14. The monument to Mathenkamen is in the shape of a pyramid resting on its square base with a side of $18 \mathrm{~m}$. Its height measures $15 \mathrm{~m}$, and the foot of the height falls at the center of the square. The pyramid is oriented so that when the sun's rays come from the south at an angle of $45^{\circ}$ to the ground, the area of the ground on which it casts its shadow is as small as possible. What is this area expressed in $\mathrm{m}^{2}$?

(Note: the ground covered by the base of the pyramid is not counted as ground in shadow.)
14. The answer is 27. Let $ABCD$ be the base of the pyramid and $V$ its vertex far from the ground. Note that the shadow $V'$ of point $V$ falls on a circle of radius $15 \, \text{m}$ centered at the center of $ABCD$, and depending on the orientation of south relative to the pyramid (which we consider fixed), it can fall on any point of the circle. The shadow of the pyramid (including this time the area of the ground under the pyramid itself) is formed by the union of the four triangles $ABV'$, $BCV'$, $CDV'$, $DAV'$.
![](https://cdn.mathpix.com/cropped/2024_04_17_74150396f6f11420587fg-17.jpg?height=438&width=1636&top_left_y=734&top_left_x=220)

This shadow can take two notably different forms. Refer to the figures: when $V'$ falls in one of the dotted parts of the circle in the top figure, the shadow is a pentagon (left figure); when $V'$ falls in the other parts, the shadow is a quadrilateral (right figure). Let $W$ (resp. $X$) be the intersection of the ray $DA$ (resp. $BA$) with the circle. We want to show that in both configurations, the area of the shadow is greater than or equal to the area of the trapezoid $WBDC$. Indeed, in the configuration of the left drawing, $S(\text{shadow}) = S(ABCD) + S(ABV')$, and $S(ABV') \geq S(ABW)$ because they are two triangles with the same base $AB$, and the height of $ABV'$ is greater than that of $ABW$. Similarly, in the configuration of the right drawing, $S(\text{shadow}) = S(BCD) + S(BDV')$, and $S(BDV') \geq S(BDW)$ because they are two triangles with the same base $BD$, and the height of $BDV'$ is greater than that of $BDW$ (to see this, note that $WX$ is parallel to $BD$). Therefore, the shadow of minimum area is obtained when $V' = W$, or in the symmetric configurations on the other sides. The area of triangle $ABW$ must then be calculated. Let $M$ be the midpoint of $AD$, so $MW \perp OM$, $OW = 15 \, \text{m}$, $OM = 9 \, \text{m}$, and by the Pythagorean theorem, $MW = \sqrt{OW^2 - OM^2} = 12 \, \text{m}$. Thus, $AW = MW - AM = 12 - 9 = 3 \, \text{m}$, and $S(ABW) = \frac{AB \cdot AW}{2} = \frac{18 \cdot 3}{2} = 27 \, \text{m}^2$.