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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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Fibonacci Numbers

 The sequence of numbers that starts as $1,1,$ and continues with each new number calculated as the sum of the previous two numbers, i.e., $1,1,2,3,5,8,13,21,34,55,89,144....$ is called the sequence of Fibonacci numbers. It was first described in 1202 by Leonardo of Pisa, better known as Fibonacci. It appeared in his book Liber Abaci as the solution to a problem involving the growing number of rabbits over time. Since then, Fibonacci numbers have seen many interesting applications and connections to a variety of topics, both within mathematics and in other disciplines, such as computer science, physics, chemistry, biology, and architecture.  This pattern of Fibonacci numbers is given by $u_1 = 1, u_2 = 1$ and the recursive formula $u_n = u_{n-1} + u_{n-2}, n > 2$. Simple Properties of Fibonacci Numbers $1.$ Sum of the Fibonacci Numbers The sum of the first n Fibonacci numbers can be expressed as $u_1 + u_2 + ... + u_{n-1} + u_n = u_{n+2} − 1.$ $2.$ Sum of Odd Terms The sum ...
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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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Triangle Similarity

Two triangles are called similar , if we can get two congruent triangles after enlarging or compressing the sides of one of them according to an equal ratio. That is, two triangles are similar means they have a same shape but may have different sizes. Criteria for Similarity of Two Triangles  (I) Each pair of corresponding angles are equal (A.A.A.);  (II) All corresponding sides are proportional (S.S.S.);  (III) Two pairs of corresponding sides are proportional, and the included corresponding angles are equal (S.A.S.);  (IV) For two right triangles, a pair of two corresponding acute angles are equal (A.A.);  (V) Among the three pairs of corresponding sides two pairs are proportional (S.S.). Basic Properties of Two Similar Triangles (I) For two similar triangles, their corresponding sides, corresponding heights, corresponding medians, corresponding angle bisectors, corresponding perimeter are all proportional with the same ratio.  (II) Consider the similarit...
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Clock Problems

When solving problems relating to clocks, keeping the following two points in mind will be helpful. The Minute hand covers $360^{\circ}$ in 60 minutes, or $6^{\circ}$ per minute. The Hour hand covers $360^{\circ}$ in 12 hours, or   $30^{\circ}$ per hour. Using these facts, the we can easily derive the results below: In a period of 12 hours, the hour and minute hand coincide 11 times. In a period of 12 hours, the hour and minute hand form a $180^{\circ}$ 11 times. In a period of 12 hours, any other angle between the minute and hour hand is formed 22 times. The time gap between any two coincidences is $\cfrac{12}{11}$ hours or $65\cfrac{5}{11}$ minutes. One common problem involving clocks is to find the angle between the minute and hour hand at a given time. To do so, one can use the formula  $$\theta = |5.5m - 30h |$$ Where $\theta =$ desired angle, and the time can be written as $h:m$. Let's use what we have learned to solve a couple problems. 1. What is the angle between the ...
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Middle School Competition Problem

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Let $ABCD$ be a convex quadrilateral with $AB=BC=AC$. Suppose that a point P lies in the interior of the quadrilateral such that $AD=AP=DP$ and $\angle PCD=30^{\circ}$. Given that $CP=2$ and $CD=3$, find that length of $AC$.  The first step in solving this question is to draw the segment $PB$. Doing this allows us to see that $\triangle APB \cong \triangle ADC$ because $AP=AD$, $AB=AC$, and $\angle DAC = \angle PAB$ because $\angle DAP = \angle CAB = 60^{\circ}$. Thus, $PB=3$.  Next, let's set $\angle ACD = x$. We see that $\angle PCA = 30-x$, and $\angle BCP = 30 +x$. Due to the triangle congruency we established earlier, $\angle PBA = x$ as well. This means that we can write $\angle PBC = 60 - x$. Now,  $$\angle CPB = 180^{\circ} - \angle BCP - \angle PBC = 180^{\circ} - (30+x) - (60-x) = 90^{\circ}$$.  Thus, $\triangle BPC$ is a right triangle, and we know that $CP=2$ and $PB=3$, so we can easily find $CB$ to be  $$CB = \sqrt{2^2+3^2} = \boxed{\sqrt{13}}$$
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Pigeonhole Principle

The Pigeon Hole principle follows a simple observation that if you have 10 pigeons and 9 pigeonholes and you distribute these pigeons randomly, its quite obvious that there will be one pigeonhole with more than 1 pigeon. In other words, there are n boxes (the Pigeonholes), into which more than n objects (the Pigeons), have been placed, then at least one of the boxes must have received more than one of the objects. This principle is frequently applied in problems where we need to determine a minimal number of objects to ensure that some integral number property is satisfied. Consider the following puzzle:  A box contains red, black and white balls. The objective is to pick balls satisfying some constraints. How many balls must be taken to ensure that there is a pair of same color? The key to solve this puzzle is pigeonhole principle.  The principle says:  If $n$ pigeons are nesting in $m$ pigeonholes, where $n > m$, then at least one pigeonhole has more than one pig...
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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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Irrational Numbers

A number which cannot be written in the form $ \cfrac {p}{q}$  where $p$ and $q$ are integers and $ q $  $ \neq 0$. Example: $ \sqrt {2}$, $ \sqrt {3}$, $ \sqrt {6}$, $ \sqrt {7}$, $ \sqrt {8}$, $ \sqrt {10}$ Theorem: If $p$ divides $x^3$, then $p$ divides $x$, where $x$ is a positive integer and $p$ is a prime number.  Proof :  Let $x = p_{1} p_{2}...p_{n}$ where $p_{1}, p_{2}, p_{3}.....p_{n}$  are primes, not necessarily distinct.  $ \rightarrow x^3 = p_{1}^3 p_{2}^3.....p_{n}^3$ Given that $p$ divides $x^3$ By fundamental theorem, $p$ is one of the primes of $x^3$.  By the uniqueness of fundamental theorem, the distinct primes of $x^3$ are same as the distinct primes of $x$.  $ \rightarrow p$ divides $x$ Similarly if $p$ divides $x^2$, then $p$ divides $x$, where $p$ is a prime number and $x$ is a positive integer. Example: Prove that  $ \sqrt {2}$ is irrational.  Solution : Let us assume that  $ \sqrt ...
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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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