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## Area of the Trapezium | PRMO-2017 | Question 30

Try this beautiful Problem from Geometry based on Area of the Trapezium from PRMO 2017.

## Area of the Trapezium – PRMO 2017, Problem 30

Consider the areas of the four triangles obtained by drawing the diagonals $\mathrm{AC}$ and $\mathrm{BD}$ of a trapezium ABCD. The product of these areas, taken two at time, are computed. If among the six products so obtained, two products are 1296 and 576 , determine the square root of the maximum possible area of the trapezium to the nearest integer.

• $9$
• $40$
• $13$
• $20$

### Key Concepts

Geometry

Triangle

Trapezium

Answer:$13$

PRMO-2017, Problem 30

Pre College Mathematics

## Try with Hints

First hint

Let $x, y, z, w$ be areas of the four triangles as shown in figure.
then area of $\triangle ADB$= Area of $\triangle ACB$
$\Rightarrow x+y=x+w \Rightarrow y=w$

Also $\frac{AE}{EC}$=$\frac{area of \triangle ADE}{area of \triangle DEF}$=$\frac{area of \triangle AEB}{area of \triangle BEC}$
$\Rightarrow \frac{y}{z}=\frac{x}{w}=\frac{x}{y} \Rightarrow y^{2}=z x$
$\Rightarrow z, y, x$ are in G.P.

Can you now finish the problem ……….

Second Hint

Let $y=z r$ and $x=z r^{2},$ where $r \geq 1$ To make area of trapezium ABCD maximum, we take $z y=z^{2} r=576$
and $y w=z^{2} r^{2}=1296$
As $( z \leq y \leq x)$
Therefore $\frac{z^{2} r^{2}}{z^{2} r}=\frac{1296}{576} \Rightarrow r=\frac{9}{4} \Rightarrow z=16$

Can you finish the problem……..

Final Step

Therefore area of trapezium $\mathrm{ABCD}$
$=x+y+z+w=z r^{2}+2 z r+z$
$=z(1+r)^{2}=16\left(1+\frac{9}{4}\right)^{2}=13^{2}$
Therefore Answer is $13 .$

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## Problem on Circle and Triangle | AMC 10A, 2016 | Problem 21

Try this beautiful problem from Geometry: Problem on Circle and Triangle

## Problem on Circle and Triangle – AMC-10A, 2016- Question 21

Circles with centers $P, Q$ and $R,$ having radii 1,2 and 3 , respectively, lie on the same side of line $l$ and are tangent to $l$ at $P^{\prime}, Q^{\prime}$ and $R^{\prime}$ respectively, with $Q^{\prime}$ between $P^{\prime}$ and $R^{\prime}$. The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of triangle $P Q R ?$

,

• $0$
• $\sqrt{6} / 3$
• $1$
• $\sqrt{6}-\sqrt{2}$
• $\sqrt{6} / 2$

### Key Concepts

Geometry

Circle

Triangle

Answer: $\sqrt{6}-\sqrt{2}$

AMC-10A (2016) Problem 21

Pre College Mathematics

## Try with Hints

First hint

We have to find out area of the Triangle PQR. But PQR is not a Standard Triangle that we can find out eassily. Join $PP^{\prime}$, $QQ^{\prime}$, $RR^{\prime}$. Now we can find out PQR such that $\left[P^{\prime} P Q R R^{\prime}\right]$ in two different ways: $\left[P^{\prime} P Q Q^{\prime}\right]+\left[Q^{\prime} Q R R^{\prime}\right]$ and $[P Q R]+\left[P^{\prime} P R R^{\prime}\right]$, so $\left[P^{\prime} P Q Q^{\prime}\right]+\left[Q^{\prime} Q R R^{\prime}\right]=[P Q R]+\left[P^{\prime} P R R^{\prime}\right]$

Can you now finish the problem ……….

Second Hint

$P^{\prime} Q^{\prime}=\sqrt{P Q^{2}-\left(Q Q^{\prime}-P P^{\prime}\right)^{2}}=\sqrt{9-1}=\sqrt{8}=2 \sqrt{2}$

$Q^{\prime} R^{\prime}=\sqrt{Q R^{2}-\left(R R^{\prime}-Q Q^{\prime}\right)^{2}}=\sqrt{5^{2}-1^{2}}=\sqrt{24}=2 \sqrt{6}$

$\left[P^{\prime} P Q Q^{\prime}\right]=\frac{P^{\prime} P+Q^{\prime} Q}{2} * 2 \sqrt{2}=\frac{1+2}{2} * 2 \sqrt{2}=3 \sqrt{2}$

$\left[Q^{\prime} Q R R^{\prime}\right]=5 \sqrt{6}$

$\left[P^{\prime} P R R^{\prime}\right]$ = $P^{\prime} R^{\prime}=P^{\prime} Q^{\prime}+Q^{\prime} R^{\prime}=2 \sqrt{2}+2 \sqrt{6}$

$\left[P^{\prime} P R R^{\prime}\right]=4 \sqrt{2}+4 \sqrt{6}$

$\left[P^{\prime} P Q Q^{\prime}\right]+\left[Q^{\prime} Q R R^{\prime}\right]=[P Q R]+\left[P^{\prime} P R R^{\prime}\right]$

$3 \sqrt{2}+5 \sqrt{6}=4 \sqrt{2}+4 \sqrt{6}+[P Q R]$

$[P Q R]=\sqrt{6}-\sqrt{2}$

Categories

## Some Classical Problems And Paradoxes In Geometric Probability||Cheenta Probability Series

This is our 6th post in our ongoing probability series. In this post, we deliberate about the famous Bertrand’s Paradox, Buffon’s Needle Problem and Geometric Probability through barycentres.

“Geometry is not true, it is advantageous.” ~ Henri Poincare

Yes , exactly… it’s time to merge the two stalwarts together : “Geometry” and “Probability”.

## The Probability Measure of Geometrical Elements

In probability theory one is usually concerned with random variables which are quantities, or sets of quantities, taking values in some set of possibilities on which there is defined a non-negative measure, satisfying certain required conditions which enable us to interpret it as a probability. In the theory of geometrical probabilities the random elements are not quantities but geometrical objects such as points, lines and rotations. Since the ascription of a measure to such elements is not quite an obvious procedure, a number of “paradoxes” can be produced by failure to distinguish the reference set. These are all based on a simple confusion of ideas but may be useful in illustrating the way in which geometric probabilities should be defined.

We consider one paradox due to J.Bertrand (1907).

The problem of interest is precisely: Determine the probability that a random chord of a circle of unit radius has a length greater than the square root of 3, the side of an inscribed equilateral triangle.

## Context of the problem

The development of the Theory of Probability has not been smooth at all. The first attempts to formalize the calculus of probability were due to Marquis De Laplace (1749-1827) who proposed to define the probability $P(A)$ of an outcome A as the ratio of the number of events that result in the outcome A to the total number of possible events. This is of course only meaningful if the number of all possible events is finite and, in addition, all the events are equi-probable. The notion which Laplace has also defined. However, in our first blog post, we addressed the fact that the definition is, in a sense, circular – a notion of equi-probable is defined prior to the introduction of probable.

Thus, at the time, the field did not seem to have a sound foundation. Attempts to extend the definition to the case of infinite number of events led to even greater difficulties. The Bertrand’s Paradox is one such discovery that made mathematicians wary of the whole notion of probability.

Apparently, this problem has more than one solution, meaning as the perspective of the reader changes, the solution also changes! Worthy of a paradox right?!

## Some of the most discussed solutions

What about the probability is $\frac{1}{3}$ ?

Yeah, this is correct! Provided your thought process follows the same lines as this proof :

Any chord of the circle intersects it in two points, and we may suppose these to be independently distributed in probability distributions which are uniform over the circumference of the circle. Without loss of generality, we can suppose one of the two points to be at a vertex of an inscribed equilateral triangle. There is then just $\frac{1}{3}$ of the circumference in which the other point can lie in order that the resulting chord has length greater than $\sqrt{3}$ so that the probability is $\frac{1}{3}$.

What about $\frac{1}{4}$ ?

Umm…sure! why not? Any chord is uniquely defined by the foot of a perpendicular on it from the centre. If this point is distributed uniformly over the circle the probability of it lying in any region of area $A$ is $A \pi^{-1}$ since
the total area of the circle is $\pi$. For the chord to have length greater than $\sqrt{3}$ the foot of the perpendicular must lie inside a circle of radius $\frac{1}{2}$ and hence the probability is $\frac{1}{4}$.

But but.. it is also $\frac{1}{2}$ !?

Try to think of a proof why this probability is also $\frac{1}{2}$ .

Based on constructing a random chord in a circle, the paradox involves a single mathematical problem with three reasonable but different solutions. It’s less a paradox and more a cautionary tale. It boils down to the same old question: “What do you mean by random?”

## Let’s hand it over to Buffon

The “Buffon needle problem” which many of us encountered in our college days has now been with us for 200 years. One major aspect of its appeal is that its solution has been tied to the value of $\pi$ which can then be estimated by physical or computer simulation today. It is interesting that in the original development, Buffon (1777) extols geometry as a companion tool to the calculus in establishing a science of probability and suggests that chance is amenable to the methods of geometry as well as those of the calculus. Buffon indicates that the human mind, because of prior mathematics, preferred numbers to measures of area but that the invention of games revolving around area and ratios of areas could rectify this. To highlight this point he investigated a game already in practice in the 18th century
known as “clean tile”.

### Clean Tile Problem

In a room tiled or paved with equal tiles, of any shape, a coin is thrown upwards; one of the players bets that after its fall the coin will rest cleanly, i.e., on one tile only; the second bets that the coin will rest on two tiles, i.e., that it
will cover one of the cracks which separate them; a third player bets the coin will rest over 3, 4, or 6 cracks: it is required to find the chances for each of these players.

This problem is regarded as the precursor of the famous “needle problem” .

Buffon in his own words states, “I assume that in a room, the floor of which is merely divided by parallel lines, a stick is thrown upwards and one of the players bets the stick will not intersect any of the parallels on the floor, whereas on the contrary the other one bets the stick will intersect some one of these lines; it is required to find the chances of the two players. It is possible to play this game with a sewing needle or a headless pin.

### Buffon’s Needle

Suppose we have a floor made of parallel strips of wood, each the same width, and we drop a needle onto the floor. What is the probability that the needle will lie across a line between two strips?

#### Uspensky’s Proof:

Let the parallel lines be separated $d$ units apart. The length of the needle is given by $l$,with the assumption $l \le d$.Uspensky (1937) provides a proof that the probability of an intersection is $p = \frac{2l}{\pi d}$ . He develops this by considering a finite number of possible positions for the needle as equally likely outcomes and then treats the limiting case as a representation of the problem. This includes a definition of randomness for the distance $x$ of the needle’s midpoint to the nearest line and the acute angle $\phi$ if formed by the needle and a perpendicular from the midpoint to the line. The solution is obtained by computing the ratio of favorable outcomes to the total set
of outcomes and passing to the limit.

A measure of the set of total outcomes is given by:

$\int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{d}{2}} dx d \phi = \frac{\pi d}{4}$

From he figure above, it’s evident that the measure of the set of intersections is:

$\int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{l}{2} \cos \phi } dx d \phi = \frac{l}{2}$

Therefore $p = \frac{l/2}{(\pi d)/4} = \frac{2l}{\pi d}$.

For knowing about the further generalizations to this problem you must go through Laplace’s Extension and Long Needle case.

Now, let’s explore something new!

## Barycentric Coordinates take guard!

What are barycentric coordinates?

For precise definiton and illustration of barycentres , you can go through Particle Geometry and Triangles.

The barycentric coordinates are generally defined in context of triangles but they can be set up in a more general space $\mathbb{R}^n$ . For n=1, it takes 2 distinct points $A$ and $B$ and two coordinates $u$ and $v$. Every point $K$ on the real line is uniquely represented as $K = uA+vB$ , where $u+v=1$. More generally, to define the barycentric coordinates in $\mathbb{R}^n$ , one need $n+1$ points that do not lie in a space of lesser dimension.

Now, let’s look at a problem:

Choose $n$ points at random on a segment of length 1. What is the probability that an $(n+1)$-gon (a polygon with $(n+1)$ sides) can be constructed from the $(n+1)$ thus obtained segments?

This is a generalization of this famous problem: “Two points are selected at random on a straight line segment of length 1. What is the probability that a triangle can be constructed out of thus obtained three segments?”

The above problem has a simple geometric proof and does not require heavy machinery , which will probably be addressed in our future posts. But the generalization makes the problem tempting enough to use barycentres.

First of all for the validity of the $(n+1)$-gon, it must satisfy:

$x_i < x_0 + x_1 +…+x_n – x_i =1- x_i \ \forall i=\{ 0,1,…,n\}$

Thus, $x_i < \frac{1}{2}$.

The object obtained by this set of inequalities is is obtained from the basic simplex by removing smaller simplexes, one at every vertex. Each of these $(n+1)$ smaller simplexes has the hypervolume equal to $(\frac{1}{2})^n$ of the hypervolume of the basis simplex.

Thus, the probability that the barycentric co-ordinates satisfy this set of inequalities is $p= 1- \frac{n+1}{2^n}$.

See that this probability goes to 1 as $n$ grows larger and larger explaining the fact that it is easier to find segments to construct a many sided polygon than it is to find the sides of a triangle, which is rather natural.

## Food For Thought

Today’s section does not contain any problem, rather I would like to share a popular research problem regarding “The extension of Buffon’s Needle Problem in three dimensions”. There have been many attempts to incorporate another dimension to the traditional needle problem for instance, If the needle were released in a weightless environment, then it wouldn’t drop down to the plane, it would float. This introduces another dimension into the problem. I would suggest some research articles in the bibliography which discusses this problem in detail. You can go through them if this problem excites you enough to dig deep into it.

Till then , stay safe!

Ciao!

## References:

1. ” The Buffon Needle Problem Extended ” – JAMES MCCARRY & FIROOZ KHOSRAVIYANI

2. “The Buffon–Laplace needle problem in three dimensions” – Zachary E Dell and Scott V Franklin

3. “Fifty Challenging Problems in Probability with Solutions” – Mosteller

4. “Geometric Probability” – Kendall, Morran

Categories

## How to Measure the Length of your Earphone from a Pic?| Cheenta Probability Series

This is our 5th post in the Cheenta Probability Series. This article teaches how to mathematically find the length of an earphone wire by its picture.

Let’s explore some truths.

• A Line is made up of Points.
• A Curve is made up of Lines.
• Earphones are as Messy.

This article is all about connecting these three truths to a topic in Probability Theory $\cap$ Geometry, which is not so famous among peers, yet its’ usefulness has led one to make an app in Google Play Store, which has only one download (by me). The sarcasm is towards the people, who didn’t download. (obviously).

Today is the day for Cauchy Crofton’s Formula, which can measure the length of any curve using random straight lines by counting points. It is all encoded in the formula.

$L(c)= \frac{1}{2} \times \iint n(p, \theta) dp d \theta$

This is the begining of a topic called Integral Geometry.

This formula intuitively speaks the truth

The Length of a Curve is the Expected Number of Times a “Random” Line Intersects it.

## Describing a Line (like seriously?).

I mean a “Random” Line.

Thus the Space of all Lines in $\mathbf{R}^2$ can be represented in “kinda” polar coordinates S = {${(\theta, p) \mid 0 \leq \theta \leq 2 \pi, p \geq 0},$} where $p, \theta$ are as indicated.

Now, we are now uniformly selecting this $(\theta, p)$ over S. This is how we select a “Random” Line.

## Understanding Crofton’s Gibberish

If C is the curve, whose length we are interested to measure. Then, $L(C)= \frac{1}{2} \times \iint n(p, \theta) dp d \theta$.

$L(C)$ = The Length of the curve

$n(\theta, p)$ = The Number of times “Random Line” corresponding to $(\theta, p)$ intersects C.

Crofton meant the following on a pretty, differentiable, and cute (regular) curve C that:

$L(C)= \frac{1}{2} \times E_{(p, \theta)} [n(p, \theta)]$

## The First Truth

Here, we will find the length of a Line using “Random Lines”. We will see, how counting points on the line, can help us find the length of a line.

Here, we choose C = Line.

A Line intersects another Line exactly once. $n(p, \theta) = 1$. Hence,

$L(C)= \frac{1}{2} \times \iint n(p, \theta) dp d \theta = \frac{1}{2} \times \iint dp d\theta = \frac{1}{2} \times \int_{0}^{2 \pi}\left(\int_{0}^{|\cos \theta|(1 / 2)} d p\right) d \theta = \frac{1}{2} \times \int_{0}^{2 \pi} l / 2|\cos \theta| d \theta= l$

## The Second Truth

Approximate Curves by Lines.

Refer to Approximation of Curves by Line Segments by Henry Stone.

## Just Limit!

Now, take apply the result on these lines. By linearity of expectation and the uniform convergence of the Lines to the Curve, we get the same for the curve.

Suppose that the curve c is parameterized on the interval [a,b]. Define a partition $P$ to be a collection of points that subdivides the interval, say { $a = a_0, a_1, a_2, …, a_n = b$ }.

Define $L_{r}(c)=\sum_{i=1}^{n} |c\left({a}_{i-1}\right)-c\left({a}_{i}\right)|$

Then $L(c)= sup_{r} L{r}(c)$

Now, each of the $|c\left({a}_{i-1}\right)-c\left({a}_{i}\right)|$ are approximated by the Crofton’s Formula. Now, taking the limit gives ust eh required result. QED.

## Measuring the Length of an Earphone

Create a Mesh of Lines (Coordinate System) – A Fixed Sample representing the Population of the Random Lines. Then count the number of intersections using those mesh of lines only and add it up.

The mathematics of the above argument is as follows.

• Construct a grid of lines parallel to the x-axis at a distance r apart.
• Rotate this family of curves about the origin by $\frac{\pi}{3}, \frac{2\pi}{3}$ to get a system of 3 parallel system of lines. (We can make this finer).
• Now if a line in the grid intersects a curve n times, there might be more intersections if we consider all lines in $\mathbf{R}^{2}$- we only know about the grid lines. The parameters of these unknown lines can range from 0 to $r$ and from 0 to $\frac{\pi}{3}$
• Therefore we can approximate the length of C by:

$L(c)= \frac{1}{2} \times \iint n(p, \theta) dp d \theta \cong \frac{1}{2} \times n r \times \frac{\pi}{3}$

An example is show below. This is done in the app named Cauchy-Crofton App, which unfortunately is not working now and I have reported.

## Future Articles and Work

There are lots of corollaries coming up like Isoperimetric Inequality, Barbier’s Result, etc. I will discuss these soon. Till then,

Stay Tuned!

Stay Blessed!

Categories

## Physics of Coin Tossing and Uncertainty | Cheenta Probability Series

This is our 4th post in the Cheenta Probability Series that deals with the physics involved in coin tossing. It reveals the true nature of uncertainty.

“It is a very tedious task !! First you have to calculate where he is and where is is not, then you must calculate where he could possibly be, then you must seek where he is at this moment, then finally you have to calculate the probability that what is the chance of finding him when you reach, where you are suspecting him to be right now . ”

Sukumar Ray in his Novel, ” Utter Nonsense”( Ho-Jo-Bo-Ro-Lo)

Uncertainty is something that has drawn attention of mathematicians ad philosophers right from the beginning of the modern civilization. There has been many school of thoughts about the true nature of uncertainty, which also changed and is keeping changing as the mathematics is getting sophisticated gradually. The perspective of uncertainty that was(is) very much appealing to me originated from the perspective of the great probabilist Bruno de Finetti is “Uncertainty is actually the quantification of our ignorance of the lack of information “. This straightaway made the idea of probability subjective, which should be a topic for discussion for another day. But one of the living legend and an ambassador of this school of thought Persi Diaconis is man who is obsessed with the way the most fundamental places where the uncertainty exists, goes further finding a physical solution to the problem of quantifying uncertainty, which our present discussion is all about.

In statistical experiments lack of knowledge do motivate statisticians to develop more sophisticated laws of predictions, and statisticians gain only that much knowledge that they can afford given a cost. So, lack of knowledge is what induces the desire of prediction in extension handling uncertainty. They basically do a mapping of knowledge gain with cost, though cost-gain of knowledge relationship is a different topic altogether and not really a core probability stuff, but from this we end up with two most important philosophical questions

• What is uncertainty ?
• Can we actually gather all the information if we have a luxury of infinite investment ?

You can see clearly that the both questions are correlated and assuming on gives you the justification for another, and I choose to discuss on the first question as that is also a more general one.

## Is Coin flipping is Random

There are quite a number of fundamental models that engage probabilists and mathematicians in their quest for the actual nature of uncertainty, coin tossing being the most basic and fundamental of them, made the great mathematician and probabilist Diaconis obsessed and he dived in the search of the true nature of uncertainty. Inspite of the simplicity of the experiment, and the number of unknowns are many and they accumulate to an ultimate uncertain system, which are subject to physical changes that one can’t possibly have the luxury to be aware of. But Diaconis being an obsessed man, follows the thoughts of de Fenetti and questions, “Is coin tossing is truly random, or its just our ignorance over the physical parameters ?” .

what if we knew about the magnitude of the force that our thumb imparted on the coin? At how many spins per second the coin spun while going up or coming down ? At what velocity the coin went up ? How high did the coin went before it succumb on the ground ? What about the collisions with the air ?

If we if can answer these questions, then Physics allows us to completely determine the outcome of coin flipping, so coin tossing is not at all random, its Physics !!

Diaconis is so obsessed by the enigma of the uncertainty, he claimed that coin tossing involves the law of mechanics( ignoring the effect of air molecules). To demonstrate this, he had had the physics department build him a coin-tossing machine. The coin starts out on a spring , the spring released, the coin spins upward and lands in a cup.(as shown in the figure). Because the force imparted are controlled he claims that the coin always lands with the same side up. Magicians and crooked gamblers (including Persi Diaconis himself) also possess such abilities. I further suggest the interested readers to read articles and watch videos of Diaconis on coin tossing and game of cards, you will be amazed. Those are the reason, I feel he is among the very few who worked extensively and mathematically over the philosophical questions about uncertainty.

The careful study of flipped coin started by Keller (in 1986). he assumed that a coin flips about an axis in its plane with spin about this axis at a rate $w$ revolutions per second. If the initial velocity in the up direction $\vec{K}$ is $v_t$ , after $t$ seconds a coin is flipped from a height $z_o$ will be at height $z_o+tv_t-\frac{g}{2}t^2$. If the coin drops at a surface the time elapsed $t*$ satisfies $t*v_t-\frac{gt*^2}{2}=0$ or $t*=\frac{2v_t}{g}$ (simple equations of motion). This coin will revolve $\frac{2wv_t}{g}$ times. If this is between $2j$ and $2j+1$ the initial side will turn up. If it is between $2j+1$ and $2j+2$ the opposite side turns up. Hyperbolas as defined by the various initial values of $w$.

The figure shows a space $(w,t)$ in the regions where the coin coes up as the same side or the opposite side. The edges of the hyperbola $\frac{2wv}{g}=j$. Visually the regions get closer together, implying that small changes in initial conditions make for the difference between heads and tails.

How then is the probabilistic treatment of coin flips so widespread and so successful ? the basic answer was given by Poincare. If a coin is flipped vigorously, with sufficient vertical and angular velocity, there is a sensitive dependence on initial conditions. Then a little uncertainty as to initial conditions is amplified to a large uncertainty about the outcome, where equiprobability of outcomes are not at all a bad assumption.

In 1992, Engel carried on the work based on Keller and fitted a probabilistic model on the velocity and spins per second and even he too ended up concluding that the chance of getting head is nearly half and inspite of accounting for the velocity and spins it seemed difficult to argue on the fact that almost with equal chances one can obtain head or tail.

Engel proposed a theorem that considered two probability distribution on velocity $v$ and spin per second $w$ and gave a bound on the chances of getting a head and found that the chance revolves around $\frac{1}{2}$.

So, even in a classical mechanical set up, the entire uncertainty can’t be wiped out completely.

### Does Fair Tosses only depends on Fair Coins ?

According to Persi Diaconis’s observation , coin tossing outcomes are heavily dependent on the way it is tossed up. The exact determination of the bias depends in a delicate way on the shape of the coin’s edges also, which are used by magicians to trick naked human eye. (We will see this fact as a consequence in a problem, which we will solve later). He even goes further to suggest that the coin tossing gives a lot more unbiased result when tossed say 100 times, but the same penny when spun showed reasonable bias towards tails. We provide the result in a histogram comparing the outcomes of coin tossing and coin spinning. Diaconis further extends his search for uncertainty in coin flipping explaining more about physical chances, which interested readers can make their way further to those.

Explaining and discussing the approaches and ideas that are associated in understanding the physics behind the coin flipping, lets shift ourselves to some beautiful problems which involves and explains the physical geometry and the classical mechanical fundamentals which influences the outcome of the toss.

## Wishful Coin Tossing

Suppose you are tossing a coin, you know nothing about its fairness, we assume that the thickness of the coin is negligible, the coin is tossed, what is the chance of getting a head ?

Now, here we really know next to nothing about the initial conditions, which we saw impacts the outcome heavily. So here we will deal with this problem geometrically, but don’t forget the fact that classical mechanics is basically originated from this kind of mathematics. Hence, lets assume that after the coin strikes the surface the vector of the normal $\vec{N}$ applied on the heads side of the coin generates a cone ( follow the figure). The axis of the cone makes an angle $\theta$ $(-\frac{\pi}{2} \le \theta \le \frac{\pi}{2})$ with the horizontal plane and $\alpha$ is the angle between the generatrix of the cone and its axis, $(0 \le \alpha \le \frac{\pi}{2})$. This figure demontrates the dynamics of the coin according to the stated problem, here the shaded sector, will be where the coin will flip over, and the darker region will be below the horizontal plane when, $|\theta| \le \alpha$.

When the coin hits the ground and starts spinning before settling finally, the normal vector spins randomly over the circumference of the base of the (imaginary) cone. Now imagine that coin is in any arbitrary position (as given in the figure).

Now observe that it is quite obvious from the figure that if $\theta < \alpha$, then the coin will surely not flip over .i.e. heads will come up.

Similarly if $\theta < – \alpha$ then the coin will surely flip and heads will never come up.

But when $|\theta| \le \alpha$, then we have to observe the rotation of the normal vector $\vec{N}$ more closely, and have to locate the regions on the circumference where if the normal vector stops will result in the flip over of the coin. Further observe that the coin will flip over whenever the normal vector will intersect(or penetrate ) the horizontal plane. It is hard to imagine, the figure and ones own imagine is what I can offer for clarification. Observe that part of the base of the cone that will be immersed within the horizontal plane will be part of the circumference that will flip over the coin.

So, observing the figure, if say $O’P=r’$ and radius of the base of the cone is $r$ and the height of the cone is $h$. Using further simple properties of the circle, we need to find the circumference of the shaded sector as in that part of the circle, the coin will flip over, following the argument we established above. Then,

$tan{\theta} = \frac{r’}{h}$ and $tan {\alpha}= \frac{r}{h}$

Let the angle of the sector be $\phi$ (angle at $O’$).

So, $Cos{\frac{\phi}{2}}=\frac{r’}{r}=\frac{h tan{\theta}}{h tan{\alpha}} \Rightarrow \phi=2cos^{-1}{(\frac{tan {\theta}}{tan {\alpha}})}$

So, $P( coin \ flips \ over)=\frac{r \phi}{2 \pi r}=\frac{1}{\pi}cos^{-1}{(\frac{tan {\theta}}{tan {\alpha}})}$

Hence, $P(Heads)= 1-\frac{1}{\pi}cos^{-1}{(\frac{tan {\theta}}{tan {\alpha}})}$.

So, we can also say that the toss will be fair when $\theta =0$ .i.e. $P(Head)=\frac{1}{2}$, which we can conclude also from the figure that if the axis of the cone becomes parallel to the horizontal plane there is a equal chance of falling on either side and expose any of the faces.

If you can toss a coin such that it first lands on its edges when it hits the ground (surface), then even with a biased coin you can perform a fair toss !!

So, as Diaconis claimed that edges play an important role in determining the outcome of a toss, this problem showed the same. Actually when you see magicians tossing a coin and claiming tat heads will land (as Diaconis does this himself) and truly ends up with a head, then you can be sure about the fact that he/she always using the “$\theta > \alpha$” case. Though in this problem I used lesser parameters ( as compared to standard mechanical models), still I claim that I ended with a quite satisfactory chance of getting, since the geometry was specific and subtle enough to cover the all the possibilities after the coin hits the surface.

Can you control the uncertainty on the edges of the coin like the magicians ? Try it !!

## Edgy Tossing

Suppose I ask you about the chance of a coin landing on one of its faces, you may think its how foolish of me to ask such a question, and smartly gave the reply that ” landing on of its faces is a sure event man !!”, and then I further argue with strong mathematical logic that we can always assign some chances on the event that the coin may sometimes land on its edges !! Edgy right !

Imagine a thick coin, (for psychological convenient), now we can actually assume the coin as a cylinder right ! a flatter one perhaps. Say the coin has radius $r$ and thickness $h$, eventhough the mass is accountable we ignore the mass for simplicity. Now you rolled the coin (cylinder), what is the probability you think that the cylinder (coin) lands on its lateral surface (edge) ?

Try this problem yourself !! Use the assumption that Keller used, that is the normal vector of the coin will rotate in a spherical space, and you have to use the idea of Solid angle of the geometric shapes you wish to use in analyzing the phenomenon. If you don’t know what is a solid angle, then give a breif read on the topic and come back to the problem !! Good luck !

I tried the problem with a standard coin of diameter ($2r$) 19.05 mm and thickness ($h$) is 1.52 mm . I ended up with the fact that there exists about 3% chance of the coin to land on its edges !! Hence landing on edges are not at all impossible, on the contrary it is edgy enough !!

## The Coin finally becomes Schordinger’s Cat !

So, till now we have been talking about how Classical Physics try to explain what are disguised as uncertainties, but at the end of every instances we failed to free ourselves from chances. Though we must admits that Laws of Mechanics explain some part of the uncertainty, leaving us to propose a “model of uncertainty” for better understanding of our readers, the model is as follows ,

$Uncertainty = “Lack \ of \ Classical \ Mechanical \ information ” + “Error’$.

Now before concluding, we will just discuss and try to argue from where this error coming and what is this error actually represents.

According to the latest researches, of two physicists Andreas Albrecht and Daniel Philips of University Of California, they argue that probabilities we use in our daily life and science do not “quantify our ignorance” but instead reflect the inherent random nature of the physical world as described by Quantum Mechanics.

They claim that the reason we fail to determine the coin toss outcome even after accounting for most of the classical parameters is because we cant anticipate the collisions of the air particles due to Brownian motion over the surface of the coin which in turn leaving its invisible impact on the instantaneous velocity and spins of the coin, Remember Heisenberg’s Uncertainty Principle !!

They infact claim that anyone who is tossing a coin is actually performing Schordinger’s Cat Experiment. But rather than a cat that is both alive and dead the quantum object in this case is a coin whose final state is here Heads or Tails (or Edge !!). hence outcome of the flip generally remains genuinely open until the upward face of the coin is looked at which the system takes a definite value of either Heads or a tails.

So, basically what uncertainty stands to be is that “Uncertainty is the manifestations of Quantum Chanciness .” Hence our intuitive model of uncertainty modifies as follows,

$Uncertainty = “Lack \ of \ Classical \ Mechanical \ information ” + “Quantum \ phenomenon”$.

This is also the instant where classical probabilities branches up to their respective directions. I hope our interested readers will find themselves hanging on any of these branches. Will you toss a coin to choose a branch ?? What do you think !!

## Food For Thought

If you have solved the problem left as an thought exercise in “Edgy Tossing ” , then take some rest and think again !!

Suppose the chance of the coin landing on its edge is $\frac{1}{3}$, How thick you think the coin should be !! Do you Have any idea !!

You can share them with us perhaps !!

Share your thoughts with us, and stay tuned as the next week is the “Geometric Probability week” , and we hope to make most of it !! Be ready to be perplexed !

## References

1. Dynamical Bias in the Coin Toss – Persi Diaconis , Susan holmes and Richard Montgomery
2. Ten Great ideas About Chances – Persi Diaconis & Brian Skyrms
3. The Quantum Coin Toss- Edwin Cartlidge, PhysicsWorld
4. Problems in The Theory of Probability – Sevastyanov, Chistyakov & Zubkov
5. Fifty Challenging Problems in Probability – Frederick Mosteller.
6. Special Thanks to my friends Avishek Dutta and Soham Ghosh ( also the co-writer of this blog), for some discussions which turned out to be productive while writing this article.

Categories

## ISI MStat PSB 2004 Problem 4 | Calculating probability using Uniform Distribution

This is a very beautiful sample problem from ISI MStat PSB 2004 Problem 4 based on finding the probability using Uniform distribution . Let’s give it a try !!

## Problem– ISI MStat PSB 2004 Problem 4

Two policemen are sent to watch a road that is $1 \mathrm{km}$ long. Each of the two policemen is assigned a position on the road which is chosen according to a uniform distribution along the length of the road and independent of the other’s position. Find the probability that the
policemen will be less than 1/4 kilometer apart when they reach their assigned posts.

### Prerequisites

Uniform Distribution

Basic geometry

## Solution :

Let X be the position of a policeman and Y be the position of another policeman on the road of 1km length .

As it is given that chosen according to a uniform distribution along the length of the road and independent of the other’s position hence we can say that $X \sim U(0,1)$ and $Y \sim U(0,1)$ and X,Y are independent .

Now we have to find the probability that the policemen will be less than 1/4 kilometer apart when they reach their assigned posts , which is

nothing but $P(|X-Y|< \frac{1}{4} )$ .

So , let’s calculate the probability $P(|X-Y|< \frac{1}{4} )$ here some sort of geometry will help to calculate it easily !

In general we have 0<X<1 and 0<Y<1 and hence the total probability is the area of the square $1 \times 1$

And in favourable case we have $|X-Y|<1/4 , 0<X<1 , 0<Y<1$ . so, it’s basically the area covered by ACBDEF = Area covered by square – area of the triangles BGD and AFH = $1 \times 1$ – $2 \times$ $\frac{1}{2}$ $\times \frac{3}{4} (1- \frac{1}{4} )$ = $1-9/16$ .

Therefore $P(|X-Y|<1/4)= \frac{1-9/16}{1} = \frac{7}{16}$

## Food For Thought

Calculate the same under the condition that road is of length (b-a) , b>a and both are positive real number .

Categories

## Median of numbers | AMC-10A, 2020 | Problem 11

Try this beautiful problem from Geometry based on Median of numbers from AMC 10A, 2020.

## Median of numbers – AMC-10A, 2020- Problem 11

What is the median of the following list of $4040$ numbers$?$

$1,2,3,…….2020,1^2,2^2,3^2………..{2020}^2$

• $1989.5$
• $1976.5$
• $1972.5$

### Key Concepts

Median

Algebra

square numbers

Answer: $1976.5$

AMC-10A (2020) Problem 11

Pre College Mathematics

## Try with Hints

To find the median we need to know how many terms are there and the position of the numbers .here two types of numbers, first nonsquare i.e (1,2,3……2020) and squares numbers i.e $(1^2,2^2,3^2……2020^2)$.so We want to know the $2020$th term and the $2021$st term to get the median.

Can you now finish the problem ……….

Now less than 2020 the square number is ${44}^2$=1936 and if we take ${45}^2$=2025 which is greater than 2020.therefore we take the term that $1,2,3…2020$ trms + 44 terms=$2064$ terms.

can you finish the problem……..

since $44^{2}$ is $44+45=89$ less than $45^{2}=2025$ and 84 less than 2020 we will only need to consider the perfect square terms going down from the 2064 th term, 2020, after going down $84$ terms. Since the $2020$th and $2021$st terms are only $44$ and $43$ terms away from the $2064$th term, we can simply subtract $44$ from $2020$ and $43$ from $2020$ to get the two terms, which are $1976$ and $1977$. Averaging the two,=$1976.5$

Categories

## Circle | Geometry Problem | PRMO-2017 | Question 27

Try this beautiful Problem from Geometry based on Circle from PRMO 2017.

## Circle – PRMO 2017, Problem 27

Let $\Omega_{1}$ be a circle with centre 0 and let $A B$ be a diameter of $\Omega_{1} .$ Le $P$ be a point on the segment $O B$ different from 0. Suppose another circle $\Omega_{2}$ with centre P lies in the interior of $\Omega_{1}$. Tangents are drawn from A and B to the circle $\Omega_{2}$ intersecting $\Omega_{1}$ again at $A_{1}$ and $B_{1}$ respectively such that $A_{1}$, and $B_{1}$ are on the opposite sides of $A B$. Given that $A_{1} B=5, A B_{1}=15$ and $O P=10,$ find the radius of $\Omega_{1}$

• $9$
• $40$
• $34$
• $20$

### Key Concepts

Geometry

Circle

Answer:$20$

PRMO-2017, Problem 27

Pre College Mathematics

## Try with Hints

Let radius of $\Omega_{1}$ be $R$ and that of $\Omega_{2}$ be $r$
From figure, $\Delta \mathrm{ADP} \sim \Delta \mathrm{AA}_{1} \mathrm{B}$
[
\begin{array}{l}
\Rightarrow \frac{D P}{A, B}=\frac{A P}{A B} \
\Rightarrow \frac{r}{5}=\frac{R+10}{2 R}
\end{array}
]

Can you now finish the problem ……….

Again, $\Delta B P E \sim \Delta B A B_{1}$
Therefore $\frac{P E}{A B_{1}}=\frac{B P}{B A}$
$\Rightarrow \frac{r}{15}=\frac{R-10}{2 R}$

Can you finish the problem……..

Dividing (1) by (2)

$3=\frac{R+10}{R-10} \Rightarrow R=20$

Categories

## Side of Square | AMC 10A, 2013 | Problem 3

Try this beautiful problem from Geometry: Side of Square.

## Sides of Square – AMC-10A, 2013- Problem 3

Square $ABCD$ has side length $10$. Point $E$ is on $\overline{BC}$, and the area of $\triangle ABE$ is $40$. What is $BE$?

,

i

• $4$
• $5$
• $6$
• $7$
• $8$

### Key Concepts

Geometry

Square

Triangle

Answer: $8$

AMC-10A (2013) Problem 3

Pre College Mathematics

## Try with Hints

Given that Square $ABCD$ has side length $10$ and area of $\triangle ABE$ is $40$.we have to find out length of $BE$ where $E$ is the point on $BC$. we know area of the $\triangle ABE=\frac{1}{2} AB.BE=40$

Can you find out the side length of $BE$?

Can you now finish the problem ……….

$\triangle ABE=\frac{1}{2} AB.BE=40$

$\Rightarrow \triangle ABE=\frac{1}{2} 10.BE=40$

$\Rightarrow \triangle ABE=\frac{1}{2} 10.BE=40$

$\Rightarrow BE=8$

Categories

## Triangle Area Problem | AMC-10A, 2009 | Problem 10

Try this beautiful problem from Geometry based on Area of Triangle.

## Area of Triangle – AMC-10A, 2009- Problem 10

Triangle $A B C$ has a right angle at $B$. Point $D$ is the foot of the altitude from $B, A D=3$, and $D C=4 .$ What is the area of $\triangle A B C$ ?

• $4 \sqrt{3}$
• $7 \sqrt{3}$
• $14 \sqrt{3}$
• $21$
• $42$

### Key Concepts

Triangle

Similarity

Geometry

Answer: $7 \sqrt{3}$

AMC-10A (2009) Problem 10

Pre College Mathematics

## Try with Hints

We have to find out the area of the Triangle ABC where $\angle B=90^{\circ}$ and $BD \perp AC$

Area of a Triangle = $\frac{1}{2}\times$ Base $\times$ Height.But we don know the value of $AB$ & $BC$. But we know $AC=7$. So if we can find out the value of $BD$ then we can find out the are of $\triangle ABC$ by $\frac{1}{2}\times AC \times BD$

Can you now finish the problem ……….

Let $\angle C=\theta$, then $\angle A=(90-\theta)$ (as $\angle B=90^{\circ}$, Sum of the angles in a triangle is $180^{\circ}$)

In $\triangle ABD$, $\angle ABD=\theta$ $\Rightarrow \angle A=(90-\theta$)

Again In $\triangle DBC$, $\angle DBC$=($90-\theta$) $\Rightarrow \angle C=\theta$

From the above condition we say that , $\triangle ABD \sim \triangle BDC$

Therefore , $\frac{BD}{CD}=\frac{AD}{BD}$ $\Rightarrow {BD}^2=AD.CD=4\times 3$

$\Rightarrow BD=\sqrt {12}$

can you finish the problem……..

Therefore area of the $\triangle ABC=\frac{1}{2}\times AC \times BD=\frac{1}{2}\times 7 \times \sqrt{12}=7 \sqrt{3}$