The side of the base of a regular triangular prism $A B C A_1 B_1 C_1$ is 4, and the lateral edge is 3. On the edge $B B_1$, a point $F$ is taken, and on the edge $C C_1$, a point $G$ is taken such that $B_1 F=1, C G=\frac{2}{3}$. Points $E$ and $D$ are the midpoints of edges $A C$ and $B_1 C_1$ respectively. Find the minimum possible value of the sum $E P+P Q$, where point $P$ belongs to the segment $A_1 D$, and point $Q$ belongs to the segment $F G$.
Note that for any fixed position of point $P$, the sum $E P+P Q$ will be the smallest if $P Q \perp F G$. Through the line $A 1 D$, we draw a plane perpendicular to the line $F G$. This can be done since the lines $A 1 D$ and $F G$ are perpendicular (the line $F G$ lies in the plane of the face $B B 1 C 1 C$, and the line $A 1 D$ is perpendicular to this plane). The drawn plane intersects the line $F G$ at the desired point $Q 0$ (Fig.1). In this case, $D Q 0 \perp F G$. Consider the rectangle $B B 1 C 1 C$ (Fig.2). Let $L-$ be the projection of point $Q 0$ on $B 1 C 1, N-$ the intersection point of the lines $F G$ and $B 1 C 1$, and $F^{\prime}$ - the projection of point $F$ on the side $C C 1$. Denote $\angle G F F^{\prime}=\alpha, L D=$ 3t. Then $\angle L Q 0 D=\angle Q 0 N D=\angle G F F^{\prime}=\alpha$. From the right triangle $G F F^{\prime}$, we find that

$$
\operatorname{tg} \alpha=\frac{G F^{*}}{F F^{*}}=\frac{G C_{1}-F^{*} C_{1}}{B_{1} C_{1}}=\frac{G C_{1}-F B_{1}}{B_{1} C_{1}}=\frac{T_{1}-1}{4}=\frac{1}{3}
$$

Then

$$
Q L=\frac{L D}{l \lg v_{2}}=\frac{\frac{1}{\frac{1}{3}}}{3}=3 t, N B 1=\frac{F^{\prime} B_{1}}{\lg m_{2}}=\frac{\frac{1}{1}}{\frac{1}{3}}=3, N D=N B 1+B 1 D=3+2=5 .
$$

In the right triangle $N Q 0 D$, the segment $Q 0 L$ is the height drawn from the vertex of the right angle, so $Q 0 L 2=N L \cdot L D$, or $9 t 2=(5-t) t$, from which we find that $t=\frac{1}{2}$. Then $L D=t=\frac{1}{2}, Q 0 L=3 t=\frac{3}{2}$. Let $Q^{\prime}-$ be the point symmetric to point $Q 0$ relative to the plane of the base $A 1 B 1 C 1$. Then $E P+P Q 0=E P+P Q^{\prime}$, so the problem reduces to finding the minimum sum $E P+P Q$ '. We will prove that the segments $E Q^{\prime}$ and $A 1 D$ intersect (then the desired point $P$ will be the intersection of these segments). Since the planes of the faces $B B 1 C 1$ and $A 1 B 1 C 1$ are perpendicular, point $L$ is the midpoint of $Q 0 Q^{\prime}$. Let $E 1$ be the projection of point $E$ onto the plane of the base $A 1 B 1 C 1$, and $T$ be the projection of point $Q 0$ onto the plane of the base $A B C$. Then $E 1$ is the midpoint of $A 1 C 1$. Consider the section of the prism by the plane passing through the parallel lines $E E 1$ and $T L$ (Fig.3). Let the segments $E Q$ ' and $L E 1$ intersect at point $O$. From the similarity of triangles $L Q O$ and $E E 1 O$, we find that

$$
\frac{L O}{O E_{1}^{\prime}}=\frac{L Q^{\prime}}{E E_{1}^{\prime}}=\frac{L Q_{0}}{E E_{1}}=\frac{3}{3}=\frac{1}{2}
$$

Consider the base $A 1 B 1 C 1$ (Fig.4). Let the segments $A 1 D$ and $L E 1$ intersect at point $O 1$. Through point $E 1$, draw a line parallel to $B 1 C 1$ until it intersects $A 1 D$ at point $S$. Then $E 1 S$ is the midline of triangle $A 1 D C 1$, so $E 1 S=\frac{1}{2} D C 1=1$. From the similarity of triangles $D O 1 L$ and $S O 1 E 1$, we find that

$$
\frac{L O_{1}}{O_{1} E_{1}}=\frac{L D}{E_{1} S}=\frac{1}{1}=\frac{1}{2}
$$

Thus, the segments $E Q$ ' and $A 1 D$ intersect the segment $L E 1$ at points $O$ and $O 1$ respectively, and $\frac{L O}{\partial E_{1}^{\prime}}=$ $\frac{L O_{1}}{O_{1}}$ $\frac{O_{1} E_{1}}{}$. Therefore, points $O$ and $O 1$ coincide. Hence, the segments $E Q^{\prime}$ and $A 1 D$ intersect. Let $P 0$ be the point of their intersection. Then

$$
E P+P Q \leqslant E P+P Q 0=E P+P Q^{\prime} \leqslant E Q^{\prime}
$$

and equality is achieved when point $P$ coincides with point $P 0$. In this case,

$$
\begin{gathered}
E P+P Q=E P 0+P Q 0=E Q^{\prime}=\sqrt{E T^{2}+Q T^{2}}=\sqrt{E_{1} L^{2}+Q^{\prime T^{2}}}= \\
=\sqrt{C_{1} E_{1}^{2}+C_{1} L^{2}-2 C_{1} E_{1} \cdot C_{1} L \cos 60^{\circ}+Q^{\prime T^{2}}}= \\
=\sqrt{2^{2}+\left(\frac{5}{2}\right)^{2}+2 \cdot 2 \cdot \frac{5}{2} \cdot \frac{1}{2}+\left(3+\frac{3}{2}\right)^{2}}=\sqrt{\frac{51}{2}}
\end{gathered}
$$

## Peтиетие methodon жоордчнит.

Choose a coordinate system as shown in Fig.5. Denote $A 1 P=x$. Then the considered points have coordinates: $P(x ; 0 ; 0), E(\sqrt{3} ; 1 ; 3), Q 0\left(2 \sqrt{3} ;-\frac{1}{2} ; 2\right)$. The problem reduces to finding the minimum value of the function

$$
f(x)=P E+P Q 0=\sqrt{(x-\sqrt{3})^{2}+1^{2}+3^{2}}+\sqrt{(x-2 \sqrt{3})^{2}+\left(\frac{1}{2}\right)^{2}+\left(\frac{3}{2}\right)^{2}}=
$$

$$
=\sqrt{(x-\sqrt{3})^{2}+10}+\sqrt{(x-2 \sqrt{3})^{2}+\frac{5}{2}}
$$

on the segment $[0 ; 2 \sqrt{3}]$. Note that for each fixed $x \in[0 ; 2 \sqrt{3}]$, the number $f(x)$ is the sum of the distances from the point $W(x ; 0)$ in the plane $x O y$ to the points $U(\sqrt{3} ; \sqrt{10})$ and $V\left(2 \sqrt{3} ; \frac{\sqrt{10}}{2}\right)$ of this plane. In turn, this problem reduces to finding on the axis $O x$ a point $X$ for which the sum of the distances to the points $U$ and $V$ is minimal. It is known that $X$ is the intersection point of the axis $O x$ with the segment $W V$, where $W$ is the point symmetric to point $U$ relative to the axis $O x$. If $U 1$ and $V 1-$ are the projections of the points $U$ and $V$ on the axis $O x$, then $\frac{U_{1} X}{X v_{1}}=\frac{\underline{U}_{1} W}{V v_{1}}=\frac{\frac{\sqrt{10}}{\sqrt{10}}}{2}=2$. Therefore,

$$
x=\sqrt{3}+\frac{2}{3}(2 \sqrt{3}-\sqrt{3})=\frac{5 \sqrt{3}}{3}
$$

Therefore,

$$
\begin{gathered}
E P 0+P Q 0=U X+X V=f\left(\frac{5 \sqrt{3}}{3}\right)= \\
=\sqrt{\left(\frac{5 \sqrt{3}}{3}-\sqrt{3}\right)^{2}+10}+\sqrt{\left(\frac{5 \sqrt{3}}{3}-2 \sqrt{3}\right)^{2}+\frac{5}{2}}=\sqrt{\frac{51}{2}}
\end{gathered}
$$

## Answer

$\sqrt{\frac{51}{2}}$