\section*{Problem 2 - 071022}

Given a rectangle \(A B C D\). The midpoint of \(A B\) is \(M\). Connect \(C\) and \(D\) with \(M\) and \(A\) with \(C\). The intersection point of \(A C\) and \(M D\) is \(S\).

Determine the ratio of the area of the rectangle \(A B C D\) to the area of the triangle \(\triangle S M C\)!
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![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1249.jpg?height=294&width=440&top_left_y=1526&top_left_x=274)

If the lengths of the sides of the rectangle \(A B C D\) are denoted by \(a\) and \(b\), then the area of the rectangle \(I(A B G B)=a \cdot b\).

Furthermore, \(\triangle A S M \sim \triangle D S C\) because \(\angle A S M \cong \angle D S C\) (vertical angles) and \(\angle S A M \cong \angle D C S\) (alternate interior angles).

Since \(M\) is the midpoint of \(A B\), it follows that \(A M: C D=1: 2\). Additionally, \(I(\triangle A M C)=\frac{1}{4} a \cdot b\), and because \(S M: S D=1: 2\) (by the intercept theorem), we have

\[
I(\triangle A M S)=\frac{1}{2} \cdot \frac{a}{2} \cdot \frac{b}{3}=\frac{1}{12} a \cdot b
\]

Thus, we obtain

\[
I(\triangle S M C)=I(\triangle A M C)-I(\triangle A M S)=\frac{1}{4} a \cdot b-\frac{1}{12} a \cdot b=\frac{1}{6} a \cdot b
\]

Therefore, \(I(A B C D): I(\triangle S M C)=6: 1\).