Let $A B C D E$ be a convex pentagon satisfying property $(P): \operatorname{area}(A B C)=\operatorname{area}(B C D)=$ area $(C D E)=\operatorname{area}(D E A)=\operatorname{area}(E A B)=1$. Calculate the area of the pentagon $A B C D E$. Is this pentagon necessarily regular?

## - Solutions of the exercises -


The triangles $BCD$ and $CDE$, having the same area, have the same base $CD$ and thus the same height relative to $CD$, which means $(BE)$ is parallel to $(CD)$. Similarly, we can prove that, for example, $(CE)$ is parallel to $(AB)$ and $(BD)$ to $(AE)$. If we call $P$ the intersection of the diagonals $BD$ and $CE$, $ABPE$ is a parallelogram. By symmetry with respect to the center of the parallelogram, the area of $(\mathrm{BPE})$ equals the area of $(EAB)=1$. It remains to determine the area $s$ of $PCD$, because the area of $(PCB) = \operatorname{area}(PDE) = 1-s$. Therefore, the area of $(ABCDE) = 4 - s$. Now, $\frac{\text{area}(\mathrm{PCB})}{\text{area}(\mathrm{PCD})} = \frac{\mathrm{PB}}{\mathrm{PD}} = \frac{\text{area}(EPB)}{\text{area}(EPD)}$ can be written as: $\frac{1-s}{s} = \frac{1}{1-s}$, which simplifies to: $1 - 3s + s^2 = 0$. The only root between 0 and 1 is: $\frac{3 - \sqrt{5}}{2}$. Therefore, the area of $(ABCDE) = \frac{5 + \sqrt{5}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_10_59677aa68548da326d08g-192.jpg?height=428&width=456&top_left_y=2079&top_left_x=802)

The pentagon is not necessarily regular. Consider, for instance, a regular pentagon, and choose an orthonormal coordinate system in which the vertices $A, B, C, D, E$ have coordinates $(a, a'), (b, b'), (c, c'), (d, d'), (e, e')$ respectively. Consider another coordinate system that is neither orthogonal nor normalized, and draw in this new system the points with the same coordinates. All areas in the first system will be multiplied by the same constant in the second system, so in this second system, the five triangles will also have the same area. However, the pentagon will no longer be regular.

## Use of Complex Numbers in Geometry

To conclude, here are two examples of using complex numbers in geometry, one classic and easy, the other non-classic and more difficult.

Recall that complex numbers are numbers of the form $x + iy$, where $i$ satisfies $\mathfrak{i}^2 = (-\mathfrak{i})^2 = -1$. Each complex number $x + iy$ can be associated with a point of coordinates $(x, y)$ or a vector with components $(x, y)$. $x + iy$ is called the affix of the point or vector. We can thus quotient two vectors by calculating the quotient of their affixes: this quotient $x + iy$ has a modulus $\rho = \sqrt{x^2 + y^2}$, which is the ratio of the moduli of the vectors, and an argument $\theta$ (such that $x + iy = \rho(\cos \theta + i \sin \theta)$, which is the angle between the denominator vector and the numerator vector. The two complex numbers $x + iy$ and $x - iy$ are called conjugates, and if a complex number has a modulus of $1 = \sqrt{x^2 + y^2}$, its conjugate is equal to its inverse.