Happy New Year!!!

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 .

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.