Name: Second Derivative Test
Uploaded: 2019-07-19
Duration: 12 min 48 s
Description: The maxima and minima of a function are the largest and smallest value of the function. Watch this amazing video to find the maxima and minima of a function.

Question

Two squares are inscribed in a circle of diameter

\sqrt{2}

units. Prove that the area of the common region of the squares is at least

2 (\sqrt{2} - 1)

.

Accepted Answer

Given a circle whose diameter is $\sqrt{2}$

So radius of circle will be $\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$

Since, the radius of circle is $\frac{1}{\sqrt{2}}$ , so side of the square $= \sqrt{2} \times \frac{1}{\sqrt{2}} = 1$ unit.

Let both the squares are overlapping in initial position.

Squares be $A B C D$ and $P Q R S .$

Circle is depicted in the above figure whose center is taken as origin.

In the below figure both the squares are rotated by same angle but in different directions to make use of symmetry.

Now, $▭ P Q R S$ is rotated anti-clockwise by an angle $\frac{θ}{2}$ and $▭ A B C D$ is rotated clockwise by an angle $\frac{θ}{2}$ .
As figure is having symmetry, so $∆ L P E ≅ ∆ E A F$ .

Again, all the corners are symmetrically placed, so the triangle thus formed will be identical or equal.

So, $∆ L P E$ , $∆ E A F, ∆ F S G, ∆ G D H, ∆ H R l, ∆ I C J, ∆ J Q K$ and $∆ K B L$ will be identical.
Since, we are looking for least area at any point of time, i.e., area of overlapped figure $E F G H I J K L E$ depends on area of triangle $E A F$ .

Area of figure $E F G H I J K L E$ will be minimum when area of $∆ E A F$ will be greatest.

Since $A E = P E$ and $A F = F S, A E + E F + F A = 1$

So, perimeter of $∆ E A F = 1$ unit, which is not changing at any point of time.
Again, $∆ E A F$ is a right angle triangle which is having constant perimeter.

For maximising area of $∆ E A F$ it should be an isosceles right angled triangle.

Let $A E = x = A F$
$\Rightarrow E M = x \cos 45^{o} = \frac{x}{\sqrt{2}}$
$\Rightarrow x + x + \frac{2 x}{\sqrt{2}} = 1$

$\Rightarrow x = \frac{1}{2 + \sqrt{2}} = \frac{2 - \sqrt{2}}{2}$

$∆ E A F = \frac{1}{2} (base) \cdot (height)$
$\Rightarrow$ Area of $∆ E A F = \frac{1}{2} A E \cdot A F$
$= \frac{1}{2} {(\frac{2 - \sqrt{2}}{2})}^{2}$

$= \frac{3 - 2 \sqrt{2}}{4}$
Hence, minimum area of overlapped figure $E F G H I J K L E =$

(Area of square) $- 8$ (Area of $∆ E A F$ )
$= 1 - 8 (\frac{3 - 2 \sqrt{2}}{4})$

$= 1 - 6 + 4 \sqrt{2}$

$= 4 \sqrt{2} - 5$

Two squares are inscribed in a circle of diameter $\sqrt{2}$ units. Prove that the area of the common region of the squares is at least $2 (\sqrt{2} - 1)$ .

Important Questions on Monotonicity, Maxima and Minima

Learn and Prepare for any exam you want !

Two squares are inscribed in a circle of diameter 2 units. Prove that the area of the common region of the squares is at least 22-1.

Important Questions on Monotonicity, Maxima and Minima

Learn and Prepare for any exam you want !

Two squares are inscribed in a circle of diameter $\sqrt{2}$ units. Prove that the area of the common region of the squares is at least $2 (\sqrt{2} - 1)$ .