Category: Geometry
The symbol π — where does it come from?
Where does the symbol π come from?
In 1652, William Oughtred used π to refer to the periphery of a circle (in his expression, the ratio of circumference-to-diameter of a circle is π ⁄ δ, the latter referring to diameter).
In 1665, Jonh Wallis used a Hebrew letter 
(mem) to equal the ratio of one-quarter of circumference to diameter of a circle. (This letter plays the role as of “M” in Latin alphabets, but look how close its shape resembles a quarter of a circle, as well as the Greek letter pi !)
In 1705, William Johns used π to represent the ratio of circumference-to-diameter of a circle (believed to be first use with exactly same meaning as in today) .
From 1736, Leonard Euler, both famous and a prolific writer in mathematics works, spread the use of π in his publications.
Counting from the first relevant use, the symbol π has already had a history of more than 360 years!
Proof: Diameter is the shortest curve that bisects circular area
The picture below actually shows the proof.
The goal is to show the red line (which is supposed to bisect circular area) is longer than the diameter AB’.
We see it from:
Length of red curve AB is greater than: AE + EB = AE + EB’
which is certainly longer than AB.
Just one more minute. Let’s review the construction-proof process.
Connect the two endpoints A, B of the curve by a line segment. Draw diameter CD // AB.
Take O (the midpoint of CD), then passing A and O we get another diameter AB’.
The two arguments needed for completing proof are:
(1) Red line must have at least one intersection with diameter CD; (think like this: if the red line resides completely at only one side, then there is no point that it can evenly divide the circular area) Suppose the intersection is at point E;
(2) B and B’ are symmetric to the diameter CD therefore EB = EB’ (think on why? using the property of circles and parallel lines)
Now the rest of the proof is straight forward.
3D objects with 3 views from top, front and side
For a 3D objects, given three views to you: one from top, one from front, and one from side, can you imagine what the original 3D objects looks like?
The question is not posed to a mechanic engineer, it would be trivial in that case. The question is raised to get a junior middle student to think a bit.
For a cylinder one of the three views is a circle, and the other two views are rectangles. For a cone one of the views is a circle, and the other two views are triangles. What if the three views given are a circle, a rectangle and a triangle? Can you figure out the original shape?
Find the Center of a Circle — Do you know how to do it?
Using a compass, you can draw a circle at any place, with any radius.
Now let’s reverse the problem. Given a circle, do you know how to find its center? (Of course, once the circle is found, there shall be no problem at all to tell its diameter, or radius.) You only see the circle itself, there is no explicit indication on where the center is.
With two set of restrictions on what kind of tools you can use, there are actually two questions. In general, you can find the center using any convenient method, including copy-and-paste the circle onto a paper, and then fold it. In particular (from classical Euclidean geometry), where it’s required to do so with a ruler (with which you are allowed to draw lines and line segments only) and a compass (with which you are allowed to draw circles only).
See the following article on how to do it in general.
How to find the center of a given circle?
If you attempt to solve this problem with a ruler and a compass, then you are required to know how to make a perpendicular bisector. This will be discussed in another posting.
Right Triangle: Find Hypotenuse Given Two Legs — Area-based Approach!
How Long Does the Hypotenuse of a Right Triangle Measure
— Find it Using Area-based Approach!
We will have an exciting journey of discovery, following the footstep of early pioneers in math.
Let the problem be raised as:
Problem Given two legs of a right triangle, what is the length of its hypotenuse?
After a bit thought, we decide to reduce the problem to the following form:
Standardized Problem Given a right triangle with one leg being 1, the other leg being x, what
length of the hypotenuse?
With familiarity with concepts of similar triangles and proportion, you will find that solution to
standardized problem leads immediately to a solution for the original one. Even without that,
the connection of the two problems can be intuitively understood. Suppose a triangle with two legs:
1 and x, and a hypotenuse of y, then we know a triangle with two legs 2, 2x will have a hypotenuse of 2y.
For this reason, below we will focus only on the standardized problem. The goal is to fill out a form
where leg one is always 1, leg two is any integer numbers: 1, 2, 3, .. etc.
This provides us with the length of the hypotenuse.
[table]
Leg-1(a),Leg-2(b),Hypotenuse(c)
1,1,?
1,2,?
1,3,?
1,4,?
[/table]
Let’s set out to work!
For row 1 – Leg two equals Leg one Equals 1
This is the case for a right isosceles triangle.
The right isosceles triangle is shown. Reflect it twice, to the horizontal leg and respectively
to vertical leg, as line symmetry. Then rotate the original triangle
around the right-angle corner for 180 degrees. In such way we obtain three new
triangles. The new three and the original one together form one square.
Since the original triangle has an area of (1/2) × 1 × 1 = (1/2), the four triangles have a total area of
4 × (1/2) = 2. While they together form one square with a side to be decided; let us suppose it to be y.
Then the area of square must equal 2. So y2 = 2.
We will simply write for this case as Leg one = 1, Leg two = 1, and the Hypotenuse is sqrt 2.
Ribbon Square is Fun
Have you ever tried to use ribbons crossing each other to enclose a square? And do it in a rectangle “something”, like a pool in fitness center?
An external link plays with fun on this.
It answers some of these questions:
What is the largest ‘ribbon square’ you can make? And the smallest? How many different squares can you make altogether?