## Coordinate system

### Line passing through an A.P.

B7, 2010

Let $$p_{1}, p_{2}, \ldots, p_{n}$$ and $$q_{1}, q_{2}, \ldots, q_{n}$$ be two different arithmetic progressions. Prove that the sequence of points $$S=\left(p_{1}, q_{1}\right),\left(p_{2}, q_{2}\right), \ldots,\left(p_{n}, q_{n}\right)$$ is collinear.

Solution

Let the common difference of the elements in the first AP and the second AP be $$h$$ and $$k$$, respectively.

Any line segment that connects two consecutive points has a slope equal to: $\frac{q_{i+1}-q_i}{p_{i+1}-p_i} = \frac{k}{h}$

The points in $$S$$ lie on the line that passes through $$\left(p_1,q_1\right)$$ with slope $$k/h$$. Therefore, all the points in $$S$$ must be collinear.

### Tangent to a cubic

A4, 2014

Find the slope of a line $$L$$ that satisfies both of the following properties: (i) L is tangent to the graph of $$y=x^{3}$$ (ii) L passes through the point $$(0,2000)$$.

Solution

The equation of the line is given by $$y=mx+2000$$. Suppose the line touches the graph at $$(x_0,{x_0}^{3})$$. The slope of the tangent at that point is $$y^{\prime} = 3{x_0}^2$$. \begin{align} y &= mx + 2000 \\ {x_0}^3 &= (3{x_0}^2)x_0 + 2000 \\ x_0 &= -10 \\ y^{\prime} &= 3{x_0}^{2} = 300 \end{align} This implies that the slope of the line $$L$$ is 300.

### Circles with Pythagoras

A6, 2015

Fill in the blanks. Let $$C_{1}$$ be the circle with center (-8,0) and radius $$6 .$$ Let $$C_{2}$$ be the circle with center (8,0) and radius 2. Given a point $$P$$ outside both circles, let $$\ell_{i}(P)$$ be the length of a tangent segment from $$P$$ to circle $$C_{i}$$. The locus of all points $$P$$ such that $$\ell_{1}(P)=3 \ell_{2}(P)$$ is a circle with radius $$\underline{\;\;}$$ and center at $$(\underline{\;\;}\;,\;\underline{\;\;})$$.

Solution

Center $$=(10,0),$$ radius $$=6$$. Using the distance formula and the Pythagorean theorem we get: $y^{2}+(x+8)^{2}-6^{2}=9\left(y^{2}+(x-8)^{2}-4\right)$ Simplifying gives $$y^{2}+(x-10)^{2}=6^{2}$$.

### Circle touching a parabola

B2, 2016

The region inside the parabola $$y=x^{2}$$ is the set of points $$(a, b)$$ such that $$b \geq a^{2}$$ We are interested in those circles all of whose points are in this region. A bubble at a point $$P$$ on the graph of $$y=x^{2}$$ is the largest such circle that contains $$P$$. Assume that there is a unique such circle at any given point on the parabola.

• (a) A bubble at some point on the parabola has radius $$1 .$$ Find the center of this bubble.
• (b) Find the radius of the smallest possible bubble at some point on the parabola. Justify.

Solution

A bubble at the point $$P=\left(a, a^{2}\right)$$ must be tangential to the parabola at $$\left(a, a^{2}\right)$$, since the circle must lie within the region of the parabola. The circle must also be symmetric with respect to $$y$$-axis, since the parabola is symmetric. So its center $$O$$ must be on the $$y$$-axis. See the figure below. The radius $$OP$$ of this bubble is perpendicular to the common tangent to the parabola and to the bubble at $$P$$. The slope of this tangent = $$2 a,$$ so the slope of radius $$O P=\frac{-1}{2 a}($$ for $$a \neq 0)$$.

Let $$Q=\left(0, a^{2}\right)$$. Using the $$\Delta O P Q$$, the slope of $$O P=\frac{-O Q}{a}=\frac{-1}{2 a}$$. Therefore $$O Q=\frac{1}{2},$$ regardless of the value of $$a$$

(a) We are given that the radius of the circle $$OP=1$$ cm. By Pythagoras: $OP^{2}=\left(\frac{1}{2}\right)^{2}+a^{2}=1$ So $$a^{2}=\frac{3}{4}$$ and $$O=\left(0, \frac{3}{4}+\frac{1}{2}\right)=\left(0, \frac{5}{4}\right)$$ Hence, the center of the bubble is at $$(0,5/4)$$.

(b) For any nonzero $$a$$, the radius of the bubble satisfies $$O P^{2}=\left(\frac{1}{2}\right)^{2}+a^{2},$$ so $$O P>\frac{1}{2}$$. The smallest bubble is at the origin and its radius is $$\frac{1}{2}$$.

### Cross-product of vectors

A8, 2021

Let $$\vec{v}_n$$ denote a vector for $$n\in \mathbb{N}$$. We define $$\vec{v}_{n+2} = \vec{v}_{n}\times \vec{v}_{n+1}$$, where $$\times$$ denotes the cross-product. Let $$\mathbb{0}$$ denote the zero vector. $$\vec{v}_{n}$$ maybe the zero vector too. Define sets $$S$$ and $$T$$ as follows: $S = \{ v_n : n\in \mathbb{N} \}$ $U = \{ \frac{\vec{v}_n}{|\vec{v}_n|} : n \in \mathbb{N} \text{ and } \vec{v}_n \neq \mathbb{0} \}$ (a) There exists $$\vec{v}_1$$ and $$\vec{v}_2$$ such that $$|S| =2$$.
(b) There exists $$\vec{v}_1$$ and $$\vec{v}_2$$ such that $$|S| =3$$.
(c) There exists $$\vec{v}_1$$ and $$\vec{v}_2$$ such that $$|S| =4$$.
(d) If $$\vec{v}_1$$ and $$\vec{v}_2$$ exist such that $$|S| = \infty$$, then $$|U|=\infty$$.

Solution (a) True. Pick $$\mathbb{1}$$ and $$\mathbb{0}$$. False, if $$v_1$$ and $$v_2$$ are non-zero.
(b) True. Pick $$\vec{v_1} = \vec{i}$$ and $$\vec{v_2}=\vec{j}$$.
(c) True. Pick $$\vec{v_1} = i+k$$ and $$\vec{v_2} = i$$.
(d) False. Pick $$\vec{v_1} = 2\vec{i}$$ and $$\vec{v_2}=3\vec{j}$$. $$|S|=\infty$$ but $$|U|=3$$.

### Vector perpendicular to a plane

A2, 2020

Let vectors $$\vec{a},\vec{b}$$ and $$\vec{c}$$ be defined as follows: \begin{align} \vec{a} &= 6i+6j+9k \\ \vec{b} &= 7i+8j+10k \\ \vec{c} &= 2i-3j+4k \end{align} Let $$P$$ be the plane containing vectors $$\vec{a}$$ and $$\vec{b}$$. Find a unit vector that lies in the plane $$P$$ and is perpendicular to vector $$\vec{c}$$.

Solution

We are looking for a vector $$\vec{u}$$ that can be written as $$t\vec{a}+\vec{b}$$ for some $$t$$. Since $$\vec{u}$$ is perpendicular to $$\vec{c}$$ we must have $$\vec{u}\cdot\vec{c} = 0$$.

\begin{align} 2(6+7t) - 3(6+8t) + 4(9+10t) &= 0 \\ t&=-1 \end{align} So $$\vec{u}=i+2j+k$$. A unit vector along $$\vec{u}$$ is $$\frac{1}{\sqrt{6}} (i+2j+k)$$.

### Find a curve given the tangent

B2, 2013

A curve $$C$$ has the property that the slope of the tangent at any given point $$(x, y)$$ on $$C$$ is $$\frac{x^{2}+y^{2}}{2 x y}$$

a) Find the general equation for such a curve. You may find it useful to substitute $$z=\frac{y}{x}$$.

b) Specify all possible shapes of the curves in this family. In particular, could an ellipse be one of the shapes?

Solution

The given property of the curve $$C$$ can be expressed as a differential equation:

$\frac{d y}{d x}=\frac{x^{2}+y^{2}}{2 x y}=\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}\right)$

It is convenient to let $$z=y / x,$$ so the equation becomes $$\frac{d y}{d x}=\frac{1}{2}\left(\frac{1}{z}+z\right).$$

To get this in terms of only $$x$$ and $$z,$$ differentiate $$z=y / x$$ with respect to $$x$$ to get: $\frac{d z}{d x}=\frac{1}{x} \frac{d y}{d x}-\frac{y}{x^{2}}=\frac{1}{x}\left(\frac{d y}{d x}-z\right)=\frac{1}{x}\left(\frac{1}{2}\left(\frac{1}{z}+z\right)-z\right)=\frac{1}{x} \frac{1-z^{2}}{2 z}$

Separating the variables and integrating, we get: \begin{align} \int \frac{d x}{x}&=\int \frac{2 z d z}{1-z^{2}}\\ \log |x|&=-\log \left|1-z^{2}\right|+ \text{some constant}\\ \log \left|1-z^{2}\right|&=-\log |x|+K=\log |x|^{-1}+K \\ 1-z^{2}&=\pm \frac{e^{K}}{x}=\frac{c}{x} \end{align}
where $$c$$ is a nonzero constant. Substituting $$z=y / x,$$ we get $$1-\frac{y^{2}}{x^{2}}=\frac{c}{x},$$ i.e., $$x^{2}-y^{2}=c x$$.

To get the shape of the curve, complete the square to get $$\left(x-\frac{c}{2}\right)^{2}-y^{2}=\frac{c^{2}}{4},$$ which is a hyperbola when $$c \neq 0$$. It could also represent two straight lines $$y=\pm x$$, when $$c=0$$.

### Mix a sin and a circle

B1, 2014

Find the area of the region in the $$X Y$$ plane consisting of all points in the set $\left\{(x, y) \mid x^{2}+y^{2} \leq 144 \text { and } \sin (2 x+3 y) \leq 0\right\}$

Solution

The area of the circular region $$S=\left\{(x, y) \mid x^{2}+y^{2} \leq 144\right\}$$ is $$144 \pi .$$ The condition $$\sin (2 x+3 y) \leq 0$$ is equivalent to $$2 x+3 y$$ being in one of the intervals $$[k \pi,(k+1) \pi],$$ where $$k$$ is an odd integer. The key point is that due to the symmetry of the circle $$S$$ about any diameter, in particular the diameter $$2 x+3 y=0,$$ the strip inside $$S$$ lying between the lines $$2 x+3 y=k \pi$$ and $$2 x+3 y=(k+1) \pi$$ is the mirror image of strip lying between the lines $$2 x+3 y=-k \pi$$ and $$2 x+3 y=-(k+1) \pi .$$ For each integer $$k,$$ precisely one of these two equal strips is included in the desired area. Thus the desired area is half that of $$S,$$ i.e., $$72 \pi$$

#### Reference

This problem appears in Chap.~2, Problem~7, 103 Problems in trigonometry Titu Andrescu.

### Intersecting planes

B2, 2017

Let $$L$$ be the line of intersection of the planes $$x+y=0$$ and $$y+z=0$$.

• (a) Write the vector equation of $$L$$, i.e., find $$(a, b, c)$$ and $$(p, q, r)$$ such that $$L=\{(a, b, c)+\lambda(p, q, r) \mid \lambda$$ is a real number. $$\}$$
• (b) Find the equation of a plane obtained by rotating $$x+y=0$$ about $$L$$ by $$45^{\circ}$$

Solution

Clearly the line $$L$$ passes through the origin. Moreover $$L$$ is in the direction perpendicular to the normals to the both the planes. The direction vector can be obtained by computing the following cross product: $(i+j) \times(j+\hat{k})=i-j+\hat{k}$ Hence $$L$$ can be written as $$L=\{(0,0,0)+\lambda(1,-1,1) \mid \lambda$$ is a real number $$\}$$. First, note that the equation of any plane that contains the line $$L$$ is given by $x+(1+\lambda) y+\lambda z=0$ Second, note that one can rotate the plane $$x+y=0$$ in either clockwise or in anticlockwise direction. Consequently there are two such planes. The normal of one of the planes makes an angle of $$45^{\circ}$$ with the normal of $$x+y=0$$ and the other normal makes an angle of $$135^{\circ}$$. $\begin{array}{c} (i+j) \cdot(i+(1+\lambda) j+\lambda \hat{k})=\pm|i+j \| i+(1+\lambda) j+\lambda \hat{k}| \cos \left(\frac{\pi}{4}\right) \\ 2+\lambda=\pm \sqrt{1+(1+\lambda)^{2}+\lambda^{2}} \\ \lambda^{2}-2 \lambda-2=0 \\ \lambda=1 \pm \sqrt{3} \end{array}$ So the equation of the plane is $x+y+(1 \pm \sqrt{3})(y+z)=0$