Question

Let a square with vertices at (1, 1), (-1, 1), (-1, -1) and (1, -1) and S be the region consisting of all points inside the square which are nearer to the origin than to any edge. If the area enclosed by region S is $\frac{4}{3}(a\sqrt{2}-b)$ then the value of a and b is equal to: $(a, b \in N)$

• A. (4,5)
• B. (5,4)
• C. (1,2)
• D. (2,1)

Answer 1

Answer: (A)

(4, 5)

Answer 2

We are given a square with vertices

$$(1,1),\;(-1,1),\;(-1,-1),\;(1,-1)$$

and we define the region $S$ as the set of points inside the square which are closer to the origin than to any edge. By symmetry it is enough to consider the first quadrant (i.e. $x\ge0,\; y \ge0$) and then multiply the computed area by 4.

A point $(x,y)$ in the first quadrant satisfies:

• Its distance from the origin is $r=\sqrt{x^2+y^2}$.
• Its distances from the edges $x=1$ and $y=1$ are $1-x$ and $1-y$ respectively.

Thus, in the first quadrant the condition becomes

$$\sqrt{x^2+y^2} < \min(1-x,\,1-y).$$

It is natural to switch to polar coordinates:

$$x=r\cos\theta,\quad y=r\sin\theta,\quad 0\le\theta\le\frac{\pi}{2},\; 0\le r.$$

The distances from the edges become:

$$1-x=1-r\cos\theta\quad \text{and}\quad 1-y=1-r\sin\theta.$$

For each $\theta$ we require:

$$r < 1-r\cos\theta \quad \text{and}\quad r < 1-r\sin\theta.$$

Solving these:

$$r(1+\cos\theta)<1 \quad\Longrightarrow\quad r< \frac{1}{1+\cos\theta},$$ $$r(1+\sin\theta)<1 \quad\Longrightarrow\quad r< \frac{1}{1+\sin\theta}.$$

Thus for a fixed angle $\theta$, the allowed $r$ runs from $0$ to

$$r_{\max}(\theta)=\min\Bigl(\frac{1}{1+\cos\theta},\,\frac{1}{1+\sin\theta}\Bigr).$$

Notice when $\theta=45^\circ$, $\cos\theta=\sin\theta$ and the two expressions are equal. Therefore, we split the integral into two parts:

1. For $\theta$ from 0 to $\pi/4$, we have $\cos\theta\ge\sin\theta$ so that $\frac{1}{1+\cos\theta}\le\frac{1}{1+\sin\theta}$. Thus: $$I_1=\int_{0}^{\pi/4}\int_{0}^{\frac{1}{1+\cos\theta}} r\,dr\,d\theta.$$
2. For $\theta$ from $\pi/4$ to $\pi/2$, we have $\sin\theta\ge\cos\theta$ so that $r_{\max}(\theta)=\frac{1}{1+\sin\theta}$: $$I_2=\int_{\pi/4}^{\pi/2}\int_{0}^{\frac{1}{1+\sin\theta}} r\,dr\,d\theta.$$

The area in the first quadrant, $A_1$, is:

$$A_1= \int_{0}^{\pi/4}\frac{1}{2}\frac{1}{(1+\cos\theta)^2}\,d\theta +\int_{\pi/4}^{\pi/2}\frac{1}{2}\frac{1}{(1+\sin\theta)^2}\,d\theta.$$

By making a substitution in the second integral, $\phi=\frac{\pi}{2}-\theta$, it turns out that both integrals become identical. Thus,

$$A_1= \int_{0}^{\pi/4}\frac{1}{(1+\cos\theta)^2}\,d\theta.$$

Multiplying by the factor $\frac{1}{2}$ from the $r$-integration, we denote:

$$I_1=\frac{1}{2}\int_{0}^{\pi/4}\frac{d\theta}{(1+\cos\theta)^2}.$$

Evaluating $I_1$:

Recall that

$$1+\cos\theta = 2\cos^2\left(\frac{\theta}{2}\right).$$

Then,

$$(1+\cos\theta)^2=4\cos^4\left(\frac{\theta}{2}\right).$$

Thus,

$$I_1=\frac{1}{2}\int_{0}^{\pi/4}\frac{d\theta}{4\cos^4(\theta/2)} =\frac{1}{8}\int_{0}^{\pi/4}\sec^4\left(\frac{\theta}{2}\right)d\theta.$$

Let $u=\frac{\theta}{2}$ so that $d\theta=2du$ and when $\theta=0,\; u=0$; when $\theta=\pi/4,\; u=\pi/8$. Then,

$$I_1=\frac{1}{8}\cdot 2\int_{0}^{\pi/8}\sec^4 u\,du =\frac{1}{4}\int_{0}^{\pi/8}\sec^4 u\,du.$$

The antiderivative is:

$$\int\sec^4u\,du = \tan u +\frac{1}{3}\tan^3u + C.$$

Thus,

$$I_1=\frac{1}{4}\left[\tan u+\frac{1}{3}\tan^3u\right]_{0}^{\pi/8} =\frac{1}{4}\left[\tan\left(\frac{\pi}{8}\right)+\frac{1}{3}\tan^3\left(\frac{\pi}{8}\right)\right].$$

The area in the first quadrant is then:

$$A_1=2I_1=\frac{1}{2}\left[\tan\left(\frac{\pi}{8}\right)+\frac{1}{3}\tan^3\left(\frac{\pi}{8}\right)\right],$$

and because the full region is symmetric in four quadrants:

$$\text{Area }S=4A_1 = 2\left[\tan\left(\frac{\pi}{8}\right)+\frac{1}{3}\tan^3\left(\frac{\pi}{8}\right)\right].$$

Using the Known Value: It is known that

$$\tan\left(\frac{\pi}{8}\right)=\sqrt{2}-1.$$

Then,

$$\tan^3\left(\frac{\pi}{8}\right)=(\sqrt{2}-1)^3.$$

Expanding $(\sqrt{2}-1)^3$:

$$(\sqrt{2}-1)^3 = (\sqrt{2})^3 - 3(\sqrt{2})^2(1) + 3\sqrt{2}\cdot1^2 - 1^3 = 2\sqrt{2}-6+3\sqrt{2}-1= 5\sqrt{2}-7.$$

Plug these back in:

$$S=2\left[(\sqrt{2}-1)+\frac{1}{3}(5\sqrt{2}-7)\right] =2\left[\sqrt{2}-1+\frac{5\sqrt{2}-7}{3}\right].$$

Write the expression with common denominator:

$$S=2\cdot\frac{3(\sqrt{2}-1)+(5\sqrt{2}-7)}{3} =\frac{2}{3}\left[3\sqrt{2}-3+5\sqrt{2}-7\right] =\frac{2}{3}\left[8\sqrt{2}-10\right] =\frac{16\sqrt{2}-20}{3}.$$

We can factor this as:

$$S=\frac{4}{3}\left(4\sqrt{2}-5\right).$$

Thus, comparing with the given form $\frac{4}{3}(a\sqrt2-b)$, we have:

$$a=4,\quad b=5.$$