# Win the relay competition by selecting the longer path for competitors [closed]

Two teams relay race: members of a team of three take turns running from the point P to a points on the circle; A for the first, B for the second, and C for the third, starting and returning at point P, transferring the baton at point P.

The points A, B, and C are set by the organizers of the competition.

Assuming that all runners have the same rate, each team selects the point P for the other team. Your task to maximize the length of the run for the competitors, by selecting the best point P to improve your probability to beat the competition.

The basic mathematical problem can be described as follows:

Given three points $A$, $B$, and $C$ and the circle that goes through them, find the point $P$ on the circle that maximizes $PA + PB + PC$.

• I assume we can assume that C and A are diametrically opposite, as appears to be indicated? – Patrick N Aug 15 '15 at 3:25
• If you want to, but actually the A, B, C, points are randomly chosen on the circle. No need for this assumption to solve the puzzle. – Moti Aug 15 '15 at 5:03
• @Moti I tried the classic trick of using the physical interpretation that the sum of the unit forces pointing at P from A, B, and C must point radially, but this too gave an algebraic equation that wasn't too nice, as did the calculus. Is there something nicer that that? – xnor Aug 15 '15 at 7:20
• Also, you can draw a circle that goes through three points in GeoGebra. No need for a center and radius. – Joe Z. Aug 15 '15 at 15:13
• Given that the accepted answer uses nothing but standard algebra and calculus methods, with nothing really nice or clever, I think this is a math problem rather than a puzzle and should be closed as such. – xnor Aug 16 '15 at 4:40

Suppose that AC is larger than AB or BC. If P is on the arc AB, then we can construct P' where PP' is parallel to BC. Then P'B=PC, P'C=PB, and P'A>PA:

Similarly, if P is on arc BC, we can also find a better point. So to maximize PA+PB+PC, P must be on arc AC.

Let the diameter of the circle be 1. PA=$\sin(θ)$, PB=$\sin(θ+γ)$, and PC=$\sin(β-θ)$. We need to find the maximum of PA+PB+PC, which is either when the derivative is zero, or at the boundaries θ=0 or θ=β. The derivative at θ=0 is $1+\cos(γ)-\cos(β)$, which is positive. The derivative at θ=β is $\cos(β)-\cos(α)-1$, which is negative. Neither of these can be the maximum, so we set the derivative equal to zero:
$\cos(θ)+\cos(θ+γ)-\cos(β-θ)=0$

Using the sum and difference formulas,
$\cos(θ)+\cos(θ)\cos(γ)-\sin(θ)\sin(γ)-\cos(θ)\cos(β)-\sin(θ)\sin(β)=0$

We can rearrange and factor to get
$\cos(θ)(1+\cos(γ)-\cos(β))=\sin(θ)(\sin(γ)+\sin(β))$

or equivalently:
$\tan(θ)=\frac{1+\cos(γ)-\cos(β)}{\sin(γ)+\sin(β)}$

Therefore P is maximized when $θ=\tan^{-1}(\frac{1+\cos(γ)-\cos(β)}{\sin(γ)+\sin(β)})$. I couldn't simplify this any further.

• I wonder if differentiation will provide a solvable system for $\gamma$ and $\beta$. – Moti Aug 15 '15 at 18:20
• @Moti I'm not sure what you mean. γ and β are fixed because they only depend on the locations of A, B, and C. – f'' Aug 15 '15 at 22:36
• My error. Will get back to you after a careful review. Aren't you looking at arcs rather than straight lines? – Moti Aug 16 '15 at 1:02
• @Moti Definitely straight lines. If I was using arcs then I would have PA=θ instead of sin(θ). – f'' Aug 16 '15 at 1:08
• Should not $PA=2\sin\theta$? – Moti Aug 16 '15 at 1:30

This answer has the assumption that runners have to move along the circle. Although not clearly stated in the puzzle I think, although the image in the puzzle has straight lines

I must say that this answer is mostly intuition so I am not sure it is proof enough and I even don't know if it is the correct answer but I have a good feeling it is.

WLOG assume A and B are the points that are closest together among the points ABC. Now take point A* and B* that are directly opposite of A and B on the circle. What you now notice is that if you pick P anywhere between A* and B* the walk from P to A plus the walk from P to B will be the same no matter where you pick it between these points, and notice that if you pick P outside these points the walk will become smaller. So you'll want to pick it between A* and B*. Now the third point C is left. Now we want the point that is farthest from C that is still between A* and B*. Logically this means that this point is either exactly on A* or on B*. So that is this is my answer:

From the two point that are closest together pick the opposite point on the circle that is farthest from the third point.

• I don't think your statement that "if you pick P anywhere between A* and B* the walk from P to A plus the walk from P to B will be the same" is correct. – f'' Aug 15 '15 at 12:14
• If A and B are 120° apart in a circle of radius 1, the distance from A* or B* is 3, but the distance from the point halfway between them is 2sqrt3. – f'' Aug 15 '15 at 12:16
• I don't understand what you're saying. But let me explain it this way. If you are between A* and B*, then the shortest path to A is clockwise and to B counterclockwise or vice versa so the total distance is the greater arc between A and B always. If you're outside this range the shortest path will be both clockwise or both counterclockwise. Maybe you are thinking of straight lines? – Ivo Beckers Aug 15 '15 at 12:53
• Or maybe I guess I misunderstood. It doesn't really indicate they should follow the circle. My answer is only correct if they move along the circle – Ivo Beckers Aug 15 '15 at 12:54
• I assumed they move in straight lines. – f'' Aug 15 '15 at 13:01

I have an okay answer, but I'm struggling with the proof, so I'll post it anyways to see if anyone has thoughts.

The ideal location for P to maximize distance is

Diametrically opposite B

My thought process:

We are given no hard information about where B is, but let's just say it's between A and C on the lower half of the circle. (This is with my assumption that A and C are opposite one another). I checked the simple cases first- if B is at the very bottom, the ideal P is at the top, and if B is at A or C, then the ideal P is at the other point. So it looks as if the best solution for any B is for P to be directly opposite it.

-

Now, I was thinking about how to prove this geometrically, and I tried doing some stuff with Ptolemy's theorem or the fact that if P and B are opposite, then PCBA forms a rectangle, but I didn't really get anywhere. And then of course I'd want to generalize for any A,B,C... But calculus is so much less fun...

• I tried this with GeoGebra and it wasn't the best point for P (at least in my testing). I think it would work in some situations, but I also think he wants a general rule or formula for the puzzle, and not what would be best in this case. – user14478 Aug 15 '15 at 15:05
• @Luxx, you are right, no interest in specific cases – Moti Aug 15 '15 at 18:17