Baudhāyana's construction: turning a rectangle into a square of matching area

0:00There is an old problem that sounds like a riddle and is not. Given a shape, build a square with the same area. That is what squaring a shape meant, and it has nothing to do with multiplying a number by itself. Three shapes were the ones people wanted: a rectangle, a triangle, and a circle. Two of those can be done with a straight edge and compasses. One of them cannot, ever, and we will come back to which.

0:30Today, the rectangle - by a construction written down around eight hundred years before the common era. It is one line of algebra, drawn. Start with the target, because it is easy to lose sight of. Here is a rectangle, long side a, short side b. Its area is a times b. So the square we want has area a b, and its side is the square root of a b.

0:58Produce a segment of length root a b, using nothing but a straight edge and a compass. You cannot measure it out: for most rectangles that length is irrational. It has to be constructed. First, one piece of algebra. Take the average of a and b, and square it. Take half their difference, and square that. Subtract the second from the first. The first expands to a squared plus two a b plus b squared, over four.

1:32The second is a squared minus two a b plus b squared, over four. The a squared cancels, the b squared cancels, and four a b over four is left. So a b is a difference of two squares. That is the identity, and every line of the construction is one of those two terms made visible. Now the drawing, in six moves. Label the corners A, B, C and D, with A at the bottom left.

2:03One. On the bottom edge mark E, so that A E equals the short side. Two. Take F as the midpoint of what is left over, the segment E D. Three. On A F build a square, A F G H, with H standing on the left edge extended upwards. Four. Put the compass at H, open it to A, and swing an arc. Five. That arc crosses the top edge of the rectangle; call the crossing K.

2:33Six. From K go straight up to the top of the square, meeting it at P. The square on H P is the answer. Six instructions, and not one of them explains itself. Start with A F. A E was set equal to the short side, so A E is b, and A D is a. What is left over, E D, has length a minus b, and F cuts it in half.

3:00So A F is b plus half of a minus b. Which is a plus b, over two. The average of the two sides - the first term of the identity, sitting on the board as a length. And notice what F is not. F is not the midpoint of A D. Halve the whole and you get a over two, and every step after comes out wrong while looking perfectly reasonable.

3:28H is a corner of that square, so A H is also a plus b over two, and A B is just b. So B H, the bit of the left edge sticking up above the rectangle, is a plus b over two, minus b. Which is a minus b, over two. Half the difference. Both halves of the identity are now lengths in the picture: the average is the arc's radius, half the difference the gap above the rectangle.

3:59Once you see that, the six instructions stop being arbitrary. Now the step where a careful reader gets stuck. Look at the triangle H K P. Where is the right angle? Not at K, and not at H. It is at P, because K P runs straight up and H P runs straight across. So the theorem gives H P squared equals H K squared minus K P squared. But the argument you will usually see writes H P squared equals H K squared minus B H squared.

4:36That is not an error. It is missing one sentence. B and K sit on the rectangle's top edge, H and P on the square's, and those two lines are parallel. So B H P K is a rectangle, and its two upright sides are equal. K P is B H. Now finish it. H K is a radius of that arc, and so are H A and H G.

5:05All radii of one circle are equal, so H K is a plus b over two. K P we have just shown is B H, which is a minus b over two. Substitute. H P squared is the square of the average, minus the square of half the difference. And that is the identity from three minutes ago. It is a b. So the square standing on H P has exactly the area of the rectangle we started with.

5:37Put numbers on it. Take a rectangle eight by two. A E is two, E D is six, so A F is five. The arc has radius five, and B H is five minus two, which is three. So H P squared is twenty five minus nine - sixteen - and H P is four. Four squared is sixteen, and eight times two is sixteen. Now try five by three.

6:04A F is four, B H is one, and H P squared is sixteen minus one. Fifteen. So H P is root fifteen. That is the ordinary case. Across all eight hundred and twenty rectangles with whole sides up to forty, only twenty eight give a square with a whole-number side. The other seven hundred and ninety two are surds, and the construction draws every one of them exactly. Could the drawing ever fail?

6:37Two things must go right, and each is one inequality. First, the arc has to reach the rectangle's top edge. Its radius is the average; the distance down to that edge is half the difference, which is smaller whenever b is positive. So the arc always crosses. Second, P must land on the square's top edge, not beyond it. That needs root a b to be no bigger than the average, and it never is.

7:08The gap between them is exactly the square on half the difference, so it closes only when a and b are equal - when the rectangle was already a square. It is tempting to add that the square then sits inside the original left edge. It does not always. Across those same rectangles it stays inside on six hundred and ninety two, and rises clear above on one hundred and twenty eight.

7:37The construction only ever promised the area. Here is why every one of those steps had to be argued. Suppose you misremember the first instruction, and mark E so that A E is a minus b instead of b. Everything else you do exactly as written. The drawing still works. The arc still reaches, P still lands, and out comes a square. On a rectangle eight by two it gives area twenty four, where the right answer was sixteen.

8:11But test it on eight by four and it gives thirty two, which is correct. That is not luck. Across the eight hundred and twenty rectangles, the misremembered construction gives the right area on exactly twenty, and those twenty are precisely the ones whose length is twice their breadth. A wrong construction, right on a whole family. So checking one example proves nothing. It is the only thing standing between you and a square of the wrong size.

8:45Two things to finish. A triangle first. Its area is base times height halved, which is the area of a rectangle with the same base and half the height. Build that rectangle, then square it by everything we have just done. And the circle. The circle cannot be squared, and the obvious reason is the wrong one. It is not because the answer is irrational. Seven hundred and ninety two of the squares we just built had irrational sides, and the compass drew every one without complaint.

9:22Irrational lengths are constructible all day long. The circle fails for a stronger reason. Pi satisfies no polynomial equation with whole-number coefficients at all, and a straight edge and compass can only ever reach lengths that do. What the drawing shows is smaller, and more useful. An identity you can prove in one line, standing up in ink, eight hundred years before the common era.