Projectile: $v^*w^*=gk$ for minimum launch velocity A projectile launched from $O(0,0)$ at velocity $v$ and launch angle $\theta$, passes through $P(k,h)$. The velocity of the projectile at $P$ is $w$. The slope of $OP$ is $\alpha$, i.e. $\tan\alpha=\frac hk$, and the length of $OP$ is $R$. 
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Let $v^*$ be the minimum launch velocity for the projectile to reach $P$, and $w^* $ the corresponding minimum terminal velocity at $P$. In the course of working out  $v^*$, I noticed this neat  relationship:
$$\color{red}{\boxed{v^*w^*=gk}}\tag{1}$$
which can be proven easily using calculus. The relationship is interesting because of its symmetry and also its independence from $\theta$ and $h$. It also helps simplify the solution of projectile problems relating to minimum velocities, e.g. this question here. 

Question 1: Is it possible to derive this relationship given by $(1)$ directly without using calculus but by exploiting some geometric or kinematic symmetry?

Separately, we know that, for a projectile for a given range $R$ (on the same level), the minimum launch velocity is given by $v^*=\sqrt{gR}$. Assume that, for another given range $r$, the minimum launch velocity is $w^*=\sqrt{gr}$. Substituting in $(1)$ gives $k=\sqrt{Rr}$, i.e. $k$ is the geometric mean of $R$ and $r$. 

Question 2: Can the relationship given by $(1)$ be derived using the relationships between minimum launch velocity and range shown above, perhaps through a geometric transform of some sort?


This is a complement to the previously posted answer. While solving this problem I realized that the set-up also serves to prove another interesting fact. That is, the flight times of the optimal trajectories to the target depend only on $R=\sqrt{h^2+k^2}$, independent of inclination.
In the second picture below, when the minimum energy projectile thrown from $A$ reaches $D$, a "projectile" released with zero velocity from $A$ will reach point $G$. In addition, projectiles thrown with parallel velocities (including zero velocity) from the point of origin at the same time remain forming a line parallel to the original velocity at all subsequent times. Therefore, $GD$ (being the line joining the projectiles at $G$ and $D$ at time $t_f$) is parallel to $AX$ (that is, $\vec{v}\parallel AX\parallel GD$), thus, $\angle{BDG}$ is a right angle and $G$ is also in the circumference. Therefore, the flight time  for all minimum energy projectiles is the same as an object falling from $A$ to $G$, given by $\sqrt{\frac{2R}{g}}$. 

Second, more explicit derivation
Using a bit of trigonometry, we show that the smallest initial velocity $\vec{v}$ of the projectile to reach a distance of $R$ in a direction forming an angle of $\alpha$ with the horizontal is attained when the $\vec{v}$ direction bisects the angle between the vertical and $\alpha$. Part of the interest of this answer is that it does not use calculus either.
First, we calculate the range of the projectile in the direction given by $\alpha$ when being thrown with an arbitrary velocity $\vec{v}$ forming an angle $\theta$ with the horizontal. Following the same reasoning as in the previous diagram, we deduce that is $KH\parallel\vec{v}$ (in the second picture below). We can immediately see that the $\angle AHK=\theta-\alpha$, and
$$ \frac{gt_f^2}{2}= \frac{R\sin(\theta-\alpha)}{\cos\theta}~~.\tag{1}$$
 
Furthermore, it is also clear that, since there is no acceleration in the $\hat{x}$ direction,
$$ R\cos\alpha=vt_f\cos\theta~~.\tag{2}$$
From (1) and (2) we deduce 
$$R=\frac{v^2}{g}\frac{2\cos(\theta)\sin(\theta-\alpha)}{\cos^2\alpha}=\frac{v^2}{g}\frac{(\sin(2\theta-\alpha)-\sin(\alpha))}{\cos^2\alpha}~~,\tag{3}$$
from where it is very clear that the maximum range is given by $2\theta-\alpha=\pi/2~~$ (or $\theta$ bisects the angle between $AH$ and the vertical).
In turn, the maximum range is given by $\frac{v^2}{g(1+\sin\alpha)}$.
Concluding, by squaring (1)
$$\begin{align}
t_f^4&= \left(\frac{2R}{g}\right)^2\frac{\sin^2((\pi/2-\alpha)/2)}{\cos^2((\pi/2+\alpha)/2)}\qquad;\theta=(\pi/2+\alpha)/2\\
&=\left(\frac{2R}{g}\right)^2 \frac{\frac{1-\cos(\pi/2-\alpha)}{2}}{\frac{1+\cos(\pi/2+\alpha)}{2}}\\
&=\left(\frac{2R}{g}\right)^2 \frac{1-\sin\alpha}{1-\sin\alpha}\\
&=\left(\frac{2R}{g}\right)^2~~,
\end{align}
$$
therefore $t_f=\sqrt{\frac{2R}{g}}$, that is, the same result as in the first answer with $t_f$ independent of $\alpha$.
