## Friday, June 29, 2012

## Thursday, June 28, 2012

### An Interesting link

I found this link when I was searching for something.

http://www.cmi.ac.in/~vipul/olymp_resources/preparingforolympiads.pdf

http://www.cmi.ac.in/~vipul/olymp_resources/preparingforolympiads.pdf

### Justifying the cost of shipping

Today I shipped some 26 kg of books to Bogota through DHL and it costed me $980 considering the cost of all the books is less than $300 because they are mostly printed from the DJVU or pdf files. I was left wondering if this was a safe decision. I think it was because to me a hard copy gives inspiration to dwel on anytime.

## Wednesday, June 27, 2012

### 2nd last day at work

Today is my 2nd last day at work here. It was a good experience. I learned a lot and when I look back I can look with satisfaction about becoming a better student and teacher over the past one year. I had the opportunity to visit Angkor Wat few times. I explored some of the great places and cities in Cambodia and Vietnam and now its time to say good bye to these amazing people.

## Monday, June 25, 2012

## Wednesday, June 20, 2012

### Is there only 1 unique triangle with given medians of triangle ? Ans: No

Just to convince myself that the order of construction doesn't matter. I drew the original triangle by extending different segement. So I have shown here all three triangles. I had to hide previous construction except the original 1/3rd median triangle (red triangle)and being anew for each original triangle.

### Drawing triangle when length of all 3 Medians of a triangle are given

To draw a triangle when medians are given. We first draw a triangle using 1/3rd of medians. Now we take one side of this triangle and extend on both sides by the same side to get the proper length of the median of original triangle. Now at the extended end we join with the remaining point of our 1/3rd triangle. The twice of this segment will give us one side of the triangle and the 2nd vertex of the original triangle. Now use this and the original extended point to draw a segment and twice of that segment will give the 2nd side of the triangle. Now the remaining two points will determine the third side.

The drawing shows the red triangle as the triangle which is formed by 1/3rd the sides of median. The blue triangle is the original triangle. We verify by measuring the length of median of the triangle.

## Monday, June 18, 2012

### 8 point Circle

While 9 point Circle is a celebrated idea and has been known for quite some time. The 8 point circle is less than 70 years old. To draw this we need a quadrilateral whose diagonals are mutually perpendicular. The 4 points are the mid points of the quadrilateral and the other 4 points are obtained by drawing perpendiculars from these mid points to opposite sides. Notice that all eight points lie on our special quadrilateral.

## Sunday, June 17, 2012

### Bhat Kot and Pandu Kholi Trip

## Friday, June 15, 2012

### Awesome Tim Parli

As you travel you meet incredible people everywhere. One of my good friend and a person very dear to me is Tim Parli. When I lived on Carico St. He was my neighbor. So why Tim is so incredible ? I think it is his genuine kindness and goodness. He walks his talk. He is a believer with no pretense or hubris about it. He is one of those person who will not come to your rescue at a moment's notice if he knows that you require his help. I am fortunate enough to have watched him closely and have spent many wonderful moments. I met him through small group I joined at Joshua and Rebekkah. I can recount many stories about how wonderful and helpful Tim is. When I moved from Southern hill apts to Shamrock apts. Tim was there along with Joshua to move my stuff. I had rack full of books and I still vividly recall how his old car sunk over 20 cm with the load. It was almost like carrying bricks and he laughed about it. Another time was when I just joined John A Logan college and my car was wrecked in the Inland storm. He gave me a ride all the way to DuQuoin. If you think that is generous. He sold me his Camry Car for a lot less price he could have earned and it was in excellent condition. There were umpteen number of times I had meal with them or given ride to his friends. You have to be really lucky to bump into such people and I am proud to call him my friend. Tim you are awesome.

## Wednesday, June 13, 2012

### Drawing an inscribed circle bounded between two arcs and a line

This is a fun problem which requires a little bit of algebra to analyze it to find the center of the inscribed circle. If the arcs of original circles is a then the center of the circle is at (a/2,3a/8). As should be evident from the figure that I used eight circles to achieve that division and than used 3 times of that to get the length 3/8a. After that I used the vector function to place that point on the perpendicular bisector.

### Line Division in Euclidean Geometry

One of the earliest problem when teaching euclidean geometry using just ruler and compass is how to divide a line into n equal parts using only straight edge and compass. The following snapshots using KSEG illustrates this idea.

It makes it clear that it doesn't matter what is the side of the division we use on the line. Also the helping line on which divisions are made can be at any angle other than 180 and 360 degrees.

## Monday, June 11, 2012

### Finding three equal segment on two intersecting lines

This is a nice problem I came across. The problem is about two intersecting lines and we have to find two points on these intersecting lines such that the resulting three segments are equal. The idea is to start backward. Think that the problem is already solved and one can see that if we can construct a parallelogram than we can find the solution. To find the parallelogram one needs to use the idea of intercept theorem. We begin with drawing a small isosceles triangle (CDF) and then we we draw a line of equal length (FH) on the side joining CA and then we make use of properties of parallelogram to find the length AJ and then JK is drawn parallel as shown in the figure below

### Finding the side of equilateral triangle when the distances from a point inside the triangle is given

Given a point inside an equilateral triangle and its distance from the three vertices. How do you find the side of the equilateral triangle ?

To find the length of the equilateral triangle

1. Construct the triangle with the sides (the distance from the vertex)

2. Draw an equilateral triangle on one of the side of this triangle. This forms a quadrilateral.

3. Measure the diagonal of this quadrilateral and that will be the side of the original equilateral triangle

I have used KSEG here to draw an equilateral triangle and chosen a point. Now I have also drawn the triangle on the same using vector function

To find the length of the equilateral triangle

1. Construct the triangle with the sides (the distance from the vertex)

2. Draw an equilateral triangle on one of the side of this triangle. This forms a quadrilateral.

3. Measure the diagonal of this quadrilateral and that will be the side of the original equilateral triangle

I have used KSEG here to draw an equilateral triangle and chosen a point. Now I have also drawn the triangle on the same using vector function

### Viviani Theorem

One of the well know theorem about equilateral triangle is the Viviani's Theorem. It states that given an equilateral triangle and any random point inside. Then sum of the distance of the perpendiculars drawn from this point on the three sides is equal to the length of the altitude of the triangle. Here are some snapshots of the theorem using KSEG. The proof of the theorem is real simple. From E if we draw join the vertices of the original triangle we get three triangles. The area of these is 1/2 base * altitude. Let the sides be a unit each (recall its an equilateral triangle) then we have

1/2*a*EG+1/2*a*EH+1/2*a*EF=1/2*a*AI

EG+EH+EF=AL

Q.E.D

## Wednesday, June 06, 2012

### Fermat point or Torricelli point

One of the fun topic I taught in last couple of days is the Fermat point and Torricelli point. This is a point inside a triangle such that its total distance from all three vertices is minimum. We sometimes also define this point as one which makes an angle of 120 degree with the sides of the triangle.

To find this point

1. Construct equilateral triangle on the sides of the original triangle.

2. Draw circumcircle of these equilateral triangles.

3. The point where these two circles intersect (ofcourse other than the common vertex) will be the Fermat point.

In my drawing that point is Q and D is a random point and you should observer that when D coincide with Q we have the least sum. Also note the angles

To find this point

1. Construct equilateral triangle on the sides of the original triangle.

2. Draw circumcircle of these equilateral triangles.

3. The point where these two circles intersect (ofcourse other than the common vertex) will be the Fermat point.

In my drawing that point is Q and D is a random point and you should observer that when D coincide with Q we have the least sum. Also note the angles

### Constructing a regular Pentagon

There are many methods to construct a regular pentagon. Here is one method I saw on a website.

These are the sequence of steps

1. Draw a circle and its diameter

2. Draw a perpendicular to its diameter

3. The perpendicular will intersect the circle at two points, find the middle point of the radius on this

perpendicular and draw a circle.

4. Now join a line through foot of the diameter to the center of the circle and let it intersect the circle at two points.

5. Use foot of the diameter and the upper part where it crosses the circle as the radius and draw an arc which crosses the original big circle at two points. The top point of diameter and the crossing on circle gives you the measure of the side of the pentagon. Use this to find other sides using compass.

I did these steps using KSEG and also measured the sides and it came out perfect

These are the sequence of steps

1. Draw a circle and its diameter

2. Draw a perpendicular to its diameter

3. The perpendicular will intersect the circle at two points, find the middle point of the radius on this

perpendicular and draw a circle.

4. Now join a line through foot of the diameter to the center of the circle and let it intersect the circle at two points.

5. Use foot of the diameter and the upper part where it crosses the circle as the radius and draw an arc which crosses the original big circle at two points. The top point of diameter and the crossing on circle gives you the measure of the side of the pentagon. Use this to find other sides using compass.

I did these steps using KSEG and also measured the sides and it came out perfect