# Best angle to attack

Suppose a circle (O,R) and a point A within its area having distance r from O (0<=r<=R). Which points X of the circle minimize the angle between the line AX and the tangent on X?

I was pondering on "realistic" (take this with many grains of salt) ways an energy shield could stop incoming projectiles without stopping outcoming ones. So I was led to think of a one-way high air drag setup. The drag would be "outwards"; thus, incoming bullets would hit it dead on and suffer an 100% penalty; but if they hit the shield at an angle, they would only be dragged back by the cos of that angle. The question then becomes, if you are an outward attacker, what is the best angle to attack someone inside.

In the following image, the question is: which are the points X that minimize the angle φ. The radius is perpendicular to the tangent. Therefore $$\angle OXA + 90 + \theta = 180$$.

Minimizing $$\theta$$ is therefore maximizing $$\angle OXA$$. We can now just concentrate on the triangle $$OXA$$ and ignore the circle. Two sides lengths are given, $$|OX|=R$$ and $$|OA|=r$$. Think of $$XA$$ as the base of the triangle. Since $$OX$$ is a fixed distance, the angle $$\angle OXA$$ is maximal if the distance between the apex $$O$$ and the base line $$XA$$ is largest, i.e. the height of the triangle when considering $$XA$$ as the base needs to be maximized.

The height of a triangle cannot be more than an adjacent side, so in this case it cannot be more than $$|OA|=r$$. It can be equal however, and this happens when $$OA$$ is that height, i.e. if $$OAX$$ is a right angle.

• It was hard to choose which to tick, I chose yours for brevity. Seems the attacker has to choose a spot so that the shield is on the side of the defender. – George Menoutis Nov 7 at 15:23
• @GeorgeMenoutis I was editing it when you gave the tick. I just felt I hadn't fully justified the last step. It is no longer as brief, so I hope you still like it. – Jaap Scherphuis Nov 7 at 15:27
• @GeorgeMenoutis: Minor comment - at 90 degrees it would not enter the circle, merely touch it at one point. The object inside would be unaffected. – Krad Cigol Nov 9 at 5:08

In that triangle $$OAX$$, angle $$OXA$$ is $$\frac\pi2-\phi$$; let's call it $$\theta$$. We want to make $$\theta$$ as large as we can. The side-lengths of that triangle are $$R$$, $$r$$, and the distance $$AX$$ which I'll call $$t$$. Then the cosine rule for angle $$X$$ of triangle $$OAX$$ says $$r^2=R^2+t^2-2Rt\cos\theta$$ so maximizing $$\theta$$ is the same as minimizing $$\frac{R^2+t^2-r^2}{2Rt}$$. $$R$$ and $$r$$ are fixed so we are trying to minimize $$\frac{R^2-r^2}t+t$$ by varying $$t$$. We can do this either by calculus or by appealing to the "AM-GM inequality"; the conclusion is that we need $$t^2=R^2-r^2$$ -- which is equivalent to requiring that $$OAX$$ be a right-angled triangle with right angle at $$A$$. So: draw a line through $$A$$ perpendicular to $$OA$$, and the places where it meets the circle are the points you want.

Construction

Construct the midpoint between $$A$$ and $$O$$, called $$E$$ in the diagram, and construct the perpendicular bisector through it. Determine a point $$Y$$ on this line such that $$|OY| = \frac{R}{2}$$ (there will be at most two such points). Extend the line $$OY$$ to meet the circle at point $$X^*$$ such that $$Y$$ is between $$O$$ and $$X^*$$ on the line. This is a point which will minimise the angle $$\phi$$. There are two such points on the circle (indicated as $$X^*$$ and $$X^{*'}$$ in the diagram) if $$A$$ is inside the circle and does not coincide with the centre. There is just one such point if $$A$$ is on the circle ($$A$$ itself) and infinite such points (all of them) if $$A$$ coincides with the centre. Proof

The angle between $$OX$$ and the tangent is always $$\frac{\pi}{2}$$ so minimizing the angle $$\phi$$ is equivalent to maximising the angle $$\angle AXO$$. Now consider a circle which passes through $$O$$ and $$A$$ and is tangent to the original circle. $$OA$$ will be a chord on this new circle and for any point $$X$$ on the original large circle, the point $$X$$ will either be on or outside the new circle. This means that the angle $$\angle OXA$$ will be maximised precisely when $$X$$ coincides with the tangent point between the new circle and the original. In our construction, the point $$Y$$ corresponds precisely to the centre of a circle with radius $$\frac{R}{2}$$ which passes through $$O$$ and $$A$$ and is tangent to the original circle at $$X^*$$ so $$X^*$$ is the point which will maximise the angle $$\angle AXO$$.

As Jaap and Gareth both discovered, the angle $$\angle OAX^*$$ must be $$\frac{\pi}{2}$$ which follows in this instance as $$OX^*$$ is a diameter of the small circle and $$A$$ is on the circle but they have found a quicker way to this end.