Fourth Regional Math Count and Third Casio MTAP-TL MATHalino Challenge

mtapAre you one of the Math wizards in your school? Math enthusiast? Winning coach of any Mathematics competition? If you think you are one, surely this update is for you.

The MTAP-TL in cooperation with Negros Oriental State University-Main Campus 1 will be holding an interscholastic contest entitled “FOURTH MTAP-TL REGIONAL Math Counts 2014″ and THIRD CASIO-MTAP-TL MATHalino CHALLENGE 2014”,which aims to promote mathematics in different fields among college students in regions outside Luzon. In connection with this, the said association is encouraging interested schools to participate in the quiz show. The elimination round will be set on March 1, 2014 at Negros Oriental State University-Main Campus 1, Dumaguete City at 8am.Regional winners will join the Math Counts 2014 to be held in Manila on March 8 2014. If you have not participated yet in competitions like this, but you know you have what it takes, then do what you need to, Join! It might be a starting step towards your journey to Math-Stardom. If in case you are already recognized in the “hall of fame” bear this in mind, competitions never end.

Kindly check this links for more details


4th Regional Math Count 2014 Rules and Regulations

3rd Casio- MTAP-TL MATHalino Challenge 2014 Rules and Regulations

To get more updates about this event feel free to visit this Facebook group. Please help us to disseminate the information by sharing it.

Number of Intersections of Lines with Given Points

This topic will enlighten the mind of students who are not aware how to solve for the maximum number of intersections are there with a given points. I will give different insights how to deal with the problem.  Let us start treating the problem geometrically.

 Remember: A line can be drawn if we have two points

line1Now, let’s add point C and draw lines and make sure that the number of intersections is maximized.


As shown in the figure we can tell that the number of intersections we can draw if we have 3 points is also 3. To investigate this further let’s put another point D and again make sure to maximize their number of intersections. As seen in the figure below we have 6 intersections.


Now, by observation the next number of intersections we can obtain if we draw another point is the number of intersections in 4 points which is 6 plus the number of intersection of 1 line to 4 lines which is 4 that is a total of 10. Add another point and the number of intersections will be 10+5 since there are already 5 lines. We can continue with the process. But to do it quickly let’s derive a formula for that.


Result by analysis:



Observe the table above. Let n be the number of points and S for the number of intersections.

Having a knowledge in generating of equation, by doing quick subtraction for S(number of intersection) we can see that the constant appeared in second degree. Meaning, the formula will follow a quadratic equation.


S = an^2 + bn +c    If n=3, S=3.

By direct substitution,

3 = a(3)^2 + b(3) +c

3 = 9a+ 3b +c     -> Eqn.1

Since we have 3 unknowns we need to have 3 equations.

If n=4, S = 6

S = an^2 + bn +c

6 = a(4)^2 + b(4) +c

6 = 16a + 4b +c  -> eqn.2


If n =5, S= 10

10= a(5)^2 + 5b +c

10= 25a+ 5b +c     -> eqn.3


Solving the equations simultaneously we have  a=1/2, b= -1/2, c=0

Going back to our equation

S = an^2 + bn +c by plugging the values of a,b, and c we have.

S = \displaystyle\frac{n(n-1)}{2}

That is the required formula. To test the formula you can try to plug in n=3, n=4, n=5, n=6. It will give you a quick answer.

Is there a quicker way to do this? The answer is yes! You don’t need to generate that formula. The concept is through combinations. That will be in a separate topic.





The Tales of Pi ( π )

This is not about the life story of the fictional character pi in the movie. This is about how the pi(π) was derived by our great mathematicians long time ago.

Pi has been known as early as 250 B.C. The first method of finding the value of π was made by Archimedes of Syracuse, Italy. He first draw a circle with an inscribed regular hexagon like as shown in the figure.

pie pix project


The radius of the circle is 1. Therefore, the length of each side of the hexagon is also 1. He approximated the value of π by the following formula,

\pi=\displaystyle\frac{Perimeter Of Hexagon}{Radius Of Circle}

\pi=\displaystyle\frac{6(1)}{2} = 3    , this was the first approximation of the value of π.

After doing the calculation he doubled the number of sides. He draw a regular dodecagon and calculated again the value of pi using the same formula.



He didn’t stop there, he continued the process up to 96 sides. You might be asking yourselves now why he used this technique. Remember that the value of π is the ratio of the circumference of a circle to diameter. As the number of sides of a polygon increases the polygon approaches to the shape of the circle and the perimeter of the polygon approaches the circumference of the circle. Take a look of the 24-sided polygon inscribed in a circle below.



The polygon in red colored side is the 24-gon. Just imagine a 96-sided polygon plotted inside the circle. We cannot even distinguish it from a circle. According to studies, the value of pi we used today is equal to the approximate value if we draw 25,165,824-sided polygon inside this circle.

The method of finding the value of pi revolutionized on 16th -17th century in the existence of infinite series. Today, we are celebrating annually March 14 as the Pi day.