## Task 6 - 310836

For a triangle $A B C$, the following are given:

The side length $a=\overline{B C}=845 \mathrm{~cm}$, the length $h_{a}=\overline{A E}=840 \mathrm{~cm}$ of the height perpendicular to $B C$, and the length $h_{c}=\overline{C D}=780 \mathrm{~cm}$ of the height perpendicular to $A B$.

Calculate the side lengths $c=\overline{A B}$ and $b=\overline{A C}$ for each triangle $A B C$ where these lengths occur!
It holds that $a \cdot h_{a}=c \cdot h_{c}$; this follows from the area equations of $\triangle ABC$ (or also from the similarity of triangles $\triangle BCD$ and $\triangle BAE$). Thus, we get

$$
c=\frac{a \cdot h_{a}}{h_{c}}=910 \text{ cm}
$$

Since $\angle BDC=90^{\circ}$, it follows from the Pythagorean theorem that

$$
BD=\sqrt{a^{2}-h_{c}^{2}}=325 \text{ cm}
$$

Depending on whether $D$ lies between $A=A_{1}$ and $B$ or $B$ lies between $A=A_{2}$ and $D$, we further obtain

$$
A_{1}D=c-BD=585 \text{ cm} \quad \text{or} \quad A_{2}D=c+BD=1235 \text{ cm}
$$

The sought side length $b$ is then in the first case

$$
b_{1}=A_{1}C=\sqrt{A_{1}D^{2}+h_{c}^{2}}=975 \text{ cm}
$$

and in the second case

$$
b_{2}=A_{2}C=\sqrt{A_{2}D^{2}+h_{c}^{2}}=1460.69 \text{ cm}
$$

Solutions from the third round of 1991 taken from [5]

# 5.34 XXXII. Olympiad 1992

### 5.34.1 First Round 1992, Class 8