Pentagon $A B C D E$ is such that all five diagonals $A C, B D, C E, D A$, and $E B$ lie entirely within the pentagon. If the area of each of the triangles $A B C, B C D, C D E, D E A$, and $E A B$ is equal to 1 , the area of the pentagon $A B C D E$ is closest to
(A) 3.62
(B) 3.64
(C) 3.66
(D) 3.68
(E) 3.70
Solution 1

We will consider an arbitrary pentagon satisfying the given condition.

First, we will show that each of the diagonals is parallel to the corresponding side.

Since Area $\triangle B C D=$ Area $\triangle C D E$ and these triangles have a common base, then their heights must be equal, that is, the points $B$ and $E$ must be the same perpendicular distance from the line $C D$, that is, the line $B E$ is parallel to the line $C D$. The same can be said for all of the other diagonals.

Now join $B D, C E$ and $B E$. Since $B D$ is parallel to $A E$

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-243.jpg?height=334&width=455&top_left_y=2034&top_left_x=1366)
and $C E$ is parallel to $A B$, then $B A E X$ is a parallelogram, and so the area of $\triangle B E X$ is equal to the area of $\triangle E A B$, which is 1.

Suppose the area of $\triangle C X D$ is $t$. Then the areas of $\triangle C X B$ and $\triangle E X D$ are both $1-t$. Since $\triangle C X D$ and $\triangle C X B$ have a common height, then the ratio of their bases is the ratio of their areas, ie.

$\frac{D X}{B X}=\frac{t}{1-t}$. Similarly, $\frac{C X}{E X}=\frac{t}{1-t}$.

But since $B E$ is parallel to $C D$, then $\triangle C X D$ is similar to $\triangle E X B$, and so the ratio of their areas is the square of the ratio of their side lengths, ie. $\frac{\text { Area } \triangle E X B}{\text { Area } \triangle C X D}=\frac{1}{t}=\left(\frac{1-t}{t}\right)^{2}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-244.jpg?height=331&width=345&top_left_y=344&top_left_x=1367)

Simplifying, $t^{2}-3 t+1=0$ or $t=\frac{3-\sqrt{5}}{2}$ (since $t$ is smaller than 1).

Therefore, the area of the entire pentagon is

$$
1+1+t+(1-t)+(1-t)=4-t=\frac{5+\sqrt{5}}{2} \approx 3.62
$$

## Solution 2

Since the problem implies that the area of the pentagon is independent of the configuration (assuming the appropriate area conditions hold), let us assume that the pentagon is a regular pentagon, with side length $s$.

We first determine the value of $s$. Consider $\triangle A B C$. Since a regular pentagon has each interior angle equal to $108^{\circ}$, then $\triangle A B C$ is isosceles with one angle equal to $108^{\circ}$. Let $X$ be the midpoint of $A C$, and join $B$ to $X$. Then $B X$ is perpendicular to $A C$ since $\triangle A B C$ is isosceles, and $\angle A B X=\angle C B X=54^{\circ}(B X$ bisects $\angle A B C)$.

Therefore, $B X=s\left(\cos 54^{\circ}\right)$ and $A X=C X=s\left(\sin 54^{\circ}\right)$.

Therefore, the area of $\triangle A B C$ is

$$
1=\frac{1}{2}\left[s\left(\cos 54^{\circ}\right)\right]\left[2 s\left(\sin 54^{\circ}\right)\right]=s^{2} \sin 54^{\circ} \cos 54^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-244.jpg?height=295&width=591&top_left_y=1300&top_left_x=1361)

which enables us to determine that $s^{2}=\frac{1}{\sin 54^{\circ} \cos 54^{\circ}}$.

Area of pentagon $=$ Area $\triangle A B C+$ Area $\triangle A C E+$ Area $\triangle C D E$

$$
=2+\text { Area } \triangle A C E
$$

So we lastly consider $\triangle A C E$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-244.jpg?height=336&width=333&top_left_y=1751&top_left_x=1492)

Joining $A$ to $Y$, the midpoint of $C E$, we see that $A Y$ is

perpendicular to $C E, C Y=Y E=\frac{1}{2} s$ and $\angle A C E=\angle A E C=72^{\circ}$.

Therefore, $A Y=\frac{1}{2} s\left(\tan 72^{\circ}\right)$, and thus

Area $\triangle A C E=\frac{1}{2} s\left[\frac{1}{2} s\left(\tan 72^{\circ}\right)\right]=\frac{1}{4} s^{2}\left(\tan 72^{\circ}\right)=\frac{\tan 72^{\circ}}{4 \sin 54^{\circ} \cos 54^{\circ}} \approx$

and so the area of the pentagon is closest to $2+1.618$ or 3.62 .

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-245.jpg?height=447&width=265&top_left_y=243&top_left_x=1491)

ANSWER: (A)