2020 MEXT Japanese Government Scholarship Undergraduate Students Natural Sciences Qualifying Examination Mathematics (B): Problem 1(5)

This problem appears at the Qualifying Examinations for Applicants for Japanese Government (MEXT) Scholarships 2020. There are two mathematics exams: one for biology-related natural sciences (Mathematics A), and another for physics- and engineering-related natural sciences (Mathematics B). This problem is from the 2020 Mathematics (B) questionnaire. The official answer key is here.

Problem 1(5)

The angle $\theta\left( 0\lt \theta \lt \frac{\pi}{2}\right)$ between the two lines $y=(2-\sqrt{3})x$ and $y=(\sqrt{3}-2)x$ on the $xy$-plane is $\fbox{ A }$.

Solution

For convenience, let us name the lines $l_1$ and $l_2$. $$\begin{align} l_1: y = (2-\sqrt{3})x\\ l_2: y = (\sqrt{3}-2)x \end{align}$$ Let us also call the angle that $l_1$ makes with the positive $x$-axis $\theta_1$, and the angle that $l_2$ makes with the positive $x$-axis $\theta_2$. The problem seeks to find the angle $\theta$ which is the smaller of $|\theta_1-\theta_2|$ and $\pi-|\theta_1-\theta_2|$.

In general, the slope of any line can be expressed as the tangent of the angle it makes with the positive $x$-axis. $$\begin{align} m_1&=\tan{\theta_1} = 2-\sqrt{3}\\ m_2&=\tan{\theta_2} = \sqrt{3}-2 \end{align}$$

We notice two things.

1. We observe that $m_1 \gt 0$ because $2 \gt \sqrt{3}$. This is even clearer when we take the square of both sides to get $4 \gt 3$. This observation means that $0 \lt \theta_1 \lt \frac{\pi}{2}$.
2. We observe that $m_2=-m_1$, so we can say that $\theta_2 = -\theta_1$.

The figure below summarises what we know.

So, now, we are looking for the smaller of $|\theta_1-\theta_2|=|\theta_1-(-\theta_2)|=2\theta_1$ and $\pi-|\theta_1-\theta_2| = \pi-2\theta_1$. The absolute value symbol could be removed because we know from the first observation that $\theta_1>0$.

We can use this with following tangent identity. $$\begin{align} \tan{2\theta_1} &= \frac{2\tan{\theta_1}}{1-\tan^2{\theta_1}}\\ & = \frac{2\left(2-\sqrt{3}\right)}{1 - (2-\sqrt{3})^2}\\ & = \frac{2\left(2-\sqrt{3}\right)}{1 - (4-4\sqrt{3}+3)}\\ & = \frac{2\left(2-\sqrt{3}\right)}{- 6+4\sqrt{3}}\\ & = \frac{2\left(2-\sqrt{3}\right)}{2(-3+2\sqrt{3})}\\ & = \frac{(2-\sqrt{3})(-3-2\sqrt{3})}{(-3+2\sqrt{3})(-3-2\sqrt{3})}\\ & = \frac{-\sqrt{3}}{-3}\\ & = \frac{1}{\sqrt{3}}\\ 2\theta_1 &= \frac{\pi}{6} \end{align}$$

Therefore, $$\begin{align} 2\theta_1 &= \frac{\pi}{6}\\ \pi - 2\theta_1 &= \frac{5\pi}{6}, \end{align}$$ and clearly, the smaller between the two values is $\frac{\pi}{6}$. So, $\theta=\frac{\pi}{6}$.■