3) Given a square $A B C D$ with a unit side and let $P$ be a point inside it such that the angle $A \hat{P} B$ measures $75^{\circ}$. The sum of the areas of triangles $A B P$ and $C D P$ is:
$\begin{array}{ll}\text { (A) } 1 & \text { (B) } \frac{\sqrt{2}}{2}\end{array}$
(C) $\frac{1}{2}$
(D) cannot be determined
(E) none of the above is correct.
3) The answer is $(\mathbf{C})$. Let $H$ be the foot of the altitude from $P$ in the triangle $A B P$ and let $K$ be the foot of the altitude from the triangle $C D P$ (i.e., $H$ and $K$ are the projections of $P$ onto $A B$ and $C D$). Then the sum of the areas of the two triangles $A B P$ and $C D P$ is

$$
\frac{A B \cdot P H}{2}+\frac{C D \cdot P K}{2}=A B \cdot \frac{P H+P K}{2}=\frac{1}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_8c48e706878e314feb6bg-1.jpg?height=300&width=303&top_left_y=1506&top_left_x=1619)

because $P H+P K=H K$ (two segments perpendicular to two parallel lines and having one endpoint in common are collinear) and $A B=C D=H K=1$. The area sought is therefore the same regardless of the position of $P$ and does not depend on the angle $A \widehat{P} B$; the fact that the angle $A \widehat{P} B$ measures $75^{\circ}$ is irrelevant.