Maximize the area of an isosceles triangle bounded by a circle

Question

Maximize the area of an isosceles triangle (let the smallest of the triangle's three angles = $2\theta$)

Triangle is bounded by a circle with radius $R$

Find angle $\theta$ which maximizes the area of the bounded triangle

Drew this out with $2\theta$ vertex pointing upward

Drew lines from center of the circle out to each vertex and noted that the angle directly below $2\theta$ (at center of circle) = $4\theta$

Next steps: possibly finding $\sin(2\theta)$ and $\cos(2\theta)?$ Not sure where to go from here. Any help welcomed.

Answer 1

Hint: Start by drawing a figure. Make a circle of radius $R$, with center at $O$. on the vertical diameter, the top point is $A$. Draw a horizontal line below the $O$ point. The intersections with the circle are points $B$ and $C$. The intersection of $BC$ with the vertical is $D$. Note that $OA=OB=OC=R$. Angle $\angle BAD=\theta$. Angle $\angle OBA=\theta$ (isosceles triangle). Angle $\angle ABD=\pi/2-\theta$. Calculate angle $\angle OBD$. Then calculate $OD$ and $BD$ in the triangle $\triangle OBD$ (right angle triangle) as a function of $R$ and $\theta$. The area of the triangle $\triangle ABC$ is given by $area=AD\cdot(2BD)/2$. This is going to be a trigonometric expression as a function of $\theta$. Take the derivative, equate it to $0$, to find $\theta$

Answer 2

Write an expression for the area of the isosceles triangle where the height is (r + y) and 1/2 the base is x. A = x(r + y).

Next substitute in for x and y, expressions in terms of $\theta. x = rsin(2\theta)$ and $y = rcos(2\theta)$......to obtain:

$$A = rsin(2\theta)(r + rcos(2\theta))$$ $$A = r^2sin(2\theta) + r^2sin(2\theta)cos(2\theta)$$ $$dA/d\theta = r^2cos(2\theta)+r^2cos^2(2\theta) - r^2(sin^2(2\theta)$$

Solve for $dA/d\theta = 0$ and you should get $\theta = pi/6 = 30\deg$