## 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)

Theorem. (Cosine Law)

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

Define be the area of the triangle ABC.

Now notice that

Since it is right angled triangle, , and therefore

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 be a point on the parabola , are points on the same parabola such that . Let G be the centroid of the triangle PQR. Let angle QPR be .

Source: Tokyo University Entrance Exam (modified)

a) Find the locus of G in parametric form.

b) Find in terms of the parameter used.

c) By (b) or otherwise, show that .

By equating , , then we have .

Moving all terms to the left and do some factorization we have . Due to the symmetricities of the parabola AND the insymmetricities between Q and R, we have , therefore .

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 , then .

Now we deal with the coordinate of G:

By substituting the value of P we have the constrain , and that gives us the parametric equation desired.

In the above expression we can clearly see that 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 where .

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 and . 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 and .

Now . It is not suprising that it is negative when t is large because they are positive-negative pairs for enough big t.

which is a bit messy. After some simplification we can get

Now we can compute the value questioned in part b.

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 , . Consider trigonometric functinos, . To make life easier we call if .

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 , i.e., exist M and k so that for all we have , 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 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, and 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: on the same parabola. Then and , and instead of ! 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 , express your answer in 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