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Prompt: In a regular hexagon $ABCDEF$, let $G$ be the intersection of $\overline{AD}$ and $\overline{CE}$, and let $H$ be the intersection of $\overline{BD}$ and $\overline{AF}$. If the side length of the hexagon is $1$, find the length of $\overline{GH}$.

To solve this problem, we can utilize the properties of regular hexagons and the symmetry within this figure.



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The diagonals $\overline{AD}$, $\overline{CE}$, and $\overline{BD}$, $\overline{AF}$ create equilateral triangles within the hexagon due to its regularity. Let's denote the side length of the hexagon as $s = 1$.



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Since $G$ is the intersection of $\overline{AD}$ and $\overline{CE}$, and given that these segments create equilateral triangles with the sides of the hexagon, we can conclude that $\overline{AG}$ is the height of an equilateral triangle with side length $1$. This means that $\overline{AG} = \frac{\sqrt{3}}{2} s = \frac{\sqrt{3}}{2}$. By symmetry, $\overline{CG} = \overline{EG} = \frac{\sqrt{3}}{2}$.



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Similarly, $\overline{BH}$ and $\overline{DH}$ are heights of equilateral triangles, making $\overline{BH} = \overline{DH} = \frac{\sqrt{3}}{2}$. Also, $\overline{FH} = \overline{AH} = \frac{\sqrt{3}}{2}$.



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We notice that $\triangle AGH$ and $\triangle BHC$ are congruent right triangles due to the side lengths found and the properties of the regular hexagon. We can apply the Pythagorean theorem to find $\overline{GH}$, but first, we need the length of the base of one of these triangles.



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Considering that the height $\overline{AG} = \frac{\sqrt{3}}{2}$, and knowing that this comes from an equilateral triangle with side $1$, we can deduce that the segment from $A$ to the point where $\overline{GH}$ intersects $\overline{AF}$ (let's call this point $I$) plus $\overline{IH}$ would give us a segment of length $1$. Since $\overline{AH} = \frac{\sqrt{3}}{2}$, we have $\overline{AI} = 1 - \frac{\sqrt{3}}{2}$. However, due to symmetry and properties of the hexagon, $\overline{AI} = \frac{1}{2}$, as the hexagon can be divided into six equilateral triangles, and each of its diagonals cuts through the midpoint of the opposite side when extended.



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Thus, we can see that $\overline{GH}$ effectively bisects the equilateral triangle it intersects, meaning its length can be found by considering the height of an equilateral triangle with side length $1$, which is a standard $\frac{\sqrt{3}}{2}$.



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This thought process shows that we effectively want to calculate the segment which is $\overline{GH}$, but from the setup provided, realizing the equilateral triangle properties simplifies this. So, we don't need the Pythagorean theorem directly for $\overline{GH}$ as initially thought; the setup directly implies $\overline{GH}$'s length due to the hexagon's geometric properties.



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Given the considerations above, $\overline{GH}$ is simply the side of the hexagon projected into its interior, a result of intersecting heights of equilateral triangles, making $\overline{GH} = \frac{\sqrt{3}}{2} \cdot 1 = \frac{\sqrt{3}}{2}$ but due to considering it as a part of a triangle formed by joining the midpoints of the sides of the hexagon, we actually look at it as a side of a smaller equilateral triangle, however this is still the same as $$\frac{\sqrt{3}}{2} \cdot \frac{2}{2}.$$



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Correct - Python is appropriate