Gravity Center(3)

The final part of explanation on the gravity center of a triangle.

Assume in the below figure that AL=LB and CN=NA.
AM is drawn to join A and G.
But BM:MC is still unknown.
We are going to prove that BM=MC.

We need a supplementary line also this time.
It is drawn from the midpoint L on AB, being parallel to AM.
Let J to denote the intersection of the line with BC.
That is, LJ//AM.

So it is obvious that BJ=JM, because BL=LA.

The we think of another triangle including LG and JM, that is, the triangle CLJ.

In the above figure, it has already been proved that CG:GL=2:1.
Then CM:MJ = 2:1 is derived, because GM, a part of AM, is parallel to LJ.

CM=2MJ
BM=BJ+MJ
and BJ=MJ

So BM=2MJ=CM

Thus it is proved that three medians cross each other with a single point G.

You might think it a meaningless trial to explain where such and such point would be located or what length a line would have. But to prove a geometrical property under a certain condition is to prove that you could control the geometry by fixing conditions. You could get the one and only center of gravity for a given triangle, and triangles are fundamental elements of all polygons.

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Gravity Center(2)

When you draw the line from A through G, the intersection of two medians from the other two vertices, the line will be the last midpoint…

To explain the above axiom, we will first explain that the intersection G of two medians BG:GN = CG:GL = 2:1.

If only the case of BG:GN is explained, the same explanation will be valid for the other.

For the discussion a supplementary line is needed. Draw it from the midpoint L to AC, parallel to BN. Let K to denote the intersection of the line and AC.

As AL=LB and LK//BN, AK=KN (This requires a proof, two, but I’d like to skip it, because it would be visibly understandable)

Then it is derived that CN=2NK=2KA, as N is the midpoint of AC.

And BN=2LK, as the triangle ALK is just 1/2 scale of the triangle ABN.

Let’s look at LK in a different triangle, CLK, including a smallaer triangle CGN.

As CN=2NK, CK=3NK.
As G is a point on BN, LK//GN. So the triangle CGN is 2/3 scale of the triangle CLK.
So, LK:GN=3:2.

As BN=2LK, BN:LK:GN=6:3:2.
As BG=BN-GN, BG:GN=4:2=2:1.

In this way, it was proved that BG:GN=2:1 .
It’s a crazy problem of geometry.

Gravity Center(1)

I have been working with a problem of explaining geometrical properties of triangles at school. Recently I met much complicated a problem of the gravity center. I would like to record what I learned, because someone (probably young people in high school) might need it.

The gravity center of a triangle is located at one and only intersection of all three medians. A median is the segment drawn from one vertex to the midpoint of its opposing side of the triangle. And a midpoint is the point on a given side of the triangle, which devides the side into two halves.
The gravity center, on the other hand, devides each medians at 2:1 ratio, the longer part being the side towards the vertex.

That is, in the figure below,
BM=MC
CN=NA
AL=LB
AG:GM=BG:GN=CG:GL=2:1

All three medians will intersect with one another at a unique point – You must prove it, however. It is obvious that two straight lines will have only one intersection, unless they are parallel. But there is no guarantee the third one should also cross the other two at the same point.

Then how can man prove it?
One of the ways is, express the assumption in another way.

Let G to denote the intersection of BN and CL, N and L being the midpoints of BC and CA, respectively.
If you draw a line from A through G to cross the BC. Then the intersection M will be the midpoint of BC.


Fisrt, forget the point M and think of only BN and CL.


Then draw a line from A through G…


That line will go through the midpoint of BC.

The above explanation is in itself complicated. It will need a preparatory explanation that the known medians BN and CL will divide each other at G, with BG:GN = CG:GL = 2:1. That will be discussed in the next post.

Items for School

One of the funs in studying is to use various tools for writing.
I often used a mechanical pencil when I was young, but now I’m fond of using a pencil.

This soft plastic grip helps me to hold a pencil tight, without pressing the wooden surface directly onto the finger. And the transparent color is beautiful.

Soft plastic grip for a pencil

A manual sharpener. I like it because it doesn’t make the tip too sharp. Of course, It’s so small that I can take it to the classroom.
Manual sharpener

I keep using each pencil, in an extender, until it becomes so short.
A short pencil in an extender

A small package of cocoa cakes with strawberry cream…
Seems like a cookie package

Indeed, it’s an eraser. It gives a sweet flavor of strawberry. Why this package simulates a cookie box? Because girls like sweets!…and I used to be a girl, too.

…but the quality as an eraser is not so good. The letters A and M are still visible. I could not use it in an examination. It’s just a toy.
Can't erase a writing so clean ...

Finally, a book band. All that I need at school is bundled with it.
A band to bundle everything

I remember in my schooldays, there was a boom for using this band. Students thought it cool to hang on the shoulder only a few books with this band, instead of carrying heavy bags. Teachers banned to the bands (banned the band…), because the corner of the books would collapse. But now I put the whole bundle in a bag and carry it. I just don’t want the each item to go somewhere.

Tomorrow I will have the last classes of the year. I’m glad now I’m pretty well to go out.

Compass and Ruler

I take an adult education in the university named Shonan Institute of Technology. It’s quite a specific system of this university. It is not a special course for adults – anyone who pays a certain charge can take the same lecture for the very students! I go there every monday, to take three classes. I wish I could take more, but I have so many other things to do.

One of the classes I take is geometry. This class is for students going to be math teachers. So the content itself is at high-school level, which they learn for much more detailed explanations. I myself don’t plan to be a teacher. I just want to learn the things again.

Today I spent much time to do the homework of geometry.

The below is an exercise to set a midpoint on each side of a triangle. Perhaps you remember your school days. This task can be done only by a compass and a straight ruler. Draw circles of the same radius at both edges of a segment as their centers, and joint the two intersections, which will cross with the segment at its midpoint.

Using compass to set the midpoint of a segment

Join each vertex and the midpoint of the opposite side. The resulting three lines will cross at a single point…oh…a little deviation.

The three lines should cross at one point...

Today’s homework was heavy. I had to write so much in the answer sheet. I wish the teacher gave us a sheet with more space.

More space in the answer sheet, sir!
Anyway it’s fun to learn again.

Matrix multiplication

For the dimension of 3 or more, multiplication process of matrices become complex. Think about multiplying two 3-by-3 matrices.
Multiplication of two 3-by3 matrice

People often say, it is convenient to divide the left Matrix in rows and the right in columns, to think of the procedure.
Left Matrix with rows, right with columns

This seemed to me difficult. Possibly I have a strong favor of seeing things from left to right.
People say up to down, I see always left to right

Then I had designed up my own way. I will divide both Matrices in each rows.
Thinking about only rows in both matrices

Denote each element of resulting Matrix as cij, i and j varying from 1 to 3.
Thinking about only rows in both matrices

For all cij, write out a calculation sheet with blanks to fill with aij or
bij at corresponding i and j.
Calculation sheet for all elements
In the sheet, there are two types of blanks: located at the right and the left to a “dot”, which is operator of multiplication.

Elements of the left Matrix, aij

Elements of the row aij are arranged in the right blanks to be summed up to c11.
Blanks at the right to dot

For all c1j the same arrangement of a11 to a13

All for c11 to c13 , elements a1j are used.

Then all c2j with a21 to a23, c3j with a31 to a33

All blanks right to dots

By these task, the rest blanks look like group of columns, don’t they? Now for me it’s much easier to arrange elements vertically.

Elements of the right Matrix, bij

Each b1j in the row is then written in a vertical arrangement of the first column-like blanks.
Since the rest blanks look like colums, it's easy to arrange the elements.

Then Each b2j in the second column. b3j to the third.
b2j to the second column, b3j to the third

Eventually, the calculated sheet is complete.
The calculation sheet complete

By this method now I scarcely make a mistake in calculation.
When I did this task, indeed, in the class, the professor looked at it and said “Good! Your work is so cautious!”
Yeah!

The last argument on isosceles

In the lecture of geometry, we had discussed various problems on proving properties of an isosceles triangle for a long time. In the last lecture we had its final argument.

(Today I was so tired of drawing pictures with a vector tool that I drew them by moving mouse on a bitmap tool).
The property to prove was the following:

If the two of three angles of a triangle are equal in measure, the triangle is an isosceles.

At first, we drew a figure:
If angle CAB=CBA then CA=CB

With the symbols above, we write the axiom and conclusion:

axiom
In the above triangle ABC, angle CAB = angle CBA

conclusion
CA=CB

[Failed Case 1] To set the midpoint M on the side AB and join M and C
With the midpoint M, you can't prove

Denote M on the side AB to be AM=BM, that is, M is the midpoint of AB.
Join C and M, then the triangle ABC is divided into two triangles, CAM and CBM.
If CAM and CBM are congruent, then CA would be equal to CB. Is it true?

AM=BM by the definition of M.
CM is identical.
But we can’t prove easily that the angle CMA would be equal to CMB.
So the proof in this way might be complicated.

[Failed Case 2] To draw the ray from C perpendicular to AB
From C to the side AB, can you really draw a perpendicular ray?

If you draw the ray from C perpendicular to the side AB….
No, it is not yet proved that you can really do that, because there might be a case that ray goes to the extension of the segment AB.
This case cannot be denied yet.

[The Forgiven Case] To draw the bisector of the angle ACB
The professor told finally, it is permitted to use without proof, to draw the bisector of the angle ACB and get its intersection on the side AB.
This would not require a proof.

Well, it was for me understandable. If the ray from C goes outside of the triangle, this ray cannot go inside the angle to divide it. That might not require a proof…
If the ray going outside, it cannot divide the angle

Therefore, denoting D as the intersection of the bisector of the angle ACB with the side AB, we could say,

AC=AC(identical)
angle ACD = angle BCD (the definition)
angle ADC = 180 degrees – angle ACD – angle CAD
= 180 degrees – angle BCD – angle CBA =angle BDC

Then we use the ASA rule to prove that the triangle ACD is congruent to the triangle BCD
So AC= BC

This problem was so difficult to me, because I had not been able to decide, which property was required a proof and which was not.

Therefore our discussion on an isosceles triangle ended.
Next time we will learn about a parallelogram.