What Is the Shortest Way to the River and Back?

Stop by stop

Our script worked out the walk for a stop every 100 m along the bank, measuring from the point straight below home. (Each part of the walk is the long side of a right triangle, so its length comes from Pythagoras’ relation: for a stop 200 m along, home to stop is $\sqrt{200^2 + 450^2} \approx 492$ m.)

Stop, along the bank	0 m	100 m	200 m	300 m	400 m	500 m	600 m	700 m	800 m
Whole walk	1,264 m	1,177 m	1,111 m	1,063 m	1,029 m	1,008 m	1,000 m	1,012 m	1,068 m

Three things to notice:

• a.The walk gets shorter and shorter, then longer again. There is one best stop, not a range of them.
• b.Near the best stop the numbers hardly change: 500 m and 700 m are only 8 m and 12 m worse. Far from it they change fast.
• c.The best stop, 600 m, splits the 800 m into 600 and 200. That is 3 to 1, and home and the field are 450 m and 150 m from the river: 3 to 1 again.

Look at the angles

There is a second pattern, easier to see than to measure. At the best stop, look at the two straight paths where they meet the bank. Home is 450 m up and 600 m back, so the first path climbs 3 m for every 4 m along; the second climbs 150 m in 200 m, also 3 for every 4. They slope equally steeply, one on each side: the path comes down to the river at the same angle as it leaves, about 37∘° each. At any other stop the two angles are different.

Where have you seen that before? A ball bouncing off a wall. A ray of light bouncing off a mirror.

Here is the trick that solves the puzzle. Imagine the field reflected in the river, as if the water were a perfect mirror: a field F′ as far below the bank as the real field F is above it, straight below it.

Now compare two lines from any stop P on the bank: the real path to the field, PF, and the dashed line to the reflection, PF′. They are always the same length. Here is why.

In Figure 19.2, M is the point of the bank straight below the field. The triangles PMF and PMF′

• share the side PM;
• have MF = MF′, because the reflection is as far below the bank as the field is above it;
• both have a right angle at M.

Two sides and the angle between them are the same, so the triangles are congruent (the same shape and size), and PF = PF′. This is what a reflection always does: it moves every point, but it keeps every length and every angle. You met that in Grade 7 when you reflected shapes in a line; here it does real work.

So every walk H → P → F is exactly as long as the walk H → P → F′, whichever stop you choose. And H and F′ are on opposite sides of the bank. The shortest way from H to F′ is the straight line, and it crosses the bank at one point, Q. Any other stop P bends the line at P, and a bent line is longer than a straight one.

Why is a bent line longer? H, P and F′ make a triangle, and in every triangle two sides together are longer than the third: the third side goes straight between its ends, and the other two make a detour through P. (Stretch a string tight between two pins: it is shortest when it is straight.) This is called the triangle inequality. Only when P is on the straight line HF′, at Q, do the two sides add up to exactly the third.

That is the whole proof. The method, reflect one end in the line you must touch and draw a straight line, is called the reflection trick, and the puzzle is known as Heron’s problem, after Heron of Alexandria, who solved it this way nearly two thousand years ago (see the story below).

Now with numbers

The straight line from home to F′ goes 800 m along the bank and 450 + 150 = 600 m down. So the shortest walk is

$$\sqrt{800^2 + 600^2} = \sqrt{1{,}000{,}000} = 1{,}000 \text{ m},$$

exactly what the table found. And where does it cross the bank? It falls 600 m in 800 m along, steadily, and it reaches the bank after falling 450 m, which is $\tfrac{450}{600} = \tfrac{3}{4}$ of the way down, so $\tfrac{3}{4}$ of the way along: $\tfrac{3}{4} \times 800 = 600$ m. That is pattern c of the table: the stop splits the distance along the bank in the same ratio as home and the field are from the river, 450 to 150. Now it isn’t a pattern any more; it is proved.

Why the angles are equal

Look at the angles at Q in Figure 19.2. The line from H through Q to F′ is straight, so the angle it makes with the bank on the left, above, equals the angle on the right, below (they are vertically opposite angles, from Grade 8’s Lines and Angles). And QF is the reflection of QF′, so it makes the same angle with the bank, above. So the path comes down to the bank and leaves it at the same angle: the equal angles of the table are no accident. The shortest walk is exactly the one where the angle in equals the angle out.

When light from a lamp bounces off a flat mirror into your eye, it obeys the law of reflection you met in Grade 7 Science: the angle of incidence equals the angle of reflection. Those angles are measured from the normal, the line at right angles to the mirror, but if the angles with the normal are equal, so are the angles with the mirror itself. So it is the shortest walk’s rule. Light going from the lamp to the mirror and then to your eye takes the shortest path that touches the mirror, just as Sunita’s walk does.

And it explains what you see in a mirror. Your eye can’t tell that the light bent at the mirror; it sees the light arriving in a straight line, as if from F′, the reflection. So the thing appears to be behind the mirror, as far behind as it is in front: the image of a plane mirror is exactly our F′.

Read more in Grade 7 Science, Laws of reflection of light, and try it with a torch in Law of Reflection.

The same trick wins games. On a carrom board, the striker bounces off the sides almost like light off a mirror (spin and friction spoil it a little: see below). Suppose the queen sits behind another coin, so you can’t hit it straight. Where on the side should you aim, so that the striker bounces off and hits the queen?

Reflect the queen in the side, and aim straight at its image. The striker’s path to the side and on to the queen is exactly as long as the straight line to the image, and it bounces at equal angles: the mirror in the river again. One detail: the striker’s center turns 2.05 cm before the frame (half its width) when its edge touches the side, so the mirror line is that far in from the frame, not the frame itself.

Off two sides

Now the queen is near the top-left corner, behind a crowd of coins, and no shot off one side gets through. Bounce off two sides: first the left, then the top. Reflect the queen in the left side, then reflect that image in the top side, and aim straight at the image of the image.

Why does that work? After the first bounce the striker moves as if it came from the mirror world of the left side; in that world, the top side has its own mirror image, and the second bounce is a reflection there. Unfold both bounces and the whole path is one straight line to the twice-reflected queen.

The river and the road

Sunita’s cousin lives between a river and a road that meet at an angle of 60∘°. Each morning he walks from home to the river, then to the road, then back home. What is his shortest round trip?

This time there are two lines to touch, and the walk starts and ends at the same place. Reflect home in the river, to H₁, and in the road, to H₂. The first part of the trip, home to the river, is as long as H₁ to that point; the last part, the road to home, is as long as that point to H₂. So the whole trip is as long as a broken line from H₁ to the river, to the road, to H₂: shortest when it is straight.

The twist: if the river and the road meet at a right angle or wider, the straight line from H₁ to H₂ passes through the corner O or outside the angle, and the best trip goes to the corner itself, touching the river and the road at the same point. With home 1 km from a right-angled corner the shortest trip is 2 km: there and back. The trick still gives the answer; it just tells you the stops have run together.

Where to build the bridge

A different kind of shortest walk. Sunita’s school is 1,200 m along the river and on the other side: home is 300 m from the near bank, the river is 50 m wide, and the school is 200 m beyond the far bank. The village will build one footbridge, straight across the river. Where should it go, so that the walk to school is as short as possible?

Every route crosses the bridge, and every bridge is 50 m long, so the bridge part doesn’t depend on where it stands. Here the trick isn’t a reflection: slide home 50 m straight across, the width of the river, to H′. The walk from home to the bridge is then exactly as long as from H′ to the bridge’s far end (they are opposite sides of a parallelogram), so the rest of the route is as long as H′ → far end of the bridge → school: shortest when straight.

That slide is a translation, another of Grade 7’s transformations. Like a reflection, it keeps every length, and that is all the trick needs.

An ant on a box

Shortest walks on the outside of a box are found by unfolding it, which is the mirror trick again: folding a face down flat is a reflection of it in the edge it hinges on.

Back at the river. Sunita doesn’t walk at the same speed both ways. Say she walks 80 m a minute with the empty gagri, but with it full only 40 m a minute. Now the shortest walk isn’t the quickest.

Every meter with the full gagri costs twice as much time as a meter with the empty one. So the best stop moves along the bank to shorten the slow part, even though the whole walk gets a little longer (about 1,023 m instead of 1,000 m). The reflection trick finds the shortest path; when the two parts are walked at different speeds, the quickest path is a different one.

• How Many Ways Can You Prove Pythagoras’ Theorem? Every length in this topic came from Pythagoras’ relation: here is why it is true.
• What Is the Quickest Way Across the Map? Shortest routes again, but along roads with a time on each, where no ruler can help.
• Why Do Flights to America Fly Near the North Pole? (Grades 9–12) The shortest way between two places on a round Earth is not the straight line on a flat map.
• How Did Eratosthenes Measure the Earth? Straight rays of sunlight and equal angles, used to measure the whole planet.
• Can a Robot Get Out of Any Maze? (Computing, Going Further 6–8) A different kind of finding a way, when you can’t see the whole map.
• Grade 7 Science, Reflection of light, and Grade 10 Optional Mathematics, Combined transformation: two reflections as one move.