What is the cosine of the angle between two non-zero vectors ?

January 29, 2024January 31, 2024 / Michael Xue / Leave a comment

If $\vec{V_1}$ and $\vec{V_2}$ are two non-zero vectors, and $\vec{V_3}$ is defined as $\vec{V_1} - \vec{V_2}$ , then on one hand,

$|\vec{V_3}|^2 = <\vec{V_3},\vec{V_3}>\;=\;<\vec{V_2}-\vec{V_1},\vec{V_2}-\vec{V_1}>\;=\;|\vec{V_1}|^2+|\vec{V_2}|^2-2<\vec{V_1},\vec{V_2}>.\quad(1)$

On the other hand, applying the law of cosine results in

$|\vec{V_3}|^2 = |\vec{V_1}|^2 + |\vec{V_2}|^2-2|\vec{V_1}||\vec{V_2}|\cos(\theta).\quad\quad\quad(2)$

Equating (1) and (2), we obtain

$\underbrace{ |\vec{V_1}|^2 + |\vec{V_2}|^2 - 2|\vec{V_1}||\vec{V_2}|\cos(\theta)}_{(2)}=\underbrace{|\vec{V_1}|^2+|\vec{V_2}|^2-2<\vec{V_1},\vec{V_2}>}_{(1)}.$

This leads to the conclusion:

$|\vec{V_1}||\vec{V_2}|\cos(\theta) = <\vec{V_1},\vec{V_2}> \implies \cos(\theta) = \frac{<\vec{V_1},\vec{V_2}>}{|\vec{V_1}||\vec{V_2}|}.$

An Area-Based Proof

January 17, 2024January 19, 2024 / Michael Xue / Leave a comment

There are numerous proofs of the Pythagorean Theorem.

The one attributed to president James Garfield is notably compelling:

Having examined his Area-Based approach, we present the following:

Consider a right triangle $\Delta ABC$ where $AC \perp BC$ and $DE \perp BC$ (Fig. 1)

Fig. 1

It can be affirmed that $\frac{BE}{BC}= \frac{DE}{AC} = \frac{BD}{AB}.$ Furthermore, $\frac{CE}{BE} = \frac{AD}{BD}.$

From Fig. 1,

$S_{\Delta ABC} = S_{\Delta BDE} + S_{\text{trapzoid}\;ACDE}$

$\implies \frac{1}{2} BC \cdot AC = \frac{1}{2}BE \cdot DE + \frac{1}{2}(DE+AC)\cdot(BC-BE)$

$\implies BC\cdot AC = BE\cdot DE + DE\cdot BC -DE\cdot BE +AC\cdot BC - AC\cdot BE$

$\implies DE\cdot BC=AC\cdot BE.$

So,

$\frac{BE}{BC} = \frac{DE}{AC}.$

Moreover, letting

$\frac{BE}{BC} = \frac{DE}{AC} = k\quad\quad\quad(1)$

gives

$BE = k \cdot BC$ and $DE=k \cdot AC.\quad\quad\quad(2)$

By Pythagorean Theorem,

$AB = \sqrt{BC^2+AC^2}\quad\quad\quad(3)$

and,

$BD = \sqrt{BE^2+DE^2}$

$\overset{(2)}{\implies} BD = \sqrt{(k\cdot BC)^2+ (k\cdot AC)^2}$

$\implies BD = k\cdot \sqrt{BC^2+AC^2}$

$\overset{(3)}{=}k\cdot AB.$

That is,

$\frac{BD}{AB} = k.$

Thus,

$\frac{BE}{BC} =\frac{DE}{AC} \overset{(1)}{=} \frac{BD}{AB}$

$\underline{\frac{BC}{BE}}= \frac{AC}{DE} = \underline{\frac{AB}{BD}}.$

It follows that

$\underline{\frac{BC}{BE}} -1 = \underline{\frac{AB}{BD}}-1 \implies \frac{BC-BE}{BE} = \frac{AB-BD}{BD} \overset{BC-BE=CE, AB-BD=AD}{\implies} \frac{CE}{BE} = \frac{AD}{BD}.$

See also “cos(x), sin(x)/a and 1 (Part 1)” and “A Proof without Calculus“.

Exercise-1 Given any triangle $\Delta ABC$ where $DE \parallel BC$ :

Show that $\frac{BC}{DE}=\frac{AC}{AE}=\frac{AB}{AD}, \frac{AD}{DB} = \frac{AE}{EC}$ and $\frac{AD}{DB} = \frac{DE}{BC}$ .

Analytic vs. Synthetic

January 15, 2024February 7, 2024 / Michael Xue / Leave a comment

An isosceles right triangle ABC is depicted blow where M is the midpoint of the hypotenuse AB, P is any point on AB, PE $\perp$ AC and PF $\perp$ BC. Show that MF $\perp$ ME.

Consider the triangle within a coordinate system, as illustrated in Fig. 1:

Fig. 1 $AC = BC, PE\perp AC$ and $PF \perp BC$

Let $k_1$ and $k_2$ denotes the slopes of $MF$ and $ME$ , respectively. We express them as follows:

$k_1= \frac{a-y}{a-0}$ and $k_2 = \frac{a-0}{a-x}$ .

This means that

$MF \perp ME \implies k_1\cdot k_2 = -1$

$\implies \frac{a-y}{a-0}\cdot \frac{a-0}{a-x} = -1$

$\implies \frac{a-y}{a-x } = -1$

$\implies a-y = x-a$

$\implies x + y =2a$

$\implies \frac{x}{2a} + \frac{y}{2a} = 1$

Certainly, this holds true for $P(x, y)$ as it lies on the hypotenuse $AB$ , which can be represented by the equation

$\frac{x}{2a} + \frac{y}{2a} = 1.\quad\quad\quad(1)$

Now, let’s provide a step-by-step proof:

$P(x,y)$ lies on the hypotenuse of $AB$

$\implies \frac{x}{2a} + \frac{y}{2a} = 1$

$\implies x + y =2a$

$\implies a-y = x-a$

$\implies \frac{a-y}{a-x}=-1$

$\implies \frac{a-y}{a}\cdot\frac{a}{a-x}$

$\implies\frac{a-y}{a-0} \cdot \frac{a-0}{a-x}\cdot=-1$

$\implies k_1\cdot k_2= -1$

$\implies ME \perp MF$ .

An alternative proof is:

$MF^2 + ME^2 = \underbrace{(a-y)^2+a^2}_{MF^2} + \underbrace{(a-x)^2+a^2}_{ME^2}$

$= 4a^2+x^2+y^2-2a(x+y)$

$\overset{(1) \implies x + y = 2a}{=} 4a^2+x^2+y^2-2a\cdot 2a$

$=x^2+y^2$ .

From the right triangle $\Delta CEF$ , we observe that

$x^2+y^2= EF^2.$

Therefore,

$MF^2+ME^2=EF^2.$

This implies that $\Delta ME$ is a right triangle. i.e.,

$MF \perp ME$ .

Treating $MF$ and $ME$ as vectors and utilizing the vector equation of line $AB$ , we have yet another proof:

Fig. 2

Let

$\overrightarrow{MF} = \begin{pmatrix} 0-a \\ y-a \end{pmatrix}$ and $\overrightarrow{ME} = \begin{pmatrix} x-a \\ 0-y \end{pmatrix}.$

The inner product of $\overrightarrow{MF}$ and $\overrightarrow{ME}$

$<\overrightarrow{MF}, \overrightarrow{ME}>=(0-a)(x-a) + (y-a)(0-a) = 2a^2-a(x+y).\quad\quad\quad(2)$

The vector equation of the line $AB$

$\overrightarrow{CP} = (1-t) \overrightarrow{CB} + t \overrightarrow{CA}, \quad t \in \mathbb{R}$

$\begin{pmatrix} x \\ y \end{pmatrix} = (1-t)\begin{pmatrix} 0 \\ 2a \end{pmatrix} + t \begin{pmatrix} 2a \\ 0 \end{pmatrix}.$

Thus, combining the components:

$x + y = \underbrace{(1-t)\cdot 0 + t\cdot 2a}_{x} + \underbrace{(1-t)\cdot 2a + t\cdot 0}_{y}= 2a.$

Submitting $x +y=2a$ into (2) gives

$<\overrightarrow{MF}, \overrightarrow{ME}> = 2a^2 - a\cdot 2a = 0 \implies MF \perp ME.$

Let’s also prove it synthetically.

Fig. 3 $MH \perp BC, MG \perp AC$

Given

$MG \perp AC, HC \perp AG \implies HC \parallel MG \implies \angle FMG = \angle MFH\quad\quad\quad(3)$

and

$\frac{MH}{AC} = \frac{BH}{BC} = \frac{1}{2}, \frac{MG}{BC} = \frac{AG}{AC} = \frac{1}{2}\implies\frac{MH}{AC} = \frac{MG}{BC} \overset{AC=BC}{\implies}NH = MG.\quad\quad\quad(4)$

Similarly,

$HF = GE.\quad\quad\quad(5)$

It follows that

$(4):MH=MG, (5):HF=GE, \Delta MHF$ and $\Delta MGE$ are right triangles

$\implies \Delta MFH \cong \Delta MGE.$

Thus,

$\angle MFH = \angle MEG.\quad\quad\quad(6)$

Consequently,

$FME = \angle FMG + \angle GME$

$\overset{(3):\angle FMG = \angle MFH}{=} \angle MFH + \angle GME$

$\overset{(6):\angle MFH=\angle MEG}{=} \angle MEG + \angle GME = \frac{\pi}{2} \implies MF \perp ME.$

See also “Deriving the vector equation of line”.

Being Effortless

January 9, 2024January 10, 2024 / Michael Xue / Leave a comment

In a triangle, the line segment that joins a vertex and the midpoint of the opposite side is called a median. Show that the medians of a triangle intersect at a common point that is two-thirds of the distance from the vertices to the midpoint of the opposite sides.

Let $\Delta ABC$ be a triangle with vertices $(x_1, y_1), (x_2, y_2)$ and $C(x_3, y_3)$ . The medians of the triangle are denoted as $AE, BF$ and $CD$ where $E(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}), F(\frac{x_1+x_3}{2}, \frac{y_1+y_3}{2})$ and $D(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$ are the midpoints.

Expressing the medians in determinant form, we have:

$AE: \begin{vmatrix} x_1 & y_1 & 1 \\ \frac{x_2+x_3}{2} & \frac{y_2+y_3}{2} & 1 \\ x & y & 1\end{vmatrix}=0\quad\quad\quad(1)$

$BF: \begin{vmatrix} x_2 & y_2 & 1 \\ \frac{x_1+x_3}{2} & \frac{y_1+y_3}{2} & 1 \\ x & y & 1 \end{vmatrix}=0\quad\quad\quad(2)$

$CD: \begin{vmatrix} x_3 & y_3 & 1 \\ \frac{x_1+x_2}{2} & \frac{y_1+y_2}{2} & 1 \\ x & y & 1 \end{vmatrix}=0\quad\quad\quad(3)$

Solving $\begin{cases} (1) \\ (2) \end{cases}$ (see Fig. 1) shows that $AE$ and $FB$ intersect at the point

$G: (\frac{x_1+x_2+x_3}{2}, \frac{y_1+y_2+y_3}{3}).\quad\quad\quad(4)$

Substitute (4) into (3), we see that (4) satisfies (3). Therefore, $AE, BF$ and $CD$ intersect at $G$ . i.e.,

the medians of a triangle intersect at a common point.

Fig. 1

Now, we define

$U = \underbrace{\sqrt{\left(x_1-\frac{x_1+x_2+x_3}{3}\right)^2+\left(y_1-\frac{y_1+y_2+y_3}{3}\right)^2}}_{AG}$

and

$V = \frac{2}{3}\cdot \underbrace{\sqrt{\left(x_1-\frac{x_2+x_3}{2}\right)^2+\left(y_1-\frac{y_2+y_3}{2}\right)^2}}_{AE}$ .

Consequently,

$U^2-V^2=0\quad\quad\quad(5)$

(see Fig. 1).

Since $U^2-V^2=(U+V)\cdot(U-V)$ and $U+V>0, (5)\implies U-V=0.$ That is,

$AG = \frac{2}{3}AE$ .

Similarly, $BG = \frac{2}{3}BF$ and $CG =\frac{2}{3}CD$ (see Exercise-1).

Exercise-1 Show that $BG = \frac{2}{3}BF$ and $CG =\frac{2}{3}CD.$

Easy Does It

January 8, 2024January 8, 2024 / Michael Xue / Leave a comment

Prove: Given any quadrilateral, the lines connecting the midpoint of two opposite sides bisect each other.

The quadrilaterals are illustrated in Fig. 1 with $E, F, G$ and $H$ representing the midpoints of the sides.

Fig. 1

In a rectangular coordinate system:

For points $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$ and $D(x_4, y_4)$ , the midpoints are:

$E(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}), F(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}), G(\frac{x_3+x_4}{2}, \frac{y_3+y_4}{2}), H(\frac{x_1+x_4}{2}, \frac{y_1+y_4}{2}).$