# Area Ratio of a Polygon

Let $A_{1}A_{2}A_{3}A_{4}A_{5}$ be a regular pentagon with side length $1$. The sides of the pentagon are extended to form the $10$-sided polygon shown in bold at right. Find the ratio of the area of quadrilateral $A_{2}A_{5}B_{2}B_{5}$ (shaded in the picture above) to the area of the entire $10$-sided polygon.

• This is from usamts.org/Tests/Problems_26_3.pdf isn't it? – Fizz Jan 20 '15 at 13:30
• @RespawnedFluff The competition ended yesterday – problemsolver Jan 20 '15 at 13:42
• @problemsolver: It's still nice to know the source of a question, in order to get an indication of its difficulty or intended toolset. (Also, it's generally preferred that questioners provide context about their own attempts to solve their problems, so that we know how best to answer.) – Blue Jan 20 '15 at 14:30

Connect, say $A_2$ to $A_4$ and argue that $\triangle A_2A_4A_5$ is congruent to $\triangle A_3A_4B_1$, hence has the same area. The same is also (obviously) true of $\triangle A_2A_3A_4$ and $\triangle A_1A_2A_5$. So there are equal amounts of white and shaded area (hence the requested ratio is $1/2$).

the lengths of the sides of the pentagram $A_3B_1 = \tau A_3A_4,$ the golden ratio. let the area of the sharper triangle $B_3A_4 = 1,$ then $A_2A_3A_4 = 1/\tau$ because the base in this ratio and has the third vertex. the area of the shaded region is $3 + 1/\tau$ made up three acute and one obtuse triangle. so is the unshaded, therefore the ratio is half.

Tangrams solution to cover the left half of the picture:

Move $\Delta A_4A_5B_2$ to $\Delta A_1A_2B4$

Rotate quadrilateral $A_2A_3A_4A_5$ to $A_1 A_2A_3A_4$

Let $M$ be the midpoint between $A_2$ and $A_3$. Rotate triangle $A_1MA_4$ to $B_1MA_3$.

Hint:

I used simple angles and splits of the diagram to come up with the following and

I get a ratio of $\dfrac{4(sin72+sin72cos72+0.5tan72)}{5(\frac{1}{tan36}+tan72)} = \frac{11.13513}{22.2703} = 0.500$