3. Given an equilateral triangle $ABC$ with side $a$, center $O$, and a point $P$ lying on the segment $OC$. Construct an equilateral triangle $XYZ$ inscribed in triangle $ABC$, such that points $X, Y, Z$ lie on sides $BC, CA, AB$ respectively, and side $XY$ contains the point $P$. When does the problem have a solution?
Solution. Analysis. Let $X Y Z$ be the desired triangle (diagram on the right). Then $X Y=Y Z$,

$$
\begin{aligned}
\measuredangle Y X C & =180^{\circ}-\measuredangle Y C X-\measuredangle X Y C \\
& =180^{\circ}-60^{\circ}-\measuredangle X Y C \\
& =180^{\circ}-\measuredangle Z Y X-\measuredangle X Y C=\measuredangle Z Y A, \\
\measuredangle X Y C & =180^{\circ}-60^{\circ}-\measuredangle Y X C \\
& =180^{\circ}-60^{\circ}-\measuredangle Z Y A=\measuredangle Y Z A,
\end{aligned}
$$

therefore, triangles $C X Y$ and $A Y Z$ are congruent. Similarly, it can be shown that each of these triangles is congruent to triangle

![](https://cdn.mathpix.com/cropped/2024_06_05_8b1793bf5544e167118bg-04.jpg?height=510&width=526&top_left_y=1187&top_left_x=958)
$B Z X$. Furthermore, we have $X A_{1}=Y B_{1}=Z C_{1}, O A_{1}=O B_{1}=O C_{1}$, so the right triangles $O A_{1} X, O B_{1} Y, O C_{1} Z$ are congruent. Therefore, $O X=O Y=O Z$, i.e., point $O$ is the center of the equilateral triangle $X Y Z$. Let $Q=Y Z \cap A O, R=Z X \cap B O$. The rotation around point $O$ by an angle of $120^{\circ}$ (or $-120^{\circ}$) maps point $P$ to point $R$ (or to point $Q$). Therefore, $O P=O Q=O R$.

Construction. On the segment $O B$, construct a point $R$ such that $O R=O P$. Then, on the side of the segment $P R$ that contains the side $B C$, construct the geometric locus of points $l$ from which the segment $P R$ is seen at an angle of
$60^{\circ}$. This is an arc of a circle. Let $X$ be the intersection point of arc $l$ and segment $B C$. Finally, construct points $Y$ and $Z$ sequentially as the intersections of lines $X P$ and $A C$, and $X R$ and $A B$, respectively. Triangle $X Y Z$ is the desired triangle.

Proof. By construction, $P \in X Y$ and $\measuredangle Z X Y=60^{\circ}$. To prove that triangle $X Y Z$ is equilateral, it suffices to prove that $X Z=X Y$. Similarly, as at the beginning, we get

$$
\measuredangle Y X C=\measuredangle X Z B, \measuredangle X Y C=\measuredangle Z X B
$$

The rotation by $120^{\circ}$ around point $O$ maps points $R, B, Z, X$ sequentially to points $P, C, Z^{\prime} \in B C, X^{\prime} \in C A$. Then $\triangle Z^{\prime} C X^{\prime} \cong \triangle Z B X$ and $\triangle Z^{\prime} C X^{\prime} \cong \triangle X C Y$, so

![](https://cdn.mathpix.com/cropped/2024_06_05_8b1793bf5544e167118bg-05.jpg?height=43&width=584&top_left_y=728&top_left_x=246)

$$
C X=C Z ', C Y=C X^{\prime}
$$

Since $X^{\prime}, Y \in C A$ and $Z^{\prime}, X \in B C$, we get $X=Z'$ and $Y=X'$. Finally,

$$
X Z=X^{\prime} Z^{\prime}=Y X
$$

Discussion. Since arc $l$ and line $BC$ can have two, one, or no intersection points, the problem has two, one, or no solutions. Let $O P=x, A B=a$, $S=P R \cap A A_{1}$ and $S_{1}=l \cap A A_{1}$. Then

$$
P S=\frac{x \sqrt{3}}{2}, O S=\frac{x}{2}, S A_{1}=\frac{a \sqrt{3}}{6}-\frac{x}{2}, S S_{1}=P S \sqrt{3}=\frac{3 x}{2}
$$

The problem has a solution if and only if $S S_{1} \geq S A_{1}$, i.e., if and only if

$$
O P=x \geq \frac{a \sqrt{3}}{12}=\frac{1}{4} \cdot \frac{2}{3} \frac{a \sqrt{3}}{2}=\frac{1}{4} O C
$$