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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Sides and Angles of a Triangle

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 the information above.

Problem 1: How many different triangles can be made with sides lengths that are prime factors of $1729$?

Solution: The prime factorization of $1729 = 7 \cdot 13 \cdot 19$. In theory, triangles of following lengths are possible

{$7, 7, 7$}                {$13, 13, 13$}            {$19, 19, 19$}             {$7, 7, 13$}                 {$7, 7, 19$}

{$13, 13, 7$}            {$13, 13, 19$}            {$19, 19, 7$}             {$19, 19, 13$}             {$7, 13, 19$}

But since we know that the sum of any two sides must be larger than the third (Triangle Inequality), some of these combinations are invalid. Can you tell which ones?

For those of you who could figure out, I'm sure you would know which one is a bogus combination.

For others who need a little bit of a reminder, let me give you another hint. Think about how the {$7, 7, 19$} wouldn't work. Post a comment if you need more explanation. You can always email me for further help.

Problem 2: (AHSME/1956) In the figure below $AB = AC$, $\angle BAD = 30^{\circ}$ , and $AE = AD$. Then what is $m\angle CDE$?


Solution: 
 Let $m\angle EDC = x$. 

$\Rightarrow$ We can say that  $x = \angle ADC - \angle ADE$.

$\Rightarrow \angle ADE = \angle AED$ (Base angles in an isosceles triangle)

$\Rightarrow x = \angle ADC - \angle AED$. (Substitution)

$\Rightarrow \angle AED = \angle C + x$ (Exterior Angle theorem)

$\Rightarrow x = \angle ADC - (\angle C + x)$ (Substitution)

Solving for $x$, we get $x = \frac{1}{2}(\angle ADC - \angle C)$

$\Rightarrow \angle ADC = \angle B + 30^{\circ}$ (Exterior Angle Theorem)

$\Rightarrow x = \frac{1}{2}(\angle B + 30^{\circ} - \angle C)$ (Substitution)

$\Rightarrow \angle B = \angle C \Rightarrow \angle B - \angle C = 0$ (Base angles in an isosceles triangle)

$\Rightarrow x = \frac{1}{2}(\angle B - \angle C+ 30^{\circ}) = \frac{1}{2} \cdot 30 ^{\circ} = \boxed{15^{\circ}}$

Problem 3: (CHINA/1997) In a right-angled $4ABC$, $\angle ACB = 90^{\circ}$, $E$ and $F$ are on $AB$ such that $AE = AC$ and $BF = BC$. Find $m\angle ECF$ in degrees.

$\angle AEC = \frac{1}{2}(180^{\circ} - \angle A) = 90^{\circ}-\frac{1}{2} \angle A$ (Base angles) $\text{Similarly, } \angle BFC = 90^{\circ}-\frac{1}{2} \angle B$ $$\text{Therefore, } \angle ECF = 180^{\circ} - \angle AEC - \angle BFC$$ $\text{Substituting our values for $\angle AEC$ and $\angle BFC$ and simplifying, we get }$ $\angle ECF = \frac{1}{2}(\angle A + \angle B) = \boxed{45^{\circ}}$ 

As always, if you have any questions on the content covered or if you are looking for additional problems for practice, feel free to reach out. 



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Equilateral and Equiangular Polygons

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