
5. Given is a regular pentagon $A B C D E$. Determine the least value of the expression

$$
\frac{P A+P B}{P C+P D+P E},
$$

where $P$ is an arbitrary point lying in the plane of the pentagon $A B C D E$.


Solution. Without loss of generality assume that the given pentagon $A B C D E$ has the side equal to 1 . Then the length of its diagonal is equal to

$$
\lambda=\frac{1+\sqrt{5}}{2} .
$$

Set $a=P A, b=P B, c=P C, d=P D, e=P E$ (Fig. 2).

![](https://cdn.mathpix.com/cropped/2024_04_17_dea7690a2ad723913678g-29.jpg?height=629&width=579&top_left_y=1513&top_left_x=750)

Fig. 2

Applying the Ptolemy inequality for the (not necessarily convex) quadrilaterals $A P D E, B P D C$ and $P C D E$ we obtain (respectively)

$$
a+d \geqslant e \lambda, \quad b+d \geqslant c \lambda, \quad e+c \geqslant d \lambda .
$$

We multiply the third inequality by $\frac{\lambda+2}{\lambda+1}$ and add together with the first and the second inequalities. As a result we obtain

$$
a+b+2 d+e \cdot \frac{\lambda+2}{\lambda+1}+c \cdot \frac{\lambda+2}{\lambda+1} \geqslant e \lambda+c \lambda+d \cdot \frac{\lambda(\lambda+2)}{\lambda+1} .
$$

Grouping the respective terms, the above inequality reduces to

$$
a+b \geqslant \frac{\lambda^{2}-2}{\lambda+1}(c+d+e)
$$

Therefore

$$
\frac{a+b}{c+d+e} \geqslant \frac{\lambda^{2}-2}{\lambda+1}=\sqrt{5}-2 .
$$

The equality holds if and only if the convex quadrilaterals $A P D E, B P D C$ and $P C D E$ are cyclic. This condition is satisfied if and only if the point $P$ lies on the minor arc $A B$ of the circumcircle of the pentagon $A B C D E$ (Fig. 3). Therefore the smallest possible value of the given expression is $\sqrt{5}-2$.

![](https://cdn.mathpix.com/cropped/2024_04_17_dea7690a2ad723913678g-30.jpg?height=591&width=579&top_left_y=1121&top_left_x=750)

Fig. 3
