Friday, June 29, 2012

Obama Care and Republicans

With passing of Obama Care, Obama has acheived the second greatest success in his political carrier. His other notable acheivements would ofcourse are getting  Bin Laden, Bringing troops back home, killing of significant number of jihadis, amnesty for latinos who behaved well and were under certain age bracket. His support behind Gay rights. Not engaging in Egypt, Syria and pulling out from Iraq and now Afghanistan. World has changed a lot. Now Romney is asking for the repeal of this historic bill. Mind you this bill is of same historical significance as women right,segregation bill, ease of immigration for asian people and it would be preposterous if republicans could garner enough vote to turn it. It will be the epitome of how corporate world can blind the eye of gullible people. Having lived for so long in United States among the conservative people I can vouch that they are some of the nicest people who meant no harm. They are just blinded by their cunning pastors and corporate czars. I would say they are pretty much like slaves to these rich people who use them for their own selfish agenda by playing with their emotions. If Americans elect Romney they would be butt of joke in the whole world and time and again the mass hysteria has been able to acheive things (extermination of jews by gullible nazis, khmer rouge in Cambodia) which totally goes against any reason. I love America because I have learned so much there and it opened my eyes to so many things, friendships, generosity of people. Its one place which has no comparision. I wish all the best for you my friends over there.

Thursday, June 28, 2012

An Interesting link

I found this link when I was searching for something.

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

17 Sided Polygon

One of the big achievement in Math is the Gauss's finding that we can inscribe a regular 17 sided polygon inside a circle. Here are couple of those I drew using Kseg and Kig and my student Sok who drew it using compass and scale right in class.

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

 Last year in March 2011 I went to Pandu Kholi with the people I was working with. We left for Dunagiri temple and from there you can go a couple of more kilometers to Kukuchina. The trek for Pandu Kholi begins there. Its only a couple of hr trek and is not so bad. Recently I went to Cat Ba island in Vietnam and the trek there though only 1 hr long was excruciating because of the stones which are hard as iron and the humidity. Pandu Kholi is at an elevation of 2800 ft whereas Rankihet is at 1800 meters. So its almost a Kilometer up in the sky. I wanted to go to Pinath and Bhat Kot but this didn't happen. Here is a memorable pic of me at Pandu Kholi and Bhat Kot mountain is in the background. Reaching Bhat Kot is not so simple as it is less traveled and the danger of encountering a live bear. Talking of live bear. I had the fortune of watching one on trek to Kafni. That experience has etched in my memory forever.

Heptagon construction

If you think all regular polygons can be circumscribed think again. The very first one to buckle this trend is heptagon. Wikipedia has an animated gif expounding its construction. I used KSEG to draw this and measured the sides of this approximate regular heptagon.

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

Pascal triangle and counting regions

One of the well know sequence is 2,4,This depicts how one can count all the regions. There is another way one can count the regions is by counting the number of lines, number of points of intersection and adding 1 to that

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

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

Wednesday, June 06, 2012

My year 10 students

I had these students for 2011-12, year 10 maths. From left its Dhalis, Sonika, Sok, Jessica, Joseph and Chris. It was a fun class to teach. They all responded very well to the task and I have faith they are on track to enjoy more maths.

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

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

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