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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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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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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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Popular posts from this blog

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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