A regular hexagon $S$ has an inscribed triangle $H$ such that one side of $H$ is parallel to one side of $S$. What is the maximum possible ratio of the area of $H$ to the area of $S$?
I. solution. Let the triangle $H=P Q R$ inscribed in the regular hexagon $S=A B C D E F$ have its side $P Q$ parallel to the side $A B$ such that $P$ is on the side $A F$, and thus $Q$ is on the side $B C$. It is obvious that the area of $H$ is maximized when $R$ is on the side $D E$, and furthermore, moving $R$ along $D E$ (while keeping $P Q$ unchanged) does not change the area. Therefore, we can fix $R$ at the midpoint of $D E$.

We will show that the area of $H$ is maximized when $P$ bisects the side $A F$: $P A=P F$. Let $P_{1}$ be on the segment $P F$, and let the line through $P_{1}$ parallel to $A B$ intersect $B C$ at $Q_{1}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_a8b3e6f1738d1d6a5f47g-1.jpg?height=585&width=637&top_left_y=434&top_left_x=728)

Figure 1

Then, denoting the area of each region by the same symbol as the region itself (Figure 1):

$$
\begin{gathered}
H_{1}=P_{1} Q_{1} R=P Q R+P P_{1} R+Q Q_{1} R-Q Q_{1} P-P P_{1} Q_{1}= \\
=P Q R-\left(P P_{1} Q_{1}-P P_{1} R\right)<P Q R=H
\end{gathered}
$$

since, by assumption, $P R \| B C$ and thus $Q Q_{1} R-Q Q_{1} P=0$, and $R Q \| A F$ and $Q_{1}$ is on the segment $Q C$, so it is farther from $P P_{1}$ than $Q$ and $R$, and thus the difference in the parentheses is positive, and by omitting the subtrahend, we have increased the expression.

Similarly, if $P_{2}$ is on the segment $P A$ and $P_{2} Q_{2} \| A B$, then $Q_{2}$ is on the segment $B Q$ and

$$
\begin{aligned}
H_{2}= & P_{2} Q_{2} R=P Q R+P P_{2} Q+Q Q_{2} P_{2}-P P_{2} R-Q Q_{2} R=P Q R- \\
& -\left(P P_{2} R-P P_{2} Q\right)-\left(Q Q_{2} R-Q Q_{2} P_{2}\right)<P Q R=H
\end{aligned}
$$

because the first parenthesis is 0, the second is positive, since $P_{2}$ is closer to $B C$ than $R$ and $P$.

![](https://cdn.mathpix.com/cropped/2024_05_02_a8b3e6f1738d1d6a5f47g-1.jpg?height=550&width=544&top_left_y=1731&top_left_x=758)

Figure 2

According to these, the triangle $H$ with the prescribed properties and the largest area constitutes $9 / 24=3 / 8$ of the area of $S$ (Figure 2), which is the largest possible value of the ratio in question.

Ágoston Péter (Budapest, Berzsenyi D. Gymnasium, 2nd grade)

II. solution. Let the distance of $P Q$ from $A B$ be $x(\leq A E / 2=A B \sqrt{3} / 2)$, and let $P Q$ intersect $A E$ at $U$ (Figure 3).

![](https://cdn.mathpix.com/cropped/2024_05_02_a8b3e6f1738d1d6a5f47g-2.jpg?height=553&width=520&top_left_y=122&top_left_x=781)

Then, considering the symmetry,

$$
\begin{gathered}
P Q R=\frac{1}{2} P Q \cdot U E= \\
=\frac{1}{2}(A B+2 P U)(A E-A U)= \\
=\frac{1}{2}\left(A B+\frac{2}{\sqrt{3}} x\right)(A B \sqrt{3}-x)= \\
=\frac{1}{\sqrt{3}}\left(\frac{A B \sqrt{3}}{2}+x\right)(A B \sqrt{3}-x) .
\end{gathered}
$$

The sum of the variable factors is constant, so - as is well known - their product is maximized when the two factors are equal, from which

$$
x=\frac{A B \sqrt{3}}{4}=\frac{A E}{4}
$$

that is, $P Q$ bisects the distance between $A B$ and $F C$: $P A=P F$.

Takács Andor (Csurgó, Csokonai Vitéz M. Gymnasium, 4th grade)