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.

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 .

Pascal's butterfly

Today we will present a beautiful union between Pascal's hexagon theorem and the butterfly theorem.

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! $$c^2 = a^2 + b^2.$$

The Fermat's point of a triangle II

In the previous post, we have analyzed the Fermat's problem, the problem of finding a point $M$ for a given triangle $ABC$ such that $MA + MB + MC$ is the minimum.

the Fermat's problem: find $M$ so that $MA + MB + MC$ is minimum

The Fermat's point of a triangle

In previous posts about modulo, we learn about the mathematician Fermat and his famous problem $$x^n + y^n = z^n.$$

Today, we will look at a geometry problem that bears his name. As we already know, Fermat was not a professional mathematician, but was a lawyer. He was doing math probably just for fun and most of his achievements that we know of today originated from his letters to his friends and also from his occasional writings on the margin of books that he read. The most famous is, of course, the problem $x^n + y^n = z^n$ and his note "I have found a beautiful proof but there is not enough space" that he wrote on the margin of the book by Diophantus.

The problem that we investigate today was raised in a letter that Fermat sent to an Italian mathematician, Torricelli. In his letter, Fermat challenged Torricelli to find a point such that the total distance from this point to the three vertices of a triangle is the minimum possible. Well, this problem was not hard for Torricelli. Since Torricelli knew how to find such a point, today some people refer to this point as the Fermat's point, and others refer it as the Torricelli's point of the triangle.

the Fermat's problem: find a point $M$ so that $MA + MB + MC$ is minimum

A problem about finding shortest path and a property of the ellipse

Today we will look at two problems that seem to be unrelated. The first one is a beautiful geometry problem about finding shortest path and the other one is about a property of an ellipse.

But first, let us introduce the ellipse. An ellipse is drawn below.

for any point $P$ on the ellipse, $PF_1 + PF_2 = \ell$

Solve for special cases first!

I would like to share with you a lesson that I have learnt. That is when facing a problem and we do not know what to do, the first thing we can do is to look at special cases of that problem. Investigating special cases can help us gain a greater understanding of the problem. To illustrate the point, let us solve some problems.

Similar triangles

Today, we are going to learn about similar triangles. We will use similar triangles to give a proof of Pythagorean Theorem. 

Morley's Theorem

In a triangle $ABC$, draw the trisectors of the angles $A$, $B$, $C$. These trisectors intersect at $A'$, $B'$ and $C'$ as in the figure below. Prove that the triangle $A'B'C'$ is an equilateral triangle.