Given two points P, Q on the same side of a line *l*, find the point X which minimises the sum of the distances from X to P, Q and *l*.

**Solution**

wlog take P to the right of Q and Q to be closer to the line than P. Take A on the line so that PA makes a 30^{o} angle with the line and A is to the left of P. Similarly, take B on the line so that QB makes a 30^{o} angle with the line and B is to the right of Q. There are three cases to consider.

(1) If the two segments PA and QB intersect, then their point of intersection X is the required point.

(2) If A is to the right of B, then join P to Q' the reflection of Q in the line. X is the point at which PQ' meets the line.

(3) If PA and QB do not meet, but A is to the left of B, then take X = Q.

We must now prove these statements. It is convenient to take the line *l* to be the x-axis. Take Q to be (a, b) and P to be (c, d). We assume that d > b > 0. Notice first that the sum of the three distances from the point X with coordinates (x, y) , which we will write as f(x, y), is a continuous and differentiable function of x and y. Also it tends to infinity as x or y tends to infinity. Hence the minimum must occur either (A) at a point where grad f = 0, or (B) on the boundary of the allowed region, in other words on y = 0, or (C) at a point where grad f does not exist.

We have f(x, y) = y + √( (x - a)^{2} + (y - b)^{2}) + √( (x - c)^{2} + (y - d)^{2}). Hence grad f = ( (x - a)/QX + (x - c)/PX, 1 + (y - b)/QX + (y - d)/PX) = **r** + **q** + **p**, where **r** is the unit vector along the perpendicular line from the line *l* to X, **q** is the unit vector from Q to X, and **p** is the unit vector from P to X. By resolving along the line perpendicular to **r** we find that the angles between **q** and **r** and between **p** and **r** must be equal. Similarly, the other pairs of angles. Hence the angle between each pair of **p**, **q**, **r** is 120^{o}. But it is clear that the only possible such point is the intersection H of the segments PA and QB (if it exists). H is thus the only candidate of type (A).

It is obvious that for points X on the line *l*, f(x, y) is minimised at K, the intersection of the line with PQ' (joining P to the reflection of Q), so that is the only candidate of type (B). Finally, the only points were the existence of grad f is doubtful are P and Q and indeed it is easy to see that grad f is not well defined there (for example, the limits as we approach parallel to the two axes are different). But clearly the distance sum is lower at Q than at P, so Q is the only candidate of type (C).

We now look separately at cases (1), (2) and (3). In case (1) all three candidates H, K and Q are available. We show first that H is better than Q. Let the perpendicular from Q to *l* meet *l* at C and the perpendicular from H to *l* meet *l* at D. Then angle HQC is 60^{o}, so QC = HD + QH/2. By the cosine rule we have PQ^{2} = PH^{2} + QH^{2} + PH·QH > (PH + QH/2)^{2}. Hence PQ + QH/2 > PH + QH and so PQ + QC = PQ + QH/2 + HD > PH + QH + HD. Next we show that H is better than K. As before let HD be the perpendicular to *l*. It is sufficient to show that PH + QH + HD < PD + QD, since we know that PD + QD <= PK + QK. Now PD^{2} = PH^{2} + HD^{2} + PH·HD > (PH + HD/2)^{2}, so PD > PH + HD/2. Similarly, QD > QH + HD/2. Adding gives the required result. That shows that H minimises the distance sum for case (1).

In case (2) only candidates K and Q are available. Take QC and PE as the perpendiculars to *l*. Then PQ^{2} = CE^{2} + (PE - QC)^{2}, PQ'^{2} = CE^{2} + (PE + QC)^{2}. We wish to show that PQ' < PQ + QC, or CE^{2} + (PE + QC)^{2} < CE^{2} + (PE - QC)^{2} + QC^{2} + 2 QC √(CE^{2} + (PE - QC)^{2}), or (4 PE - QC)^{2} < 4(CE^{2} + PE^{2} - 2 PE QC + QC^{2}), or 4 CE^{2} > 12 PE^{2} - 3 QC^{2}. That is certainly true since we have CE > (PE + QC) √3.

Finally, in case (3) the only candidates are K and Q. So as in the previous paragraph we wish to show that 4 CE^{2} < 12 PE^{2} - 3 QC^{2}. Now we have CE < √3 (PE - QC) from which the inequality follows.

*This is a messy problem. It is fairly well known that in the general case one wants 120 deg angles and that for points on a line one looks at the straight line to the reflection, so it is fairly clear what the solution is. But there are lots of messy details.*

© John Scholes

jscholes@kalva.demon.co.uk

15 Feb 2002