Selected Problems from Open Contests

Let $H$ be the foot of the altitude drawn from $C$. First prove that $|AH|$ and $|BH|$ are rational numbers. For that, use the Pythagorean theorem for triangles $ACH$ and $BCH$ to obtain $|AH|^2 + |CH|^2 = |AC|^2$ and $|BH|^2 + |CH|^2 = |BC|^2$. Therefore $|AC|^2 - |BC|^2 = |AH|^2 - |BH|^2 = (|AH| - |BH|) \cdot (|AH| + |BH|) = (|AH| - |BH|) \cdot |AB|$. We see that $|AH| - |BH| = \frac{|AC|^2 - |BC|^2}{|AB|}$ is rational and so are $|AH| = \frac{|AC|^2 + |BC|^2}{|AB|}$

Fig. 6
and $|BH| = |AH| - (|AH| - |BH|)$. Let now $K$ be the projection of $P$ to $BC$ (see Fig. 6). As $P$ lies on the angle bisector of $B$, it is equidistant from both $AB$ and $BC$, i.e., $|PH| = |PK|$. Consequently,

$$\frac{S_{APB}}{S_{BPC}} = \frac{|AB| \cdot |PH|}{|BC| \cdot |PK|} = \frac{|AB|}{|BC|} \cdot \frac{|BH|}{|AH|} \quad \text{As } CH \perp AB, \text{ also } \frac{S_{BPC}}{S_{APC}} = \frac{|CP| \cdot |BH|}{|CP| \cdot |AH|} = \frac{|BH|}{|AH|}$$
Thus, $\frac{S_{APB}}{S_{APC}} = \frac{|AB|}{|BC|} \cdot \frac{|BH|}{|AH|}$ is rational as a product of two rational numbers.

Let $H$ be the foot of the altitude drawn from $C$ and let $L$ be the projection of $P$ to $AC$ (see Fig. 6). Now $\angle CPL = 90^\circ - \angle ACH = 90^\circ - (90^\circ - \angle CAB) = \angle CAB$, giving $\frac{|PH|}{|PL|} = \frac{|PH|}{|PC| \cdot \cos \angle CAB}$. The angle bisector theorem gives $\frac{|PH|}{|PC|} = \frac{|BH|}{|BC|} = \cos \angle ABC$. Consequently,
$$\frac{S_{APB}}{S_{APC}} = \frac{|AB| \cdot |PH|}{|AC| \cdot |PL|} = \frac{|AB|}{|AC|} \cdot \frac{\cos \angle ABC}{\cos \angle CAB}$$
As the side lengths of the triangle $ABC$ are integers, $\frac{|AB|}{|AC|}$ is rational. By the cosine law, the cosines of the angles of triangle $ABC$ are rational, whence $\frac{\cos \angle ABC}{\cos \angle CAB}$ is rational. Altogether, $\frac{S_{APB}}{S_{APC}}$ is rational.