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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Linear Equations - II

Cross multiplication is an important technique for solving equations that contain fractions. In this technique, we multiply both sides of an equation by the denominators of the fractions within it. Doing so will remove the denominators and make the equation easier to solve. 
Cross multiplication is based on the principle that for any nonzero values $a$ and $b$, 
$  a \cdot \cfrac{b}{a} = b$ 
In other words, multiplying a fraction by its denominator leaves just its numerator.

 Example 1 : $ \cfrac {2}{x} = \cfrac{1}{5}$.

  • First, we multiply both sides by $5$, obtaining $ \cfrac{10}{x} = 1$.
  • Then, we multiply both sides by $x$. This cancels the denominator of the left side, leaving  $x = 10$


Example 2 : Simplify $ \cfrac {a}{1 + \cfrac {4}{5}a} $

  • Here we must first collapse the denominator into a single term. This can be done using common denominators.
     $ \Rightarrow$ $ 1 + \cfrac {4}{5}a = \cfrac {5+4a}{5}$

     $ \Rightarrow$ $  \cfrac {a}{ \cfrac {5 + 4a}{5}} = a \cdot \cfrac{5}{5 + 4a} = \boxed{\cfrac{5a}{4a + 5}} $

Your turn problems: 

$(1)$ Simplify $ \cfrac {2x}{ \cfrac {x}{3} + \cfrac{y}{2} + \cfrac{1}{5}} $

$(2)$ Simplify $ \cfrac {4(3+x)}{3(9-x)} = \cfrac{8}{3} $. Find $x$.

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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

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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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