The use of Pi in geometry is very common. It is seen in formulas for area and circumference of circles, volume of cones and cylinders and more. Most of us know the value of Pi to be around 3.14, with many people memorizing Pi up to many digits. But where does the value of Pi come from?

The distance around the outside a circle is its circumference. Consider the formula for the circumference of a circle, Circumference = Pi x Diameter. Solving for Pi we get, Pi = Circumference divided by Diameter. Therefore, for a circle of any size, the circumference divided by the diameter is approximately 3.14. But how do we know this?

We can start approximating Pi but considering the perimeter of an regular n-sided figure and dividing that by the length of a diagonal. Take a square with side equal to one. By using the Pythagorean Theorem, or the knowledge of 45-45-90 triangles, the diagonal is approximately 1.41. Therefore the perimeter divided by the diagonal is approximately 2.82.

Now take an octagon, with sides of length equal to one. The length of the diagonal found by taking 2 times the apothem is approximately 2.61. Taking the perimeter of 8 divided by 2.61, we get an approximation for Pi to be 3.06. Notice that this is fairly close to the value of Pi or 3.14.

As the number of sides of a regular polygon increases, the shape more closely approximates that of a circle. When considering a regular polygon with 500 sides, each side of length one, we see calculate the diagonal to be approximately 159.155 and the approximation for Pi to be approximately 3.14157. This is extremely close to the value of Pi at 5 digits, which is 3.14159.

One can continue this process by increasing the number of sides of a regular polygon to get as close of an approximation as possible to the actual value of Pi. I have a book which has the value of Pi calculated out to 1 million digits. Why someone needs to know that, I don't know. For all practical purposes, 3.14 works fine. Also, the fraction 22/7 is often used in calculations involving Pi.

I was working with a student on completing the square and it tends to be a more tricky method of factoring than using the quadratic formula.  There are several steps to remember.

Recall how we factored and solved quadratic equations using the reverse FOIL method. An example of this type of factoring is x^2 + 5x + 4 = 0 factored is (x +4)(x + 1) = 0. Sometimes it's quite difficult to solve quadratic equations using this method, so we can solve by a method known as completing the square.
The idea behind completing the square is to turn a binomial into a perfect square trinomial. For example, consider the binomial x2 + 6x. The perfect square trinomial with the first two terms x2 + 6x is x2 + 6x + 9 because (x + 3)(x + 3) = x^2 + 6x + 9. Notice how we added a 9 to x^2 + 6x. The question we ask ourselves is, “What number squared equals 9?” We know that 3^2 = 9. Also notice that 3 is half of the coefficient of the middle term 6x. We take half of the middle term, square it and
add it to form the perfect square trinomial. Then we factor the trinomial.

Examples: Complete the square and factor the perfect square trinomial.
1. x^2 + 8x

Step one: Take half the coefficient of the middle term. (1/2)(8) = 4.
Step two: Square the result in step one. 4^2 = 16.
Step three: Add the result in step two to the binomial to form the trinomial x^2 + 8x + 16.
Step four: Factor the trinomial. (x + 4)(x + 4) or (x + 4)^2.

 x^2 – 7x
Step one: Take half of the coefficient of the middle term. (1/2)(-7) = -7/2.
Step two: Square the result in step one. (-7/2)^2 = 49/4.
Step three: Add the result in step two to the binomial to form the trinomial x^2 – 7x + 49/4.
Step four: Factor the trinomial. (x – 7/2)(x – 7/2) or (x -7/2)^2


To solve quadratic equations by completing the square we must remember the following:
1. Make sure the coefficient of the squared term is one. If it is not one, we must divide both sides of the equation by the coefficient of that term. For example, if the term is 2x2, we must divide both sides of the equation by 2.
2. Get all variables on one side of the equation and the constants on the other side. This makes sure we have a binomial in the form x2 + bx.
3. Take half of the coefficient of the middle term, square it and add it to both sides of the equation.
4. Factor the perfect square trinomial.
5. Solve the equation using the square root property and check answers by substituting into the original equation.

Note if the coefficient in front of the x^2 term is not 1, must divide the equation by the coefficient before completing the score.

For B to be the multiplicative inverse of A, AB = I and BA = I. So we begin by multiplying the first row of A with the
first column of B to get
(-3)(-5/27) + (4)(1/9) = 15/27 + 4/9

For those having difficulty finding the inverse of a matrix, here's part of a chapter from my book on this topic.

Multiplicative Inverses of Matrices

Recall that the multiplicative inverse of any real number x is the number when multiplied by x equals 1. In this case the multiplicative inverse of x is 1/x. Suppose we have matrix A of the form








 If we multiply A by what is known as the identity matrix, we still get A. We will call the identity matrix I, note the illustration below of AI = A.

 
If A is an n by n matrix, then there exists another matrix A-1, called A inverse, so that AA-1 = I and A-1A = I. So we basically are looking for a matrix that when multiplied by the original matrix, equals the identity matrix and vice versa. The identity matrix is always a matrix with 1's along the diagonal from upper left to lower right and 0's everywhere else.

Example: Show that B is the multiplicative inverse of A where
 







 For B to be the multiplicative inverse of A, AB = I and BA = I. So we begin by multiplying the first row of A with the first column of B to get

(-3)(-5/27) + (4)(1/9) = 15/27 + 4/9
= 15/27 + 12/27
= 1

Next we multiply the first row of A with the second column of B to get
(-3)(4/27) + (4)(1/9) = -12/27 + 4/9
= -12/27 + 12/27
= 0

Now proceed to multiply the second row of A with the first row of B to get
(3)(-5/27) + (5)(1/9) = -15/27 + 5/9
= -15/27 + 15/27
= 0

Finally, multiply the second row of A by the second column of B to get
(3)(4/27) + (5)(1/9) = 12/27 + 5/9
= 12/27 + 15/27
= 1

Therefore B is the multiplicative inverse of A. Notice the illustration of this below.

Here's a little review on graphing and finding equations of perpendicular lines.

Example: Find the equation of the line that passes through the point (2, 3) and is perpendicular to the line
y = 3x - 5.

Solution:
The slope of the line perpendicular to y = 3x – 5 must multiply by 3 equal -1. We think of the slope of the first line as m1 and the slope of the second line as m2, therefore 3(m1) = -1, so m1= -1/3.
Now we know that m = – 1/3, x = 2 and y =3.
Use the formula y = mx + b
3 = (-1/3)(2) + b
3 = -2/3 + b
3 2/3 = b
11/3 =b
The equation of the line perpendicular to y = 3x – 5 that passes through the point (2, 3) is y = (-1/3)x + 11/3.
Notice the graph of both lines below.

Here's a math puzzle


Suppose you go to a hotel and ask for a room and the clerk says that there are no room available.  The clerk then states that if I ask everyone to move to the room next them, then a room will become available. For example, the people in room 1 will move to room 2, from room 2 to room 3 and so on. The people in the last room will move to room 1.  How does a room open up???

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This is a hotel with an infinite number of rooms, therefore noone ever moves into room 1.  It's kind of an absurd concept, which leads to the concept of infinity. It's a very uncertain concept in and of itself.

Suppose we wish to graph the equation y = -3(x - 3)^2 + 5.   We recognize this is a parabola with the vertex at (3, 5) because we can rewrite this as y - 5 = -3(x - 3)^2, which is in the form y - k = a(x - h)^2 (vertex at (h, k)).

The parabola opens up or down, and since a is negative, the parabola opens down. 

Interesting, we can use calculus to determine whether (3, 5) is a maximum or a minimum. If it's a maximum, the parabola opens down, if it's a minimum, the parabola opens up.

y' = -6(x - 3) = -6x + 18

We now find the derivative for a value of x < 3 and one for a value for x > 3

for x = 2, y' = -6(2-3) = -6(-1) = 6

for x = 4, y' = -6(4-3) = -6(1) = -6

Since the slope of the curve at x = 2 is positive and the slope of the curve at x = 4 is negative we have a maximum at x = 3.  Therefore (3, 5) is a maximum and the parabola opens downward.

Just another way to determine how a parabola opens.

I am noticing a disturbing trend in schools today regarding mathematics and teaching of mathematics.  The majority of students I work with tell me how poor their teachers are. Many basically just read things from the book, don't take time to really explain concepts in class or take the time to work with student after school. I even heard of one teacher that simply posts youtube videos on the topic and hands out worksheets with the answer key.  Making matters worse, the school has a tutoring center, but that has been deemed pretty useless as well. When I was in school, teachers seemed more apt and wanting to help students.  Are teachers, in general, getting more lazy? I know there are still plenty of excellent teachers and certainly want to give credit to those fine educators that take their jobs seriously. But I can't overlook what seems to be a higher percentage of poor teachers today than 25 to 30 years ago.