# Solution to the [[A Triangle Inside a Hexagon]] Puzzle +-- {.image} [[ATriangleInsideAHexagon.png:pic]] > An equilateral triangle inside a regular hexagon. What’s the angle? =-- ## Solution by [[Congruent Triangles]], [[Angles in a Regular Hexagon]], and [[Angles in an Equilateral Triangle]] +-- {.image} [[ATriangleInsideAHexagonAnnotated.png:pic]] =-- In the above diagram, $O$ is the centre of the hexagon. Although $O H F$ looks like a straight line - and will turn out to be a straight line - that is not a given from the diagram. Consider triangles $A G O$ and $A H F$. Side lengths $A G$ and $A H$ have the same length, as do $A O$ and $A F$. Triangles $B A O$, $G A H$, and $O A F$ are all [[equilateral triangles]], so angles $G \hat{A} H$ and $O \hat{A} F$ are both $60^\circ$. Therefore, angles $G \hat{A} O$ and $H \hat{A} F$ are equal. This establishes that triangles $A G O$ and $A H F$ are [[congruent]]. Hence angle $A \hat{F} H$ is the same as angle $A \hat{O} G$, which is $60^\circ$. Since the interior angle in a [[regular hexagon]] is $120^\circ$, angle $H \hat{F} E$ is $60^\circ$. ## Solution by [[Rotation]] This is an application of the [[weak invariance principle]]. The point $G$ is not fixed on the line $B E$ and as $G$ moves along this line, so then $H$ moves as well. To identify the possible positions of $H$, consider a rotation of angle $60^\circ$ anticlockwise about point $A$. This takes point $B$ to $O$, $G$ to $H$, and $O$ to $F$. Therefore, since $B G O$ is a straight line, so also is $O H F$. Hence, $H$ always lies on the line $O F$. This establishes that angle $H \hat{F} E$ is the same as angle $O \hat{F} E$, which is $60^\circ$. ## Solution by [[Invariance Principle]] +-- {.image} [[ATriangleInsideAHexagonInvariance.png:pic]] =-- Since $G$ is free to move along the line segment $B E$, there are configurations where the angle is straightforward to see. In the above, $G$ coincides with $B$ which brings $H$ to $O$ and hence angle $H \hat{F} E$ is $60^\circ$.