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Category: Exercises

Partial Fraction Integrals

Problem 1. Given a real number $\alpha$ , evaluate the integrals

$\displaystyle \int \frac{1}{x-\alpha}\, \mathrm dx \quad \text{and} \quad \int \frac{1}{(x-\alpha)^2}\, \mathrm dx.$

(Click for Solution)

Solution. By the chain rule,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}\ln|x-\alpha| &= \frac 1{x-\alpha} \\ \frac{\mathrm d}{\mathrm dx} \left(\frac 1{x-\alpha}\right) &= \frac{\mathrm d}{\mathrm dx} \left( (x-\alpha)^{-1} \right) \\ &= (-1)(x-\alpha)^{-2} \\ &= -\frac 1{(x-\alpha)^2}.\end{aligned}$

By linearity,

$\displaystyle \frac{\mathrm d}{\mathrm dx} \left(-\frac 1{x-\alpha}\right) = - \frac{\mathrm d}{\mathrm dx} \left(\frac 1{x-\alpha}\right) = \frac 1{(x-\alpha)^2}.$

Therefore,

$\begin{aligned} \int \frac{1}{x-\alpha}\, \mathrm dx &= \ln|x-\alpha| + C_1, \\\int \frac{1}{(x-\alpha)^2}\, \mathrm dx &= -\frac 1{x-\alpha} + C_2. \end{aligned}$

Problem 2. Given distinct real numbers $\alpha,\beta$ , determine constants $A, B$ such that

$\displaystyle \frac{ 1 }{(x-\alpha)(x-\beta)} = \frac A{x-\alpha} + \frac B{x-\beta}.$

Hence, evaluate $\displaystyle \int \frac{ 1 }{ (x-\alpha)(x-\beta) } \, \mathrm dx$ .

(Click for Solution)

Solution. Firstly, multiply both sides by $x - \alpha$ :

$\displaystyle \frac 1{x-\beta} = A + \frac{B}{x-\beta} \cdot (x-\alpha).$

Setting $x = \alpha$ ,

$\displaystyle \frac 1{\alpha-\beta} = A + \frac{B}{\alpha-\beta} \cdot 0.$

Therefore, $A = 1/(\alpha - \beta)$ .

Similarly, multiply both sides by $x - \beta$ :

$\displaystyle \frac 1{x-\alpha} = \frac{ A }{ x - \alpha } \cdot (x-\beta) + B.$

Setting $x = \beta$ ,

$\displaystyle \frac 1{\beta-\alpha} = \frac{ A }{ \beta - \alpha } \cdot 0 + B.$

Therefore, $B = 1/(\beta-\alpha) = -A$ .

Consolidating,

$\begin{aligned} \frac{ 1 }{(x-\alpha)(x-\beta)} &= \frac {\frac 1{\alpha - \beta}}{x-\alpha} - \frac {\frac 1{\alpha - \beta}}{x-\beta} \\ &= \frac 1{\alpha - \beta}\left( \frac {1}{x-\alpha} - \frac {1}{x-\beta} \right). \end{aligned}$

Problem 3. Use Problems 1 and 2 to evaluate the integral

$\displaystyle \int \frac{ x^2 + x + 1 }{(x-2)(x-3)^2} \, \mathrm dx.$

(Click for Solution)

Solution. Write

$\displaystyle \frac{ x^2 + x + 1 }{(x-2)(x-3)^2} = \frac{x^2 + x+ 1}{x-2} \cdot \frac 1{(x-3)^2}.$

Write $x = (x-2)+2$ and expand the fraction therein:

$\begin{aligned} \frac{ x^2 + x + 1 }{x-2} &= \frac{ ((x-2)+2)^2 + ((x-2)+2) + 1}{x-2} \\ &= \frac{(x-2)^2 + 2 \cdot (x-2) \cdot 2 + 2^2 + (x-2) + 2 + 1}{x-2} \\ &= \frac{ (x-2)^2 + 5(x-2) + 7 }{x-2} \\ &= x-2 + 5 + \frac 7{x-2} \\ &= x+3 + \frac{7}{x-2} \\ &= (x-3)+6 + \frac{7}{x-2}. \end{aligned}$

Therefore, by multiplying excess terms,

$\begin{aligned} \frac{ x^2 + x + 1 }{(x-2)(x-3)^2} &= \left( (x-3)+6 + \frac{7}{x-2} \right) \cdot \frac 1{(x-3)^2} \\ &= \frac 1{x-3} + \frac{6}{(x-3)^2} + \frac 7{(x-2)(x-3)^2}. \end{aligned}$

By Question 2,

$\begin{aligned} \frac 1{(x-2)(x-3)} &= \frac 1{2-3} \left(\frac 1{x-2} - \frac 1{x-3}\right) \\ &= \frac 1{x-3} - \frac 1{x-2}. \end{aligned}$

Therefore,

$\begin{aligned} \frac 1{(x-2)(x-3)^2} &= \frac 1{x-3} \cdot \frac 1{(x-2)(x-3)} \\ &= \frac 1{x-3} \cdot \left(\frac 1{x-3} - \frac 1{x-2}\right) \\ &= \frac 1{(x-3)^2} - \frac 1{(x-2)(x-3)} \\ &= \frac 1{(x-3)^2} - \left(\frac 1{x-3} - \frac 1{x-2}\right) \\ &= \frac 1{x-2} - \frac 1{x-3} + \frac 1{(x-3)^2}. \end{aligned}$

Combining the displays,

$\begin{aligned} \frac{ x^2 + x + 1 }{(x-2)(x-3)^2} &= \frac 1{x-3} + \frac{6}{(x-3)^2} + \frac 7{(x-2)(x-3)^2} \\ &= \frac 1{x-3} + \frac{6}{(x-3)^2} + 7 \cdot \left( \frac 1{x-2} - \frac 1{x-3} + \frac 1{(x-3)^2} \right) \\ &= \frac{7}{x-2} - \frac 6{x-3} + \frac{13}{(x-3)^2}.\end{aligned}$

By the linearity of integration and Question 1,

$\begin{aligned} &\int \frac{ x^2 + x + 1 }{(x-2)(x-3)^2}\, \mathrm dx \\ &= 7 \cdot \int \frac 1{x-2}\, \mathrm dx - 6 \cdot \int \frac 1{x-3}\, \mathrm dx + 13 \cdot \int \frac 1{(x-3)^2}\, \mathrm dx \\ &= 7 \ln|x-2| - 6 \ln |x-3| - \frac{13}{x-3} + C. \end{aligned}$

Remark 1. In the process of evaluating the integral, we have uncovered the partial fraction decomposition for the following expression with distinct $\alpha,\beta$ :

$\displaystyle \frac{px^2 + qx + r}{(x-\alpha)(x-\beta)^2} = \frac A{x-\alpha} + \frac {B}{x-\beta} + \frac C{(x-\beta)^2}.$

We leave it as an exercise to check that

$\begin{aligned} A &= \frac{ p\alpha^2 + q\alpha + r }{(\alpha - \beta)^2}, \\ B &= p -A = p - \frac{ p\alpha^2 + q\alpha + r }{(\alpha - \beta)^2}, \\ C &= \frac{p\beta^2 + q\beta + r }{\beta - \alpha}. \end{aligned}$

Problem 4. Similarly, evaluate

$\displaystyle \int \frac{ x^2 + x + 1 }{(x-2)(x-3) (x-4)} \, \mathrm dx.$

(Click for Solution)

Solution. Using the simplification in Problem 2,

$\begin{aligned} \frac{ x^2 + x + 1 }{(x-2)(x-3) (x-4)} &= \frac{ x^2 + x + 1 }{(x-3) (x-4)} \cdot \frac 1{(x-3)(x-4)} \\ &= \left((x-3)+6 + \frac{7}{x-2}\right) \cdot \frac 1{(x-3)(x-4)} \\ &= \frac{1}{x-4} + \frac 6{(x-3)(x-4)} + \frac 7{(x-2)(x-3)(x-4)}. \end{aligned}$

By Question 2 applied twice,

$\begin{aligned} \frac 1{(x-3)(x-4)} &= \frac 1{3-4} \left( \frac 1{x-3} - \frac 1{x-4} \right) \\ &= \frac 1{x-4} - \frac{1}{x-3} \end{aligned}$

so that

$\begin{aligned} \frac 1{(x-2)(x-3)(x-4)} &= \frac 1{x-2} \cdot \frac 1{(x-3)(x-4)} \\ &= \frac 1{x-2} \left( \frac 1{x-4} - \frac{1}{x-3} \right) \\ &= \frac 1{(x-2)(x-4)} - \frac{1}{(x-2)(x-3)} \\ &= \frac 1{2-4} \left(\frac 1{x-2} - \frac 1{x-4}\right) - \frac 1{2-3} \left(\frac 1{x-2} - \frac 1{x-3}\right) \\ &= \frac 12\left(\frac 1{x-4} - \frac 1{x-2}\right) + \left(\frac 1{x-2} - \frac 1{x-3}\right) \\ &= \frac{\frac 12}{x-2} - \frac 1{x-3} + \frac {\frac 12}{x-4}. \end{aligned}$

Combining the displays,

$\begin{aligned} \frac{ x^2 + x + 1 }{(x-2)(x-3) (x-4)} &= \frac{ x^2 + x + 1 }{(x-3) (x-4)} \cdot \frac 1{(x-3)(x-4)} \\ &= \frac{\frac 72}{x-2} - \frac {13}{x-3}+ \frac{\frac{21}2}{x-4} .\end{aligned}$

By the linearity of integration and Question 1,

$\begin{aligned} \int \frac{ x^2 + x + 1 }{(x-2)(x-3) (x-4)}\, \mathrm dx &= \frac 72 \ln|x-2| - 13 \ln|x-3| + \frac{21}2 \ln|x-4| + C.\end{aligned}$

Remark 2. The corresponding partial fraction decomposition with distinct $\alpha,\beta,\gamma$ is as follows:

$\displaystyle \frac{px^2 + qx + r}{(x-\alpha)(x-\beta)(x-\gamma)} = \frac A{x-\alpha} + \frac {B}{x-\beta} + \frac C{x-\gamma}.$

We leave it as an exercise to check that

$\begin{aligned} A &= \frac{ p\alpha^2 + q\alpha + r }{(\alpha - \beta)(\alpha - \gamma)}, \\ B &= \frac{ p\beta^2 + q\beta + r }{(\beta - \alpha)(\beta - \gamma)}, \\ C &= \frac{ p\gamma^2 + q\gamma + r }{(\gamma - \alpha)(\gamma - \beta)}. \end{aligned}$

Problem 5. Given distinct real numbers $\alpha,\beta$ , determine constants $A,B,C$ such that

$\displaystyle \frac{ \alpha^2+\beta^2 }{(x-\alpha)(x^2+\beta^2) } = \frac A{x-\alpha} + \frac {Bx+C}{x^2 + \beta^2}.$

Hence, evaluate

$\displaystyle \int \frac{ x^2 + x + 1 }{(x-2)(x^2+9) } \, \mathrm dx.$

(Click for Solution)

Solution. By observing that $0 = x^2 -x^2$ ,

$\begin{aligned} \frac{\alpha^2 + \beta^2}{(x-\alpha)(x^2 +\beta^2)} &= \frac{x^2 + \beta^2 - (x^2 - \alpha^2) }{(x-\alpha)(x^2+\beta^2)} \\ &= \frac{x^2 + \beta^2}{(x-\alpha)(x^2+\beta^2)} - \frac{x^2 -\alpha^2}{(x-\alpha)(x^2+\beta^2)} \\ &= \frac 1{x-\alpha} - \frac{(x-\alpha)(x+\alpha)}{(x-\alpha)(x^2+\beta^2)} \\ &= \frac 1{x-\alpha} - \frac{x+\alpha}{x^2+\beta^2}, \end{aligned}$

so that $A = 1, B = -1, C = -\alpha$ . Dividing both sides by $\alpha^2 +\beta^2$ for the next part,

$\displaystyle \frac 1{(x-\alpha)(x^2+\beta^2)} = \frac 1{\alpha^2+\beta^2} \left( \frac 1{x-\alpha} - \frac{x+\alpha}{x^2+\beta^2} \right).$

Using the solution in Problem 3,

$\begin{aligned} \frac{ x^2 + x + 1 }{(x-2)(x^2+9) } &= \frac{ x^2 + x + 1 }{x-2 } \cdot \frac 1{x^2+9} \\ &= \left(x+3 + \frac 7{x-2}\right)\cdot \frac 1{x^2+9} \\ &= \frac{x+3}{x^2+9} + \frac 7{(x-2)(x^2+9)}. \end{aligned}$

Using the first part with $\alpha = 2$ and $\beta = 3$ ,

$\begin{aligned} \frac 1{(x-2)(x^2+9)} &= \frac 1{2^2 + 3^2} \left(\frac 1{x-2} - \frac{x + 2}{x^2 + 3^2}\right) \\ &= \frac 1{13} \left(\frac 1{x-2} - \frac{x+2}{x^2+9}\right). \end{aligned}$

Combining the displays,

$\begin{aligned} \frac{ x^2 + x + 1 }{(x-2)(x^2+9) } &= \frac{x+3}{x^2+9} + \frac 7{(x-2)(x^2+9)} \\ &= \frac{x+3}{x^2+9} + \frac 7{13} \left(\frac 1{x-2} - \frac{x+2}{x^2+9}\right) \\ &= \frac 1{13} \left( \frac{7}{x-2} + \frac{13(x+3) - 7(x+2)}{x^2+9} \right) \\ &= \frac 1{13} \left( \frac{7}{x-2} + \frac{ 6x+ 25}{x^2+9} \right). \end{aligned}$

Using the chain rule,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx} \ln(x^2 + 9) &= \frac{2x}{x^2+9}, \\ \frac{\mathrm d}{\mathrm dx} \tan^{-1}(x/3) &= \frac{1}{1 + (x/3)^2} \cdot \frac 13 = \frac{3}{x^2+9}. \end{aligned}$

Therefore,

$\begin{aligned} \int \frac{x}{x^2+9}\, \mathrm dx &= \frac 12 \ln(x^2+9) + C_1, \\ \int \frac{1}{x^2+9}\, \mathrm dx &= \frac 13 \tan^{-1}(x/3) + C_2, \end{aligned}$

By the linearity of integration,

$\begin{aligned} \int \frac{x^2+x+1}{(x-2)(x^2+9)}\, \mathrm dx &= \frac 1{13} \int \left( \frac{7}{x-2} + \frac{ 6x}{x^2+9}+ \frac{ 25}{x^2+9} \right)\, \mathrm dx \\ &= \frac 1{13}\left( \ln|x-2| + \frac 62 \ln(x^2 + 9) + \frac{25}{3} \tan^{-1}(x/3) \right) + C \\ &= \frac 1{13} \ln|x-2| + \frac{3}{13} \ln(x^2+9) + \frac{25}{39} \tan^{-1}(x/3) + C. \end{aligned}$

Remark 3. The corresponding partial fraction decomposition with distinct $\alpha,\beta$ is as follows:

$\displaystyle \frac{px^2 + qx + r}{(x-\alpha)(x^2+\beta^2)} = \frac A{x-\alpha} + \frac {Bx+C}{x^2 + \beta^2}.$

—Joel Kindiak, 29 Mar 26, 1546H

June 2, 2026
Imaginary Numbers

Define the 2 × 2 matrices $\mathbf I$ and $\mathbf J$ as follows:

$\mathbf I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \quad \mathbf J = \begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix}.$

Problem 1. Evaluate $\mathbf J^2$ .

(Click for Solution)

Solution. Using matrix multiplication,

$\begin{aligned} \mathbf J^2 &= \begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix} \\ &= \begin{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} & \begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix} \begin{bmatrix} -1 \\ 0 \end{bmatrix} \end{bmatrix} \\ &= \begin{bmatrix} 0\begin{bmatrix} 0 \\ 1\end{bmatrix} + 1\begin{bmatrix} -1 \\ 0\end{bmatrix} & (-1)\begin{bmatrix} 0 \\ 1\end{bmatrix} + 0\begin{bmatrix} -1 \\ 0\end{bmatrix} \end{bmatrix} \\ &= \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = (-1) \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} \\ & = (-1)\mathbf I = -\mathbf I. \end{aligned}$

Let $a,b,c,d$ be real numbers.

Problem 2. Show that

$a \mathbf I + b \mathbf J = c \mathbf I + d \mathbf J$

implies that $a = c$ and $b = d$ .

(Click for Solution)

Solution. Using the definition of $\mathbf I$ and $\mathbf J$ ,

$\begin{aligned} a\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} + b \begin{bmatrix}0 & -1 \\ 1 & 0 \end{bmatrix} &= c \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} + d \begin{bmatrix}0 & -1 \\ 1 & 0 \end{bmatrix} \\ \begin{bmatrix}a & 0 \\ 0 & a \end{bmatrix} + \begin{bmatrix}0 & -b \\ b & 0 \end{bmatrix} &= \begin{bmatrix}c & 0 \\ 0 & c \end{bmatrix} + \begin{bmatrix}0 & -d \\ d & 0 \end{bmatrix} \\ \begin{bmatrix}a & -b \\ b & a \end{bmatrix} &= \begin{bmatrix}c & -d \\ d & c \end{bmatrix}.\end{aligned}$

Comparing the entries of the matrices on both sides yields $a = c$ and $b = d$ .

Define $\mathbf Z := a \mathbf I + b \mathbf J$ and $\mathbf W := c \mathbf I + d \mathbf J$ .

Problem 3. Show that $\mathbf Z \mathbf W = \mathbf W \mathbf Z$ .

(Click for Solution)

Solution. We observe that $\mathbf I \mathbf J = \mathbf J = \mathbf J \mathbf I$ . By Problem 1,

$\begin{aligned}\mathbf Z \mathbf W &= (a \mathbf I + b \mathbf J)(c \mathbf I + d \mathbf J) \\ &= (ac) \mathbf I^2 + (ad) \mathbf I \mathbf J + (bc) \mathbf J \mathbf I + (bd) \mathbf J^2 \\ &= (ac) \mathbf I +(ad)\mathbf J + (bc)\mathbf J + (bd)(-1)\mathbf I \\ &= (ac - bd) \mathbf I + (ad + bc) \mathbf J.\end{aligned}$

Similarly, $\mathbf W \mathbf Z = (ca - db) \mathbf I + (da + cb) \mathbf J$ . Therefore, $\mathbf Z \mathbf W = \mathbf W \mathbf Z$ .

Define $\mathbf Z^* := a \mathbf I - b \mathbf J$ .

Problem 4. Evaluate $\mathbf Z \mathbf Z^*$ . Hence, if $\mathbf Z \neq \mathbf 0$ , construct a matrix $\mathbf Z^{-1}$ with the property that

$\mathbf Z \mathbf Z^{-1} = \mathbf Z^{-1} \mathbf Z = \mathbf I.$

(Click for Solution)

Solution. Using the multiplication in Problem 3 but setting $c = a$ and $d = -b$ ,

$\mathbf Z \mathbf Z^* = (a^2 - b(-b))\mathbf I + (a(-b) + ba)\mathbf J = (a^2 + b^2) \mathbf I.$

If $\mathbf Z \neq \mathbf 0$ , then either $a \neq 0$ or $b \neq 0$ , so that $a^2 + b^2 \neq 0$ . Therefore,

$\begin{aligned} \mathbf Z\left( \frac 1{a^2 + b^2} \mathbf Z^* \right) &= \frac 1{a^2 + b^2} \mathbf Z \mathbf Z^* \\ &= \frac 1{a^2 + b^2} (a^2 + b^2) \mathbf I \\ &= \mathbf I. \end{aligned}$

Denoting $\alpha := (\alpha^2 + \beta^2)^{-1}$ , define $\mathbf Z^{-1} := \alpha \mathbf Z^*$ , so that

$\mathbf Z \mathbf Z^{-1} = \mathbf Z (\alpha \mathbf Z^*) = \mathbf I.$

By Problem 3, $\mathbf Z^{-1} \mathbf Z = \mathbf Z \mathbf Z^{-1} = \mathbf I$ .

Problem 5. Determine the two possible matrices $\mathbf Z = a\mathbf I + b \mathbf J$ such that

$\mathbf Z^2 - 4\mathbf Z + 13\mathbf I = \mathbf 0.$

(Click for Solution)

Solution. Write $\mathbf Z = a\mathbf I + b \mathbf J$ . Using the multiplication in Problem 3 with $c = a$ and $d = b$ ,

$\mathbf Z^2 = (a^2 - b^2) \mathbf I + (2ab) \mathbf J.$

Therefore,

$((a^2 - b^2) \mathbf I + (2ab) \mathbf J) - 4(a\mathbf I + b\mathbf J) + 13\mathbf I = \mathbf 0.$

Grouping the terms together,

$(a^2 - 4a - b^2 + 13) \mathbf I + 2b (a - 2) \mathbf J = \mathbf 0 = 0 \mathbf I + 0\mathbf J.$

Using Problem 2,

$a^2 - 4a - b^2 + 13 = 0\quad \mathbf{and} \quad 2b (a - 2) = 0.$

Using the second equation, either $b = 0$ or $a = 2$ . If $b = 0$ , then substituting into the first equation,

$a^2 - 4a+ 13 = 0.$

However, this equation has discriminant $(-4)^2 - 4 \cdot 1 \cdot 13 = - 36 < 0$ , and so there are no real roots to the equation, a contradiction.

Therefore, we must have $a = 2$ . Substituting into the first equation again,

$2^2 - 4 \cdot 2 - b^2 + 13 = 0 \quad \Rightarrow \quad b^2 = 9.$

Therefore, $b = \pm 3$ . Hence, the two possible matrices for $\mathbf Z$ are

$\mathbf Z = 2\mathbf I +3\mathbf J \quad \text{or} \quad \mathbf Z = 2\mathbf I - 3 \mathbf J.$

We can condense them to the expression $\mathbf Z = 2\mathbf I \pm 3 \mathbf J$ .

Remark 1. By denoting $a \mathbf I \equiv a$ and $b\mathbf J \equiv bi$ , we have created a model for the complex numbers $a + bi$ , where $i^2 = -1$ by Problem 1. In particular, numbers of the form $bi$ are called purely imaginary. The solution to Problem 5 would then look like

$z^2 - 2z + 5 = 0 \quad \iff \quad z = 2 \pm 3i.$

The letter ‘z‘ is used to denote a complex number by convention. Furthermore, the calculation $i^2 = -1$ motivates the (somewhat debatable) notation $\sqrt{-1} := i$ . For more information, see this post.

—Joel Kindiak, 25 Mar 26, 0056H

June 1, 2026
Baby Quantitative Finance

In this post, we will explore some basic notions in quantitative finance.

More specifically, buying and selling stocks.

Suppose 1 unit of a stock KMATH costs $1 at time t = 0. Assume negligible trading fees.

Problem 1. At time t = 0, you buy 200 units of KMATH. What is the value of your position?

(Click for Solution)

Solution. The value of $200$ units of KMATH is

$200 \times \$1 = \$200.$

Problem 2. Suppose at time t = 1, the price per unit of KMATH increased by 10%. What is the value of your position at t = 1?

(Click for Solution)

Solution. The value of the position has increased by $10\%$ , that is,

$10\% \times \$200 = 0.1 \times \$200 = \$20.$

Therefore, the position has a new value of

$\$200 + \$20 = \$220.$

Alternate Solution. If the initial position has a value of $\$P$ and the price increased by a percentage of $r_1$ , then the position increases by the value

$r_1 \times \$P = \$(P \times r_1).$

Therefore, the position would have a new value of

$\begin{aligned} \$P + \$(P \times r_1) &= \$P + r_1 \times \$P \\ &= (1 + r_1) \times \$P \\ &= \$((1 + r_1) \times P).\end{aligned}$

In particular, setting $P = 200$ and

$\displaystyle r_1 = 10\% = \frac{ 10 }{ 100 } = 0.1$

in Problem 2 yields a new value of

$\$((1 + 0.1) \times 200) = \$220.$

Problem 3. Suppose at time t = 2, the price per unit of KMATH decreased by 10%. What is the overall change in your position from t = 0 to t = 2? How about its overall percentage change? Is your position in a profit or a loss?

(Click for Solution)

Solution. We will use the calculation in Remark 1. Let

$\displaystyle r_2 = -10\% = -\frac{ 10 }{ 100 } = -0.1$

denote the percentage decrease of the price per unit of KMATH from $t = 1$ to $t = 2$ .

By Remark 1, the new position at $t = 1$ is $\$( (1+r_1) \times 200 )$ . Therefore, the new position at $t = 2$ has a value of

$\$ ( (1 + r_2) \times ((1 + r_1) \times 200) ) = \$ ( (1 + r_1) \times (1+r_2) \times 200 ).$

Substituting $r = 0.1$ , the new position has a value of

$\begin{aligned} \$ ( (1 + r_1) \times (1+r_2) \times 200 )&= \$ ( (1 + 0.01) \times (1-0.01) \times 100 ) \\ &= \$ ( (1 - 0.1^2) \times 200 ) \\ &= \$ ( (1 - 0.01) \times 200 ) \\ &= \$ 198. \end{aligned}$

The overall percentage change is

$\displaystyle -0.01 \times 100\% = -1\%.$

Since the percentage change is negative, our position currently sits in a loss.

Problem 4. For any positive integer n, let r_n denote the percentage change in your position from t = n – 1 to t = n. Show that the overall percentage change $r$ between t = 0 and t = n is calculated by

1 + r = (1 + r₁) × (1 + r₂) × … × (1 + r_n).

(Click for Solution)

Solution. Let $P_n$ denote the value of the position at time $t = n$ . Applying the alternate solution in Problem 2 repeatedly,

$\begin{aligned} \$P_n &= (1 + r_n) \times \$P_{n-1} \\ &= (1 + r_n) \times (1 + r_{n-1}) \times P_{n-2} \\ &= \, \vdots \\ &= (1 + r_n) \times (1 + r_{n-1}) \times \cdots \times (1 + r_1) \times \$P_0 \\ &= \$ ( (1 + r_n) \times (1 + r_{n-1}) \times \cdots \times (1 + r_1) \times P_0 ). \end{aligned}$

On the other hand, denoting the overall percentage change by $r$ , we have

$\$P_n = \$((1+r) \times P_0)$

Equating the two sides,

$\$ ((1 + r) \times P_0) = \$ ( (1 + r_n) \times (1 + r_{n-1}) \times \cdots \times (1 + r_1) \times P_0 ).$

Dividing by $P_0$ on both sides yields the desired result:

$1 + r = (1 + r_n) \times (1 + r_{n-1}) \times \cdots \times (1 + r_1).$

Problem 5. What is the minimum percentage increase of the price per unit of KMATH from t = 2 to t = 3 required for you to not incur loss?

(Click for Solution)

Solution. Let $r_3$ denote the required percentage increase of the price per unit of KMATH from $t = 2$ to $t = 3$ . By Problem 3, the overall percentage change $r$ is given by

$1 + r = (1 + r_1) \times (1 + r_2) \times (1 + r_3).$

Substituting $r_1 = 0.1$ and $r_2 = -0.1$ , since $1+r_1 > 0$ and $1+r_2 > 0$ , we can divide on both sides to obtain

$\displaystyle 1 + r_3 = \frac{ 1 + r }{ (1 + 0.1) \times (1 - 0.1) } = \frac{ 1 + r }{ 1 - 0.01}.$

Subtracting by $1$ on both sides,

$\displaystyle r_3 = \frac{ 1 + r }{ 1 - 0.01 } - 1.$

Since we do not want to incur loss, the overall percentage change must be non-negative, that is to say, $r \geq 0$ :

$\begin{aligned} r_3 = \frac{ 1 + r }{ 1 - 0.0001 } - 1 \geq \frac{ 1 + 0 }{ 1 - 0.01 } - 1 = \frac{ 1 }{ 99 } > 0.01 = 1\%. \end{aligned}$

In particular, we need more than $1\%$ increase in order to compensate for an overall decrease of $1\%$ .

—Joel Kindiak, 22 Feb 26, 1324H

May 18, 2026
Solving Trigonometric Equations
Problem 1. Determine the value(s) of $x$ such that $\sin(x) = 1/2$ , where
- $0 < x < \pi/2$ ,
- $0 \leq x < 2\pi$ ,
- $x$ is a real number.
We call this process solving a trigonometric equation.

(Click for Solution)

Solution. Using special angles, we recall that for $0 < x < \pi/2$ ,

$\sin(x) = 1/2 \quad \iff \quad x = \pi/6.$

Denote $\alpha := \pi/6$ . Using the extended definitions of $\sin(x)$ , for $0 \leq x < 2\pi$ , we have

$\begin{aligned} \sin(\alpha) &= 1/2,\\ \sin(\pi - \alpha) &= \sin(\alpha) = 1/2,\\ \sin(\pi + \alpha) &= -{ \sin(\alpha) } = -1/2,\\ \sin(2\pi - \alpha) &= -{ \sin(\alpha) } = -1/2. \end{aligned}$

Since only the first two equations work, we have $x = \pi/6$ or

$x = \pi - \pi/6 = 5\pi/6.$

Graphically, we have the following.

Finally, for general $x$ , we recall that $\sin(x)$ is $2\pi$ -periodic:

$\sin(x) = \sin(x+2\pi).$

By repeating this process, given any integer $k$ , we have

$\sin(x + 2k\pi) = \sin(x).$

In particular,

$\sin(\pi/6 + 2k\pi) = \sin(\pi/6) = 1/2.$

Furthermore, for any $x_0 \neq \pi/6$ and $x_0 \neq 5\pi/6$ ,

$\sin(x_0 + 2k\pi) = \sin(x_0) \neq 1/2.$

Therefore, we must have $x = \pi/6 + 2k\pi$ or $x = 5\pi/6 + 2k\pi$ , where $k$ is some integer.

Problem 2. Solve the equation $\cos(x) = 1/2$ , where
- $0 < x < \pi/2$ ,
- $0 \leq x < 2\pi$ .
(Click for Solution)

Solution. Using special angles, we recall that for $0 < x < \pi/2$ ,

$\cos(x) = 1/2 \quad \iff \quad x = \pi/3.$

Using the extended definitions of $\cos(x)$ , for $0 \leq x < 2\pi$ , we have

$\begin{aligned} \cos(\pi/3) &= 1/2,\\ \cos(\pi - \pi/3) &= -{ \cos(\pi/3) } = -1/2,\\ \cos(\pi + \pi/3) &= -{ \cos(\pi/3) } = -1/2,\\ \cos(2\pi - \pi/3) &= \cos(\pi/3) = 1/2. \end{aligned}$

Since only the first and last equations work, we have $x = \pi/3$ or

$x = 2\pi - \pi/3 = 5\pi/3.$

Graphically, we have the following.

Problem 3. Solve the equation $\tan(x) = 1$ , where $0 \leq x < 2\pi$ .

(Click for Solution)

Solution. Using special angles, we recall that for $0 < x < \pi/2$ ,

$\tan(x) = 1 \quad \iff \quad x = \pi/4.$

Using the extended definitions of $\tan(x)$ , for $0 \leq x < 2\pi$ , we have

$\begin{aligned} \tan(\pi/4) &= 1,\\ \tan(\pi - \pi/4) &= -{ \tan(\pi/4) } = -1,\\ \tan(\pi + \pi/4) &= \tan(\pi/4) = 1, \\ \tan(2\pi - \pi/4) &= -{ \tan(\pi/4) } = -1. \end{aligned}$

Since only the first and third equations work, we have $x = \pi/4$ or

$x = \pi + \pi/4 = 5\pi/4.$

Graphically, we have the following.

Problem 4. For each of the cases $r=-1, 0, 1$ , solve the equation $\sin(x) = r$ , where $0 \leq x \leq 2\pi$ .
(Click for Solution)

Solution. Firstly, we work with the case $r = 1$ . Using special angles, we recall that for $0 \leq x \leq \pi/2$ ,

$\sin(x) = 1 \quad \iff \quad x = \pi/2.$

Using the extended definitions of $\sin(x)$ , for $0 \leq x \leq 2\pi$ , we have

$\begin{aligned} \sin(\pi/2) &= 1,\\ \sin(\pi - \pi/2) &= \sin(\pi/2) = 1,\\ \sin(\pi + \pi/2) &= -{ \sin(\pi/2) } = -1, \\ \sin(2\pi - \pi/4) &= -{ \sin(\pi/2) } = -1. \end{aligned}$

Since only the first and third equations work, we have $x = \pi/2$ or

$x = \pi - \pi/2 = \pi/2.$

In fact, we get one and only one solution: $x = \pi/2$ .

Secondly, we work with the case $r = 0$ . Using special angles, we recall that for $0 \leq x \leq \pi/2$ ,

$\sin(x) = 0 \quad \iff \quad x = 0.$

Using the extended definitions of $\sin(x)$ , for $0 \leq x \leq 2\pi$ , we have

$\begin{aligned} \sin(0) &= 0,\\ \sin(\pi - 0) &= \sin(0) = 0,\\ \sin(\pi + 0) &= -{ \sin(0) } = 0, \\ \sin(2\pi - 0) &= -{ \sin(0) } = 0. \end{aligned}$

Now, all equations work, so that we have exactly three solutions: $x = 0, \pi, 2\pi$ .

Finally, we work with the case $r = -1$ . Using the same equations as the case for $r = 1$ , the solutions must be
- $x = \pi + \pi/2 = 3\pi/2$ , or
- $\quad x = 2\pi - \pi/2 = 3\pi/2$ .
In fact, we get one and only one solution: $x = 3\pi/2$ .

Notice that the $x$ -values of interest take the form $k\pi/2$ , where $0 \leq k \leq 4$ is an integer.

Graphically, we have the following.
Problem 5. For each of the cases $r=-1, 0, 1$ , solve the equation $\cos(x) = r$ , where $0 \leq x \leq 2\pi$ .
(Click for Solution)

Solution. We can either adopt the strategy in Problem 4, or simply focus on the special values $k\pi/2$ , where $0 \leq k \leq 4$ is an integer. Observe that

$\begin{aligned} \cos(0) &= \cos(2\pi) = 1, \\ \cos(\pi/2) &= \cos(3\pi/2) = 0, \\ \cos(\pi) &= -1. \end{aligned}$

Therefore, for $0 \leq x \leq 2\pi$ , we must have:
- $\cos(x) = 1 \iff x = 0, 2\pi$ ,
- $\cos(x) = 0 \iff x = \pi/2, 3\pi/2$ ,
- $\cos(x) = -1 \iff x = \pi$ .
Graphically, we have the following.
Problem 6. Fix $-1 \leq r \leq 1$ . Define $0 \leq \alpha \leq \pi/2$ such that

$\sin(\alpha) = |r|.$

Solve the equation $\sin(x) = r$ in terms of $\alpha$ , where $0 \leq x \leq 2\pi$ .
(Click for Solution)

Solution. Using the extended definitions of $\sin(x)$ , for $0 \leq x \leq 2\pi$ , we have

$\begin{aligned} \sin(\alpha) &= |r|,\\ \sin(\pi - \alpha) &= \sin(\alpha) = |r|,\\ \sin(\pi + \alpha) &= -{ \sin(\alpha) } = -|r|,\\ \sin(2\pi - \alpha) &= -{ \sin(\alpha) } = -|r|. \end{aligned}$

We consider the three cases:
- If $r > 0$ , then only the first two equations work: $x = \alpha, \pi -\alpha$ .
- If $r = 0$ , then $\alpha = 0$ : $x = 0, \pi, 2\pi$ by Problem 4.
- If $r < 0$ , then only the last two equations work: $x = \pi + \alpha, 2\pi -\alpha$ .
Graphically, we have the following.
Problem 7. Fix $-1 \leq r \leq 1$ . Define $0 \leq \alpha \leq \pi/2$ such that

$\cos(\alpha) = |r|.$

Solve the equation $\cos(x) = r$ in terms of $\alpha$ , where $0 \leq x \leq 2\pi$ .
(Click for Solution)

Solution. Using the extended definitions of $\cos(x)$ , for $0 \leq x \leq 2\pi$ , we have

$\begin{aligned} \cos(\alpha) &= |r|,\\ \cos(\pi - \alpha) &= -{ \cos(\alpha) } = -|r|,\\ \cos(\pi + \alpha) &= -{ \cos(\alpha) } = -|r|,\\ \cos(2\pi - \alpha) &= { \cos(\alpha) } = |r|. \end{aligned}$

We consider the three cases:
- If $r > 0$ , then only the first and last equations work: $x = \alpha, 2\pi -\alpha$ .
- If $r = 0$ , then $\alpha = 0$ : $x = \pi/2, 3\pi/2$ by Problem 5.
- If $r < 0$ , then only the middle two equations work: $x = \pi - \alpha, \pi +\alpha$ .
Graphically, we have the following.
Problem 8. Fix any real number $r$ . Define $0 \leq \alpha \leq \pi/2$ such that

$\tan(\alpha) = |r|.$

Solve the equation $\tan(x) = r$ in terms of $\alpha$ , where $0 \leq x \leq 2\pi$ .
(Click for Solution)

Solution. Using the extended definitions of $\tan(x)$ , for $0 \leq x \leq 2\pi$ , we have

$\begin{aligned} \tan(\alpha) &= |r|,\\ \tan(\pi - \alpha) &= -{ \tan(\alpha) } = -|r|,\\ \tan(\pi + \alpha) &= { \tan(\alpha) } = |r|,\\ \tan(2\pi - \alpha) &= { \tan(\alpha) } = -|r|. \end{aligned}$

We consider the three cases:
- If $r > 0$ , then only the first and last equations work: $x = \alpha, \pi + \alpha$ .
- If $r < 0$ , then only the middle two equations work: $x = \pi - \alpha, 2\pi -\alpha$ .
- If $r = 0$ , then $\alpha = 0$ : $x = 0, \pi, 2\pi$ .
Graphically, we have the following.
Remark 1. Problems 6–8 detail the ASTC strategy used to solve trigonometric equations. Given a trigonometric equation $f(x) = r$ , the solutions for $0 \leq x \leq 2\pi$ depend on
- the type of trigonometric function $f(x)$ , and
- the sign of $r$ .
Example 1. Given the equation $\sin(x) = r$ , by calculating $0 \leq \alpha \leq \pi/2$ such that $\sin(\alpha) = |r|$ , the candidate solutions in the domain $0 \leq x < 2\pi$ for the equation are

$x = \alpha, \quad x = \pi-\alpha,\quad x = \pi+ \alpha,\quad x = 2\pi -\alpha.$

If $r \geq 0$ , we choose the first two candidates (i.e. ASTC), yielding

$x = \alpha, \quad x = \pi-\alpha.$

If $r < 0$ , we choose the latter two candidates (i.e. ASTC), yielding

$x = \pi + \alpha, \quad x = 2\pi-\alpha.$

Graphically, we have the following.

We then add and subtract enough angles by multiples of $2\pi$ until we recover all desired solutions, as per the solving strategy in Problems 1 and 3.

Example 2. Given the equation $\cos(x) = r$ , by calculating $0 \leq \alpha \leq \pi/2$ such that $\cos(\alpha) = |r|$ , the candidate solutions in the domain $0 \leq x < 2\pi$ for the equation are

$x = \alpha, \quad x = \pi-\alpha,\quad x = \pi+ \alpha,\quad x = 2\pi -\alpha.$

If $r \geq 0$ , we choose the first and last candidates (i.e. ASTC), yielding

$x = \alpha, \quad x = 2\pi-\alpha.$

If $r < 0$ , we choose the middle two candidates (i.e. ASTC), yielding

$x = \pi - \alpha, \quad x = \pi + \alpha.$

We then add and subtract enough angles by multiples of $2\pi$ until we recover all desired solutions, as per the solving strategy in Problems 2 and 5.

Graphically, we have the following.

Example 3. Given the equation $\tan(x) = r$ , by calculating $0 \leq \alpha \leq \pi/2$ such that $\tan(\alpha) = |r|$ , the candidate solutions in the domain $0 \leq x < 2\pi$ for the equation are

$x = \alpha, \quad x = \pi-\alpha,\quad x = \pi+ \alpha,\quad x = 2\pi -\alpha.$

If $r \geq 0$ , we choose the first and third candidates (i.e. ASTC), yielding

$x = \alpha, \quad x = \pi+\alpha.$

If $r < 0$ , we choose the second and fourth candidates (i.e. ASTC), yielding

$x = \pi - \alpha, \quad x = 2\pi - \alpha.$

Graphically, we have the following.

We then add and subtract enough angles by multiples of $2\pi$ until we recover all desired solutions, as per the solving strategy in Problems 2 and 5.

Problem 9. Solve the equation $\sin(2x) = 1/2$ for $0 \leq x \leq 2\pi$ .

(Click for Solution)

Solution. Write $u = 2x$ so that $0 \leq u \leq 4\pi$ . To solve the equation

$\sin(u) = 1/2,$

we appeal to Problem 1 (or ASTC techniques) to conclude that for $0 \leq u \leq 2\pi$ ,

$u = \pi/6,\quad 5\pi/6.$

To account for all solutions $0 \leq u \leq 4\pi$ , we add $2\pi$ to each solution:

$u = \pi/6,\quad 5\pi/6,\quad 13\pi/6,\quad 17\pi/6.$

Since $u = 2x$ , we divide by $2$ to obtain all solutions in $x$ :

$x = \pi/12,\quad 5\pi/12,\quad 13\pi/12,\quad 17\pi/12.$

Problem 10. Solve the equation $\cos(4x) - 3\sin(2x) + 1 = 0$ for $0 \leq x \leq 2\pi$ .

Hint. Use the double-angle formula $\cos(2A) = 1 - 2 \sin^2(A)$ .

(Click for Solution)

Solution. Using the double-angle formula,

$\cos(4x) = \cos(2 \cdot 2x) = 1 - 2 \sin^2(2x).$

Hence, we can simplify the given equation in terms of $\sin(2x)$ :

$\begin{aligned} \cos(4x) - 3\sin(2x) + 1 &= 0 \\ (1 - 2 \sin^2(2x)) - 3\sin(2x) + 1 &= 0 \\ - 2 \sin^2(2x) - 3\sin(2x) + 2 &= 0 \\ 2\sin^2(2x) + 3\sin(2x) - 2 &= 0 \\ (\sin(2x) + 2)(2\sin(2x) - 1) &= 0. \end{aligned}$

Hence, $\sin(2x) + 2 = 0$ or $2 \sin(2x) - 1 = 0$ . In the former,

$\sin(2x) = -2.$

However, since for any $x$ we have $-1 \leq \sin(2x) \leq 1$ , this equation has no solutions. Therefore, $2 \sin(2x) - 1 = 0$ , yielding:

$\sin(2x) = 1/2,\quad 0 \leq x \leq 2\pi.$

We appeal to Problem 9 (or ASTC techniques) to conclude that

$x = \pi/12,\quad 5\pi/12,\quad 13\pi/12,\quad 17\pi/12.$

—Joel Kindiak, 2 Feb 26, 2302H
May 8, 2026
Multiplying Vectors
Problem 1. Illustrate the two vectors $\mathbf u = \begin{bmatrix} a \\ b \end{bmatrix} , \mathbf v = \begin{bmatrix} c \\ d \end{bmatrix}$ in the diagram below.

Show that the angle $\theta$ between $\mathbf u, \mathbf v$ is given by

$ac + bd = \| \mathbf u \| \| \mathbf v \| \cos(\theta).$

(Click for Solution)

Solution. Observe that

$\mathbf u - \mathbf v = \begin{bmatrix} a \\ b \end{bmatrix} - \begin{bmatrix} c \\ d \end{bmatrix} = \begin{bmatrix} a-c \\ b-d \end{bmatrix}.$

Using the law of cosines,

$\begin{aligned} \| \mathbf u - \mathbf v \|^2 &= \| \mathbf u \|^2 + \| \mathbf v \|^2 - 2 \cdot \| \mathbf u \| \cdot \| \mathbf v \| \cdot \cos(\theta).\end{aligned}$

Expanding the display on the left-hand side by Pythagoras’ theorem,

$\begin{aligned} \| \mathbf u - \mathbf v \|^2 &= (a-c)^2 + (b-d)^2 \\ &= (a^2 - 2ac + c^2) + (b^2 - 2bd + d^2) \\ &= (a^2 + b^2) + (c^2 + d^2) - 2(ac + bd) \\ &= \|\mathbf u\|^2 + \|\mathbf v\|^2 - 2(ac + bd) .\end{aligned}$

Comparing both sides of the expression $\frac 12 (\|\mathbf u\|^2 + \|\mathbf v\|^2 - \| \mathbf u - \mathbf v \|^2)$ ,

$ac + bd = \| \mathbf u \| \cdot \| \mathbf v \| \cdot \cos(\theta),$

as required.

Remark 1. The left-hand side is called the dot product of two vectors, defined by

$\displaystyle \begin{bmatrix} a \\ b \end{bmatrix} \cdot \begin{bmatrix} c \\ d \end{bmatrix} := ac+ bd.$

Then the result of Question 1 reduces to the dot product equation

$\mathbf u \cdot \mathbf v = \| \mathbf u \| \| \mathbf v \| \cos(\theta).$

Let $\alpha$ denote a real constant and $\mathbf w$ denote another two-dimensional vector.

Problem 2. Using Remark 1, show that the following equations always hold:
- $\mathbf v \cdot \mathbf v \geq 0$ .
- $\mathbf v \cdot \mathbf v = 0 \Rightarrow \mathbf v = \mathbf 0$ .
- $\mathbf u \cdot \mathbf v = \mathbf v \cdot \mathbf u$ .
- $\mathbf u \cdot (\alpha \mathbf v) = (\alpha \mathbf u) \cdot \mathbf v = \alpha (\mathbf u \cdot \mathbf v)$ .
- $\mathbf u \cdot (\mathbf v + \mathbf w) = \mathbf u \cdot \mathbf v + \mathbf u \cdot \mathbf w$ .
(Click for Solution)

Solution. Write $\mathbf u = \begin{bmatrix} a \\ b \end{bmatrix}$ and $\mathbf v = \begin{bmatrix} c \\ d \end{bmatrix}$ . The first result is almost immediate since $c,d$ are real numbers:

$\begin{aligned} \mathbf v\cdot\mathbf v &= c^2+d^2 \\ &\ge 0 + 0 \\ &= 0. \end{aligned}$

For the second result

$\mathbf v\cdot\mathbf v=0 \quad \Rightarrow \quad c^2+d^2=0.$

Furthermore,

$\begin{aligned} 0 \leq c^2 &= c^2 + 0 \\ &\leq c^2 + d^2 = 0. \end{aligned}$

Hence, $c = 0$ . Similarly, $d = 0$ . Hence $\mathbf v= \begin{bmatrix} 0 \\ 0 \end{bmatrix}= \mathbf 0$ .

The third property is straightforward:

$\begin{aligned} \mathbf u\cdot\mathbf v &= ac+bd \\ &=ca+db=\mathbf v\cdot\mathbf u. \end{aligned}$

Recall that $\alpha \mathbf v = \alpha \begin{bmatrix} c \\d \end{bmatrix} = \begin{bmatrix} \alpha c \\ \alpha d\end{bmatrix}$ . Then

$\begin{aligned} \mathbf u\cdot(\alpha\mathbf v) &= a(\alpha c)+b(\alpha d) \\ &= \alpha(ac+bd) \\ &= \alpha(\mathbf u\cdot\mathbf v), \end{aligned}$

Define $\mathbf w=\begin{bmatrix} p \\ q \end{bmatrix}$ . Then the fifth property is immediate:

$\begin{aligned} \mathbf u\cdot(\mathbf v+\mathbf w) &= a(c+p)+b(d+q) \\ &= (ac+bd)+(ap+bq) \\ &= \mathbf u\cdot\mathbf v+\mathbf u\cdot\mathbf w. \end{aligned}$

Problem 3. Explain why $\| \mathbf v \| = \sqrt{ \mathbf v \cdot \mathbf v }$ . Deduce the following:
- $\|\mathbf v \| \geq 0$ .
- $\|\mathbf v \| = 0 \Rightarrow \mathbf v = \mathbf 0$ .
- $\| \alpha \mathbf v \| = |\alpha| \|\mathbf v \|$ .
- $| \mathbf u \cdot \mathbf v | \leq \| \mathbf u \| \| \mathbf v\|$ .
- $\| \mathbf u + \mathbf v \| \leq \| \mathbf u \| + \| \mathbf v \|$ .
(Click for Solution)

Solution. Using Remark 1 and Pythagoras’ theorem,

$\| \mathbf v \| = \sqrt{c^2 + d^2} = \sqrt{ \mathbf v \cdot \mathbf v }.$

Therefore, we obtain the properties rather straightforwardly:

$\|\mathbf v \| = \sqrt{\mathbf v \cdot \mathbf v} \geq \sqrt{0} = 0$

with equality if and only if $\mathbf v \cdot \mathbf v = 0 \iff \mathbf v = \mathbf 0$ . Next,

$\begin{aligned} \| \alpha \mathbf v \| &= \sqrt{(\alpha \mathbf v) \cdot(\alpha \mathbf v)} \\ &= |\alpha | \sqrt{\mathbf v \cdot \mathbf v} \\ &= |\alpha| \|\mathbf v \|. \end{aligned}$

The fourth property, known as the Cauchy-Schwarz inequality, follows from Problem 1 and the observation that $|{\cos(\theta)}| \leq 1$ :

$\begin{aligned} |\mathbf u \cdot \mathbf v| &= \| \mathbf u \| \|\mathbf v\| |{\cos(\theta)}| \\ &\leq \| \mathbf u \| \|\mathbf v\| \cdot 1 \\ &=\| \mathbf u \| \|\mathbf v\|. \end{aligned}$

The fifth property follows from the fourth:

$\begin{aligned} \| \mathbf u + \mathbf v \|^2 &= (\mathbf u + \mathbf v) \cdot (\mathbf u + \mathbf v) \\ &= (\mathbf u \cdot \mathbf u) + (\mathbf u \cdot \mathbf v) + (\mathbf v \cdot \mathbf u) + (\mathbf v \cdot \mathbf v) \\ &= \| \mathbf u \|^2 + 2(\mathbf u \cdot \mathbf v) + \| \mathbf v \|^2 \\ &\leq \| \mathbf u \|^2 + 2 \| \mathbf u \| \| \mathbf v \| + \| \mathbf v \|^2 \\ &= (\| \mathbf u \| + \| \mathbf v \|)^2 \end{aligned}$

and taking square roots on both sides.

Problem 4. Define $d(\mathbf u, \mathbf v) := \| \mathbf u - \mathbf v\|$ . Show that the following hold:
- $d(\mathbf u, \mathbf v) \geq 0$ .
- $d(\mathbf u, \mathbf v) = 0 \Rightarrow \mathbf u = \mathbf v$ .
- $d(\mathbf u, \mathbf v) = d(\mathbf v, \mathbf u)$ .
- $d(\mathbf u, \mathbf v) \leq d(\mathbf u, \mathbf w) + d(\mathbf w, \mathbf v)$ .
(Click for Solution)

Solution. The results in Problem 4 comes from Problem 3:

$d(\mathbf u, \mathbf v) = \| \mathbf u - \mathbf v \| \geq 0,$

then

$\begin{aligned} d(\mathbf u, \mathbf v) = 0 \quad &\Rightarrow \quad \| \mathbf u - \mathbf v \| = 0 \\ &\Rightarrow \quad \mathbf u - \mathbf v = \mathbf 0 \\&\Rightarrow \quad \mathbf u = \mathbf v. \end{aligned}$

Furthermore,

$\begin{aligned} d(\mathbf u, \mathbf v) &= \| \mathbf u - \mathbf v \| \\ &= \| (-1)(\mathbf v - \mathbf u) \| \\ &= |{-1}| \| \mathbf v - \mathbf u \| \\ &= \| \mathbf v - \mathbf u \| \\ &= d(\mathbf v , \mathbf u) \end{aligned}$

and

$\begin{aligned} d(\mathbf u, \mathbf v) &= \| \mathbf u - \mathbf v \| \\ &= \| (\mathbf u - \mathbf w) + (\mathbf w - \mathbf v) \| \\ &\leq \| \mathbf u - \mathbf w \| + \| \mathbf w - \mathbf v \| \\ &= d(\mathbf u, \mathbf w) + d(\mathbf w, \mathbf v). \end{aligned}$

Remark 2. In the language of linear algebra, Remark 1 and Problem 2 defines a “multiplication” on the set of two-dimensional vectors, turning it into a real inner product space. Problem 3 show that inner product spaces are normed spaces (in that two-dimensional vectors have lengths or norms), while Problem 4 shows that normed spaces are metric spaces (i.e. the notion of distance is a reasonable one).

Problem 5. Given that the point $(x_0, y_0)$ lies on the line $ax+by = c$ , show that any other point $(x,y)$ lies on the line if and only if

$\begin{bmatrix} x - x_0 \\ y - y_0 \end{bmatrix} \cdot \begin{bmatrix} a \\ b \end{bmatrix} = 0.$

(Click for Solution)

Solution. Since $(x_0, y_0)$ lies on the line, we have

$ax_0 + by_0 = c.$

The point $(x,y)$ lies on the line if and only if

$ax + by = c.$

Subtracting the equations yields the equation

$a(x-x_0) + b(y-y_0) = 0.$

On the other hand,

$\begin{aligned} \begin{bmatrix} x - x_0 \\ y - y_0 \end{bmatrix} \cdot \begin{bmatrix} a \\ b \end{bmatrix} &= (x-x_0)a + (y-y_0)b \\ &= a(x-x_0) + b(y-y_0). \end{aligned}$

Then $(x,y)$ lies on the line if and only if the right-hand side equals $0$ , that is,

$\begin{bmatrix} x - x_0 \\ y - y_0 \end{bmatrix} \cdot \begin{bmatrix} a \\ b \end{bmatrix} = 0.$

Remark 3. Problem 5 generalises to an $(n-1)$ -dimensional hyperplane: Given that the point $(v_1, \dots, v_n)$ lies on the $(n-1)$ -dimensional hyperplane with equation

$a_1 x_1 + a_2 x_2 + \cdots + a_n x_n = c,$

any other point $(u_1,\dots, u_n)$ lies on the hyperplane if and only if

$\begin{bmatrix} u_1 - v_1 \\ \vdots \\ u_n - v_n \end{bmatrix} \cdot \begin{bmatrix} a_1 \\ \vdots \\ a_n \end{bmatrix} = 0.$

—Joel Kindiak, 30 Jan 26, 1810H
May 6, 2026
The Cubic Formula

Let $b,c,d$ be constants with $a \neq 0$ .

In this post, we develop a formula to solve the cubic equation

$x^3 + bx^2 + cx + d = 0.$

Problem 1. Using the substitution $x = y + h$ , determine the constants $B,C,D$ in terms of $b,c,d,h$ such that

$x^3 + bx^2 + cx + d = y^3 + By^2 + Cy + D.$

Deduce the value of $h$ such that the right-hand side has no $y^2$ term.

(Click for Solution)

Solution. Making the substitution on the left-hand side,

$\begin{aligned} x^3 + bx^2 + cx + d &= (y+h)^3 + b(y+h)^2 \\ &\phantom{=.} + c(y+h) + d \\ &= (y^3 + 3y^2h + 3yh^2 + h^3) \\ &\phantom{=.} + b(y^2 + 2yh + h^2) \\ &\phantom{=.} + c(y+h) + d \\ &= y^3 + (3h + b)y^2 \\ &\phantom{=.} + (3h^2 + 2bh + c)y \\ &\phantom{=.} + (h^3 + bh^2 + ch + d). \end{aligned}$

Therefore,

$\begin{aligned} B &= 3h + b, \\ C &= 3h^2 + 2bh + c, \\ D &= h^3 + bh^2 + ch + d. \end{aligned}$

Hence, we set $B = 0 \iff h = -b/3$ .

Remark 1. Substituting $x = y - b/3$ ,

$x^3 + bx^2 + cx + d = 0 \quad \iff \quad y^3 + Cy + D = 0.$

Therefore, solving the right-hand side, called the depressed cubic, allows us to solve the left-hand side.

Problem 2 (Viète’s Substitution). Making the substitution

$\displaystyle y = z - \frac C{3z}$

into the right-hand side, obtain an equation that is quadratic in $z^3$ .

(Click for Solution)

Solution. By performing painful algebra,

$\begin{aligned} y^3 + Cy &= y(y^2 + C) \\ &= \left( z - \frac C{3z} \right) \left( \left( z - \frac C{3z} \right)^2 + C \right) \\ &= \left( z - \frac C{3z} \right) \left( z^2 - 2 \cdot z \cdot \frac{C}{3z} + \frac{C^2}{(3z)^2} + C \right) \\ &= \left( z - \frac C{3z} \right) \left( z^2 + \frac{C^2}{9z^2} + \frac C3 \right) \\ &= z^3 -\frac{C^3}{27z^3}. \end{aligned}$

Therefore,

$\begin{aligned} y^3 + Cy + D &= 0\\ z^3 -\frac{C^3}{27z^3} + D &= 0 \\ (z^3)^2 + Dz^3 - \frac{C^3}{27} &= 0. \end{aligned}$

Assume $D^2/4 + C^3/27 \geq 0$ .

Problem 3. Using Problems 1 and 2, solve the equation

$x^3 + bx^2 + cx + d = 0.$

(Click for Solution)

Solution. Using Problem 2, we solve for $z^3$ using the quadratic formula: $m := -D/2$ and $p := -C^2/27$ , so that

$\begin{aligned} m^2 - p &= \left( -\frac D2 \right)^2 - \left( - \frac{C^3}{27} \right) \\ &= \frac{D^2}{4} + \frac{C^2}{27}. \end{aligned}$

yields

$\begin{aligned} z^3 &= m \pm \sqrt{m^2 - p} \\ z^3 &= -\frac D2 \pm \sqrt{\frac{D^2}{4} + \frac{C^3}{27}} \\ z_{\pm} &:= \sqrt[3]{ -\frac D2 \pm \sqrt{\frac{D^2}{4} + \frac{C^3}{27}} }. \end{aligned}$

Substituting, we get two roots $y_{\pm} := z_{\pm}^3 - C^3/(27z_{\pm}^3)$ .

Denote the final root by $y_0$ . By comparing the constant term (i.e. set $y = 0$ ) on both sides of the equation

$\displaystyle (y-y_+)(y-y_-)(y-y_0) = y^3 + Cy + D,$

we get $y_0 = -D/(y_+ y_-)$ . The final solution is then recovered by the substitution in Problem 1:

$\displaystyle x = y_+ - \frac b3,\quad y_- - \frac b3,\quad y_0 - \frac b3.$

—Joel Kindiak, 24 Jan 26, 2056H

April 16, 2026
The Shoelace Formula

Problem 1. Consider the triangle below.

Show that the triangle has area $\frac 12 (x_1 y_2 - x_2 y_1)$ .

(Click for Solution)

Solution. Add dashed lines to draw in “phantom” triangles.

We subtract the area of the three smaller triangles from the larger rectangle:

$\begin{aligned} &\textstyle x_1 y_2 - \frac 12 x_1 y_1 - \frac 12 (x_1 - x_2)(y_2 - y_1) - \frac 12 x_2 y_2 \\ & = \textstyle x_1 y_2 - \frac 12 (x_1 y_1 + (x_1 - x_2)(y_2 - y_1) + x_2 y_2) \\ & = \textstyle x_1 y_2 - \frac 12 (x_1 y_1 + (x_1 y_2 - x_1 y_1 - x_2 y_2 + x_2 y_1) + x_2 y_2) \\ & = \textstyle x_1 y_2 - \frac 12 ( x_1 y_2 + x_2 y_1 ) \\ &= \textstyle \frac 12 (2x_1 y_2 - (x_1 y_2 + x_2 y_1)) \\ &= \textstyle \frac 12 (x_1 y_2 - x_2 y_1). \end{aligned}$

Remark 1. There are many other kinds of triangles, but with enough patience, they can all be shown to yield the same area formula.

Definition 1. Define

$\begin{aligned} \left| \begin{matrix} x_1 & x_2 \\ y_1 & y_2 \end{matrix} \right| &:= x_1 y_2 - x_2 y_1. \end{aligned}$

In particular, the area of the triangle in Problem 1 can be written as $\displaystyle \frac 12 \left| \begin{matrix} x_1 & x_2 \\ y_1 & y_2 \end{matrix} \right|$ .

Problem 2. Show that $\left| \begin{matrix} x_2 & x_1 \\ y_2 & y_1 \end{matrix} \right| = -\left| \begin{matrix} x_1 & x_2 \\ y_1 & y_2 \end{matrix} \right|$ . In particular,

$\left| \begin{matrix} x_2 & x_1 \\ y_2 & y_1 \end{matrix} \right| + \left| \begin{matrix} x_1 & x_2 \\ y_1 & y_2 \end{matrix} \right| = 0.$

(Click for Solution)

Solution. By definition,

$\begin{aligned} \left| \begin{matrix} x_2 & x_1 \\ y_2 & y_1 \end{matrix} \right| &= (x_2y_1 - x_1y_2) \\ &= -(x_1 y_2 - x_1 y_2) \\ &= -\left| \begin{matrix} x_1 & x_2 \\ y_1 & y_2 \end{matrix} \right|. \end{aligned}$

Remark 1. In particular, a polygon is said to have positive orientation if its coordinates are traversed in anti-clockwise direction, and have negative orientation if its coordinates are traversed in clock-wise direction.

Problem 3. Consider the triangle below with positive orientation.

Show that the triangle has area

$\displaystyle \frac 12 \left(\left| \begin{matrix} x_1 & x_2 \\ y_1 & y_2 \end{matrix} \right| + \left| \begin{matrix} x_2 & x_3 \\ y_2 & y_3 \end{matrix} \right| + \left| \begin{matrix} x_3 & x_1 \\ y_3 & y_1 \end{matrix} \right|\right).$

(Click for Solution)

Solution. Draw some “phantom” triangles:

Denote the area of the triangle by $A$ . By Problem 1,

$\begin{aligned}A + \frac 12 \left| \begin{matrix} x_1 & x_3 \\ y_1 & y_3 \end{matrix} \right| &= \frac 12 \left| \begin{matrix} x_1 & x_2 \\ y_1 & y_2 \end{matrix} \right| + \frac 12 \left| \begin{matrix} x_2 & x_3 \\ y_2 & y_3 \end{matrix} \right|. \end{aligned}$

By Problem 2,

$\begin{aligned}A &= \frac 12 \left(\left| \begin{matrix} x_1 & x_2 \\ y_1 & y_2 \end{matrix} \right| + \left| \begin{matrix} x_2 & x_3 \\ y_2 & y_3 \end{matrix} \right| - \left| \begin{matrix} x_1 & x_3 \\ y_1 & y_3 \end{matrix} \right| \right) \\ &= \frac 12 \left(\left| \begin{matrix} x_1 & x_2 \\ y_1 & y_2 \end{matrix} \right| + \left| \begin{matrix} x_2 & x_3 \\ y_2 & y_3 \end{matrix} \right| + \left| \begin{matrix} x_3 & x_1 \\ y_3 & y_1 \end{matrix} \right| \right). \end{aligned}$

Remark 2. In a similar manner, for triangles with other orientations, as long as the vertices are traversed in an anti-clockwise direction, we still recover the same formula as in Problem 3.

Definition 2. Make the notation

$\begin{aligned} \left| \begin{matrix} x_1 & x_2 & \cdots & x_n \\ y_1 & y_2 & \cdots & y_n\end{matrix} \right| &:= \left| \begin{matrix} x_1 & x_2 \\ y_1 & y_2 \end{matrix} \right| + \left| \begin{matrix} x_2 & x_3 \\ y_2 & y_3 \end{matrix} \right| + \cdots + \left| \begin{matrix} x_{n-1} & x_n \\ y_{n-1} & y_n \end{matrix} \right|. \end{aligned}$

This notation is commonly referred to as the shoelace formula. In particular, the area in Problem 3 can be expressed as

$\displaystyle \frac 12 \left| \begin{matrix} x_1 & x_2 & x_3 & x_1 \\ y_1 & y_2 & y_3 & y_1 \end{matrix} \right|.$

Problem 4. Show that

$\left| \begin{matrix} x_1 & \cdots & x_k \\ y_1 & \cdots & y_k\end{matrix} \right| + \left| \begin{matrix} x_k & \cdots & x_n \\ y_k & \cdots & y_n\end{matrix} \right| = \left| \begin{matrix} x_1 & x_2 & \cdots & x_n \\ y_1 & y_2 & \cdots & y_n\end{matrix} \right|.$

(Click for Solution)

Solution. Patiently expand the right-hand side:

$\begin{aligned} &\left| \begin{matrix} x_1 & \cdots & x_k & \cdots & x_n \\ y_1 & \cdots & y_k & \cdots & y_n\end{matrix} \right| \\ &= \left| \begin{matrix} x_1 & x_2 \\ y_1 & y_2 \end{matrix} \right| + \cdots + \left| \begin{matrix} x_{k-1} & x_k \\ y_{k-1} & y_k \end{matrix} \right| + \left| \begin{matrix} x_k & x_{k+1} \\ y_k & y_{k+1} \end{matrix} \right| + \cdots + \left| \begin{matrix} x_{n-1} & x_n \\ y_{n-1} & y_n \end{matrix} \right| \\ &= \left| \begin{matrix} x_1 & \cdots & x_k \\ y_1 & \cdots & y_k\end{matrix} \right| + \left| \begin{matrix} x_k & \cdots & x_n \\ y_k & \cdots & y_n\end{matrix} \right| . \end{aligned}$

Problem 5. Consider the quadrilateral below.

Show that the quadrilateral has area

$\displaystyle \frac 12 \left| \begin{matrix} x_1 & x_2 & x_3 & x_4 & x_1 \\ y_1 & y_2 & y_3 & y_4 & y_1 \end{matrix} \right|.$

(Click for Solution)

Solution. Connect $(x_1,y_1)$ to $(x_3,y_3)$ .

The quadrilateral consists of two triangles, and therefore its area is given by the sum of both triangles, whose formulas are given by Problem 4:

$\begin{aligned} & \left| \begin{matrix} x_1 & x_2 & x_3 & x_1 \\ y_1 & y_2 & y_3 & y_1 \end{matrix} \right| + \left| \begin{matrix} x_1 & x_3 & x_4 & x_1 \\ y_1 & y_3 & y_4 & y_1 \end{matrix} \right| \\ &= \left| \begin{matrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \end{matrix} \right| + \underbrace{ \left| \begin{matrix} x_3 & x_1 \\ y_3 & y_1 \end{matrix} \right| + \left| \begin{matrix} x_1 & x_3 \\ y_1 & y_3 \end{matrix} \right| }_0 + \left| \begin{matrix} x_3 & x_4 & x_1 \\ y_3 & y_4 & y_1 \end{matrix} \right| \\ &= \left| \begin{matrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \end{matrix} \right| + \left| \begin{matrix} x_3 & x_4 & x_1 \\ y_3 & y_4 & y_1 \end{matrix} \right| \\ &= \left| \begin{matrix} x_1 & x_2 & x_3 & x_4 & x_1 \\ y_1 & y_2 & y_3 & y_4 & y_1 \end{matrix} \right|, \end{aligned}$

where the $0$ comes from Problem 2. Multiplying both sides by $1/2$ yields the formula

$\displaystyle \frac 12 \left| \begin{matrix} x_1 & x_2 & x_3 & x_4 & x_1 \\ y_1 & y_2 & y_3 & y_4 & y_1 \end{matrix} \right|.$

Remark 2. Using mathematical induction and the same strategy in Problem 5, the signed area of an $n$ -gon with vertices

$(x_1, y_1), (x_2,y_2), \dots, (x_n, y_n)$

can be shown to equal

$\displaystyle \frac 12 \left| \begin{matrix} x_1 & x_2 & \cdots & x_n & x_1 \\ y_1 & y_2 & \cdots & y_n & y_1 \end{matrix} \right|.$

Furthermore, this quantity is positive if the vertices are traversed in an anti-clockwise direction and negative if the vertices are traversed in a clockwise direction.

—Joel Kindiak, 24 Jan 26, 1722H

April 15, 2026
Quadri-literacy
Definition 1. A quadrilateral is a four-sided shape.

For example, a rectangle is a quadrilateral whose internal angles are all $90^\circ$ .

Definition 2. A rhombus is a quadrilateral with equal side lengths.

Problem 1. Show that a quadrilateral is a square if and only if it is both a rectangle and a rhombus.
(Click for Solution)

Solution. $(\Leftarrow)$ Suppose a quadrilateral is both a rectangle and a rhombus.
- As a rectangle, all of its angles are $90^\circ$ .
- As a rhombus, all of its side lengths are equal.
Therefore, it must be a square.

$(\Rightarrow)$ Trivial.
Definition 3. A parallelogram is a quadrilateral with two pairs of parallel lines.

Problem 2. Show that the following are equivalent for a given quadrilateral $Q$ :
1. $Q$ is a parallelogram,
2. opposite sides in $Q$ are equal,
3. opposite angles in $Q$ are equal.
(Click for Solution)

Solution. $(1 \Rightarrow 2)$ Consider the parallelogram $Q = ABCD$ .

Draw the diagonal $BD$ . Since $AB \parallel DC$ , by alternate angles,

$\angle ABD = \angle CDB.$

As a common side, $BD = DB$ . Since $AD \parallel BC$ , by alternate angles,

$\angle ADB = \angle CBD.$

By the ASA Criterion, $\Delta ABD \equiv \Delta CDB$ . In particular, $AB = CD$ and $AD = CB$ , so that its opposite sides are equal.

$(2 \Rightarrow 3)$ Draw the diagonal and use the SSS Criterion to conclude that opposite angles equal each other.

$(3 \Rightarrow 1)$ Let $\alpha,\beta,\alpha,\beta$ denote the angles of the quadrilateral.

Since angles in a quadrilateral sum to $360^\circ$ ,

$2(\alpha + \beta) = 360^\circ \quad \Rightarrow \quad \alpha + \beta = 180^\circ.$

Since this equality holds for any pair of interior angles, the parallelogram must have two pairs of parallel sides.

Problem 3. Show that a rectangle is always a parallelogram. Furthermore, show that a parallelogram is a rectangle if and only if it has at least one interior right angle.

(Click for Solution)

Solution. Since all angles in a rectangle is $90^\circ$ , opposite pairs of angles are equal.

By Problem 2, a rectangle is a parallelogram.

$(\Leftarrow)$ Let $90^\circ, \alpha, \beta, \gamma$ denote the interior angles of the parallelogram, labelled anti-clockwise.

By Problem 2, $\alpha = \gamma$ and $\beta = 90^\circ$ . Since angles in a quadrilateral sum to $360^\circ$ ,

$\alpha + 90^\circ + \alpha + 90^\circ = 360^\circ \quad \Rightarrow \quad \alpha = 90^\circ.$

Therefore, all angles equal $90^\circ$ , and the parallelogram is a rectangle.

$(\Rightarrow)$ Trivial.

Problem 4. Show that a rhombus is always a parallelogram. Furthermore, show that a parallelogram is a rhombus if and only if it has at least one pair of equal adjacent sides.

(Click for Solution)

Solution. Since opposite sides in a rhombus are equal, by Problem 2, a rhombus is a parallelogram.

$(\Leftarrow)$ Let $w,x,y,z$ denote the sides of a parallelogram, labelled anti-clockwise.

By Problem 2, $w=y$ and $x=z$ . By hypothesis, suppose $w=x$ without loss of generality. Then $z=x=w=y$ . Therefore, all side lengths are equal, and the parallelogram is a rhombus.

$(\Rightarrow)$ Trivial.

Problem 5. Show that a rectangle is a square if and only if it has at least one pair of equal adjacent sides. Likewise, show that a rhombus is a square if and only if it has at least one interior right angle.
(Click for Solution)

Solution. We first prove the rectangle claim:
- $(\Leftarrow)$ By Problem 3, a rectangle is a parallelogram. By hypothesis and Problem 4, it is a rhombus. By Problem 1, it is a square.
- $(\Rightarrow)$ Trivial.
The rhombus claim follows similarly:
- $(\Leftarrow)$ By Problem 4, a rhombus is a parallelogram. By hypothesis and Problem 2, it is a rectangle. By Problem 1, it is a square.
- $(\Rightarrow)$ Trivial.
Definition 4. A trapezium is a quadrilateral with at least one pair of opposite sides that are parallel.

Problem 6. Show that a parallelogram is always a trapezium. Furthermore, show that a trapezium is a parallelogram if and only if it has at least one pair of equal opposite angles.

(Click for Solution)

Solution. Since a parallelogram has two pairs of equal and parallel sides, it has at least one pair of parallel sides, and is thus a trapezium.

$(\Leftarrow)$ Denote the angles of the trapezium by $\alpha, \beta, \alpha, \theta$ .

Given the pair of parallel sides, interior angles are supplementary, so that

$\alpha +\beta = 180^\circ = \alpha + \theta \quad \Rightarrow \quad \beta = \theta.$

By Problem 2, the trapezium is a parallelogram.

$(\Rightarrow)$ Trivial.

Definition 5. A kite is a quadrilateral with two pairs of adjacent sides that are equal in length.

Problem 7. Show that a kite has at least one pair of equal opposite angles and that its diagonals are perpendicular to each other.

(Click for Solution)

Solution. Consider the kite $ABCD$ below.

As base angles of isosceles triangles,

$\angle CBD = \angle CDB, \quad \angle ABD = \angle ADB.$

Hence,

$\begin{aligned} \angle ABC &= \angle ABD + \angle CBD \\ &= \angle ADB + \angle CDB \\ &= \angle ADC, \end{aligned}$

as required. Using the SAS Criterion, $\Delta ABC \equiv \Delta ADC$ . In particular, $\angle BAR = \angle DAR$ . As a kite, $AB = AD$ . As base angles of an isosceles triangle,

$\angle ABR = \angle ABD = \angle ADB = \angle ADR.$

Using the ASA Criterion, $\Delta ABR \equiv \angle ADR$ , so that $\angle ARB = \angle ARD$ . Since adjacent angles on a straight line are supplementary, $\angle ARB + \angle ARD = 180^\circ$ . Solving, $\angle ARB = 90^\circ$ .

Problem 8. Show that a quadrilateral is a rhombus if and only if it is both a kite and a trapezium.

(Click for Solution)

Solution. $(\Leftarrow)$ Suppose the quadrilateral is both a kite and a trapezium. By Problem 7, it has at least one pair of equal opposite angles. By Problem 6, it is a parallelogram. As a kite, it has at least one pair of equal adjacent sides. By Problem 4, it is a rhombus.

$(\Rightarrow)$ Trivial.

—Joel Kindiak, 24 Jan 26, 1651H
April 15, 2026
Quadratic Identities

Problem 1. Verify the following quadratic identities:

$\begin{aligned} (a+b)^2 &= a^2 + 2ab + b^2, \\ (a-b)^2 &= a^2 - 2ab + b^2, \\ a^2 - b^2 &= (a+b)(a-b). \end{aligned}$

(Click for Solution)

Solution. By the rainbow method,

$\begin{aligned} (a+b)^2 &= (a+b)(a+b) \\ &= a(a+b) + b(a+b) \\ &= a^2 + ab + ba + b^2 \\ &= a^2 + ab + ab + b^2 \\ &= a^2 + 2ab + b^2. \end{aligned}$

Replacing $b$ with $-b$ ,

$\begin{aligned} (a-b)^2 &= (a+(-b))^2 \\ &= a^2 + 2a(-b) + (-b)^2 \\ &= a^2 - 2ab + b^2. \end{aligned}$

Finally,

$\begin{aligned}(a+b)(a-b) &= a(a-b) + b(a-b) \\ &= a^2 - ab + ba - b^2 \\ &= a^2 - ab+ab - b^2 \\ &= a^2 - b^2.\end{aligned}$

Let $b, c$ be integers.

Problem 2. Define $m := -b/2$ and $p:= c$ . Show that the quadratic equation

$x^2 + bx + c = 0$

has solutions given by $x = m \pm \sqrt{m^2 - p}$ .

(Click for Solution)

Solution. Write $b = -2m$ . Using the quadratic formula, the solutions are given by

$\begin{aligned} x &= \frac{-b \pm \sqrt{b^2 - 4 \cdot 1 \cdot c}}{2 \cdot 1} \\ &= \frac{2m \pm \sqrt{(2m)^2 - 4p}}{2} \\ &= \frac{2m \pm \sqrt{4m^2 - 4p}}{2} \\ &= \frac{2m \pm \sqrt 4 \cdot \sqrt{m^2 - p}}{2} \\ &= \frac{2m \pm 2 \sqrt{m^2 - p}}{2} \\ &= m \pm \sqrt{m^2 - p}.\end{aligned}$

Problem 3. Given that the quadratic equation

$x^2 + bx + c = 0.$

has roots $r, s$ , evaluate $b,c$ in terms of $r,s$ . Deduce that

$x^2 + bx + c = (x-r)(x-s),$

and show that $\Delta = 4(s-r)^2$ .

(Click for Solution)

Solution. Using Problem 2, suppose

$r = m - \sqrt{m^2 - p},\quad s = m + \sqrt{m^2 - p}.$

Then $r+s = 2m = -b$ and by Problem 1,

$\begin{aligned} rs &= (m - \sqrt{m^2 - p}) \cdot (m + \sqrt{m^2 - p})\\ &= m^2 - (m^2-p) \\ &= p = c. \end{aligned}$

Therefore, $b = -(r+s)$ and $c = rs$ . Hence,

$\begin{aligned} x^2 + bx + c &= x^2 -(r+s)x + rs \\ &= x^2 -rx-xs + rs \\ &= (x-r)x-(x-r)s \\ &= (x-r)(x-s). \end{aligned}$

It follows that $\Delta = 4 (m^2 - p) = 4(s-r)^2$ .

Now suppose $c$ is a positive integer.

Problem 4. Suppose a quadratic equation of the form

$x^2 - x - c = 0$

has integer solutions with larger solution $r$ . Evaluate $c$ in terms of $r$ .

(Click for Solution)

Solution. By Problem 3,

$-(r+s) = -1\quad \Rightarrow \quad s = 1-r = -(r-1)$

and $c = -rs = -r(1-r) = r(r-1)$ .

Problem 5. Show that a quadratic equation of the form

$x^2 - x + c = 0$

has no real roots.

(Click for Solution)

Solution. Since $c \geq 1$ , by calculating the discriminant of the quadratic equation,

$\begin{aligned} \Delta &= (-1)^2 - 4 \cdot 1 \cdot c \\ &= 1 - 4c \leq 1 - 4 \cdot 1 \\ &= -3 < 0. \end{aligned}$

Therefore, the quadratic has no real roots.

Problem 6. Determine the values of $r > 0$ such that

$x^2 - x + r = 0$

has real roots.

(Click for Solution)

Solution. We require $\Delta \geq 0$ :

$(-1)^2 - 4 \cdot 1 \cdot r \geq 0 \quad \iff \quad r \leq 1/4.$

Therefore, we require $0 < r \leq 1/4$ .

Problem 7. Given non-negative numbers $a, b$ , show that

$\displaystyle \sqrt{ab} \leq \frac{a+b}{2}.$

This inequality is a special case of the arithmetic mean-geometric mean (AM-GM) inequality.

(Click for Solution)

Solution. Using Problem 1,

$\begin{aligned} 0 &\leq (\sqrt{a}-\sqrt{b})^2 \\ &= (\sqrt{a})^2 - 2\sqrt{a}\sqrt{b} + (\sqrt{b})^2 \\ &= a - 2\sqrt{ab} + b. \end{aligned}$

By algebruh, $2\sqrt{ab}\leq a+b$ . Dividing by $2$ yields the desired result.

—Joel Kindiak, 24 Jan 26, 0213H

April 14, 2026
An Absurd Puzzle

Problem 1. Evaluate

$\displaystyle S := \frac 1{\sqrt 1 + \sqrt 2} + \frac 1{\sqrt 2 + \sqrt 3} + \cdots + \frac 1{\sqrt{2024} + \sqrt{2025}}.$

(Click for Solution)

Solution. By rationalising the denominator,

$\begin{aligned} \frac 1{\sqrt n + \sqrt{n+1}} &= \frac 1{\sqrt n + \sqrt{n+1}} \cdot \frac{ \sqrt n - \sqrt{n+1} }{ \sqrt n - \sqrt{n+1} } \\ &= \frac{ \sqrt n - \sqrt{n+1} }{ (\sqrt n)^2 - (\sqrt{n+1})^2 } \\ &= \frac{ \sqrt n - \sqrt{n+1} }{ n - (n+1) } \\ &= \frac{ \sqrt n - \sqrt{n+1} }{ -1 } \\ &= \sqrt{n+1} - \sqrt{n}. \end{aligned}$

Hence, summing the terms,

$\begin{aligned} S &=\frac 1{\sqrt 1 + \sqrt 2} + \frac 1{\sqrt 2 + \sqrt 3} + \cdots + \frac 1{\sqrt{2023} + \sqrt{2024}} + \frac 1{\sqrt{2024} + \sqrt{2025}} \\ &= \sqrt 2 + \sqrt 3 + \cdots + \sqrt{2024} + \sqrt{2025} \\ &\phantom{=} - \sqrt{1} - \sqrt 2 - \cdots - \sqrt{2023} - \sqrt{2024} \\ &= \sqrt{2025} - 1 \\ &= 45 - 1 = 44. \end{aligned}$

Problem 2. Evaluate

$\displaystyle T := \frac 1{1 \times 2} + \frac 1{2 \times 3} + \cdots + \frac 1{n(n+1)}$

in terms of $n$ .

(Click for Solution)

Solution. For any $k$ ,

$\begin{aligned} \frac 1{k(k+1)} &= \frac {(k+1)-k}{k(k+1)} \\ &= \frac{k+1}{k(k+1)} - \frac k{k(k+1)} \\ &= \frac 1k - \frac 1{k+1}. \end{aligned}$

Hence,

$\begin{aligned} T &= \frac 1{1 \times 2} + \frac 1{2 \times 3} + \cdots + \frac 1{(n-1)n} + \frac 1{n(n+1)} \\ &= \frac 11 + \frac 12 + \cdots + \frac 1{n-1} + \frac 1n \\ &\phantom{=} - \frac 12 - \frac 13 - \cdots - \frac 1n - \frac 1{n+1} \\ &= 1 - \frac 1{n+1} = \frac{n}{n+1}. \end{aligned}$

Remark 1. In both problems, we used the method of differences or a telescoping sum (or more fancifully, the fundamental theorem of discrete calculus), which states that if $u_n = v_{n+1} - v_n$ , then

$u_1 + u_2 + \cdots + u_n = v_{n+1} - v_1.$

—Joel Kindiak, 19 Jan 26, 2204H

April 9, 2026

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