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,

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

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Alternative solution.

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.