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Geometry Difficulty 3.0 AMC 10/12 Prove it Canada

A regular hexagon is a polygon that has six sides with
equal length and six interior angles with equal measure. In Figure 1,
regular hexagon $ABCDEF$ has side
length $2x$ and its vertices lie on
the circle with centre $O$ . The
diagonals $AD$ , $BE$ and $CF$ divide $ABCDEF$ into six congruent equilateral
triangles.

In terms of $x$ , what is the radius of the
circle?
The midpoint of side $AB$ is labelled $M$ , as shown in Figure 2. In terms of
$x$ , what is the length of $OM$ ?
In terms of $x$ , what is the area of hexagon $ABCDEF$ ?
The region that lies inside the circle
and outside hexagon $ABCDEF$ is
shaded, as shown in Figure 3. The area of this shaded region is 123. Rounded to the nearest tenth,
determine the value of $x$ .

Solution

Regular hexagon $ABCDEF$ has
side length $2x$ , and so $AB=2x$ .

Since $\triangle OAB$ is equilateral,
then $OA=OB=AB=2x$ .

The radius of the circle is equal to $OA$ and thus is $2x$ .
Since $M$ is the midpoint of
$AB$ and $OA=OB$ , then $OM$ is perpendicular to $AB$ .

Since $M$ is the midpoint of $AB$ , then $AM=\frac12AB=x$ .

Using the Pythagorean Theorem in right-angled $\triangle OAM$ , we get $OA^2=OM^2+AM^2$ or $(2x)^2=OM^2+x^2$ , and so $OM^2=3x^2$ or $OM=\sqrt{3}x$ (since $OM>0$ ).

Alternatively, notice that $\$\triangle$
OAM $is a$ $30^{\circ}$ $-$ $60^{\circ}$ $-$ $90^{\circ}$ $triangle, and so$ $AM:OA:OM=1:2:\sqrt{3}=x:2x:\sqrt{3}x\$.$

[[IMAGE0]]

The diagonals $AD$ , $BE$ and $CF$ divide $ABCDEF$ into six congruent equilateral
triangles.

Thus the area of $ABCDEF$ is six
times the area of $\triangle OAB$
(having base $AB$ and height $OM$ ), or $6\times\tfrac12\times AB\times OM=3\times 2x\times \sqrt{3}x=6\sqrt{3}x^2$
The area of the shaded region is determined by subtracting the
area of $ABCDEF$ from the area of the
circle with centre $O$ and radius
$2x$ .

Thus, the area of the shaded region is $\pi(2x)^2-6\sqrt{3}x^2=4\pi x^2-6\sqrt{3}x^2=(4\pi-6\sqrt{3})x^2$ The area of this shaded
region is 123, and so $(4\pi-6\sqrt{3})x^2=123$ or $x^2=\dfrac{123}{4\pi-6\sqrt{3}}$ .

Since $x>0$ , we get $x=\sqrt{\dfrac{123}{4\pi-6\sqrt{3}}}$ and
so $x=7.5$ when rounded to the
nearest tenth.

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.