How to show that a triangle of maximum area inscribed in a given circle is an equilateral triangle?

1 Answer
Jun 15, 2018

As the radius of the circle is not relevant we can assume we have a circle of unitary radius centered in the origin, whose equation is:

x^2+y^2=1x2+y2=1

The triangle will then have three vertices A,B,CA,B,C lying on the circle. Because of the circular symmetry we can assume without loss of generality that the side ABAB is parallel to the yy axis and intersecting the positive xx axis, so that the coordinates of AA and BB are:

{(x_A = cos(alpha)),(y_A = sin(alpha)):}

{(x_B = cos(alpha)),(y_A = -sin(alpha)):}

where 2alpha = angle (AOB) so that the length of AB is:

bar(AB) = 2sin(alpha)

Now, as a first step let's consider for any possible choice of AB the triangle ABC letting C be the third vertex with coordinates:

{(x_C = cos(theta)),(y_C=sin(theta)):}

where theta is the angle between the x axis and OC as we generally define in trigonometry.

The height of the triangle is going to be parallel to the x axis.
As x_A is positive by construction the height is:

h = x_A -x_C = cosalpha - cos theta

and the area of the triangle is:

S = (h bar(AB))/2 = sinalpha(cosalpha-costheta)

and as 1- <= cos theta <=1 clearly the are is maximum when:

cos theta = -1

theta = pi

that is when the triangle ABC is isosceles and then the area is:

S(alpha) = sinalpha(cosalpha+1)

Maximize now this function with alpha in (0,pi/2):

(dS)/(d alpha) = cosalpha(cos alpha +1) -sin^2 alpha

(dS)/(d alpha) = cos^2alpha-sin^2 alpha +cos alpha

(dS)/(d alpha) = cos^2alpha -(1-cos^2 alpha)+cos alpha

(dS)/(d alpha) = 2cos^2alpha +cos alpha -1

The critical points are then:

cos alpha = (-1+-sqrt(1+8))/2 = (-1+-3)/2 = {(-1),(1/2):}

Excluding cos alpha = -1 because by construction cos alpha >0 we have:

alpha = pi/3

where:

(d^2S)/(dalpha^2) = -4sin alpha cos alpha -sin alpha = -sin alpha(1+4 cosalpha)

(d^2S)/(dalpha^2)|_(pi/3) = -sqrt3/2(1+4 (1/2)) = -(3sqrt3)/2 < 0

so that in fact alpha = pi/3 is a local maximum.

Remembering that:

angle(AOB) = 2alpha = (2pi)/3

as the angle angle(ACB) = 1/2 angle(AOB) we have:

angle(ACB) = pi/3

and as the triangle is isosceles the angles at the base are each:

angle (BAC) = angle(ABC) = 1/2 (pi -pi/3) = pi/3

which means the triangle is equilateral.