Question

Show that the triangle of maximum area that can be inscribed in a given circle is an equilateral triangle.
A) -1
B) 1
C) 2
D) Zero

Answer 1

Hint:
Let R be the radius of the circle and h is the height of the triangle and 2r be the base of the triangle. Then In right triangle $\Delta OBD$ $O{B^2} = O{D^2} + B{D^2}$ then find the equation of ${r^2}$ in terms of h. We know that the area of the triangle is $\dfrac{1}{2} \times base \times height$ for maximizing the area the differential equation will be equated to zero and double differentiation will be negative at the zeros of the differential equation. If the corresponding value of h area will be negative then we can say area is maximum at $h = \dfrac{{3R}}{2}$ and then we will calculate r in term in R and find the length of BC. Now, again in the right triangle ABD $A{B^2} = A{D^2} + B{D^2}$ from this we can calculate the value of side AB and similar with AC hence we can prove ABC is an equilateral triangle.

Complete step by step solution:

Let R be the radius of the circle and h is the height of the triangle and 2r be the base of the triangle.
Let AD be the height, it is perpendicular to BC
Therefore, OD be perpendicular to chord BC
We know that perpendicular to the center bisects the chord.
Then, $BD = \dfrac{{BC}}{2} = r$
Now,
In right triangle $\Delta OBD$
$\Rightarrow O{B^2} = O{D^2} + B{D^2}$
$\Rightarrow {R^2} = {r^2} + {\left( {h - R} \right)^2}$
$\Rightarrow {R^2} = {r^2} + {h^2} + {R^2} - 2Rh$
$\Rightarrow {h^2} + {r^2} - 2Rh = 0$
Now, we have to maximize area of triangle
Area of triangle =$\dfrac{1}{2} \times base \times height$
$A = \dfrac{1}{2} \times 2r \times h = r \times h$
Now, we know the value of ${r^2}$
Let, $Z = {A^2}$=${r^2} \times {h^2}$
Now, $Z = \left( {2hR - {h^2}} \right) \times {h^2}$
For maximizing Z differentiating Z w.r.t h
$\dfrac{{dZ}}{{dh}} = 2R \times 3{h^2} - 4{h^3}$
$\dfrac{{dZ}}{{dh}} = 6R{h^2} - 4{h^3}$
Putting $\dfrac{{dZ}}{{dh}} = 0$
$3R{h^2} - 2{h^3} = 0$
$\Rightarrow {h^2}\left( {3R - 2h} \right) = 0$
Therefore \[h = 0,h = \dfrac{{3R}}{2}\]
h cannot be equal to zero so $h = \dfrac{{3R}}{2}$.
Now, again differentiating z with respect to h
$\dfrac{{{d^2}Z}}{{d{h^2}}} = 12Rh - 12{h^2}$
Substituting $h = \dfrac{{3R}}{2}$ to check the sign of $\dfrac{{{d^2}Z}}{{d{h^2}}}$
$\dfrac{{{d^2}Z}}{{d{h^2}}} = 12R\left( {\dfrac{{3R}}{2}} \right) - 12{\left( {\dfrac{{3R}}{2}} \right)^2}$
$\Rightarrow \dfrac{{{d^2}Z}}{{d{h^2}}} = 18{R^2} - 27{R^2} = - 9{R^2}$
2nd differentiation of Z is negative hence Z is maximum at $h = \dfrac{{3R}}{2}$
Now,
Solving for h and r
${h^2} + {r^2} - 2Rh = 0$
Substituting $h = \dfrac{{3R}}{2}$
${\left( {\dfrac{{3R}}{2}} \right)^2} + {r^2} - 2R\left( {\dfrac{{3R}}{2}} \right) = 0$
$\Rightarrow {r^2} = 3{R^2} - \dfrac{{9{R^2}}}{4}$
So $r = \dfrac{{\sqrt 3 R}}{2}$
so, side BC=$2r = \sqrt 3 R$
Now,
In right angle triangle ABD
$\Rightarrow A{B^2} = A{D^2} + B{D^2}$
$\Rightarrow A{B^2} = {h^2} + {r^2}$
$\Rightarrow A{B^2} = {\left( {\dfrac{{3R}}{2}} \right)^2} + {\left( {\dfrac{{\sqrt 3 R}}{2}} \right)^2}$
$\Rightarrow A{B^2} = \dfrac{{9{R^2}}}{4} + \dfrac{{3{R^2}}}{4}$
So $AB = \sqrt 3 R$
Similarly,
$AC = \sqrt 3 R$
Since $AB = BC = AC = \sqrt 3 R$

Hence, $\Delta ABC$ is an equilateral triangle.

Note:
In equilateral triangles all sides are equal. for maximizing the area, the differential equation will be equated to zero and double differentiation will be negative at the zeros of the differential equation. If at the corresponding value of h area will be negative then we can say area is maximum. Don’t consider h to be 0.