In the convex pentagon $A B C D E$, $\angle A B C = \angle C D E = 90^{\circ}, B C = C D = A E = 1, A B + E D = 1$. What is the area of the pentagon?
I. solution. Let $P$ be the point on segment $AE$ such that $AP = AB$ and thus $PE = ED$. Let $CP = d$, $\angle APC = \alpha$, and $\angle EPC = \beta$.

![](https://cdn.mathpix.com/cropped/2024_05_02_c4526140a2957f1a6b70g-1.jpg?height=480&width=528&top_left_y=267&top_left_x=793)

In triangles $ABC$ and $APC$, two pairs of sides are equal, so $d > 1 \Leftrightarrow \alpha > 90^\circ$. Similarly, from the comparison of triangles $CPE$ and $CDE$, $d > 1 \Leftrightarrow \beta > 90^\circ$. Thus, if $d > 1$, then $180^\circ = \alpha + \beta > 90^\circ + 90^\circ = 180^\circ$; both are impossible, so $d = 1$, and thus triangles $ABC$ and $APC$, as well as triangles $CPE$ and $CDE$, are congruent. The area of the pentagon is thus twice the sum of the areas of triangles $ABC$ and $CDE$; if $AP = AB = x$, then

$$
2\left(\frac{x \cdot 1}{2} + \frac{(1-x) \cdot 1}{2}\right) = 1
$$

Remark. The following solution shows that the regularity of the pentagon can be used more elegantly than above.

II. solution. Rotate triangle $CDE$ around point $C$ so that vertex $D$ moves to point $B$.

![](https://cdn.mathpix.com/cropped/2024_05_02_c4526140a2957f1a6b70g-1.jpg?height=542&width=417&top_left_y=1366&top_left_x=843)

The resulting triangle $CBE'$, together with triangle $ABC$, forms triangle $ACE'$. This triangle is congruent to triangle $ACE$, since side $AC$ is common to both, $CE = CE'$, and $AE = 1 = AB + ED = AB + BE' = AE'$. Therefore, the area of the pentagon is exactly twice the area of triangle $ACE'$. Since in this triangle the height from $C$ to the unit-length side $AE'$ is also of unit length, the area of the triangle is $\frac{1}{2}$, and thus the area of the pentagon is exactly 1 square unit.

Remark. We can arrive at the solution without preconceiving the symmetry hidden in the pentagon. The lengths of segments $AC$ and $CE$ are, by the Pythagorean theorem, $AC = \sqrt{1 + x^2}$ and $CE = \sqrt{1 + (1-x)^2} = \sqrt{2 - 2x + x^2}$. Applying the cosine rule to side $CE$ in triangle $ACE$:

$$
2 - 2x + x^2 = 1 + (1 + x^2) - 2 \cdot 1 \cdot \sqrt{1 + x^2} \cdot \cos \angle EAC,
$$

from which $\cos \angle EAC = \frac{x}{\sqrt{1 + x^2}} = \cos \angle BAC$. Thus, due to the convexity of the pentagon, $\angle EAC = \angle BAC$, which implies that triangle $ABC$ is congruent to triangle $APC$. Similarly, the congruence of triangles $CPE$ and $CDE$ can also be shown.