# Two parts of five hundred **Teaches:** `sum-fixed-product-varies`, `half-sum-and-deviation`, `square-is-never-negative`, `impossible-deviation` · **Assumes:** `species-of-the-unknown` · **~35 min** --- ## One question, asked for four thousand years > Five hundred has been cut into two parts. What is the product of the parts? The question has no answer, and finding out *why* it has no answer is the first real thing that happens in this library. Fill this in. No algebra yet — just multiply. | first part | second part | product | |---:|---:|---:| | 100 | 400 | | | 200 | 300 | | | 250 | 250 | | | 300 | 200 | | | 400 | 100 | | | 50 | 450 | | | 490 | 10 | | Two things should bother you. **The sum was fixed and the product was not.** Everyone knows this once they have seen it and almost nobody believes it in advance. Ask a class whether two fields with the same distance around them hold the same amount of land and a large fraction will say yes. They are in old company. Polybius, writing history in the second century BC, breaks off to complain about it (*Histories* IX.21). Megalopolis, he says, has a circuit of fifty stades and Sparta only forty-eight — and yet Sparta is **twice the size**. People find this incredible. Tell them a city with a circuit of forty stades can be double one with a circuit of a hundred and they are, he says, utterly puzzled; and it is not only ordinary people but statesmen and generals, who have not remembered their geometry. That is a professional historian, complaining that the men running armies could not do your table. It is worth letting the table be a refutation before you explain it. **The table reads the same forwards and backwards.** $100 \times 400$ and $400 \times 100$ are the same product, which is only multiplication being multiplication. But look at the *cut* rather than the parts: cutting at 100 and cutting at 400 are different cuts of the same rope, and they give the same product. The table has a mirror in it, and the mirror is at 250. Nobody told you to look at 250. The arithmetic did. ## Stop naming the parts and name the deviation Here is the move. It is the whole lesson, and it is about four thousand years old. You have been naming the parts $x$ and $500-x$, which treats the two parts as unrelated strangers who happen to sum to 500. Name them instead by **how far the cut sits from the middle**: $$\text{first part} = 250 + d, \qquad \text{second part} = 250 - d$$ Their sum is $500$ no matter what $d$ is — the $d$'s cancel, which is the point: the constraint is now built into the names, and $d$ is free. Now multiply: $$(250+d)(250-d) = 62500 - d^2$$ Check it against your table. $d = 150$ gives parts 400 and 100 and product $62500 - 22500 = 40000$. It does. Look at what that expression is. It is a fixed number, minus a square. Read it in species, as in `quad-01-species`: *sixty-two thousand five hundred numbers, less a square.* One species is being taken away from another and nothing is being added across. ## Three facts fall out, and none of them needs a graph ### The product can never exceed 62500. $62500 - d^2$ is $62500$ with something removed. The something removed is $d^2$, and a square is never negative. So the product is at most $62500$, always. That is a complete proof of a maximum. Read it again if it went by too fast, because a great deal of school mathematics is arranged to make you believe that finding a maximum requires either a picture or calculus, and here is one that required neither. There was no searching. We did not try values and take the biggest. We showed that **62500 is a ceiling**, by exhibiting the only thing that separates any product from it and observing that that thing can never be an asset. ### The ceiling is reached exactly once. $d^2 = 0$ only when $d = 0$, and $d = 0$ means both parts are 250. So the largest product is $250 \times 250 = 62500$, at the even cut, and every other cut is strictly worse. Not "worse or equal" — strictly worse, because a non-zero square is strictly positive. ### Two different cuts give the same product, and you can say which two. The expression $62500 - d^2$ does not distinguish $d$ from $-d$. So the cut at $250+d$ and the cut at $250-d$ always tie. That is the mirror you spotted in the table, now proved rather than noticed, and it is the reason quadratic equations have two solutions that straddle a middle. ## Now run it backwards: this is the quadratic formula > Cut 500 into two parts whose product is 40000. Do not set up $x(500-x) = 40000$ and expand. Use the names you just built: $$62500 - d^2 = 40000 \;\Longrightarrow\; d^2 = 22500 \;\Longrightarrow\; d = 150$$ and the parts are $250 \pm 150$, that is, **400 and 100**. Check: $400 \times 100 = 40000$. Done, in one line, with one square root and no formula. This is not a trick and it is not a shortcut for special numbers. It is the Old Babylonian method, and scribes were running it on clay in the early second millennium BC. Given a sum $S$ and a product $P$, take half the sum, square it, subtract the product, take the root — that is your deviation — then add and subtract it from the half-sum. Written out for a general sum and product it says $$\text{parts} \;=\; \frac{S}{2} \;\pm\; \sqrt{\left(\frac{S}{2}\right)^{2} - P}$$ which, if you have met the quadratic formula, you should now stare at until you see that it *is* the quadratic formula, wearing different clothes. The $b/2a$ is the half-sum. The thing under the root is the ceiling minus what you asked for. The $\pm$ is the mirror. > **Do these.** Same sum of 500 every time. Find both parts. > > | product | | product | | product | > |---:|---|---:|---|---:| > | 60000 | | 46875 | | 22500 | > | 61600 | | 52500 | | 62500 | ## The wall Now the question that makes this lesson worth teaching before any graph exists. > Cut 500 into two parts whose product is 70000. Run the method. $62500 - d^2 = 70000$, so $d^2 = -7500$. **Stop.** Do not reach for a root. Look at what the method just told you. It did not fail. It did not run out of technique. It answered the question, and the answer is *there is no such cut*. A square is never negative, so no number $d$ does this, so no pair of parts does this. You asked for a product above the ceiling and the arithmetic reported back that the ceiling is real. That negative number is the discriminant, met four hundred years before anyone drew a parabola, and met as a fact about rope rather than as a sign test to memorize. When a student later learns that $b^2 - 4ac < 0$ means "no real solutions," the sentence should land as something they already know: you asked the impossible and the square told you so. > **Do these, and say for each whether the cut exists. Sum is 500.** > Products: 62499 · 62501 · 40000 · 62500 · 0 · 100000 · 62500.25 Two of those are worth arguing about in a group. ## What has and has not happened You can now find the largest product of two parts of a fixed total, prove that it is the largest, find both cuts giving any achievable product, and detect when a demanded product is impossible. That is the substantive content of a first course in quadratics. You have not drawn anything, you have not been given a formula to memorize, and you have not used the word "parabola." The only tools were: name the deviation instead of the parts, and remember that a square is never negative. --- ## Exit 1. The parts sum to 500 and the product is 52500. Find them without writing $x$. 2. Someone claims two parts of 500 with product 63000. Refute them in one sentence. 3. A different total: two parts of 90 with product 1800. Find them. What is the ceiling for a total of 90, and how did you get it without a table? 4. Prove, in general, that of all the ways to split a total $S$ into two parts, the even split gives the largest product. Your proof should be about three lines and should not mention graphs, derivatives, or inequalities you have not proved. 5. Why does naming the parts $250 \pm d$ work better than naming them $x$ and $500-x$, when the two are obviously the same thing? Answers: [`ANSWERS.md`](ANSWERS.md). ## Related - An alternate explanation of the same two core concepts, numeric-first and interactive: `quad-02b-the-table`. - Where this gets a stake attached to it: `quad-03-assessors-rule`. - The same identity, met as a physical dissection instead of an algebraic re-parameterization: `quad-05-completing-the-square`.