# The assessor's rule **Teaches:** `gross-vs-exact`, `deviation-square-is-the-gap`, `square-is-never-negative` · **Assumes:** `sum-fixed-product-varies`, `half-sum-and-deviation` · **~40 min** --- ## A rule that was actually used On the walls of the temple of Horus at Edfu, carved around 100 BC, there is an inventory of land belonging to the priesthood, with the area of each parcel worked out. Many of the parcels are four-sided and not rectangular. The scribes computed their areas like this: > **The assessor's rule.** Average one pair of opposite sides. Average the other pair. > Multiply the two averages. In symbols, for a field with sides $a, b, c, d$ taken in order around the boundary, $$\text{assessed area} \;=\; \frac{a+c}{2} \cdot \frac{b+d}{2}$$ It is a sensible rule. For a rectangle it is exactly right. For a field that is nearly a rectangle it is nearly right, and most cultivated fields are nearly rectangles. It is easy to carry out with a rope and no geometry. It was in use, in one place or another, for something like two thousand years. It is also wrong, and this lesson is about the fact that *wrong* is not the interesting part. The interesting part is **by how much, and in which direction.** ## The field Take a field whose boundary measures 1000 rods, laid out like this: two of its corners are square, and the four sides, going around, are $$x, \quad 500-x, \quad 500-x, \quad x$$ so that the boundary total is $1000$ for any $x$ you like. Think of it as two identical right-angled parcels sharing a long diagonal boundary, each with legs $x$ and $500-x$. As $x$ moves from $0$ to $500$, you get a whole family of fields, all with the same distance around them, ranging from a long thin splinter to a square and back again. > **Do this first, and do it by hand.** For $x = 100$: draw the two right parcels, each > with legs 100 and 400. What is the area of one? Of the field? Now do $x = 250$. What > shape is the field then? The true area is not hard, because the field is two right triangles: $$\text{true area} \;=\; 2 \cdot \tfrac{1}{2} \cdot x \cdot (500-x) \;=\; x(500-x)$$ which is the quantity you spent all of `quad-02-parts-of-500` on. It has walked back in wearing a hat. ## What the assessor charges Now apply the rule. The opposite pairs are $(x,\; 500-x)$ and $(500-x,\; x)$. So: $$\frac{x + (500-x)}{2} \cdot \frac{(500-x) + x}{2} \;=\; \frac{500}{2}\cdot\frac{500}{2} \;=\; 250 \cdot 250 \;=\; 62500$$ Read that again, because it is the whole lesson in one line. **The $x$ canceled.** The assessor charges 62500 for every field in the family. The splinter and the square, the field you can barely stand in and the field that feeds a household — same assessment, every time. The rule cannot see the shape at all. It sees only the boundary, and the boundary was the same all along. > **Do this.** Fill the ledger. The middle column is what you actually own; the right > column is what you are taxed on. > > | $x$ | true area $x(500-x)$ | assessed | overcharge | > |---:|---:|---:|---:| > | 50 | | 62500 | | > | 100 | | 62500 | | > | 150 | | 62500 | | > | 200 | | 62500 | | > | 250 | | 62500 | | > | 300 | | 62500 | | > | 400 | | 62500 | | ## The overcharge is a square Look at your last column before reading on. Those numbers are 40000, 22500, 10000, 2500, 0, 2500, 22500. If they look familiar it is because they are $200^2, 150^2, 100^2, 50^2, 0^2, 50^2, 150^2$. The overcharge is always a perfect square, and it is the square on your distance from 250: $$\underbrace{62500}_{\text{assessed}} \;-\; \underbrace{x(500-x)}_{\text{true}} \;=\; x^2 - 500x + 62500 \;=\; (x-250)^2$$ Verify the middle step by multiplying out. It is one line, and it is the single most useful line in elementary algebra: $$\boxed{\;x(500-x) \;=\; 62500 - (x-250)^2\;}$$ This is the identity you already found in `quad-02-parts-of-500` by naming the deviation $d = x - 250$. There it was a change of variables. Here it is a bill. ## Four things you now know, none of which needed a graph **1. The assessor never undercharges.** The gap is $(x-250)^2$, and a square is never negative. Whatever your field, you are paying for at least as much land as you have. Not because the scribes were corrupt — the rule simply has a one-way error, and nobody in two thousand years was in a position to notice, because noticing requires exactly the algebra you have just done. **2. The rule is exact for one field only.** The gap is zero only when $x = 250$, the square field. Every other landowner in the family is overpaying, by an amount that grows as the square of how far from square their field is. A field at $x=200$ — only a fifth off — overpays by 2500, which is 4% of its true area. A field at $x=50$ overpays by 40000, which is nearly **twice** what it actually holds. **3. The largest field in the family is the square one, and its area is 62500.** This follows from the same line with nothing added. The true area is $62500$ minus a square, so it cannot exceed $62500$, and it reaches $62500$ only at $x = 250$. Notice what just happened to the assessor's rule. It has stopped being a wrong answer and become a correct one to a different question: **the assessed value is the true area of the best field you could have had with that boundary.** The scribes were not computing your land. They were computing the most land anyone could hold for that much boundary — and charging everyone that. **4. Two fields are taxed fairly-unfairly alike.** $x$ and $500-x$ give the same overcharge, because $(x-250)^2 = (250-x)^2$. The field at 100 and the field at 400 are the same field turned on its side, which the arithmetic knew before you did. ## Where the fourth power comes from Everything above used the true area, $x(500-x)$, which we got by cutting the field into two right triangles. That was available because we insisted the corners be square. In general a four-sided field is not so obliging, and the exact-area problem for quadrilaterals is genuinely hard. In 628 CE, in the *Brāhmasphuṭasiddhānta*, Brahmagupta gave a rule for it. He gave two rules, in fact, and he named them: the **gross** area (*sthūla*) — which is the assessor's rule, the same average-of-opposite-sides recipe — and the **exact** area (*sūkṣma*): $$(\text{area})^2 \;=\; (s-a)(s-b)(s-c)(s-d), \qquad s = \frac{a+b+c+d}{2}$$ Apply it to our field. The half-perimeter is $s = 500$, and the four sides are $x$, $500-x$, $500-x$, $x$, so the four factors are $500-x$, $x$, $x$, $500-x$: $$(\text{area})^2 \;=\; x^2(500-x)^2$$ There it is. A fourth-degree quantity, handed to you in factored form, by a rule from the seventh century. Take its square root and you get $x(500-x)$, agreeing with the triangles — which is the check that the rule is doing what it claims. **Do not expand it.** That is the subject of the next lesson, and it is worth arriving there with the question already live: you are holding $x^2(500-x)^2$, you could multiply it out into $x^4 - 1000x^3 + 250000x^2$, and the question is what on earth that would buy you. ## An honest complication Brahmagupta stated the exact rule without a restriction, and it is not unrestricted. Four side lengths do not determine a quadrilateral — the figure is floppy, and you can flex it through a whole range of shapes without changing a single side. So "the area of a quadrilateral with sides $a,b,c,d$" is not a well-posed question, and no formula in $a$, $b$, $c$, $d$ alone can answer it. What Brahmagupta's rule actually gives is the area of the **cyclic** one — the flexing position in which all four corners lie on a circle — and that position happens to be the one with the *largest* area. Bhāskara II, five centuries later, attacked him for the omission in memorable language. Our field is safe, because we pinned it down: two square corners force the four vertices onto a circle (the shared diagonal is the diameter), so it is the cyclic one and the rule is exact for it. But the complication is worth an argument in class rather than a footnote: > **Discuss.** A deed records four boundary lengths and nothing else. Two people claim > different acreages and both compute correctly. Who is right? What is the smallest extra > thing the deed could record to settle it? Does the assessor's rule have this problem > too — and if it does not, is that a point in its favor? --- ## Exit 1. A field in the family has true area 46875. Find $x$. There are two answers; explain why, in terms of the field rather than the algebra. 2. By what percentage is the $x=125$ field overcharged? 3. A landowner is overcharged by exactly 3600. What is their $x$? 4. Prove that no field in the family holds more than 62500, and say precisely which step of your proof would fail if the overcharge were not a square. 5. The assessor's rule multiplies averages of *opposite* sides, so its answer depends on the order the sides are written down. Brahmagupta's does not — it is symmetric in $a,b,c,d$. Take sides 10, 20, 30, 40 and compute the assessor's rule for two different orderings. What does this tell you about the rule? 6. *Harder.* Over all four-sided fields with boundary 1000 — not just our family — the largest is the square. Prove it from Brahmagupta's rule. (Hint: the four factors $s-a, s-b, s-c, s-d$ are non-negative and their sum is fixed. What is it?) Answers: [`ANSWERS.md`](ANSWERS.md). ## Related - The identity as a physical dissection: `quad-05-completing-the-square`. - Why you should not expand $x^2(500-x)^2$: `quad-04-do-not-expand`. - Historical sourcing, and which claims here are verified: [`../../SOURCES.md`](../../SOURCES.md).