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...
Post a Comment

Age Problems

Age problems is another commonly tested concept on competitive exams. Solving many practice problems is the key to mastering this topic. Interpreting the language of what is given and what is asked of is the biggest challenge you would need to overcome. Lets look at some commonly used phrases throughout these problems. 
  • If the present age is $a$, then $n$ times the present age = $an$.
  • If the present age is $a$, then $n$ years later, age = $a+n$.
  • If the present age is $a$, then age $n$ years ago = $a-n$.
The next step would be to organize the given information. What usually works for me is putting a chart together and to keep plugging pieces of data that are given. Let's walk through a few examples to see how this works. 

Example 1: Tim's age is three times his son, Alex's age. After $10$ years, he would only be twice Alex's age. What is Alex's present age? 

Solution:

Current AgeAge $10$ years later
Tim$3x$$3x+10, 2(x+10)$
Alex$x$$x+10$

I usually like to keep time periods going across on top, and my characters going down. In this example, we have current ages, and ages $10$ years later. We now start populating the cells with what we know. Lets assign Alex's current age to be $x$. The problem tells us that Tim's age is $3$ times that, so his current age must be $3x$ years. $10$ years later, Alex would be $x+10$ years old, and Tim will be $3x+10$ years old. We also have another clue on Tim's age - Tim will be twice Alex's age in $10$ years, leading us to $2(x+10)$. 
This gives us $2$ ways to represent Tim's age $\Rightarrow 3x+10$ and $2(x+10)$. We can now set these expressions equal and solve for $x$. 
$$3x+10 = 2(x+10) \Rightarrow 3x+10 = 2x+20 \Rightarrow x = 10$$ 
We thus have Alex's current age, $\boxed{10 \text{ years}}$

Example 2: Jack is twice as old as his friend Pam. Pam is $5$ years older than Austin. In $5$ years, Jack will be three times as old as Austin. How old is Pam now?

Solution

Current AgeAge $5$ years later
Jack$2x$$2x+5, 3(x-5+5)$
Pam$x$$x+5$
Austin$x-5$$x-5+5$

Here we can start by calling Pam's current age as $x$. (It often helps to set $x$ as the value the problem is asking you for, in this case, Pam's age). We can fill out this chart in a similar fashion as we did in Example 1. In this problem, we come across two ways to write Jack's age in $5$ years $\Rightarrow 2x+5$ and $3(x-5+5)$. Equating the two and solving for $x$ gives us our answer. 
$$2x+5 = 3(x-5+5) \Rightarrow 2x+5 = 3x \Rightarrow x=5$$
Therefore, Pam's current age is $\boxed{5 \text{ years}}$

Comments

Post a Comment

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...
Read more

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 ...
Read more

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...
Read more