compared to total possible outcomes. The "total" possible outcomes here is represented by the area of the square. We are interested in the likelihood of the random point to be inside the circle, hence the area of the circle.

I was minded to do some animation, so bear with it and endure lol

I hope you all got it :D

First we defined that the probability to be inside a square is: A. Circle / A. Square. A. Circle = πR^2 and A. Square = 4R^2, -> πR^2/4R^2 -> π/4 (also, R is the radio of the circle of 1: π(1*1) and in the square case we can represent as 4 = 2*2 with 1 unit R)

Any point in the square is equally likely. Some of those points are in the inscribed circle. The probability of a point being in a region is proportional to the area of that region. Therefore he desired probability is equal to the ratio of the area of the circle to the area of the square. Assume R is the radius of the circle. The areas are $\pi R^2$ and $(2 R )^2$ Therefore the ratio is $\pi/4$

Suppose that the circle has radius $r$ , and the square has side length $2r$ . The area of the circle will be $πr^{2}$ , and the area of the square will be $4r^{2}$ . Since the point has an equal probability of landing anywhere in the square, the probability that it is also in the circle is $\dfrac{\small\text{the area of the circle}}{\small\text{the area of the square}}$ which is $\frac{πr^{2}}{4r^{2}}$ . The $r^{2}$ 's cancel, and we are left with $\boxed{\frac{π}{4}}$ .

Let the $\red{Square}$ be the universal set.

• Now let P(C) be the probability of the point lying inside the $\blue{Circle}$
• Also let P(NC) be the probability of of the point lying outside the $\blue{Circle}$

Now we know that P(C) + P(NC) = 1 since they are complementary and we require P(C) = 1 - P(NC).

• Let the side length of the $\red{Square}$ be 2 units

• Now the radius of the $\blue{Circle}$ = 1 unit

• Therefore Area of the $\red{Square}$ = $2*2$ = 4

• And the Area of the $\blue{Circle}$ = $\pi$ ( $1*1$ ) = $\pi$
• P(NC) = $\frac{4-\pi}{4}$ = 1 - $\frac{\pi}{4}$

Therefore our required answer P(C) = 1 - P(NC) = $\frac{\pi}{4}$

We have to calculate the probability that a random point selected in the interior of the square is also inside the circle.

Let r be the radius of the inscribed circle. If we increase r to kr (k is constant) the area of the circle as well as the square will increase by the same proportion that is k^2.

From this it is clear that for any value of 'r' the probability will be the same.

So lets take r=1

• area of circle = πr^2 = π* 1 *1 = π
• area of square = side^2 = (2r)^2 = (2*1)^2 = 4

Therefore, the probability of required event = $\dfrac{π}{4}$

Let a be the side of the square

Then the area of square would be $a^2$ And since the diameter of the circle = the side of the square=a the radius of the circle,which is half times diameter will be $\frac{a}{2}$ Now we can substitute the value of r in $π*r^2$ (which gives us the area of a circle)

$\frac{π*a^2}{4}$

Now, the probability of a point lying in the square:

p(s)=1(because there can't be any point outside it ,assuming that it was our universal set)

The probability of a point lying in both square and circle

p(s∆c)=p(c)(because c is a subset of s)

And clearly,

p(c)= $\frac{π*a^2}{a^2}$ =π/4

Hence option b is correct!

Exotic explanation looking for ingenuity points :

I took a a quadrant inscribed in a unit square.
I uniformly (randomly) scattered a given number of points over the square.
I counted the number of points inside the quadrant.
Then I evaluated the ratio : $\boxed{ pi/4 }$ .

Not the quickiest way but it works !

The same approach has been used to evaluate $pi$ and is known as the Monte Carlo method.

Probability = $\dfrac{\text{what you want to get}}{\text{what you could get}}$ . In this case, it is the ratio of $S_\circ$ to $S_\square$ which is just $\dfrac{\pi}{4}$ .

Most Common Solution -

The probability that a point in the square is also a point in the circle is nothing but the ratio of the circle's area to the square's area.

Let the radius of the circle be $r$ . Then the side of the square is of length $2r$ . Then, the areas of the circle and square are $\pi r^{2}$ and $4r^{2}$ respectively.

Thus, the answer is $\dfrac{\pi r^{2}}{4r^{2}} = \boxed{\pi / 4}$