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Equilateral and Equiangular Polygons

A polygon is a 2 dimensional geometric figure bound with straight sides. A polygon is called Equilateral if all of its sides are congruent. Common examples of equilateral polygons are a rhombus and regular polygons such as equilateral triangles and squares.  Now, a polygon is equiangular if all of its internal angles are congruent.  Some important facts to consider The only equiangular triangle is the equilateral triangle If P is an equilateral polygon that has more than three sides, it does not have to be equiangular. A rhombus with no right angle is an example of an equilateral but non-equiangular polygon.  Rectangles, including squares, are the only equiangular quadrilaterals Equiangular polygon theorem. Each angle of an equiangular n-gon is  $$\Bigg(\frac{n-2}{n}\Bigg)180^{\circ} = 180^{\circ} -   \frac{360^{\circ}}{n} $$ Viviani's theorem   Vincenzo Viviani (1622 – 1703) was a famous Italian mathematician. With his exceptional intelligence in math...
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Angles

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An angle is formed when two rays share an origin. There are many different types of angles formed when two lines/rays/line-segments intersect. The point of origin/intersection is called the  vertex , and the two rays are called the  sides . When we name an angle, we always put the vertex in the middle. For example, in the image on the right, $\angle ABC$ is an angle with vertex at point $B$. Angles: All Different Kinds A $90^{\circ}$ angle is a Right Angle . Lines, segments, or rays that form a right angle are said to be Perpendicular . An angle smaller than $90^{\circ}$ is an Acute Angle . An angle between $90^{\circ}$ and $180^{\circ}$ is an Obtuse Angle . An angle that measures $180^{\circ}$ is a Straight Angle . An angle of more than $180^{\circ}$ is a Reflex Angle . Two angles that add up to $180^{\circ}$ are known as Supplementary Angles . Two angles that add up to $90^{\circ}$ are known as Complementary Angles . When two lines intersect, they ...
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Sides and Angles of a Triangle

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Lets look at some basic facts related to sides and angles in triangles. For any triangle, the sum of any two sides will always be larger than the third side. We could also say that the difference of lengths of any two sides must be less than the length of the third side. (Triangle Inequality Theorem) The sum of the three interior angles of a triangle is $180^{\circ}$. The sum of all exterior angles of an n-sided polygon is $360^{\circ}$. The two base angles of an Isosceles triangle are congruent. The sum of all interior angles of an n-sided polygon is $(n − 2) \cdot 180^{\circ}$ . An exterior angle of a triangle is equal to the sum of the two opposite interior angles. (Exterior Angle Theorem) For a triangle, the opposite side of a bigger interior angle is longer than that of a smaller angle, and vice versa. In a right triangle, the sum of the squares of the two legs equals the square of the hypotenuse. (Pythagorean Theorem) Here are some problems that you would solve using...
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Number of Trailing 0's

Today, we will be looking at problems that ask you to find the number of trailing $0$'s in any factorial. Say for example, we are asked to find the number of trailing zeroes of $101!$ Reminder: $n! = n \cdot (n-1) \cdot (n-2) \cdot ... \cdot 3 \cdot 2 \cdot 1$ Simply put, trailing zeroes are zeroes at the end of the number without any non zero digits to the right of them. Ex. - there are $4$ trailing zeroes in the number $20340000$. Trailing $0$'s are formed when a multiple of $2$ is multiplied by a multiple of $5$. So, all we have to do is count the number of $5$'s and $2$'s in $101!$. Let's start with counting the number of $5$'s. The numbers $5, 10, 15, 20, 25, ... 95, 100$ all contribute one factor of $5$ to the factorial, so we have $20$ factors of $5$. However, some numbers have more than one multiple of $5$. For example, the numbers $25 = 5 \cdot 5$ $50 = 2 \cdot 5 \cdot 5$  $75 = 3 \cdot 5 \cdot 5$ $100 = 4 \cdot 5 \cdot 5$   all have an ex...
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Train Problems

The key to solving moving train problems is to understand the distance that the needs to be covered or the relative speeds at which the objects are commuting in the problems. Lets take a look at both of these cases and talk about how to solve these types of questions.  If the train is passing a Stationary object, think Distance  - If the problem asks for the time taken by a moving train to pass a pole or standing man or anything similar that basically is a point, you will need to find the time taken by the train to cover the length of the train itself.   - If the problem asks for the time taken by a moving train to pass a bridge or a tunnel or anything that has a length of itself, then the time taken by the train to pass that object will be the length of the train $+$ the length of the object. If the train is passing a Moving object, think Speed  - If the problem involves $2$ moving objects, you would usually want to calculate the net speed betwee...
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Geometric Sequences

Let's take a look at geometric sequences today. In a way, they are similar to Arithmetic Sequences , but just think common ratio instead of common difference. Let's begin by taking a look at the geometric sequence $2, 6, 18, 54...$. The initial term , or the first term is $2$. The common ratio  is defined as the number you multiply to each term to get the next one. In this case, the common ratio would be $3$, because $2 \cdot 3 = 6, 6 \cdot 3 = 18, 18 \cdot 3 = 54$, and so on. Common problems include finding the $n$th term, or finding the sum of such sequences. To find the $n$th term of a geometric sequence, we use the formula: $$a_n = ar^{n-1}$$ Where $a_n =$ the $n$th term in the sequence $a =$ the initial term $r =$ the common ratio Your Turn! $1.$  Find the $7$th term in the sequence $5, 10, 20, 40...$ (Scroll down to the bottom of the page for the answer) Next, let's take a look at how to find the sum of such a sequence. Let's find the sum of the seque...
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Age Problems

Age problems is another commonly tested concept on competitive exams. Solving many practice problems is the key to mastering this topic. Interpreting the language of what is given and what is asked of is the biggest challenge you would need to overcome. Lets look at some commonly used phrases throughout these problems.  If the present age is $a$, then $n$ times the present age = $an$. If the present age is $a$, then $n$ years later, age = $a+n$. If the present age is $a$, then age $n$ years ago = $a-n$. The next step would be to organize the given information. What usually works for me is putting a chart together and to keep plugging pieces of data that are given. Let's walk through a few examples to see how this works.  Example 1:  Tim's age is three times his son, Alex's age. After $10$ years, he would only be twice Alex's age. What is Alex's present age?  Solution: Current Age Age $10$ years later Tim $3x$ $3x+10, 2(x+10)$ Alex $x$ $x+10$ ...
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Distance and Work Problems: Basics

Distance and Work problems are another type of problems that you can almost be sure will appear on all math competitions at every level. It is imperative for young olympians to understand them well and get comfortable as early as you can. This post will touch upon the very basics of Speed/Distance/Time and Work problems. There will be future posts where I will discuss different variations and complexity levels within this category. Distance Problems The key in solving Speed/Distance/Time problems is to understand the relationship between each of these concepts. The basic formula to understand here is Distance = Rate $\cdot$ Time $D = R \cdot T$ It is highly recommended to make use of various tools like diagrams and charts to organize the given information leading up to the solution. The formula can be applied to different flavors of such problems. For example, if you know the time and rate a person is traveling on a bus, you can quickly calculate how far he traveled. And if...
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Equilateral and Equiangular Polygons

A polygon is a 2 dimensional geometric figure bound with straight sides. A polygon is called Equilateral if all of its sides are congruent. Common examples of equilateral polygons are a rhombus and regular polygons such as equilateral triangles and squares.  Now, a polygon is equiangular if all of its internal angles are congruent.  Some important facts to consider The only equiangular triangle is the equilateral triangle If P is an equilateral polygon that has more than three sides, it does not have to be equiangular. A rhombus with no right angle is an example of an equilateral but non-equiangular polygon.  Rectangles, including squares, are the only equiangular quadrilaterals Equiangular polygon theorem. Each angle of an equiangular n-gon is  $$\Bigg(\frac{n-2}{n}\Bigg)180^{\circ} = 180^{\circ} -   \frac{360^{\circ}}{n} $$ Viviani's theorem   Vincenzo Viviani (1622 – 1703) was a famous Italian mathematician. With his exceptional intelligence in math...
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Incenter/Excenter Lemma

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Given $\triangle ABC$ with incenter $I$, extend $AI$ to meet the circumcircle of $\triangle ABC$ at $L$. Then, reflect $I$ over $L$, and name this point $I_a$. Then:  $(i)$ $BICI_a$ is a cyclic quadrilateral with point $L$ as the center of the circumscribed circle.  $(ii)$ $BI_a$ and $CI_a$ bisect the exterior angles of $\angle B$ and $\angle C$,  respectively.  Proof $(i)$ Notice that the the question is equivalent to proving the distance from $L$ to points $B, I, C, I_a$ are equal. In other words, we need to show that  $$LB = LI = LC = LI_a$$ Firstly, it is obvious that $LI = LI_a$ because by definition, $I_a$ is the reflection of $I$ over $L$, and reflections preserve the length of the segment.  Now, we are left with proving that $LB=LI$, because similar calculations will provide $LI=LC$.  We see that most of our given information involves angles, so we focus on proving $\angle LBI = \angle LIB$. Firstly,  $$\angle LBC = \angl...
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Sequences and Series

 Let's take a look at some interesting sequences and series problems. 1. (USAMTS) Evaluate the value of  $$S = \sqrt{1+\cfrac{1}{1^2}+\cfrac{1}{2^2}} + \sqrt{1+\cfrac{1}{2^2}+\cfrac{1}{3^2}} + ... + \sqrt{1+\cfrac{1}{1999^2}+\cfrac{1}{2000^2}}$$ Solution Notice that $$1+\cfrac{1}{n^2}+\cfrac{1}{(n+1)^2} = \cfrac{n^4+2n^3+3n^2+2n+1}{n^2(n+1)^2} = \cfrac{n^2+n+1)^2}{n^2(n+1)^2} = \Bigg(1+\cfrac{1}{n(n+1)} \Bigg)^2$$ Thus, $$S= \Bigg(1+\cfrac{1}{1 \cdot 2} \Bigg) + \Bigg(1+\cfrac{1}{2 \cdot 3} \Bigg) + \Bigg(1+\cfrac{1}{3 \cdot 4} \Bigg) + ... + \Bigg(1+\cfrac{1}{1999 \cdot 2000} \Bigg)$$ $$= 1999 + \Bigg(1 - \cfrac{1}{2} \Bigg) + \Bigg(\cfrac{1}{2} - \cfrac{1}{3} \Bigg) + ... + \Bigg(\cfrac{1}{1999} - \cfrac{1}{2000} \Bigg) = 1999 + 1 - \cfrac{1}{2000} = \boxed{\cfrac{3,999,999}{2000}}$$ 2. (HMMT) Find the value of  $$S = \cfrac{1}{3^2+1} + \cfrac{1}{4^2+2} + \cfrac{1}{5^2+3} + ...$$ Solution For this problem, we can use partial fraction decomposition and try to find an alt...
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