\section*{Problem 6 - 141036}

Given is a parallelogram \(O P Q R\). We are looking for all points \(X\) on the extension of \(O P\) beyond \(P\) that have the following property:

}

If the line parallel to \(X R\) through \(Q\) intersects the extension of \(O R\) beyond \(R\) at \(Y\), then \(P Y \|\)

\(X Q\).

Determine whether such points \(X\) exist!

If this is the case, describe and justify a construction of all such points and investigate whether there is only one such point \(X\)!
}

(I) Suppose \(X\) is a point of the required kind. Then, if \(O P=a, O R=b, P X=x\), \(R Y=y\) are set:

\[
\frac{y}{a}=\frac{b}{a+b}
\]

because due to \(\angle R O P=\angle Y R Q\) and \(\angle O R X=\angle R Y Q\), \(\triangle O R X \sim \triangle R Y Q\). Furthermore,

\[
\frac{x}{b}=\frac{a}{b+y}
\]

because due to \(\angle O P Y=\angle O X Q\) and \(\angle O Y P=\angle P Q X\), \(\triangle O Y P \sim \triangle P Q X\) holds.

From (1) and (2) it follows

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1350.jpg?height=591&width=671&top_left_y=1009&top_left_x=267)

\[
\begin{gathered}
\frac{x}{b}=\frac{a}{b+\frac{a b}{a+x}} \\
x^{2}+a x-a^{2}=0
\end{gathered}
\]

and from this, because \(x>0\),

\[
x=\frac{a}{2}(\sqrt{5}-1)
\]

If now \(\triangle O M P\) is a right triangle with the legs \(O P\) and \(O M=\frac{a}{2}\), \(N\) is the intersection of \(M P\) with the circle around \(M\) with the radius \(\frac{a}{2}\), and \(X\) is the intersection of the circle around \(P\) with the radius \(P N\) that is not on \(O P\), then \(P X=\frac{a}{2}(\sqrt{5}-1)\).

Therefore, the point \(X\) satisfies all the requirements of the problem only if it can be constructed in the following way:

(a) Construct the perpendicular \(s\) to \(O P\) at \(O\).

(b) Measure a segment \(O M\) of length \(\frac{a}{2}\) from \(O\) on \(s\).

(c) Draw the circle \(k\) around \(M\) with the radius \(M O\). If \(N\) is the intersection of \(k\) with \(P M\), then

(d) draw the circle \(k^{\prime}\) around \(P\) with the radius \(P N\). The intersection of \(k^{\prime}\) with the line through \(O\) and \(P\) that is not on \(O P\) is the point \(X\).

(III) Every point \(X\) constructed in this way satisfies the conditions of the problem.

Proof:

By the Pythagorean theorem,

\[
M P=\sqrt{O M^{2}+O P^{2}}=\frac{a}{2} \sqrt{5}
\]

Therefore, by construction,

\[
P X=P N=P M-M N=P M-M O=\frac{a}{2}(\sqrt{5}-1)
\]

Thus, \(x=P X=\frac{a}{2}(\sqrt{5}-1)\) and therefore (4) and (3) hold.

If now \(Y\) is located outside \(O P\) on the line containing \(O R\) such that \(Y Q \| R X\), then \(\triangle O Y P \sim \triangle P Q X\) and from this \(\angle O P Y=\angle P X Q\). Therefore, \(\angle P X Q+\angle Y P X=180^{\circ}\).

Hence, the lines containing \(P Y\) and \(X Q\) cannot intersect according to the angle sum theorem for triangles, so they are parallel.

(IV) The given construction is always executable in exactly two ways, both leading to the same point \(X\).

Adapted from [5]

\subsection*{7.16.4 IV. Round 1974, Class 10}