2017 UNCO Math Contest II Problems/Problem 2


[asy]   pair A=dir(60),B=dir(120),C=dir(180),D=dir(240),E=dir(300),F=dir(360),O=(0,0); pair G=(2/sqrt(3))*A,H=(2/sqrt(3))*B,I=(2/sqrt(3))*C,J=(2/sqrt(3))*D,K=(2/sqrt(3))*E,L=(2/sqrt(3))*F; draw(circle(O,1),black); draw(A--B--C--D--E--F--A); draw(G--H--I--J--K--L--G);  [/asy]

Find the ratio of the area of a regular hexagon circumscribed around a circle to the area of a regular hexagon inscribed inside the same circle. (A polygon is called regular if all its sides are the same length and all its corner angles have the same measure. A hexagon is a polygon with six sides.)


Since the ratio of the areas of two similar hexagons is the square of the ratio between the side lengths of the hexagons. If we let the radius of the circle be $r$, then the side length of the smaller hexagon is $r$ and the side length of the larger hexagon is $\frac{2}{\sqrt{3}}\cdot r$. The ratio of the lengths is $\frac{2}{\sqrt{3}}$, which squared equals $\boxed{\frac{4}{3}}$

See also

2017 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All UNCO Math Contest Problems and Solutions