\section*{Problem 5 - 311035}

Let \(A B C D\) be a given parallelogram. For each point \(P\) on the segment \(A B\), let \(S\) be the intersection of the segments \(A C\) and \(P D\).

a) For which position of \(P\) on \(A B\) is the area of triangle \(S P C\) equal to one-sixth of the area of parallelogram \(A B C D\)?

b) For which position of \(P\) on \(A B\) is the area of triangle \(S P C\) maximized?
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![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1601.jpg?height=257&width=488&top_left_y=246&top_left_x=790)

a) A point whose position can be described on a line segment is the midpoint. And indeed, this point is the desired point, as can be easily seen.

Let the length of the parallelogram be \(a\) and the height \(b\). The parallelogram then has an area of \(A = ab\). Clearly, the area of \(\triangle ASD\) is the same as that of \(\triangle SPC\), since \(\triangle APD\) is equal in area to \(\triangle APC\) (shearing).

Let the area of the triangles be \(x\). The triangles \(\triangle APS\) and \(\triangle DSC\) are similar, as all angles are equal. According to the intercept theorem, the scaling factor is 2, so the area of \(\triangle DSC\) is four times that of \(\triangle ASP\). Let \(A_{\triangle ASP} = y\), then we have:

(1) : \(\quad x + 4y = \frac{ab}{2}\)

(2) : \(\quad x + y = \frac{ab}{4}\)

(1) \(-(2): \quad 3y = \frac{ab}{4}\) and thus \(y = \frac{ab}{12}\)

Therefore, we get \(A_{\triangle SPC} = \frac{ab}{4} - \frac{ab}{12} = \frac{ab}{6}\).

b) Now - if \(P\) is moved to the left, the area will always get smaller, and if \(P\) is moved to the right, it will get larger. If \(P\) finally lies on \(B\), the area is maximal, namely \(A = \frac{ab}{4}\).