60. The diameter of the circle inscribed in triangle $ABC$ is 1. What can the side $d$ of the square inscribed in the triangle be, with two vertices lying on the base $AB$ of the triangle (not on its extensions!); and the other two on sides $AC$ and $BC?$
60. Since angles \( A \) and \( C \) at the base of the triangle must be acute (otherwise one of the vertices of the square would fall on the extension of the base), the points \( D \) and \( E \) of tangency of the sides \( AB \) and \( BC \) with the inscribed circle \( s \) are located on the "upper" semicircle (Fig. 142a), and therefore, the segment \( MN \parallel AC \) tangent to \( s \) inside the triangle will be less than the diameter \( I \) of the circle \( s \). By inscribing a rectangle \( MNQP \) with side \( MN \) in the triangle, we can see that its side \( MP = 1 \), and \( MN < 1 \); hence, the side of the inscribed square \( M_0N_0P_0Q_0 \) in \( \triangle ABC \) is less than 1. On the other hand, \( d \) can be arbitrarily close to 1; to verify this, it is sufficient to consider an isosceles triangle circumscribed around the circle \( s \) with diameter 1 and a very small angle at the vertex \( B \).

Let \( M_1N_1 \parallel AC \) be a chord of the circle \( s \) such that

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-172.jpg?height=276&width=555&top_left_y=308&top_left_x=578)
the rectangle \( M_1N_1P_1Q_1 \) is

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-172.jpg?height=278&width=522&top_left_y=596&top_left_x=612)

Fig. 142. a square (Fig. 142b); in this case, for the corresponding inscribed rectangle \( M_2N_2Q_2P_3 \) in the triangle, we have \( M_2N_3 \geq M_1N_1 = M_1P_1 = M_2P_2 \), which implies that \( d \geq M_1P_1 = M_1N_1 \), because the side \( M_0N_0 = d \) of the inscribed square in \( \triangle ABC \) is higher than the segment \( M_2N_2 \) (i.e., farther from the base \( AB \)) and thus \( M_0P_0 = d \geq M_1P_1 \). Therefore, to estimate the value of \( d \) from below, it is sufficient to find what \( M_1N_1 \) is; denoting \( M_1N_1 = x \), we find from the right triangle \( KM_1L \) depicted in Fig. 142b that

\[
M_1R^2 = KR \cdot RL,
\]

or

\[
\left(\frac{x}{2}\right)^2 = x(1 - x), \quad \text{i.e.,} \quad 5x^2 - 4x = 0
\]

(because in the notation of Fig. 142, \( \sigma \) we have \( RM_1 = \frac{x}{2} \), \( KR = M_1P_1 = x \), \( KL = 1 \)) and

\[
x = \frac{4}{5}.
\]

In this case, the equality \( d = M_1N_1 = \frac{4}{5} \) is possible: it is satisfied, for example, when the sides \( AB \) and \( BC \) touch the inscribed circle \( s \) in the points \( M_1 \) and \( N_1 \).

Thus,

\[
\frac{4}{5} \leq d < 1.
\]