Non-trig solution for AMC 10 Question

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 $r$? AMC 10A 2019#17, AMC 12A 2010#17

Since this was asked in AMC 12, Law of Cosines is an appropriate tool. And the video solution at the link above does it beautifully. However, since it was asked in AMC 10 as well, there (usually) should be a non-trig solution as well.

Can anyone help with a geometrical solution? (Note: I understood the trig solution, but can't think of a geometrical solution).

Answer 1

Let's use the property: "similar triangles are to one another in the squared ratio of (their) corresponding sides" ^[1].

Follow the image.

As $\small \triangle ABC$ and $\small \triangle BMC$ have equal height, $\mathsf{ratio\ of\ their\ areas=ratio\ of\ bases}$.

Let $\Delta$=big equilateral triangle with side-length $(2r+1)$.

$$\frac{\color{black}\Delta}{\color{blue}\Delta}=\frac{(10+3r)A}{rA}=\frac{(2r+1)^2}{r^2}$$ $$\implies r^2-6r+1=0$$

Now we can use Vieta's formula to get the answer.

Answer 2

It is elementary geometry that

• $ABCDEF$ concyclic
• $ABCD$ is an isoceles trapezium
• Construct point $P$ on $AD$ such that $ABCP$ is a parallelogram. Then
  • $\triangle CDP$ is equilateral of side length $CD=1$
  • The "main" diagonals $AD=BE=CF=r+1$.
  • The distance between $AD$ and $BC$ is $\frac{\sqrt3}2$.
  • The area of parallelogram $ABCP=\frac{\sqrt3}2r$.
  • Ptolemy's theorem gives $AC^2=AC\cdot BD = AD\cdot BC+AB\cdot CD=(r+1)r+1=r^2+r+1$
• So area of $\triangle ABC = \frac{\sqrt{3}}4 r$, and area of equilateral $\triangle ACE$ is $\frac{\sqrt3}4AC^2=\frac{\sqrt3}4(r^2+r+1)$.

Since the area of triangle $ACE$ is 70% of the hexagon, each of the smaller triangles $ABC,CDE,EFA$ are 10% of the hexagon. So $r$ must satisfy $$\require{cancel} 7 {\cancel{\frac{\sqrt3}4}}r = \cancel{\frac{\sqrt3}4}(r^2+r+1) $$ and now Vieta formula gives the answer.