Math for the curious
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:
Today, we will use Menelaus' theorem to prove Pascal's theorem for the circle case.
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 $P$.
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.
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| 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 $(O)$.
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.
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! $$c^2 = a^2 + b^2.$$
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| 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
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.
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