We wanted to announce that over the past weeks the publication capabilities of Brilliant.org have been transferred from this wordpress blog to our own main site. Desiring to integrate our blog posts with the problem solving activity done on our weekly challenges, we have transferred our entire blog archives over to our main site. Eventually each blog post will evolve to have sets of practice problems to illustrate the concepts covered in each post. To check it out click here:

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When students are first exposed to trigonometric identities, they are often given a list of formulas, which they are asked to memorize. Here is a way for you to remember many of these ideas:

Start by drawing a regular hexagon and connect each of the vertices to the center. In the left most vertex, label it . In the bottom left vertex, label it , in the bottom right vertex, label it . In the center, label it 1. Now, to figure out what to label the remaining vertices, simply look at the diagonally opposite vertex and label it as the reciprocal. You should get the following diagram:

[Note: , which is sometimes denoted as cosec . Either version is valid, and we will be using in this post.]

Now, let’s figure out some properties of this hexagon.

Property 1: How do the labels on the endpoints of a diameter relate? Recall that to establish the labels, we labelled the diagonally opposite vertex as the reciprocal. Hence, the product of the labels on a diameter is 1, which corresponds to the center vertex.

Property 2: How do we relate vertices that are connected by edges? Consider the central vertex 1. Moving directly to the left is equivalent to multiplying by , moving to the lower right is equivalent to multiplying by , moving to the lower left is equivalent to multiplying by . This seems obvious when we’re at vertex 1, and in fact holds true for any other vertex. For example, moving left from brings us to , which corresponds to . There are similar statements for moving to the right, upper left and upper right (see Test Yourself 1).

Property 3: What property do we get when reading off the 3 vertices in an anti-clockwise manner? With our 3 starting vertices, we know that . This behavior holds true for the rest, that the first vertex is equal to the second divided by the third. For example, another set of 3 vertices is , and you should be able to verify that .

Property 4: Recall the Pythagorean identity which states that . How is this expressed in the hexagon? If we look at the bottom upright triangle, we see it it has vertices of at the base, and at the top. In fact, given any other upright triangle, a similar relation holds. For example, with the right upright triangle, we get , and with the left upright triangle, we get .

Worked Examples

1. What is ?

Solution: By property 3, we know that .

2. Why does property 4 hold?

Solution: We already know that . Consider what happens when we shift this triangle to the upper right. We are simply multiplying each vertex by . Hence, the equivalent identity is that , which becomes .

3. What other properties are there?

Solution: This is an open ended question. There are as many properties as you can find for yourself. I’d state one more, which is related to property 3 (and actually is simply a different way of expressing the same idea).

Property 5: If we read a set of 3 vertices off in a clockwise-order, we get that the first vertex is equal to the second over the third. For example, are 3 consecutive vertices in clockwise order, so this property gives us that .

The principles of parity can be used to understand the potential chess moves of a Knight.

[This post is targeted at a Level 4 student. You should be familiar with Parity.]

We will further explore the idea of parity, and its various uses. Apart from simply determining if a number is odd or even, parity can be used as a simple counting argument in many cases. In Parity Test Yourself 1, we gave a question where Mary had an even number of sheep at the start, and an odd number of sheep at the end of the day. This implies that she had a different amount of sheep, and hence didn’t do her job properly.

[This is targeted at a level 3 student. This is a continuation of the Probability blog post.]

In our first blog about probability, we focused on sets of events that are mutually exclusive. In this blog, we will expand to look at events which are not mutually exclusive, and some techniques we can use to solve various types of probability questions.

One of our best tools to solve questions dealing with events that are not mutually exclusive is the Principle of Inclusion and Exclusion. We can formulate a probabilistic version of this principle.

And in general, techniques that we use when we solve counting problems are also what we often use when we solve probability problems. If we can find nice descriptions of the set we are interested in, and of the whole set of possibilities, we can use counting techniques to count both. We have seen examples of how the Rule of sum, Rule of Product have been used to calculate probabilities.

When solving problems about coin flips, the number of ways to get a certain number of heads or tails is always a binomial coefficient. We can often use properties of binomial coefficients to help solve problems of this type.

As in counting problems, it is sometimes easier to determine the probability of the complement of what you are looking for. If you roll two dice and want to know what the probability of rolling a sum that is larger than 3, you could add the probabilities of getting together, or you could add the probabilities of getting (the complement) and subtract this value from 1.

The trigonometric functions are functions of an angle, the most prominent of which are sine, cosine and tangent. These are best understood by considering the line segments from a unit circle.

Given an angle , construct the half ray, , that is an anti-clockwise rotation of from the positive x-axis. The intersection of this ray with the unit circle, labelled as is given by , which gives the definition of . We further introduce that .

From this point , let’s drop a perpendicular to the x-axis, intersecting at . Then, is a right angled triangle with . Hence, if we are given any triangle where and , then by similarity with triangle , we get that

If you need a mnemonic to remember these formulas, most english speakers use “Soh Cah Toa”. [Personally, I remembered it as "Toa Cah Soh", which means "Pig trotters women" in Cantonese.] In either case, the first letter stands for the trigonometric function, the second letter stands for the numerator, and the third letter stands for the deonominator. We refer to as the Hypotenuse, as the Opposite side and as the adjacent side, and the mnemonic states that

There are certain values of these functions which are useful to remember. They are:

The reason for writing them in this way, is to aid remembering these terms. For example, the numerator for is simply the square root of 0, 1, 2, 3, 4.

A version of the Mandelbrot set colored in relation to the distance of each iterate from the nearest gaussian integer (Image courtesy of wikimedia creative commons).

[This post is targeted at a Level 5 student. This is a continuation of the blog post Gaussian integers II.]

You are probably used to the fact that every positive integer can be uniquely expressed as a product of positive primes, up to the order of multiplication. This property is called the Fundamental Theorem of Arithmetic. While this seems like an obvious fact, can we create a system where it is not true?

Example. Consider the set of all positive integers of the form , where is a positive integer. A product of any two such integers in is again an integer of this form. We will call an integer from prime if it is not a product of two smaller integers from , and the integer is called composite otherwise. By definition, every integer from is a product of primes from . However, the product may not be uniquely expressed. For example, we can see that are all primes in , and we have .

Of course, is very different from integers or natural numbers, but this showcases the fact that there must be a reason for why the prime decomposition is unique. The deep reason is the existence of the division algorithm that produces a remainder that is strictly smaller than the divisor. Interestingly, the same property holds for the Gaussian integers, with respect to the norm:

Lemma 1. (Division algorithm) Suppose and are Gaussian integers, . Then there exist Gaussian integers and such that and

Proof. Dividing by , we get a number where and are rational (not necessarily integers). Suppose is the closest integer to and is the closest integer to . Denote . Then

So if then

This property is called the Euclidean domain property. In fact, in every Euclidean domain the factorization into a product of primes is unique. We present an argument for the Gaussian integers, which can be easily adapted for the regular integers. The following lemma is needed in the proof, and is known as Euclid’s Lemma when restricted to integers.

Lemma 2. Suppose is a prime Gaussian integer and divides for some Gaussian integers and . Then or .

Proof. Consider the set of all Gaussian integers of the form for arbitrary Gaussian integers and . This is also known as the ideal generated by and . It has the property that any sum or difference of two elements from is in , and any Gaussian integral multiple of an element from is in . Suppose is the element of with the smallest possible non-zero absolute value. From Lemma 1, every element of is a multiple of , otherwise the non-zero remainder, which also lies in , will have a smaller absolute value. In particular, . Because is prime, either is a unit or is a unit. In the first case, is a multiple of , so it is a multiple of . In the second case, multiplying by , we express as . Multiplying by , we get . Because this implies that .

Proof. First we prove that a factorization into a product of primes always exists. If not, take to be the Gaussian integer with smallest absolute value which is not a product of primes. Then is not a prime, so a product of two non-zero, non-unit Gaussian integers. Since norms are multiplicative and positive integers, must have smaller norm than , so they are both products of primes. If so, is also the product of primes, hence a contradiction.

Now we will prove the uniqueness. Suppose that is the Gaussian integer with smallest norm that admits two substantially different decompositions into a product of primes:

Applying lemma 2 to , , we get that either or . In the second case we apply it again, and again… as a result we get that must divide for some . Because the number is prime, this means that is up to a unit . Then we can cancel them out, and get a new with two different prime decompositions and smaller norm. This contradicts our choice of .

Another property of the Euclidean domain, very much related to the Unique Decomposition, is the existence of the Euclidean Algorithm for finding the greatest common divisor. This algorithm, as well as the uniqueness of prime decomposition for the usual integers goes all the way back to Euclid’s Elements!

The Euclidean Algorithm works as follows: Suppose and are (Gaussian) integers. We order the pair so that and call it . Then we use it to get new pairs, as follows. At every step, we divide by . If the remainder is zero, we stop, and claim that the greatest common divisor of and , denoted . If there is a non-zero remainder, i.e. we take and continue.

Theorem 2.
1) For every initial pair , the Euclidean algorithm terminates after a finite number of steps.
2) The number divides both and , and every (Gaussian) integer that divides both and must divide .

Proof. 1) For the Gaussian integers, note that , and the norm, being a positive integer, cannot decrease indefinitely.

2) Suppose . Then it clearly divides and . Because and it divides and . Continuing in this manner, we prove that divides and .

If divides both and , then it divides and . Continuing in this manner, we prove that divides .

We finish this post by reproducing one of the recent proofs of the Fermat Two Squares Theorem, due to Don Zagier. An involution is such map from a set to itself, that for all elements .

[This post is targeted at a Level 2 student. You should be familiar with the Rule of sum, Rule or product.]

Probability is a measure of how likely it is that a certain event will occur. Probability is a concept that turns up frequently in the real world. It has applications in finance, meteorology, marketing, and many other areas.

The probability of an event occurring is always a number between 0 and 1 inclusive. If the event will not happen the probability is 0, if the event will happen the probability is 1, and the higher the probability is, the more likely it is that the event will occur. Whenever we have a set of events and exactly one of them will occur, we say that these events are mutually exclusive, and the sum of the probabilities for these events is 1.

We will start by looking at some simple examples of probability, where our events are equally likely to occur, and exactly one of them will occur at a time. This could happen for example if we roll a single die, flip a single coin, or choose a ball at random from a bag. Suppose we have a set of equally likely, mutually exclusive, events, and we know that one of them will occur. Let be the probability that a particular event occurs. Then since the probability of one of the events occurring is 1, we have , so . If and we want to know what the probability that an event in occurs, we use the formula

Probability of equally likely outcomes:

Let’s consider an example that illustrates why “equally likely” is important. When a single 6-sided die is rolled, there are 6 possible results – the roll could be any integer from 1 to 6 inclusive. Each different number is equally likely to occur, so the probability of each number being rolled is . If we instead roll two six-sided dice, the smallest sum we can get is 2 when we roll a 1 on each die, and the largest sum we can get is 12 when we roll a 6 on each die. It is easy to see how we could get any integer between 2 and 12, so the possible outcomes for the sum of the numbers are integers from 2 to 12. If these events were each equally likely, they would each have a probability of , since there are 11 possible sums. However, this is not the case.

If you’ve ever played a game that involved rolling dice, you’ll know that some totals occur more often than others. So instead of looking at the sum of the dice, if we look at the ordered pairs of numbers that occur, then by the rule of product, there are possible outcomes. Of these outcomes, one will give a sum of 2: , and six will give a sum of 7: . So the probabilities of getting the different sums from 2 to 12 are not equal. However, in this case, each of the 36 possible ordered pairs is equally likely. So the probability of getting a sum of 2 is , and the probability of getting a sum of 7 is . The important distinction is that, while each of the events of ordered pairs of dice rolls are equally likely, the events of sum of dice are NOT equally likely.

When the probabilities of the events are not equal, we can still calculate the probability that a subset of the events occur. If we wanted to know the probability of getting a sum of , , or on a roll of two dice, we can just add the respective probabilities. So the total probability would be . This is the rule of sum applied to the numerator of the probability formula.

The above example also illustrates the concept of independence of events, which will be elaborated in a future post. Two events are independent if the outcome of one event does not affect the outcome of the other event. For example, rolling two dice, flipping a coin twice, choosing balls from different bags, and rolling a single die and then flipping a coin are all independent because the result on one of the actions is not affected by the result of the other action.

Worked Example

1. In the board game monopoly, you may end up “in jail”.  To get out of jail, you roll two six-sided dice, and need to get the same number on both of them. What is the probability that on a particular roll you will get out of jail?

Solution: We use the formula to calculate the probability. We worked out above that there are 36 possible ordered pairs that can occur when we roll two six-sided dice. Six of those: have the same number showing on both dice, so the probability is .

2. A bag contains 4 red balls, 5 blue balls, and 3 black balls.  If three balls are drawn at random from the bag, what is the probability they are all different colours?

Solution: There are balls in the bag, so there are ways to choose three balls from the bag. If we want one ball of each colour, by the rule of product there are ways we can choose them. So the probability is .

3. 6 coins are flipped. Which result is more likely to occur, 3 heads and 3 tails, or 2 of one and 4 of the other?

Solution: When we flip 6 coins, there are different possible outcomes. There are ways to get 3 heads and 3 tails. There are ways to get 4 heads and 2 tails, and the same number of ways to get 2 heads and 4 tails. So the probability of getting 3 of each is and the probability of getting 2 of one and 4 of the other is . So it is more likely that we will get 2 of one and 4 of the other than it is that we will get 3 of each.

On our discussions page, folks have been wondering how many others are using Brilliant? Currently, there are 10,000 monthly active users, and 40,000 registered users on Brilliant.org. We are steadily growing, and will keep you updated in the coming weeks and months as the community evolves. From the graph above, you will all notice an impressive international distribution of users. In the coming weeks, we will be exploring ways to permit all of you to better meet and benefit from the diverse collection of avid thinkers on our site. If you have friends, family, teachers, or meet a stranger who you think would enjoy Brilliant, spread the word. The more the merrier!

Two new features have appeared in the last week. Last Sunday(Monday for many of you), we unveiled a “Curriculum Math” category of problem sets(see picture below). Our original sections of Algebra and Number Theory, and Geometry and Combinatorics, have always aimed to pose problems that are more challenging and fascinating than what most people are offered in school. Our user feedback over the past couple of months has told us that many people find these problem sets uncannily addictive, and that their appetite for our Olympiad problems will never be sated. Many of these same people expressed a desire for additional problem sets that would more directly help them practice for school, and potentially serve as a bridge to improvement on our “Olympiad Math” problem sets.

For these reasons, we opened up an Algebra and Trigonometry section to provide more practice on the basics for those that desire it. Thinking people might enjoy a taste of calculus with their weekly problem sets, we also opened up a calculus category. We are aware that many of you are in calculus courses right now, and possibly preparing for high stakes calculus intensive exams. We wish all of you the best of luck and hope our problem sets can be both a helpful study tool and a fun diversion from your coursework.

The curriculum math section appears right below the traditional categories Algebra/Number theory and Geometry/Combinatorics on the “Challenges” page.

At the end of this last week, we unveiled a Key Techniques Trainer. Over the past several months, this blog has published a regular series of Key Techniques posts that highlight essential concepts and methods of mathematics. These posts introduce and illustrate a topic to show  how the concepts can be applied to a wide range of thought and problem solving.

The Key Technique Trainer button is currently visible to those who are levels 1-3 in Algebra/Number Theory.

Many of these posts pertain directly or indirectly to the problems on our site. If you get two wrong attempts on a question in your problem set that pertains to a Key Technique we have written about, you will be offered a link to the relevant Key Technique post. Reading through this post might offer you insights on how to solve the problem before making your last attempt. These Key Technique posts in the Trainer, now have “Test Yourself” questions that can accept numerical answers and tell you if you are right or wrong. These questions are not worth any points, but accept unlimited attempts. They allow you to verify whether you understand the concept, and to keep trying until you figure it out.

 You can also go directly to the Key Technique Trainer main page and browse all the Key Techniques available, to see which ones you can apply in your current problem set. The Trainer tracks how many problems you have solved that use each technique, so you can track your progress and gauge your mastery of a topic. If people enjoy the Key Technique Trainer, we will migrate more of the Key Technique posts from the blog into the Trainer, and expand it into other problem sections and more levels..

Both Curriculum Math and the Key Techniques Trainer are works in progress, and will evolve in response to your feedback and use patterns. Please give feedback on the Key Technique Trainer here, and give feedback on the Curriculum Math sections here. It is our hope that they are both steps toward making Brilliant the most stimulating place on the web to explore your passions for math and physics.

Happy problem solving and have a great week!

Graph of the inequality in worked example number 1

[This is targeted at a Level 2 student. Unless otherwise mentioned, all quantities are real numbers.]

In an algebra course, students are introduced to the concept of order on the real numbers. This is a relation between the real numbers, denoted by the symbol , which is often read as “greater than”. We have the following 4 order axioms for real numbers:

1. [Trichotomy] If are real numbers, then one and only one of the following is true: , , .
2. [Transitivity] If are real numbers and , then .
3. If are real numbers and , then .
4. If are real numbers such that , then .

With this, we can formally define a positive and negative number. A positive number is a number that satisfies , while a negative number is a number that satisfies . By the trichotomy axiom, a real number is either positive, negative, or zero. However, this does not appear to tell us much. In particular, is 1 a positive number?

An important inequality is the trivial inequality which states that the square of any real number is non-negative. This is easily shown by considering Trichotomy cases:
1. If , then .
2. If , then multiplying both sides by gives , where we keep the sign since is positive.
3. If , then (axiom 3) and multiplying both sides by gives (Axiom 4), and adding to both sides yields (Axiom 3).
With the trivial inequality, we now know that is a positive number.

Polynomial inequalities can be solved by understanding how the graph behaves. Because the graph is continuous (naively this means that the graph can be drawn without lifting off your pen), we only need to find the roots of the polynomial and test one value in each of the corresponding regions to determine if they are positive or negative.

Worked Examples

1. Determine the region in which .

Solution: The proper approach would be to shift the terms to one side, and then factorize, to get . This has roots and , so we test the 3 different regions ( i.e. , to conclude that the inequality is satisfied for .

Note: Similar to the equality case , the first instinct of many students is to simply take square roots on the inequality and conclude that . This likewise leads to the wrong answer.

2. Determine the region in which .

Solution: To avoid dividing by 0, we want to ensure that the denominator is never 0. As such, we have to exclude as a possibility. From Axiom 4, we can only multiply by a quantity with a fixed sign. Since is negative when , we cannot simply multiply by . Instead, we can multiply by , which we know is always non-negative, to obtain

At this point, it is tempting to cancel the term on both sides, but this is equivalent to multiplying by . We may not do so, we choose to factorize instead. Shifting all the terms to one side and factorizing, we get

We easily understand the graph of this quadratic equation, and (by using the number line) can conclude that the inequality is satisfied when (Recall that is excluded) or .

3. Show that for all real values of .

Solution: If we tried to calculate the roots to the quadratic equation , since the discriminant is negative, we know that the roots are complex numbers. Hence there is only 1 region on the number line. Testing shows that the quantity is always positive.

Another approach is to rewrite by Completing the square. Since both terms are squares, they are non-negative, and hence their sum is non-negative (axiom 3).

4. Show that the product of 2 positive numbers is positive.

Solution: If and are positive numbers, then . Using axiom 4, (with ), we get that . Hence, the product is positive.

Though named after Blaise Pascal(1623-1662), the triangular array of binomial coefficients had been studied for centuries before him by Indian, Arabic, Chinese, German, and Italian mathematicians. The above drawing is a 14th century Chinese depiction of Pascal’s Triangle attributed to the mathematician Yang Hui in 1303.

[This is targeted at a Level 2 student.]

In the blog post about Combinations, we introduced the concept of binomial coefficients. We showed that the binomial coefficient counts the number of ways to choose objects from a set of size . Binomial coefficients also arise naturally when raising binomial expressions to powers, as seen in the Binomial Theorem:

Binomial Theorem
If are variables and is a positive integer, then .

The two different ideas mentioned above are actually very closely related to each other. When we expand , each term that we get comes from choosing either or from each of the binomials in the product. This is why the terms in our expression are of the form . To get the term , we have to multiply copies of and copies of . If we choose of the binomials to use the term from, then we are left with of the binomials to use the term from. The total number of ways to choose of the binomials is , which is why this is the coefficient in the term .

Note: The Binomial Theorem is still true when is not a positive integer. In a later post, we will learn how to extend the definition of , and apply the Binomial Theorem accordingly.

When working with binomial coefficients, we can arrange them nicely into a chart called Pascal’s Triangle that uses the relation . The th row of Pascal’s triangle has the binomial coefficients . The first row is considered row 0 and contains the coefficient .

If we expand these out into their values, we get Pascal’s Triangle:

When working with binomial coefficients, things can generally be proved two different ways, either algebraically (by manipulating the expressions), or combinatorially (by interpreting the expressions). In the following worked example, we will solve the same problem both ways.

Worked Example

1. Show algebraically that .

Solution: We can expand the expression on the left hand side in terms of factorials.

2. Show combinatorially that .

Solution: The right hand side counts the number of ways to choose people from people. If we consider one of the people, then either they are chosen in the or they are not. The number of ways to choose people which includes this person is (we need to choose of the other people). The number of ways to choose people which do not include this person is (we need to choose of the other people).

3. Show that .

Algebraic solution: Recall from the Binomial Theorem that we have the relationship . If we let , this expression becomes .

Combinatorics solution: We use the technique of Double Counting. Consider the number of ways to choose a team out of players. From Combinations, we know that there are ways to choose a team of exactly players, hence the LHS represents the number of ways to choose a team according to the number of players. On the other hand, we can choose a team by deciding if a specified player is in the team or not. By the Rule of Product, there are ways to do so, which is the RHS. Hence, we have LHS = RHS.