Applying the theorem: the lotus-in-the-lake problem from the Līlāvatī

0:00A lotus grows up out of a lake and stands one unit clear of the water. A breeze comes across and tips it over, until the flower is lying on the surface. It touches the water three units from where it had been standing. The question is: how deep is the lake? One and three. Those two numbers are everything you are given. Look at what has actually been measured, and it seems hopeless.

0:29Nobody measured the depth. That is the thing being asked for. Nobody measured the stem either. And there was no second reading taken. Two numbers, and one of them is about the air above the water rather than the water at all. So the first honest reaction is that there is not enough information here. Let us find out whether that is true. Try the obvious thing. Call the depth x, and call the whole stem y.

1:00The stem, the water surface, and the line down to the lake bed make a right triangle, so three squared plus x squared is y squared. One equation. Two unknowns. Watch what that costs you. Suppose the lake is one deep. Nine plus one is ten, so the stem would have to be whatever has square ten. That is allowed. Suppose it is two deep. Nine plus four is thirteen. Also allowed. Three deep, four deep, five deep, all the way up.

1:34Every single depth you try is permitted, because y is free to be whatever it needs to be. The equation rules nothing out. That is not a hard problem. That is no problem at all. Here is the move, and it is the whole video. The swayed stem is the same stem. Nobody cut it. It did not grow. So the stem is not a second unknown. The stem is the depth plus the one unit that was standing clear.

2:06The whole stem is x plus one. One name, doing two jobs: x is the depth, and x plus one is the stem. Now try those eight depths again. Nine plus one is ten, but a stem of two has square four, and ten is not four. Out. Two deep: nine plus four is thirteen, and a stem of three has square nine. Out. Eight candidates, and the relation on its own let every one of them through. Add that one fact and exactly one survives.

2:41Before going further, say out loud what this rests on, because two things are being assumed. The first is that the lotus stood straight up at rest, at right angles to the water. Without that there is no right triangle and the theorem has nothing to grip. The second one is quieter, and it is the one people forget. The stem is the same length bent as it was upright. It does not stretch.

3:10That is the assumption doing the real work here, because it is what lets one unknown cover two lengths. If the stem stretched, the whole method collapses. Now find the right angle, and be careful, because getting this backwards is the commonest way to lose the problem. The right angle is where the upright stem met the water surface. So the two short sides are the depth and the three across, and the side facing the right angle is the swayed stem.

3:44It is the stem that faces it, not the three. The three is horizontal, and horizontal has nothing to do with it. Put the three facing the right angle instead and you get a depth of minus five. That is worth keeping. A negative depth is not a near miss. It is the equation telling you the picture was wrong. Write it down: three squared plus x squared equals x plus one, all squared.

4:15Now expand the right side, slowly, because this is where the other wreck happens. x plus one, squared, is not x squared plus one. It is x squared, plus two x, plus one. That two x in the middle is not decoration. At a depth of four, the whole stem squared is twenty-five, and x squared plus one is only seventeen. The difference is eight, and it never goes away at any depth at all.

4:45Hold on to that middle term. It is the only place the unknown is going to survive what happens next. So the equation reads: nine plus x squared, equals x squared plus two x plus one. There is an x squared in it, which usually means methods nobody has yet. But look where they are. There is one on the left and one on the right, and they are identical.

5:14Take x squared off both sides and both of them are gone. What is left is nine equals two x plus one. That is linear. The hard part vanished. Take the one off both sides: eight equals two x. So x is four. The lake is four units deep. Put that back into the picture and look at what the triangle turned out to be. The depth is four. The three across is three. The whole stem is four plus one, which is five.

5:47Three, four, five. That is the first whole-number triple anyone ever wrote down, arriving at the end of a puzzle about a flower in a lake. Nobody planned that. It fell out of the arithmetic. And notice how little the answer cost once the naming was right. Two lines, after the naming. All the difficulty was in the naming. The same tool opens two more things immediately. A square of side five: how long is the corner-to-corner line?

6:21It cuts the square into two right triangles with short sides five and five, so the diagonal squared is fifty. Fifty is not a whole number squared, so do not pretend it is one. Trap it instead. Seven squared is forty-nine and eight squared is sixty-four, so it is between seven and eight. Tighter: seven point one squared is fifty point four one, which is over. So it is between seven point zero and seven point one, and both ends are exact.

6:57Now a rhombus, with its two crossing lines measuring twenty-four and seventy. Those two cut each other exactly in half, and they cross at right angles. So each side of the rhombus faces a right angle with short sides twelve and thirty-five. Twelve squared plus thirty-five squared is thirteen hundred and sixty-nine, and that is thirty-seven squared exactly. Twelve, thirty-five, thirty-seven. Another triple, arriving uninvited. Here is the habit worth building. Find the right angle before you write anything.

7:38Take six right triangles, drawn at six different angles on the page, each with two sides labelled. In three of the six, the labelled side is the one facing the right angle, so you subtract. In the other three it is a short side, so you add. Adding when you should subtract is not a small slip. It is a different triangle. Two short sides of seven and nine give a facing side whose square is a hundred and thirty. But a short side of forty with a facing side of forty-one gives nine, and nine, forty, forty-one is a triple.

8:18And if there is no right angle in the figure at all, make one. An equilateral triangle of side six has no right angle anywhere. Drop a line from the top corner straight down to the base and there are two now. That line cuts the base exactly in half, so it makes a triangle with a facing side of six and a short side of three. Its height squared is thirty-six take away nine, which is twenty-seven.

8:50One last thing, on a grid of dots. Draw a square with its corners on dots. If you go one across and one up to get from corner to corner, the area comes out two. Two across and none up gives four. Two across and one up gives five. So two, four and five are all drawable. Now try to draw one of area three. You cannot. Go across and up by whole numbers and the area is always one square plus another square, and three is not two squares added.

9:25Search every area up to sixteen and nine of them can be drawn and seven cannot. Which brings it back round. The answer was never the point. The naming was. One unknown covering two lengths turned a problem with no way in into two lines of arithmetic. Look for the quantity that is already something you have.