3. In triangle $ABC$ with side $BC$ of length $2 \text{ cm}$, point $K$ is the midpoint of side $AB$. Points $L$ and $M$ divide side $AC$ into three equal segments. Triangle $KLM$ is isosceles and right-angled. Determine the lengths of sides $AB$ and $AC$ for all such triangles $ABC$.
SOLUTION. Points $L$ and $M$ on side $AC$ are chosen such that $|AM|=|ML|=|LC|$. The median $KO$ of triangle $KLM$ is the midline of triangle $ABC$, so $|KO|=\frac{1}{2}|BC|$, $|AC|=6|MO|$, and $|AB|=2|AK|$. We distinguish three cases:

(a) Let $|KL|=|KM|$ (Fig.2). Then $|\angle MKL|=|\angle MOK|=90^{\circ}$ and $|MO|=$

![](https://cdn.mathpix.com/cropped/2024_04_17_f9c69bf7084d05f276acg-4.jpg?height=323&width=805&top_left_y=2100&top_left_x=631)

Fig. 2
$=|KO|$. From the Pythagorean theorem for triangle $AKO$, we get

$$
|AK|=\sqrt{(3|MO|)^{2}+|KO|^{2}}=\sqrt{10|KO|^{2}}=\sqrt{10}|KO|=\frac{1}{2} \sqrt{10}|BC|,
$$

so

$$
\begin{gathered}
|AB|=2|AK|=\sqrt{10}|BC|=2 \sqrt{10} \text{ cm} \\
|AC|=6|MO|=6|KO|=3|BC|=6 \text{ cm}
\end{gathered}
$$

(b) Let $|ML|=|MK|$ (Fig.3). Then $|\angle KML|=90^{\circ}$ and $|AM|=|ML|=$

![](https://cdn.mathpix.com/cropped/2024_04_17_f9c69bf7084d05f276acg-5.jpg?height=335&width=339&top_left_y=752&top_left_x=867)

Fig. 3

$=|MK|=2|MO|$. From the Pythagorean theorem for triangle $KMO$, we get

$$
|KO|=\sqrt{|MO|^{2}+(2|MO|)^{2}}=\sqrt{5}|MO|,
$$

so

$$
|AC|=6|MO|=\frac{3}{\sqrt{5}}|BC|=\frac{6 \sqrt{5}}{5} \text{ cm}
$$

From the Pythagorean theorem for triangle $AKM$, we get

$$
|AK|=\sqrt{|AM|^{2}+|MK|^{2}}=\sqrt{2}|MK|=2 \sqrt{2}|MO|=\frac{2 \sqrt{10}}{5}|KO|=\frac{\sqrt{10}}{5}|BC|,
$$

so

$$
|AB|=2|AK|=\frac{2 \sqrt{10}}{5}|BC|=\frac{4 \sqrt{10}}{5} \text{ cm}
$$

(c) Let $|ML|=|KL|$ (Fig. 4). Then $|\angle MLK|=90^{\circ}$. Thus, $|KL|=|ML|=$

![](https://cdn.mathpix.com/cropped/2024_04_17_f9c69bf7084d05f276acg-5.jpg?height=248&width=531&top_left_y=1986&top_left_x=771)

Fig. 4

$=2|LO|=2|MO|$ and $|AL|=|AM|+|ML|=4|MO|$. From the Pythagorean theorem for triangle $KLO$, we get

$$
|KO|=\sqrt{|LO|^{2}+(2|LO|)^{2}}=\sqrt{5}|LO|
$$

so

$$
|AC|=6|MO|=6|LO|=\frac{3}{\sqrt{5}}|BC|=\frac{6 \sqrt{5}}{5} \text{ cm}.
$$

From the Pythagorean theorem for triangle $AKL$, we get

$$
\begin{aligned}
|AK| & =\sqrt{|AL|^{2}+|LK|^{2}}=\sqrt{(4|LO|)^{2}+(2|LO|)^{2}}= \\
& =2 \sqrt{5}|LO|=2|KO|=|BC|=2 \text{ cm}
\end{aligned}
$$

so

$$
|AB|=2|AK|=2|BC|=4 \text{ cm}.
$$