In triangle $ABC$, $AB = AC$. The distances from the two trisection points of the side $AB$ to the vertex $C$ are 17 cm and 20 cm, respectively. What could be the distance from vertex $A$ to the trisection points of the base?
Let the base of the isosceles triangle be denoted by $a$, the height corresponding to the base by $m$, the foot of the height by $F$, the points dividing the base into thirds from $B$ to $A$ by $H_{1}, H_{2}$, the orthogonal projections of these points onto the base by $H_{1}^{\prime}, H_{2}^{\prime}$, the distances from point $C$ to $H_{1}$ and $H_{2}$ by $p$ and $q$, and the sought distance by $x$.
![](https://cdn.mathpix.com/cropped/2024_05_02_181e962d83110129f95bg-1.jpg?height=688&width=1148&top_left_y=302&top_left_x=476)

From the theorem of parallel chords, it follows that $B H_{2}^{\prime}=\frac{2}{3} B F$, and by symmetry, $B F=\frac{1}{2} B C$, so $B H_{2}^{\prime}=\frac{1}{3} B C$, hence $A H_{2}^{\prime}=x$.

Similarly, from the theorem of parallel chords, it follows that $B H_{1}^{\prime}=\frac{1}{6} a, F H_{2}^{\prime}=\frac{1}{6} a, H_{1} H_{1}^{\prime}=\frac{1}{3} m, H_{2} H_{2}^{\prime}=\frac{2}{3} m$. Applying the Pythagorean theorem to the right triangles $\mathrm{H}_{1} \mathrm{CH}_{1}^{\prime}, \mathrm{H}_{2} \mathrm{CH}_{2}^{\prime}, \mathrm{H}_{2}^{\prime} \mathrm{AF}$:

$$
p^{2}=\left(\frac{1}{3} m\right)^{2}+\left(\frac{5}{6} a\right)^{2}, \quad q^{2}=\left(\frac{2}{3} m\right)^{2}+\left(\frac{2}{3} a\right)^{2}, \quad x^{2}=m^{2}+\left(\frac{1}{6} a\right)^{2}
$$

We simplify these equations:

$$
\begin{aligned}
36 p^{2} & =4 m^{2}+25 a^{2} \\
9 q^{2} & =4 m^{2}+4 a^{2} \\
36 x^{2} & =36 m^{2}+a^{2}
\end{aligned}
$$

Since on the right-hand side of all three equations, only $m^{2}$ and $a^{2}$ terms appear, we can look for numbers $\lambda$ and $\mu$ such that by multiplying (1) and (2) and adding them, we get exactly the right-hand side of (3). Based on the coefficients of $m^{2}$ and $a^{2}$:

$$
\begin{aligned}
4 \lambda+4 \mu & =36 \\
25 \lambda+4 \mu & =1
\end{aligned}
$$

By subtracting, we easily find $\lambda=-\frac{35}{21}=-\frac{5}{3}$, and from the first equation, $\mu=9-\lambda=\frac{32}{3}$. These are the multipliers. Indeed, by multiplying (1) by $-\frac{5}{3}$ and (2) by $\frac{32}{3}$ and adding them:

$$
96 q^{2}-60 p^{2}=36 m^{2}+a^{2}=36 x^{2}
$$

From which we immediately get $x^{2}=\left(8 q^{2}+5 p^{2}\right) / 3$.

The problem does not specify the role of the distances 17 and 20 more precisely, so both cases $p=17, q=20$ and $p=20, q=17$ are possible. In the first case, $x_{1}=\sqrt{585} \approx 24.2 \mathrm{~cm}$, and in the second case, $x_{2}=\sqrt{104} \approx 10.2 \mathrm{~cm}$. Both results satisfy the desired conditions.