Consider a semicircle. Let's look at the triangles that have one side containing the diameter of the semicircle, and the other two sides touching the semicircle. Among these triangles, which one has the smallest area?
I. solution. Let the sides of the circumscribed triangle that are tangent to the semicircle be denoted by $a$ and $b$. The segment connecting the center $O$ of the semicircle to the opposite vertex of the triangle divides the triangle into two smaller triangles, each having a height of $r$, the radius of the semicircle (Figure 1). Accordingly, the double area of the triangle is:

$$
2 T = (a + b) \cdot r
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_8e12bc553a8f18383db6g-1.jpg?height=339&width=469&top_left_y=400&top_left_x=817)

Figure 1

By the arithmetic and geometric mean inequality, $a + b \geq 2 \sqrt{a b}$; using this, we obtain the following estimate from the above equality:

$$
T \geq r \sqrt{a b}
$$

Note that equality holds here precisely when $a = b$, i.e., the triangle is isosceles.

Remark. It is evident that it suffices to find the triangle with the smallest area among the isosceles circumscribed triangles, but this path may lead to unexpected problems. From the above, it can be precisely concluded that for a circumscribed triangle with $a \neq b$, the isosceles circumscribed triangle with sides $\frac{a+b}{2}$ definitely has a smaller area, provided the latter exists. A fortunate further estimate can avoid this investigation.

From the relationship $2 T = a b \sin \gamma$ for the double area of the triangle, and since $0 < \sin \gamma \leq 1$, we get $a b \geq 2 T$, and equality holds precisely when the angle between sides $a$ and $b$ is a right angle. Therefore, the right-hand side of inequality (1) can be estimated from below: $r \sqrt{a b} \geq r \sqrt{2 T}$, and thus we obtain the inequality:

$$
T \geq r \sqrt{2 T}
$$

Squaring and dividing by $T$, we get $T \geq 2 r^2$, and we have seen that equality holds precisely when the triangle is isosceles and right-angled.

![](https://cdn.mathpix.com/cropped/2024_05_02_8e12bc553a8f18383db6g-1.jpg?height=431&width=607&top_left_y=1546&top_left_x=756)

Figure 2

II. solution. Let $T$ be the area of the triangle. Reflect the triangle across the diameter of the semicircle (Figure 2). The resulting kite is a tangential quadrilateral, and its area is the product of the radius of the inscribed circle and half the perimeter. Using the notation in the figure, the lengths of the tangent segments are:

$$
\begin{gathered}
x = r \cdot \operatorname{ctg} \gamma ; \quad y = r \cdot \operatorname{ctg} \alpha \\
z = r \cdot \operatorname{ctg} \beta
\end{gathered}
$$

Thus,

$$
2 T = r \frac{4 x + 2 y + 2 z}{2} = r^2 (2 \operatorname{ctg} \gamma + \operatorname{ctg} \alpha + \operatorname{ctg} \beta)
$$

$\alpha, \beta$, and $\gamma$ are half the angles of a convex quadrilateral, so each is an acute angle. The cotangent function is convex on the interval $\left(0 ; \frac{\pi}{2}\right]$, so by Jensen's inequality:

$$
\frac{2 \operatorname{ctg} \gamma + \operatorname{ctg} \alpha + \operatorname{ctg} \beta}{4} \geq \operatorname{ctg} \frac{2 \gamma + \alpha + \beta}{4} = \operatorname{ctg} \frac{\pi}{4} = 1
$$

Thus, $2 T = r^2 (2 \operatorname{ctg} \gamma + \operatorname{ctg} \alpha + \operatorname{ctg} \beta) \geq 4 r^2$, and equality holds precisely when $\alpha = \beta = \gamma = 45^\circ$, so the circumscribed triangle with the minimum area is isosceles and right-angled.

András Konfár (Szeged, Radnóti Miklós Gymnasium, 10th grade)