Saturday, February 26, 2011

Interesting Facts about Triangles - Quantitative Relations

In the January Metroplex Math Circle presentation, the speaker also showed the equation to calculate the area (A) of a triangle from the length of its sides a, b, c. It is simple, clean and frankly quite elegant:

A = sqrt(s (s-a) (s-b) (s-c)), where s= 0.5 (a+b+c) half perimter of the triangle.

The derivation of the formula is not hard at all. Continuing from my proof that the altitudes of a triangle is concurent (orthocenetr):



m^2+ h^2 = c^2

so h^2 = 0.5[ c^2+ b^2 - 0.5h (a^2 +((c^2-b^2)/a)^2)]

A = 0.5 h a =0.5 a sqrt{0.5[ c^2+ b^2 - 0.5h (a^2 +((c^2-b^2)/a)^2)]}

using the relation (a+b+c)^2 = a^2+b^2+c^2+2ab + 2bc + 2ca, the above equation is simplified as

A =sqrt(s (s-a) (s-b) (s-c))

or h= 2 A/a = 2 sqrt(s (s-a) (s-b) (s-c))/a

Radii of circumcircle, 9 point circle and incircle

I showed in part 1, that radius of circumcircle R = b c /(2h), so

R= abc/(4 sqrt(s (s-a) (s-b) (s-c))) = abc/(4A).

The radius of 9 point circle R9 = 0.5 R

R9= 0.5 R = abc/(8 sqrt(s (s-a) (s-b) (s-c))) = abc/(8A)

The radius of inccirlce r, and it is easy to show that A = 0.5 s

r = 2A/s = 2 sqrt(s (s-a) (s-b) (s-c)))/s

All quatities are uniquely determined by the sides of the triangle, and the formulas are all clean and simple! There are many more interesting relations exist in a triangle - as a saying goes - the more you know, the more you don't know!

Exploration is fun.

Saturday, February 19, 2011

Pictures of Snow Days

Heavy snow in the night of February 3rd foiled my plan to go to office the next day.

But the road condition was actually getting better - it was not as slippery as previous several days! The Sun came out Friday (Feb. 4) afternoon. The boys were allowed to go out play. Before dinner, Lily and I went out as well for a stroll in the community.
On Saturday morning, I got up ~ 7:30am. With a camera in hand, I went out to enjoy what nature offered: clear icicles, white snow, blue sky, bright sunshine and chirping birds.

What shown below are the pictures I took Friday (Feb. 3) afternoon and Saturday (Feb. 4) morning.
Playing in the Snow

Nick enjoyed snow

Lily's collection of icicles from cars


See the very long icicle?

first ray of sunshine in 3 days

A bright Saturday morning

Street Scenes - 1

The morning of 2/4/2011 - I could not see the road

A snowy Texas

a bit color in black, white and grey


Gravity and Heat at work

Snow covered street

A snow capped bird nest

Street Scenes - 2

The front yard of a Steelers' fan

footprints and bushes

Snow siblings

In the cold

a yellow bench

Enjoying the solitude


A cardinal

an orange breasted finch

A bird in my front yard Saturday morning

Saturday, February 12, 2011

February Freeze

When it was forecasted that there would be a strong Arctic blast coming to North Texas on February 1, I did not pay much attention, and expected a one day inconvenience. It turned out to be a 4 day spell of ice and extreme cold.

Ice Covered

We were awakened midnight by thunder on Monday night - January 31st, and it was raining, right before the strong cold front reached North Texas. The next morning, when I looked outside, the street in front our house was covered with ~ 1 inch thick ice, my north facing patio was covered by even thicker ice! One can skate in the streets if he inclines to. In fact, TV news showed video of a girl skating in her street in a neighbouring city!

It was frozen everywhere. As I tried to open our garage door, the garage door motor had a real hard time to lift it open because the garage door was frozen to the floor. The melted snow from the roof was quickly frozen into icicles large and small while dripping down. Of course the pond in the park was frozen too.

Frozen In

With all the ice on the roads, and constant TV reports of traffic accidents everywhere, we did not go anywhere, not even step out of the front door for the first day of the freeze. I had to go out the second day since a van skidded into our yard from the T intersection by our house!! We quickly spread salt and potting soil at the intersection as well as the alley way entrance to help cars to maneuver at the intersection. Per Justin's suggestion, I sent out recommendations to Crime Watch captains about spread salt/sand/soil into the slippery streets in our community. I also inquired with our HOA board president about hiring someone to sand our streets.

We were trapped inside by ice and the boys were trapped by computer games. We did a few things to relax from work (at home) and to get kids off the computers.

Removing Ice on Drive Way

The first two days, we went out to clean our driveway 30 to 45 minutes at a time, during our morning and afternoon breaks, hoping to go to work the next day. We used a shovel, and a digging hoe to remove ice, but it was too hard and thick. Justin and I were tired quickly. Instead of clean the whole driveway, we dug four troughs.

Dodge Ball

It was too cold and slippery. We had to limit the the time that the boys could be outside. So Lily introduced "dodge ball" game to them. She sewed four small bean bags as the "balls". The game was simple, two player spaced out, each has two bean bags, trying to hit the other with the ball while dodging the ball from his/her opponent.

First Lily and Nick played. To avoid the balls from his Mom, Nick ran, jumped up, twisted, or just dug like an ostrich. His postures made mom laugh. I could hear Nick shout in excitement when he hit mom. Justin was quickly attracted into the game. For a while, the boys played dodge ball whenever they were asked to time out from computers. Their laughing, jumping and excitement were really contagious. I eventually agreed to play dodge ball with Nick when he could not find others to play with him.

Dodge ball made the dull, uneventful snow days a bit exciting!

To make the most of the snow days, we also had some other quieter family activities. One night we had a reading by the fireplace. On Saturday night, we had a family Wii tournament. The time flied as we had fun together.

So one upside of the February Freeze was that it allowed us more family time.

Saturday, February 5, 2011

Interesting Facts about Triangles

Last Saturday I took Nicholas to a Metroplex Math Circle presentation about “Important points in a triangle: centroid, orthocenter, incenter, circumcenter, …” by a TI computer scientist - Branislav Kisańćanin.

The presentation

The concept is simple and straight forward.

Centroid - the intersection of medians ( a line between vertex and middle point of the opposite side) of a triangle, orthocenter - the intersection of the altitudes (heights), incenter - the intersection of angle bisectors, and circumcenter - the intersection of perpendicular bisectors.

Then he raised the question: How do you prove that the medians, altitudes, angle bisectors and perpendicular bisectors are concurrent? The proof for medians -centroid, angle bisector - incenter, and angle bisectors - incenter, and perpendicular bisectors - circumcenters are not hard; but he used the seldom used Ceva theorem to prove the orthocenter.

I was literally amazed when he mentioned that centroid, circumcenter and orthocenter are collinear, and the line they are on is called Euler Line. And there is a nine point circle - which pass the middle point of each side, and foot of each altitude and mid-point between orthocenter and each vertex

The speaker let students participated in the process of his presentation; many students - either answered questions or went to the white board to give a proof. The interactive approach really helped keep students' focused and interested in the subject. One time when he asked an alternate proof to his proof that the medians divided a triangle into six smaller triangle which have equal area, no students could provide one, so he asked parents to help - I volunteered an alternate proof :). Near the end of his presentation, the speaker somehow stuck on a proof, he remembered what to do, but he could not make a logic derivation of the result he wanted. I volunteered again to help him figure things out, and I provided an alternate proof one more time!

Nicholas was very much interested in the presentation until the last quarter of the presentation, when methods used to solve some problems involved trigonometry.

The interesting facts and my proofs

Due to time limit, the speaker did not show the proof of Euler line and 9-point circle. Fascinated by these facts, I decided to try my capability of geometry proof. I was mentally as excited as when I was at the peak of my Ph.D. thesis research! I spent several hours on Sunday and proved that 1) altitudes are concurrent witout using Ceva's theorem; 2) Euler line: centroid, orthorcenter and circumcenter are collinear; and 3) nine point circle: the middle point of each side, and foot of each altitude and mid-point between orthocenter and each vertex are on the same circle.

1) Altitudes are concurrent

Assume altitude AA' and CC' intersect at O, and OA'=X, then we have

m/h=X/n, thus X = mn/h

Similarly assume that AA' and BB' intersects at O', O'A' =Y

n/h= Y/m, thus Y=mn/h = X. i.e. A'O=A'O'.So O and O' coincides, altitudes are concurrent.

Note: this is much easier and simpler than using Ceva's theorem

2) Euler Line

I used algebra to prove the Euler Line. A triangle is placed in a Cartesian coordinate as shown.

Write out the linear equation for two altitudes, one can solve for the orthocenter, which is at
(0, mn/h) . Similarly the centroid is at ((n-m)/3, h/3) and the circumcenter is located at
(0.5(n-m), (h^2-mn)/(2h)).

The line passing orthocenter and centroid is

y-mn/h = (h-3mn/h)/(n-m) x.

It is easily verified that the circumcenter is on this line. So orthorcenter, centroid and circumcenter are on the same line - Euler line.

Note: it is easy to determine that the radius of circumcircle is AB*AC/(2h)

3) 9 - point circle

Using the same coordinates to define a triangle as before, it is found the that the center of the circle that passes the mid-points of each side of the triangle is at ((n-m)/4, (h+mn/h)/4). The radius of this circle is sqrt{[(m-n)/4]^2+[(h+mn/h)/4]^2} = AB*AC/(4h).

It turns out that the distance between this center and the foot of a altitude (0,0) is the same as above. Similarly one can prove that the feet of other two altitudes are on the circle as well.

The remaining three points are mid-points between orthocenter and each vertex. One such point is on y axis, (0, 0.5(h+mn/h)). Its distance to the center of the circle is exactly the radius!

So all nine points are on the same circle - illustration of a nine point circle is shown below (chart from

9 - point circle has additional interesting properties: it is tangent to incircle, its center is on Euler line and is midway between Orthocenter and Circumcenter.

In additional to these interesting geometric relations, there are also many elegant and clean quantitative relations, e.g. the radius of 9-point circle is half of that of circumcircle (which has been shown above).

(to be continued ...)