## Friday, 11 May 2012

### M2: Algebra and geometry Part 1

In HKCEE A-maths or HKDSE M2 paper, the topic "rate of change" as the application of differentiation always appear in the paper and sometimes it has been modelled into a geometric question like the velocity of a certain position of a pivot. What I want to illustrate is that geometric question can be modelled into a algebra question in many ways like vectors, coordinate geometry, and this time we can deal with calculus.

With the assistance of trigonometry we built up a strong relationship between segment lengths and angles, allowing us to relate different quantity, then the rate of change of a certain quantity can be transformed easily. We will demonstrate four questions, one unrelated to calculus, one A-math, one M2 and one Tokyo U questions, showing that differentiation can deal with meaningful conclusion instead of simply finding some derivatives.

Recall: Sine law and Cosine law.

Theorem. (Sine Law) $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

Theorem. (Cosine Law) $c^2=a^2+b^2-2ab\cos C$

Question 1. (Incircle) Let a,b,c be three sides of right-angled triangle ABC where c is the hyp. side. Show the incircle radius $r = \frac{a+b-c}{2}$

Define $S_{ABC}$ be the area of the triangle ABC.

Now notice that $S_{ABC}=rs=r\frac{a+b+c}{2}=\frac{ab}{2}$

$r=\frac{ab}{a+b+c}=\frac{ab(a+b-c)}{(a+b+c)(a+b-c)}=\frac{ab(a+b-c)}{(a+b)^2-c^2}$

Since it is right angled triangle, $a^2+b^2=c^2$, and therefore $(a+b)^2-c^2=2ab$

$r=\frac{ab(a+b-c)}{2ab}=\frac{a+b-c}{2}$

Comparing with the x-y-z approach of incircle (Fact: AF = AG, CF = CE, BG = BE due to tangent properties or congruent triangle.) this is completed in a more algebrical way. (Using the x-y-z approach requires a cartesian plane which the right angled properties is shown in a geometrical simulation, but in this approach it is shown in a algebrical way).

Question 2. (Triangle on parabola) Let $P(\frac{1}{2},\frac{1}{4})$ be a point on the parabola $y=x^2$, $Q(\alpha ,\alpha ^2),R(\beta ,\beta ^2)$ are points on the same parabola such that $PQ=PR$. Let G be the centroid of the triangle PQR. Let angle QPR be $\theta$.

Source: Tokyo University Entrance Exam (modified)

a) Find the locus of G in parametric form.

b) Find $\frac{PG}{QR}$ in terms of the parameter used.

c) By (b) or otherwise, show that $\theta \in (0,\pi)$.

By equating $PQ^2=PR^2$, $(\alpha-\frac{1}{2})^2+(\alpha^2-\frac{1}{4})^2=(\beta -\frac{1}{2})^2+(\beta ^2-\frac{1}{4})^2$, then we have $2\alpha ^4+\alpha ^2-2\alpha = 2\beta ^4 +\beta ^2 -2\beta$.

Moving all terms to the left and do some factorization we have $(\alpha -\beta )[(\alpha +\beta )(2\alpha ^2+2\beta ^2+1)-2]=0$. Due to the symmetricities of the parabola AND the insymmetricities between Q and R, we have $\alpha -\beta \neq 0$, therefore $(\alpha +\beta)(2\alpha ^2+2\beta ^2 +1)=2$.

Now it is a suitable time to parametize the equation. By simple trial and error, we know that the x-value of the locus tends to a certain finite value while the y-value converges. We want to make the point further from the origin for larger t. Therefore we let $2\alpha ^2+2\beta ^2+1=t$, then $\alpha +\beta =\frac{2}{t}$.

Now we deal with the coordinate of G:

$G_x = \frac{1}{3}(P_x+Q_x+R_x)=\frac{1}{3}(\alpha +\beta +\frac{1}{2})=\frac{2}{3t}+\frac{1}{6}$

$G_y = \frac{1}{3}(P_y+Q_y+R_y)=\frac{1}{3}(\alpha ^2+\beta ^2 +\frac{1}{4})=\frac{t}{6}-\frac{1}{12}$

By substituting the value of P we have the constrain $t\in (2,\infty)$, and that gives us the parametric equation desired.

In the above expression we can clearly see that $\alpha +\beta$ tends to 1/6 as t gets larger (which is not zero and is a bit out of expectation). If you want the explicit equation it is $y = \frac{2}{18x-3}-\frac{1}{12}$ where $x\in (\frac{1}{6},\frac{1}{2})$.

Now when we proceed to part b, there are two quantity that we avoided in part a must be solved in this part, they are $\alpha - \beta$ and $\alpha \beta$. In DSE cirriculum we mainly derive the symmetric functions by Viete's theorem (or known as relationship between sum and products) and now, we start with two symmetric functions $\alpha +\beta$ and $2\alpha ^2+2\beta ^2 +1$.

Now $\alpha \beta = \frac{1}{4}(2(\alpha +\beta)^2-2\alpha ^2-2\beta ^2-1)+\frac{1}{4}=\frac{1}{4}(2(\frac{2}{t})^2-t)+\frac{1}{4}=\frac{2}{t^2}-\frac{t}{4}+\frac{1}{4}$. It is not suprising that it is negative when t is large because they are positive-negative pairs for enough big t.

$\alpha - \beta=\sqrt{(\alpha +\beta)^2-4\alpha \beta}=\sqrt{\frac{4}{t^2}-4(\frac{2}{t^2}-\frac{t}{4}+\frac{1}{4})}=\sqrt{\frac{(t-2)(t^2+t+2)}{t^2}$

$PG=\sqrt{(G_x-\frac{1}{2})^2+(G_y-\frac{1}{4})^2}=\sqrt{(\frac{2}{3t}+\frac{1}{6}-\frac{1}{2})^2+(\frac{t}{6}-\frac{1}{12}-\frac{1}{4})^2$ which is a bit messy. After some simplification we can get $PG=\sqrt{\frac{(t-2)^2(t^2+4)}{36t^2}$

$QR=\sqrt{(\alpha -\beta)^2+(\alpha ^2-\beta ^2)^2}=(\alpha - \beta)\sqrt{1+(\alpha +\beta)^2}=\sqrt{\frac{(t-2)(t^2+t+2)(t^2+4)}{t^4}}$

Now we can compute the value questioned in part b. $\frac{PG}{QR}=\sqrt{\frac{(t-2)^2(t^2+4)}{36t^2}}\sqrt{\frac{t^4}{(t-2)(t^2+t+2)(t^2+4)}}=\sqrt{\frac{t^2(t-2)}{36(t^2+t+2)}}$

For part c, we need a bit basic geometry knowledge. The median PX (where X on QR) is the altitude, angle bisector and perpendicular bisector of the triangle as well because PQ=PR. Also, since PX is the median, we have PG = 2/3 (PX) by properties of median. Therefore $QX=\frac{QR}{2}$, $PX=\frac{3PG}{2}$. Consider trigonometric functinos, $\tan \frac{\theta}{2} = \frac{QX}{PX}=\frac{QR}{3PG}=2\sqrt{\frac{t^2+t+2}{t^2(t-2)}}$. To make life easier we call if $f(t)$.

The numerator of f is always positive, but the denominator tends to zero when it approach to 2 (which is the boundary case of the original figure), therefore f tends to infinity at t=2. Since $f(t)=O(\frac{1}{\sqrt{x}})$, i.e., exist M and k so that for all $t\geq k$ we have $|f(t)|\leq M\frac{1}{\sqrt{t}}$, by sandwich or whatever it tends to zero. Therefore f tends to zero when t tends to infinity. The function is everywhere continous at defined t so that taking arctan gives $\frac{\theta}{2}\in (0,\frac{\pi}{2})$ which finishes the proof.

Of course there are some basic approach for part c (directly leaving part a and b unanswered) since the result is a bit stupid. We can do the approximation that when t is large enough, $PG\approx \alpha ^2 \approx \beta ^2$ and $QR\approx |\alpha |+|\beta |$ which clearly leads to the limit when t tends to infinity. When t tends to 2, all points tends to point T, then QPR tends to be the tangent of the parabola, hence tending to 180 degrees.

I must admit that this is hardcore coordinate geometry but it's not the traditional pure maths questions (tangents involving conic sections), instead it emphasizes pretty much algebric skills in modelling a geometry problem.

Foods for thought. 1) (Rethinking parametization) When t is large enough, one suggests the approximation: $P(0,0),Q(t,t^2),R(-t,t^2)$ on the same parabola. Then $PG = \frac{2t^2}{3}$ and $QR=2t$, and $\frac{QX}{PX}=O(t^{-1})$ instead of $O(t^{-0.5})$! The above calculation is correct but the degree from approximation differs. What is happening?
2) (Generalization)
a) Instead of a fixed point P, find the locus of G if P is a varying point on the parabola $P(\gamma ,\gamma ^2)$, express your answer in $\gamma$ and t.
b) Do the same generalization for part b and finish the proof in part c. (Well, calculuating the stuffs in part b is extremely complicated and is a bit pointless, but it is a good arithmetic exercise.)

Algebra and Geometry Part II