# Thomas Simpson and Maxima and Minima - Least isosceles triangle circumscribing a circle

Author(s):
Michel Helfgott

To determine the dimensions of the least isosceles triangle ACD that can circumscribe a given circle (Example VI, p. 18).

Let $O$ be the center of the circle, with radius $a$, inscribed in the isosceles triangle ACD. Draw the altitude $\overline{DB}$ and let $x = OD$. We have $\triangle DSO$ is similar to $\triangle DBC$. Thus $\frac{DS}{DB} = \frac{SO}{BC}$. Hence

$BC = \frac{(x + a)a}{\sqrt{x^2 - a^2}}$,   so   $area(x) = \frac{(x + a)^2 a}{\sqrt{x^2 - a^2}}$.

Define

$f(x) = \frac{(x + a)^4}{x^2 - a^2}$,

which is the square of the area multiplied by $\frac{1}{a^2}$. Note that

$f(x) = \frac{(x + a)^3}{x - a}$   so   $f'(x) = \frac{3(x + a)^2 (x - a) - (x + a)^3}{(x - a)^2}$.

The numerator will be zero when $x = 2a$. Therefore, the area attains its minimum at $x = 2a$. From the similarity of triangles $\triangle DSO$ and $\triangle DBC$, we get $\frac{DC}{OD} = \frac{BC}{SO}$. Thus $DC = \frac{x \cdot BC}{a} = \frac{2a \cdot BC}{a} = 2BC = AC$. In other words, the solution is to construct the equilateral triangle that circumscribes the given circle.

Remark: It is interesting to observe that Simpson implicitly assumes that the altitude $BD$ passes through the center of the circle. Undoubtedly, he believes that his readers are knowledgeable in geometry and will realize that the center of the inscribed circle is at the point of intersection of the angle bisectors of the given triangle. Because the latter is isosceles, $\overline{BD}$ is both an altitude and an angle bisector; thus $\overline{BD}$ goes through $O$.