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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Arithmetic Sequences

Let's take a look at Arithmetic Sequences today!

What is an Arithmetic Sequence?
A set of terms where every sequential term is created by adding or subtracting a common difference from the previous term. For example, let's look at the sequence $2, 5, 8, 11, ...$ The initial term, or the first term of the sequence here is $2$. The common difference is the number that is added to each term to get to the next term. So in this case, we add $3$ to get from $2 \to 5, 5 \to 8, 8 \to 11$ and so on. To find the fifth term, I could keep adding $3$ to the first term $4$ times. Sounds easy, right! Now try finding the $25$th term. You would be here all day long trying to add $3$, $24$ times. Or, you can think smart and add $(3 \cdot 24)$ to the first term, $2$ which brings us to our first formula.
$$a_n = a_1 + (n-1)d$$
Where
$a_n =$ the $n$th term in the sequence
$a_1 =$ the first term in the sequence
$d =$ the common difference
So, going back to our example, we know that we need to find the twenty-fifth term, or $a_{25}$, having the first term as $a_1 = 2$, and the common difference $d=3$. Plugging these values into our formula give us our answer.
$$a_{25} = 2 + (25-1)3 = 2 + (24 \cdot 3) = 2 + 72 = \boxed{74}$$
Your Turn!
1. Find the twenty-third term in the sequence $3, 7, 11, 15, ...$
(Scroll down to the bottom of the page for the answer)

Let's go one step further and talk about how to find the sum of such sequences. For instance, find the sum of the first $8$ terms in the sequence $2, 4, 6, ...$. What we can do is find the sum of the first and last terms, second and second-last terms, and keep working our way inwards, as shown in the image. Anything interesting?



All the sums we get equal the same, $18$ in this case. This sum should then be multiplied by the number of pairs that we just added which is $4$ in this example.
We thus get $18 \cdot 4 = \boxed{72}$
This brings us into our next formula to find the sum of the first $n$ terms in a given arithmetic sequence. $$S_n = \frac{n(a_1+a_n)}{2}$$
Where
$S_n =$ the sum of the first $n$ terms
$a_1 =$ the first term in the sequence
$a_n =$ the $n$th term in the sequence

Your Turn!
2. Find the sum of the first $12$ terms of the sequence $1, 3, 5, ...$.
(Scroll down to the bottom of the page for the answer)









Answers
1. $91$
2. $144$

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Equilateral and Equiangular Polygons

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Irrational Numbers

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Incenter/Excenter Lemma

Image
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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