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Problem 675

AMC 10/12, early questions

Geometry Difficulty 3.0 Prove it CEMC Hypatia · Canada · 2022

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 $x$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official 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

Figure for this problem

Figure for this problem ABCDEF is six times the area of

Figure for this problem

Figure for this problem $\triangle$ 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

Figure for this problem

Figure for this problem ABCDEF from the area of the circle with centre

Figure for this problem

Figure for this problem O $and radius$ 2x. Thus, the area of the shaded region is

Figure for this problem

Figure for this problem $\pi(2x)^2-6\sqrt{3}x^2=4\pi x^2-6\sqrt{3}x^2=(4\pi-6\sqrt{3})x^2$ The area of this shaded region is 123, and so

Figure for this problem

Figure for this problem $(4\pi-6\sqrt{3})x^2=123$ $or$ $x^2=\dfrac{123}{4\pi-6\sqrt{3}}$ $. Since$ x>0 $, we get$ $x=\sqrt{\dfrac{123}{4\pi-6\sqrt{3}}}$ $and so$ x=7.5$ when rounded to the
nearest tenth.

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