Two isosceles triangles have their inscribed circles touching the legs at the points closer to the base in one case, and at the points farther from the base in the other case. In which case does the inscribed circle cover a larger fraction of the triangle's area?
First, we examine the case where the inscribed circle touches the base at the point one-third of the way from the base. We denote the vertices of the triangle as $A, B, C$, where $A B = A C$, the center of the inscribed circle as $O$, and its radius as $\varrho$. The altitude from $C$ to $B$ intersects the base at the midpoint $F$, the perpendicular from $O$ to $A B$ intersects $A B$ at $D$, and the length of segment $O A$ is denoted by $y$. (Figure 1)

$B F = B D = x$, the lengths of the tangents from an external point, and according to the condition, $A D = 2 x$.

The right triangles $A O D$ and $A B F$ are similar (they share the angle at vertex $A$), and the ratio of their corresponding sides is:

$$
y: 3 x = \varrho: x, \quad \text { hence } \quad y = 3 \varrho
$$

From the right triangle $A O D$, by the Pythagorean theorem:

$$
\varrho^{2} + 4 x^{2} = 9 \varrho^{2}, \quad \text { hence } \quad x = \varrho \sqrt{2}
$$

The height of the triangle $A B C$ is: $\varrho + y = 4 \varrho$. Now we can write the areas:

$$
T_{\triangle} = x \cdot m = \varrho \sqrt{2} \cdot 4 \varrho = 4 \sqrt{2} \varrho^{2}, \quad T_{\bigcirc} = \varrho^{2} \pi
$$

Thus,

$$
\frac{T_{\triangle}}{T_{\bigcirc}} = \frac{4 \sqrt{2}}{\pi} \approx 1.8006
$$

In the other case, using a similar approach, we find that $y = \frac{3}{2} \varrho$ (the length of the tangents is now denoted by $2 x$ according to Figure 2), and $\varrho^{2} + x^{2} = \frac{9}{4} \varrho^{2}$, hence $x = \frac{\sqrt{5}}{2} \varrho$.

$$
T_{\triangle} = 2 x (\varrho + y) = 5 \cdot \frac{\sqrt{5}}{2} \varrho^{2}, \quad T_{\bigcirc} = \varrho^{2} \pi
$$

The ratio of the areas in this case is:

$$
\frac{T_{\triangle}}{T_{\bigcirc}} = \frac{\frac{5 \sqrt{5}}{2} \varrho^{2}}{\varrho^{2} \pi} = \frac{5 \sqrt{5}}{2 \pi} \approx 1.7794
$$

Thus, in the second case, the circle covers a larger fraction of the triangle's area.

![](https://cdn.mathpix.com/cropped/2024_05_02_242b97626d24675c73a0g-1.jpg?height=835&width=561&top_left_y=1588&top_left_x=760)