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Divide a quadrilateral into equal areas

Today let us consider the following interesting compass-and-straightedge construction problem:
Dividing quadrilateral problem. Given a quadrilateral $ABCD$ and four points $M_1$, $M_2$, $M_3$, $M_4$, in this order, on $AB$. Using compass and straightedge, show how to construct four points $N_1$, $N_2$, $N_3$, $N_4$ on $CD$ so that the four line segments $M_1 N_1$, $M_2 N_2$, $M_3 N_3$ and $M_4 N_4$ divide the quadrilateral into 5 small quadrilaterals of equal area.

Compass-and-straightedge construction


Today we will start a series of posts on compass-and-straightedge construction. Believe it or not, there are a few construction problems that sound simple but it had required more than two thousand years to settle! The most famous ones are the problem of regular polygon construction and the problem of angle trisection. These problems were known in ancient times but it was not until the late 18th-19th centuries that mathematicians could finally solve them using the most modern tools of mathematics.

In this first post, we will learn about some basic compass-and-straightedge constructions. These constructions are assumed as obvious and can be employed in more complex construction problems.


James' question

My son James wanted to ask you a question:
how many triangles can you see in this picture?

The butterfly problem

In the previous post, we show a simple proof of the butterfly problem by using Pascal's hexagon theorem. Today, we will list a few more proofs.  These proofs will be presented in the form of exercises so that interested readers can brush up on their problem-solving skills .


Pappus' Theorem

Today we will learn about Pappus' theorem. This theorem states that if we take three points $1$, $3$, $5$ on a line, and another three points $2$, $4$, $6$ on another line, then the three intersection points of the following line pairs $$\{12, 45\}, ~\{23, 56\}, ~\{34, 61\}$$ are collinear.


Pascal's Theorem

In a previous post, we were introduced to Pascal's Hexagrammum Mysticum Theorem - a magical theorem - which states that if we draw a hexagon inscribed in a conic section then the three pairs of opposite sides of the hexagon intersect at three points which lie on a straight line.

For example, as in the following figure we have a hexagon inscribed in a circle and the intersection points of the three pairs of the opposite sides of the hexagon $\{12, 45\}$, $\{23, 56\}$, $\{34, 61\}$ are collinear.

There is a useful tool to prove the collinearity of points - the Menelaus' Theorem - which states as follows:

Menelaus' Theorem: Given a triangle $ABC$ and three points $A'$, $B'$, $C'$ lying on the three lines $BC$, $CA$, $AB$, respectively. Then the three points $A'$, $B'$, $C'$ are collinear if and only if $$\frac{\vec{A'B}}{\vec{A'C}} \times \frac{\vec{B'C}}{\vec{B'A}} \times \frac{\vec{C'A}}{\vec{C'B}} = 1.$$


Today, we will use Menelaus' theorem to prove Pascal's theorem for the circle case.

Radical axis and radical center

In previous post, we have learned about the concept of power of a point with respect to a circle. The power of a point $P$ with respect to a circle centered at $O$ and of radius $r$ is defined by the following formula $${\cal P}(P, (O)) = \vec{PU} \times \vec{PV} = PO^2 - r^2 = (P_x - O_x)^2 + (P_y - O_y)^2 - r^2,$$ here, $U$ and $V$ are two intersection points of the circle $(O)$ with an arbitrary line passing though $P$.
Power of a point: ${\cal P}(P, (O)) = \vec{PU} \times \vec{PV} = PO^2 - r^2 = (P_x - O_x)^2 + (P_y - O_y)^2 - r^2$.

The value of the power of a point gives us information about relative position of the point with respect to the circle. If the power of the point $P$ is a positive number then $P$ is outside the circle, if it is a negative number then $P$ is inside the circle, and if it is equal to zero then $P$ is on the circle.

Today, we will look at application of the power of a point concept. We will use two main tools: radical axis and radical center. Radical axis is often used to prove that a certain number of points lie on the same straight line, and radical center is used to prove a certain number of lines meet at a common point. 

Power of a point to a circle

Today we will learn about the power of a point with respect to a circle.

Suppose on a plane we have a point $P$ and a circle $(O)$. Draw a line through $P$ which intersects with the circle at two points $U$ and $V$. Then the value of $$PU \times PV$$ is independent of the choice of the line $PUV$.

This means that if we draw another line through $P$ which cuts the circle at two other points $A$ and $B$ then $$PA \times PB = PU \times PV.$$
This constant value is called the power of the point $P$ with respect to the circle $(O)$.


Hexagrammum Mysticum Theorem

The mathematician Pascal was most famous for Pascal's triangle, a number pattern that used in the binomial expansion formula. Today, we will introduce a theorem in geometry that bears his name. The Pascal's Hexagrammum Mysticum Theorem states that if we draw a hexagon inscribed in a circle then the three pairs of opposite sides of the hexagon intersect at three points which lie on a straight line.

Ceva's Theorem and Menelaus' Theorem

Today we will learn about two well-known theorems in geometry,  Ceva's Theorem and Menelaus' Theorem. These two theorems are very useful in plane geometry because we often use them to prove that a certain number of points lie on a straight line and a certain number of lines intersect at a single point. Both of the theorems will be proved based on a common simple principle. We also generalize the theorems for arbitrary polygons.



Pythagorean Theorem

Pythagorean Theorem is surely one of the most popular theorems in mathematics, which says that in a right triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides.

Pythagorean Theorem: $c^2 = a^2 + b^2$.

This theorem has so many proofs. One of the proofs is rather interesting because it was due to James Abram Garfield - the 20th President of the United States.

President Garfield's proof is very simple. It relies on finding the area of the following trapezoid in two different ways.

This trapezoid has two bases of length $a$ and $b$. The height of the trapezoid is $a + b$. So the area of the trapezoid is $$\frac{1}{2}(a+b)(a+b) = \frac{1}{2}(a^2 + b^2 + 2ab).$$

The other way to calculate the area of the trapezoid is to sum up the three areas of the triangles, which is $$\frac{1}{2}ab + \frac{1}{2} c^2 + \frac{1}{2}ab = \frac{1}{2}(c^2 + 2ab).$$

Comparing the two results we obtain the Pythagorean Theorem! $$c^2 = a^2 + b^2.$$

Sequence - Part 9


This is the last post of our series; here is the link to "Sequence - Part 1" if you haven't read it. Today we will do some more exercises on sequence. We will prove some interesting identities. For the Pell sequence $$P_0=0, ~~P_1 = 1, ~~P_n = 2 P_{n-1} + P_{n-2},$$ and the companion Pell sequence $$H_0=1, ~~H_1 = 1, ~~H_n = 2 H_{n-1} + H_{n-2},$$ we will show that $$H_n^2 - 2 P_n^2 = (-1)^n.$$
For the Fibonacci sequence $$F_0 = 0, ~~F_1 = 1, ~~F_n = F_{n-1} + F_{n-2},$$ we will prove the following identity $$\frac{F_{2013(n+1)} - F_{2013 (n−1)}}{F_{2013 n}} = \frac{F_{2013(n^{2013}+1)} - F_{2013 (n^{2013}−1)}}{F_{2013 n^{2013}}}.$$

Sequence - Part 8


In the last post, we learn how to determine a trigonometric formula for a sequence in the case when the characteristic equation has complex roots. Today we will solve more exercises for this case.

Sequence - Part 7


This is the 7th part of our series on sequences. Today we will learn how to solve a linear recurrence equation in the case when the characteristic equation has complex roots. In this case, we can express the complex roots of the characteristic equation in trigonometric form and then use de Moivre's identity to obtain a trigonometric formula for the sequence.


Sequence - Part 6


Today we will learn about difference operator and use it to prove a fundamental theorem for linear recurrence equations.
A fundamental theorem for linear recurrence equations. Suppose that the characteristic equation can be factored as $$f(x) = a_k x^k + a_{k-1} x^{k-1} + \dots + a_0 = (x - z)^j (b_s x^s + b_{s-1} x^{s-1} + \dots + b_0)$$ and $$f_n = p(n)~z^n,$$ where $p(n)$ is a polynomial of degree less than $j$. Then the sequence $f_n$ satisfies the recurrence equation 
$$a_k f_{n} + a_{k-1} f_{n-1} + \dots + a_1 f_{n-k+1} + a_0 f_{n-k} = 0.$$


Sequence - Part 5


Today we will go through some examples. The first type of examples involves solving a linear recurrence equation to derive a general formula for a sequence. The second type of examples does the reverse process, in which, we are given a formula of a sequence and asked to determine its recurrence equation.


Sequence - Part 4


Today we will learn how to solve a linear recurrence equation and derive a general formula for a sequence. This method will work for all cases, including the case when the characteristic equation has multiple roots.  

Sequence - Part 3


This is the third part of our series on sequences. There will be no new material in this post. The main purpose of this post is to go through some examples so we can get used to the method of solving recurrence equations. If you have not read the previous parts of our series, please read them here: Part 1, Part 2.


Sequence - Part 2


This is the second part of our series on sequences. Please have a careful read of Part 1 before you continue to this post. Today, we will introduce some more terminologies about sequences and we will present a general method to solve a linear recurrence equation.

Sequence - Part 1


Today we will begin our first post of a series on recurrence sequences. The purpose of this series is to show you how to derive a general formula for a sequence that satisfies a linear recurrence equation. We start with some basic operations on sequences. We will learn about addition of sequences and multiplication of a sequence with a number.

Fibonacci sequence and Pascal triangle


In previous post, we used the result of a tile matching puzzle to prove an identity for Fibonacci sequence. Today, we will continue on and prove another identity. We will show that $${2011 \choose 0} + {2010 \choose 1} + {2009 \choose 2}+ {2008 \choose 3}+ \dots + {1007 \choose 1004}+ {1006 \choose 1005} = F_{2012},$$ $${2012 \choose 0} + {2011 \choose 1} + {2010 \choose 2}+ {2009 \choose 3}+ \dots + {1007 \choose 1005}+ {1006 \choose 1006} = F_{2013}.$$
In general, we have the following identity $$\sum_{v+u=n}{v \choose u} = F_{n+1}.$$ Through this identity, we will see an interesting connection between the Fibonacci sequence and the Pascal triangle.


Fibonacci identities


In previous post, we have learnt about Fibonacci sequence and a tile matching puzzle. Today, we will use the result of the tile matching puzzle to prove the following identity on Fibonacci sequence $$F_{n+m+2} = F_{n+1} F_{m+1} + F_{n+1} F_m + F_n F_{m+1}.$$

Like an art painting, a mathematical problem reflects different beauties if we view it at different angles. Take, for example, the Fibonacci identity that we are considering, we can prove this identity by different methods. One method is to use the formula $$F_n = \frac{1}{\sqrt{5}} \left[ \left( \frac{1 + \sqrt{5}}{2} \right)^n - \left( \frac{1 - \sqrt{5}}{2} \right)^n \right]$$ to calculate the two sides of the identity and make them equal by straightforward algebraic manipulations. Another method is to make use of the tile matching puzzle that we learned in the previous post. I hope that you will find this "combinatorial" method more interesting.


Fibonacci numbers and a tile matching puzzle


Today we will learn about Fibonacci sequence and a tile matching puzzle. We will soon see that the tile matching puzzle seems to be unrelated to the Fibonacci sequence, but it turns out that the solution to the puzzle is, in fact, the Fibonacci sequence!

Let us first introduce the Fibonacci sequence. The Fibonacci sequence $\{F_n\}$ is determined by the following rule: $$F_0 = 0, ~F_1 = 1, ~F_{n+1} = F_n + F_{n-1}.$$

Thus, $$F_0 = 0, ~F_1 = 1, ~F_2 = 1, ~F_3 = 2, ~F_4 = 3, ~F_5 = 5, ~F_6 = 8, ~F_7 = 13, ~F_8 = 21, \dots$$

de Moivre's formula


In previous post we have learned about complex numbers. Today, we will learn about the trigonometric form of a complex number and the famous de Moivre's formula.


Complex number


Today we will learn about the complex numbers. The crucial point about complex numbers is that we accept a very special number that we will denote it by $i$. This number $i$ is very special because it satisfies the following identity $$i^2 = -1.$$

So a complex number will have the form $$a + ib$$ where $a$ and $b$ are two real numbers. When $b=0$ then $a + ib = a$ is just a normal real number, and when $a=0$ then $a + ib = ib$ is called a pure imaginary number. Here are some examples of complex numbers: $$1+ i, ~~ 2 - 3i, ~~ -\sqrt{3} + 4i, ~~5i - 4, ~~6, ~~i, ~~-3i, ~~4 + 2i, \dots$$