0
$\begingroup$

Let $A_1A_2A_3A_4A_5$ be a regular pentagon with side length $1$. The sides of the pentagon are extended to form the $10$-sided polygon shown in bold in the picture that I have attached. Find the ratio of the area of quadrilateral $A_2A_5B_2B_5$ to the area of the entire $10$-sided polygon.

Here I have attached a diagram of what I have understood so far about this problem. Pentagon Problem

$\endgroup$
2
  • $\begingroup$ This question comes from the current USAMTS Round 3 problem set (problem 2). This question will remain locked until after the submission deadline of 19 Jan 2015. $\endgroup$ – user642796 Dec 15 '14 at 7:53
  • $\begingroup$ This was a problem asked during an open mathematical competition for the purposes of gaining an unnatural and unfair advantage. It had been locked and hidden. Although it is now unlocked, since the competition has passed, this is abusive and bad; and I downvoted. It is unfortunate, as the question is pleasant and the answer good. $\endgroup$ – davidlowryduda Jan 20 '15 at 7:27
1
$\begingroup$

Join

  1. $A_2$ and $A_5$.

  2. $A_3$ and $A_5$.

Note that $A_2A_3A_5$ is congruent to any of the triangle that are formed by extending the sides of the triangle. We denote the area of any such triangle by $S$.

Notice that the pentagon is partitioned into three part. One of them is the triangle $A_2A_3A_5$ with area $S$ and the other two parts, namely $A_2A_1A_5$ and $A_3A_4A_5$ are congruent (easy to check!). Let us say each of them covers an area $T$. Hence, area of the pentagon is $S+2T$.

Area of the shaded region : $2S+(S+T)=3S+T$

Area of the star: $5S+(S+2T)=6S+2T$

Clearly, the ratio is $\frac{1}{2}$

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.