Optimal shot on goal

We consider the triangle formed by the point S (striker) and the points A and B (goal post). The best chance to score has the striker when the opening angle $\epsilon$ becomes maximal. This is true even if the goalkeeper tries to cover the space between striker and goal, because at maximum angle $\epsilon$ also the distance between the goalkeeper and the ball's path becomes maximum. (This is especially true when the ball is played against the running direction of the keeper.)

The length $s = 16\,\text{m}$ und $w = 7.3\,\text{m}$ are the distance to goal post and the width of the goal, respectively. The angles are determined by $\begin{aligned} \tan(\alpha) &= \frac{s}{x} \\ \tan(\alpha+\varepsilon) &= \frac{s+w}{x} \\ \Rightarrow \quad \varepsilon &= \arctan\left(\frac{s+w}{x}\right) - \arctan\left(\frac{s}{x}\right) \end{aligned}$ With $\arctan'(y) = \frac{1}{1+y^2}$ the derivative results $\begin{aligned} & & \frac{d\varepsilon}{dx} = \frac{- \frac{s+w}{x^2}}{1 + \left(\frac{s+w}{x}\right)^2} - \frac{- \frac{s}{x^2}}{1 + \left(\frac{s}{x}\right)^2} &\stackrel{!}{=} 0 \\ \Rightarrow & & \frac{s+w}{x^2 + (s+w)^2} &= \frac{s}{x^2 + s^2} \\ \Rightarrow & & s (x^2 + (s+w)^2) &= (s+w)(x^2 + s^2) \\ \Rightarrow & & - w x^2 &= - w(w + s) s \\ \Rightarrow & & x &= \sqrt{s(w +s)} \approx 19.3\,\text{m} \end{aligned}$ Therefore, the result is the geometric mean between the distances $s$ and $s+w$ between the striker and the goal posts along the goal line.

This solution has a similar basis as the one Markus provided. I did not want to take derivatives of trigonometric functions, though, so I used some tricks to get rid of them. I assume $\theta_1$ , $\theta_2$ are the angles to the corners of the goal with respect to the side of the pitch and $L_1$ , $L_2$ are the distances to the same corners. The quantity that needs maximized is $\theta_2 - \theta_1$ . Because $\sin(\theta_2 - \theta_1)$ follows the same trend over the domain of interest, that function can be used instead. Employing the rules of addition/subtraction for the sine function,

$\begin{aligned} \sin( \theta_2 - \theta_1) &= \sin(\theta_2) \cos(\theta_1) - \cos(\theta_2) \sin(\theta_1) \\ &= \frac{23.3}{L_2} \cdot \frac{x}{L_1} - \frac{x}{L_2} \cdot \frac{16}{L_1} \\ &= \frac{7.3 x}{L_1 L_2} \end{aligned}$

$L_1 L_2 = \sqrt{ (x^2 + 16^2) (x^2+23.3^2)} = \sqrt{x^4 + 2(16^2 + 23.3^2) x^2 + (16 \cdot 23.3)^2}$

We need to maximize this quantity with respect to $x$ , so taking its derivative and solving for zero. Using a generalized equation,

$\frac{d}{dx} \left( \frac{ a x }{ \sqrt{x^4 + b x^2 + c} } \right) = \frac{ a (x^4 - c) }{(x^4 + b x^2 + c)^{3/2} }$

$x$ can be solved for by setting $x^4 - c = 0$ .

$\begin{aligned} & x^4 - (16 \cdot 23.3)^2 = 0 \\ & x = (16 \cdot 23.3)^{1/2} \\ & x = 19.308... \end{aligned}$

Draw any circle through the two goals posts ( $P_1, P_2$ ), and put the striker $S$ on that circle (to the right of the goal line). The view angle of the goal for the striker ( $\angle P_1SP_2$ ) is independent of the location of the striker on that circle (a geometry theorem). Furthermore, it varies inversely with the radius of the circle: smaller circle gives larger view angle. So, we are looking for the smallest circle through the goal posts and the striker. This is a circle that touches the sideline on which the striker is located, and its center is located on the perpendicular bisector of the goal posts. Hence, the radius of that circle equals $16+7.3/2 = 19.65$ . See diagram:

Therefore, we have $x^2 + 3.65^2 = 19.65^2$ , and thus $x = \sqrt{19.65^2 - 3.65^2} \approx \fbox{19.308}$ .

I did y= (tan^-1(16/x))-(tan^-1(23.3/x)) and looked for the max value when it was graphed which turned out to be 19.311 Done by a highschool kid with a C+ in trig

We want the angle from the player to the goal to be largest, which can be written as

$θ = tan^{-1}(\frac{23.3}{x}) - tan^{-1}(\frac{16}{x})$

Now combine the two arctans:

$tan(θ)=tan(tan^{-1}(\frac{23.3}{x}) - tan^{-1}(\frac{16}{x}))$

$=\frac{\frac{23.3}{x}-\frac{16}{x}}{1+\frac{23.3}{x}\frac{16}{x}}$

$=\frac{7.3x}{x^2+(23.3 \times 16)}$

If we want to maximise θ, we just maximise tan(θ). Therefore we just solve

$\frac{d}{dx} \frac{7.3x}{x^2+(23.3 \times 16)}=0$

$⇒ x = \boxed{19.3080...}$

If we let $a=16,b=16+7.3$ and the angle be $\theta$ combining trig and Pythagoras we can get: $\cos{\theta}=\frac{x^2+ab}{\sqrt{(x^2+a^2)(x^2+b^2)}}=\frac{x^2+ab}{\left( \sqrt{ab} \right) \left (\sqrt{\left(\frac{x^2}{a}+a \right) \left(\frac{x^2}{b}+b \right)} \right)}$

And using AM-GM: $\cos{\theta} \geq \frac{2}{\sqrt{ab}} \cdot \frac{x^2+ab}{\frac{x^2}{a}+\frac{x^2}{b}+a+b}= \frac{2}{\sqrt{ab}} \cdot \frac{ab}{a+b}=\frac{2\sqrt{ab}}{a+b}$

As $\cos{\theta}$ is decreasing as $\theta$ is increasing this corresponds to the maximum value of $\theta$ . Equality is when $x=\sqrt{ab}=\boxed{19.3}$

Using the Law of Sines, we have

$\dfrac { \sqrt { { x }^{ 2 }+{ 16 }^{ 2 } } }{ Sin\left( ArcSin\left( \dfrac { x }{ \sqrt { { x }^{ 2 }+{ \left( 16+7.3 \right) }^{ 2 } } } \right) \right) } =\dfrac { 7.3 }{ Sin\left( \theta \right) }$

where $\theta$ is the angle the goal posts present to the striker, which we wish to maximize. Since for small $\theta$ the right side is a monotonic, we only need to find $x$ such that the following has the minimum value for the maximum $\theta$

$\dfrac { 1 }{ x } \sqrt { { x }^{ 2 }+{ 16 }^{ 2 } } \sqrt { { x }^{ 2 }+{ \left( 16+7.3 \right) }^{ 2 } }$

which occurs when $x=19.308...$