We first find that, in problem 15, each of the circles of radius $\sqrt{Z}$ is the incircle of a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle formed by cutting the equilateral one in half. The equilateral triangle itself has sidelength $\frac{2 \sqrt{Y}}{\sqrt[4]{3}}$, so the said inradius is $$\sqrt{Z}=\frac{1+\sqrt{3}-2}{2} \cdot \frac{1}{2} \cdot \frac{2 \sqrt{Y}}{\sqrt[4]{3}}$$ so that $$Z=\frac{(-1+\sqrt{3})^{2}}{4 \sqrt{3}} Y=\frac{4-2 \sqrt{3}}{4 \sqrt{3}} Y=\frac{2 \sqrt{3}-3}{6} Y$$ Now we guess that $X=2$ and see that, miraculously, everything works: in the problem 14, say a crimson flower is placed first. Then there are 2 possibilities for $C_{-} C_{-{ }_{-1}}, 4$ for $C_{-} C_{--}, 2$ for $C_{-\ldots} C_{-}$, and 2 for $C_{\ldots_{-}} C$, giving a total of 10. Of course, the first flower can be of any of the three hues, so $Y=3 \cdot 10=30$. We compute $Z$ and check $X$ in a straightforward manner. If $X>2$, then $Y>30$, and $Z>10 \sqrt{3}-15$, with the result that $X \leq 2$, a contradiction. Assuming $X<2$ results in a similar contradiction.