## Task 11/70

In which triangle are the measurements of the height $h_{b}$ and the sides $a, b$, and $c$ four consecutive whole numbers in this order?
Let $\beta$ be the angle of the triangle enclosed by the sides $a$ and $c$. Then, according to the sine rule and the cosine rule,

$$
b h_{b}=a c \sin \beta \quad \text { and } \quad \cos \beta=\frac{a^{2}+c^{2}-b^{2}}{2 a c}
$$

From this it follows that

$$
b^{2} h_{b}^{2}=a^{2} c^{2} \sin ^{2} \beta=a^{2} c^{2}\left(1-\cos ^{2} \beta\right)=a^{2} c^{2}\left(1-\frac{\left(a^{2}+c^{2}+b^{2}\right)^{2}}{4 a^{2} c^{2}}\right.
$$

Now, let $a=h_{b}+1, b=h_{b}+2, c=h_{b}+3$. Substituting these values and simplifying the equation, we get (since $h_{b} \neq 0$) $h_{b}^{4}+8 h_{b}^{3}-44 h_{b}^{2}-48 h_{b}=0=h_{b}^{3}+8 h_{b}^{2}-44 h_{b}-48$.

By sensible trial, we first find the solution $h_{b}=-2$, which, however, makes no geometric sense since $h_{b}>0$. This allows us to further reduce the equation.

$$
h_{b}^{2}-10 h_{b}-24=0 \rightarrow h_{b}=-2 \text { (double solution!) }, h_{b}=12
$$

Thus, the sought triangle has sides of lengths $a=13 \text{ units}, b=14 \text{ units}, c=15 \text{ units}$ and the height $h_{b}=12 \text{ units}$.