\section*{Problem 14 - V01114}

In a coordinate system, the points \(P_{1}(1 ; 1), P_{2}(4 ; 2), P_{3}(3 ;-2), Z(-1 ; 4)\) are given. A triangle similar to \(\triangle P_{1} P_{2} P_{3}\) is to be drawn using the similarity point \(Z\) and the similarity ratio \(2: 3\).
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![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1664.jpg?height=796&width=859&top_left_y=493&top_left_x=593)

If the areas of the similar triangles \(P_{1} P_{2} P_{3}\) and \(P_{1}^{\prime} P_{2}^{\prime} P_{3}^{\prime}\) are to be in the ratio \(2: 3\), then the side lengths must be in the ratio \(\sqrt{2}: \sqrt{3}\).

1. In a side construction (see figure), by applying the Pythagorean theorem twice, two segments of lengths \(\sqrt{2}\) and \(\sqrt{3}\) are determined.

2. From \(Z\), these segments are marked on any straight line, yielding the points \(Z_{2}\) and \(Z_{3}\).

3. From \(Z\), rays are drawn through the points \(P_{1}, P_{2}, P_{3}\).

4. \(P_{1}, P_{2}, P_{3}\) are connected to \(Z_{1}\) by straight lines.

5. The parallel shifts of these lines through \(Z_{2}\) intersect the corresponding rays through \(Z\) at the desired points \(P_{1}^{\prime}, P_{2}^{\prime}\) and \(P_{3}^{\prime}\).

According to the intercept theorem, we then have

\[
\frac{Z Z_{1}}{Z Z_{2}}=\frac{\sqrt{2}}{\sqrt{3}}=\frac{Z P_{1}}{Z P_{1}^{\prime}}=\frac{Z P_{2}}{Z P_{2}^{\prime}}=\frac{Z P_{3}}{Z P_{3}^{\prime}}
\]

and again by the intercept theorem

\[
\frac{P_{1} P_{2}}{P_{1}^{\prime} P_{2}^{\prime}}=\frac{P_{1} P_{3}}{P_{1}^{\prime} P_{3}^{\prime}}=\frac{P_{2} P_{2}}{P_{2}^{\prime} P_{3}^{\prime}}=\frac{\sqrt{2}}{\sqrt{3}}
\]

so that the triangle \(P_{1}^{\prime} P_{2}^{\prime} P_{3}^{\prime}\) is the desired triangle.