2010 AMC 10A Problem 19

Question

Equiangular hexagon

ABCDEF

has side lengths

AB=CD=EF=1

and

BC=DE=FA=r.

The area of

\triangle ACE

is

70\%

of the area of the hexagon. What is the sum of all possible values of

r?

\dfrac{4\sqrt{3}}{3}

\dfrac{10}{3}

4

\dfrac{17}{4}

6

Split the hexagon into

\triangle ACE

and three corner triangles

Express both areas using

r^2+r+1

and

r

Note that

\triangle ACE

is equilateral. Using the Law of Cosines in

\triangle ABC,

we get

AC^2=r^2+1^2

-2r\cos120^\circ

=r^2+r+1.

The area of

\triangle ACE

is then

\dfrac{\sqrt3}{4} (r^2 + r + 1).

The three corner triangles

\triangle ABC,

\triangle CDE,

and

\triangle EFA

each have area

\frac12\cdot1\cdot r\cdot\sin120^\circ=\frac{r\sqrt3}{4}.

Thus the hexagon has area

\dfrac{\sqrt3}{4}(r^2+r+1)

+3\cdot\dfrac{r\sqrt3}{4}

=\dfrac{\sqrt3}{4}(r^2+4r+1).

The condition

[ACE]=70\%\cdot[ABCDEF]

gives

r^2+r+1=\dfrac{7}{10}(r^2+4r+1),

so

r^2-6r+1=0.

By Vieta’s formulas, the sum of the possible values of

r

is

6.

Thus, E is the correct answer.

Problem 19 in other years