ABC is a right-angled triangle of given area k. Find the sides of the triangle for which the area of the circumscribed circle is least.

The triangle is isosceles (two sides are equal).

The correct answer is Option (1) → The triangle is isosceles (two sides are equal).

Let ABC be a right-angled triangle at B.

Let hypotenuse $AC = l$ and side $AB = x$, then

$BC^2=l^2-x^2⇒ BC= \sqrt{l^2 - x^2}$   ...(i)

Area of $ΔABC =\frac{1}{2} AB × BC = k$ (given)

$⇒\frac{1}{2}x \sqrt{l^2 - x^2} = k⇒ l^2 − x^2 =\frac{4k^2}{x^2}$

$⇒l^2=x^2+\frac{4k^2}{x^2}$   ...(ii)

Since $∠ABC =\frac{\pi}{2}$, therefore, AC is a diameter of the circumcircle of ΔABC.

∴ Radius of circumcircle = $\frac{1}{2}AC = \frac{l}{2}$.

Let A be the area of the circumcircle of ΔABC, then $A = π (\frac{l}{2})^2$

$⇒ A= \frac{1}{4}πl^2=\frac{1}{4}π\left(x^2 +\frac{4k^2}{x^2}\right)$   (using (ii))

$⇒\frac{dA}{dx}=\frac{π}{4}\left(2x-\frac{8k^2}{x^3}\right)$ and $\frac{d^2A}{dx^2}=\frac{π}{4}\left(2+\frac{24k^2}{x^4}\right)$

Now $\frac{dA}{dx}=0⇒2x-\frac{8k^2}{x^3}=0 ⇒x^4 = 4k^2⇒x^2 = 2k⇒x=\sqrt{2k}$   $(∵x>0)$

$\left(\frac{d^2A}{dx^2}\right)_{x=\sqrt{2k}}=\frac{π}{4}\left(2+\frac{24k^2}{x^4}\right)=2\pi>0$ ⇒ A is least when $x=\sqrt{2k}$

From (ii), when $x = \sqrt{2k}, l^2 = 2k +\frac{4k^2}{2k}=4k⇒l=2\sqrt{k}$.

From (i), $BC =\sqrt{l^2 - x^2} = \sqrt{4k - 2k} =\sqrt{2k}$.

Hence, the sides of the triangle are $\sqrt{2k}$, \sqrt{2k}$, 2\sqrt{k}$.