Saturday, 1 March 2014

Triangle centres and choice of coordination

Recently I have a few tutorial sessions with others on NSS maths and olympiad maths and there are a few questions on geometry which is not very easy to be solved by analytical method. With or without extra restrictions we may be able to solve them in coordinate system, but sometimes we don't know where should we start our calculation. What should we do to make these calculations much simplified?

Before doing this we can review a few operations in coordinate geometry in senior secondary school level.

1) Straight lines, in general form $y = mx+b$, or point slope form $(y-y_0) = m(x-x_0)$.

For parallel translation consider $y= mx + b + \epsilon $ or $(y- y_0- \epsilon _y ) = m(x-x_0 - \epsilon)$.

For perpendicular lines consider lines with slope $-m^{-1}$.

2) Centres of triangle. They are discussed in the previous entry.

3) Locus. Common locus includes straight line(s), circle and conic sections. The standard form can be generalized by $a(x-x_0)^2 + b(y-y_0)^2 = r^2$.

Mathematics is consistent (at this level) in the sense that using different methods will give you the same answer. However there are many approaches, in particular there are many method to define a geometry problem in the coordinate plane.

To define a problem in the coordination system first we define the coordinate system to be used. For 2D geometry the three choices are Cartesian, polar and Argand coordinate. Cartesian is the obvious choice as polar produces ALOT of trouble when you operate points besides the origin (we will see this later), and argand deals more with dynamic stuffs like transitions, but we don't want things to move. In the static case Argand is similar to Cartesian but the later can be much easier.

The second step is to put the points into the coordinate. You can rotate and transform freely as long as this does not violate the original given information. A very common transformation is to put two points of the triangle at (0,0) and (0,1), then the last point can be placed at $(x,y) \in {\mathbb{R}^+}^2$. (This could lead to a paradox, but this is beyond our scope.)

The third step is to deduce the transformed relation in the new coordinate, which is technically the most complicated part. You will find angle bisector giving you so much pain. Consider the following example.

Example 1. Given that $\tan (\theta ) = k$. Show that $\tan (\frac{\theta}{2}) = \frac{k}{\sqrt{1+k^2}+1}$.

This can be performed on a right-angled triangle, then the general case immediately follows. (why?)

Consider three points $A(0,0)$, $B(1,0)$ and $C(0,k)$ (since this is a right-angled triangle) and $D$ be the point of intersection of $AC$ and the angle bisector of $B$.

By angle bisection theorem, $\frac{DA}{DC} = \frac{BA}{BC}$ so that $DA = \frac{k}{\sqrt{1+k^2}+1}$ and the result follows.

Example 2. Verify that $r = \frac{a+b-c}{2}$ where $a,b,c,r$ is the sidelengths and inradius respectively if (and in fact, only if), $ABC$ is a right-angled triangle.

We had a proof using Pyth. them to get the result directly, in a much easier way, but this time we try to use a purely coordinate-geom approach.

Here we take the same set up with example 1. Then the angle bisector of $A$ is given by $y=x$ (if the triangle is not right-angled the formulation of this one will be complicated too). The slope $m_{CB} = -k$ and the angle bisector of $B$ has slope $m_{B'} = \frac{-k}{\sqrt{1+k^2}+1}$. Then we have two equations to find the incenter (two is enough, assuming the fact that angle bisectors are concurent): $y = x$ and $y = m_{B'}(x-1)$, which gives us $y = x = r = \frac{k}{1+k+\sqrt{1+k^2}}$. $r=y$ because the foot from the incenter to $AB$ is vertical.

Notice that all right angled triangle can be transformed into the above form, then $r =  \frac{k}{1+k+\sqrt{1+k^2}} = \frac{1+k-\sqrt{1+k^2}}{2}$ can be shown by rationalization.

When you have an isosceles triangle you may want it to be symmetrical in your coordinate system.

Example 3. (PCIMC 2014 Heat SS Q17) In triangle ABC, $AB=AC$. $D$ is on $AB$ so that $AD:DB = 2:5$. Extend $BC$ to $E$ so that $DE = EC$. If the area of triangle DEB is 10, find the area of triangle ABC.

Note that in this question transformation does not change the given information. Notice that there is a ratio $2:5$ so we may want to put $A$ at $(0,7)$ and do the rest of the calculations. However this question can be solved easily with rotation only (rotation do not alter ANY information, unless they give you a pre-assigned coordinate with slopes, etc.). Put $A$ at $(0,a)$ with $C(0,x)$, $B(0,-x)$. Then $D$ is at $(-\frac{2}{7}x, \frac{5}{7}a)$ (why?) and by symmetry $E$ is at $(\frac{-11}{7}x,0)$.

Now consider the area of triangle DEB we have $\frac{1}{2} \frac{4x}{7} \frac{5a}{7} = 10$, which gives us the area of triangle ABC $\frac{(2x)(a)}{2} = 49$.

Readers may want to try the case when $a$ is fixed. Now we perform one classic example: the collinearity of centres of triangle. It may look hard but this is not that hard as one imagined as: (1) it does not involve the stupid incenter, with all those complicated trigonometric functions (2) we use a nice coordination to make calculation easier.

By arbitrary transformation consider the triangle $A(0,0)$, $B(1,0)$ and $C(a,b)$ at the first quadrant.

The centroid is easy: $G(\frac{a+1}{3}, \frac{b}{3})$.

For the orthocenter, notice that $m_{BC} = \frac{y}{x-1}$ so that the altitude from $A$ to $BC$ has slope $\frac{1-x}{y}$. Moreover the altitude from $C$ to $AB$ is vertical so that we have two equations: $x=a$ and $y = \frac{1-a}{b}x = \frac{a(1-a)}{b}$ so that the orthocenter is at $H(a,\frac{a(1-a)}{b})$.

For the circumcenter, note that the perpendicular bisector of $AB$ is $x=\frac{1}{2}$. Now the perpendicular bisector of $BC$ has equation $y = \frac{b}{2} + \frac{1-a}{b}(x-\frac{a+1}{2})$. Solving the equation we have the circumcenter $O(\frac{1}{2}, \frac{b^2 - a(1-a)}{2b})$.

The coordinate of the circumcenter may not look good, but we can verify that the slope of the three points are equal to $\frac{-3a^2+3a-b^2}{(2a-1)b}$ so that they are collinear. Are there any better coordination? Please don't hesitate to tell me.

Food for thoughts:
1. (PCIMC 2014 Heat SS Q12) In triangle ABC, D is the mid-point of BC. E and F are on AB, AC respectively so that BC//EF. If $S_{AEF}:S_{DEF} = 7:3$ (area ratio) and $S_{ABC} = 100$, find $S_{DFC}$.

2. (PCIMC 2013 Final SS Q14) For a kite ABCD, $AB = AD = 50$, $CB = CD = 75$. E,F,G,H are on AB, BC, CD, DA respectively so that EFGH is a square. Moreover let angle $ABC, ADC$ be right angle and EF//AC. Find the area of the square.

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