Three-Ordinate Curves

Summary: §§374–379 of chapter-16-on-finding-curves-from-properties-of-the-ordinate. The two-ordinate analysis of §§364–373 extends directly to curves with three ordinates per abscissa, $y^3-P{y}^2+Qy-R=0$. Vieta gives sum $P$, pair-product $Q$, triple-product $R$, and Newton’s identities express any symmetric polynomial in the three ordinates as a polynomial in $P,Q,R$. Concrete applications: constant power sums (§375), constant sums of square or cube roots (§§376–377), the §378 general solution via the quadratic-in-$y$ resolvent, and the §379 constant-area-triangle problem solved by Heron’s formula.

Sources: chapter16 (§§374–379). No figure references.

Last updated: 2026-05-11.

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The setup (§374)

For a three-valued curve

$y^3-P{y}^2+Qy-R=0$

with $P,Q,R$ non-irrational in $x$, each abscissa has three ordinates $p,q,r$ (some possibly complex; we focus on the locus where all three are real). Vieta gives

$p+q+r=P,pq+pr+qr=Q,pqr=R.$

So $P=$ const means constant sum, $Q=$ const constant pair-product sum, $R=$ const constant triple-product. Demanding any one of these gives a one-parameter family with the other two coefficients free.

Power sums via Newton’s identities (§375)

For $p^n+q^n+r^n$, Newton’s identities give the explicit table:

$n$	$p^n+q^n+r^n$ in terms of $P,Q,R$
1	$P$
2	$P^2-2Q$
3	$P^3-3PQ+3R$
4	$P^4-4P^{2}Q+2Q^2+4PR$
5	$P^5-5P^{3}Q+5P{Q}^2+5P^{2}R-5QR$

For negative integer $n$, substitute $z=1/y$ in the original equation. Then $z^3-(Q/R)z^2+(P/R)z-1/R=0$ has roots $1/p,1/q,1/r$, so

$n$	$1/p^n+1/q^n+1/r^n$ in terms of $P,Q,R$
1	$Q/R$
2	$(Q^2-2PR)/R^2$
3	$(Q^3-3PQR+3R^2)/R^3$
4	$(Q^4-4P{Q}^{2}R+4Q{R}^2+2P^2{R}^2)/R^4$

Demanding any of these equal $a^n$ pins one relation, leaving two free. For example, $p^3+q^3+r^3=a^3$ requires $P^3-3PQ+3R=a^3$. Since $R=y^3-P{y}^2+Qy$ from the curve equation, substituting gives the explicit curve

$3y^3-3P{y}^2+3Qy+P^3-3PQ=a^3,$

a third-order line with two free functions $P,Q$ of $x$.

Root sums — the difficult case (§376)

When $n$ is fractional the algebra is messier. For $\sqrt{p}+\sqrt{q}+\sqrt{r}=\sqrt{a}$, square both sides to get

$P+2(\sqrt{pq}+\sqrt{pr}+\sqrt{qr})=a,$

so $\sqrt{pq}+\sqrt{pr}+\sqrt{qr}=(a-P)/2$. Square again, using $pq+pr+qr=Q$ and $\sqrt{p^{2}qr}+\sqrt{q^{2}pr}+\sqrt{r^{2}pq}=(\sqrt{p}+\sqrt{q}+\sqrt{r})\sqrt{pqr}=\sqrt{aR}$:

$\frac{(a-P{)}^2}{4}=Q+2\sqrt{aR}.$

Solve for $Q$:

$Q=\frac{(a-P{)}^2}{4}-2\sqrt{aR}.$

Substituting into the curve and clearing the remaining radical (since $R=((a^2-2aP+P^2-4Q{)}^2)/(64a)$):

$y^3-P{y}^2+Qy-\frac{(a^2-2aP+P^2-4Q{)}^2}{64a}=0.$

A complicated but explicit curve, with $P,Q$ still free functions of $x$.

Cube-root sums via auxiliary variable (§377)

For $p^{1/3}+q^{1/3}+r^{1/3}=a^{1/3}$, introduce $v=(pq{)}^{1/3}+(pr{)}^{1/3}+(qr{)}^{1/3}$. Since $(pqr{)}^{1/3}=R^{1/3}$ and Newton-style manipulation gives

$p+q+r=a-3v{a}^{1/3}+3R^{1/3}=P,pq+pr+qr=v^3-3v(aR{)}^{1/3}+3R^{2/3}=Q.$

The curve equation is then

$y^3-(a-3v{a}^{1/3}+3R^{2/3})y^2+(v^3-3v(aR{)}^{1/3}+3R^{2/3})y-R=0$

with $v$ any function of $x$. The auxiliary-variable trick is the general workaround when direct Newton-identity expansion is too irrational to handle.

The general solution via quadratic resolvent (§378)

Despite the case-by-case difficulties, a single closed-form solution covers all $n$. Take one of the three ordinates as $y$, say $p=y$; then $q+r=P-y$ and $qr=Q-y(q+r)=Q-y(P-y)=Q-Py+y^2$. So $q,r$ are roots of the quadratic

$t^2-(P-y)t+(Q-Py+y^2)=0,$

with discriminant $(P-y{)}^2-4(Q-Py+y^2)=P^2+2Py-3y^2-4Q$. Hence

$q=\frac{1}{2}(P-y)+\frac{1}{2}\sqrt{P^2+2Py-3y^2-4Q},r=\frac{1}{2}(P-y)-\frac{1}{2}\sqrt{\cdots }.$

The condition $p^n+q^n+r^n=a^n$ then becomes the single equation

$y^n+{\left[\frac{1}{2}(P-y)+\frac{1}{2}\sqrt{P^2+2Py-3y^2-4Q}\right]}^n+{\left[\frac{1}{2}(P-y)-\frac{1}{2}\sqrt{P^2+2Py-3y^2-4Q}\right]}^n=a^n.$

The sum of the two conjugate $n$-th powers is rational in the square root squared (binomial expansion drops the odd terms), giving a rational equation in $y$ once expanded. This works for any $n$ — integer or fractional.

Constant-area triangle from three ordinates (§379)

A pretty geometric application: pick three ordinates $p,q,r$ at the same abscissa and form a triangle with sides $p,q,r$. By Heron’s formula, the squared area is $(2p^2{q}^2+2p^2{r}^2+2q^2{r}^2-p^4-q^4-r^4)/16$. Setting this equal to $a^4$ and using $p^4+q^4+r^4=P^4-4P^{2}Q+2Q^2+4PR$, $p^2{q}^2+p^2{r}^2+q^2{r}^2=Q^2-2PR$:

$16a^4=4P^{2}Q-8PR-P^4,$

so

$R=\frac{PQ}{2}-\frac{P^3}{8}-\frac{2a^4}{P}.$

The curve

$y^3-P{y}^2+Qy-\frac{1}{2}PQ+\frac{1}{8}{P}^3+\frac{2a^4}{P}=0$

has the property that the three ordinates form a triangle of constant area $a^2$ at every abscissa. Taking $P=2b$ constant adds the further property that the perimeter is constant. With $P=2b$ and $Q=m{x}^2+nx+k{a}^2$, the explicit cubic is

$y^3+m{x}^{2}y-2b{y}^2+nbxy-mb{x}^2+(a^4/b-k{a}^{2}b)y-n{b}^{2}x+a^4/b-k{a}^{2}b+b^3=0,$

three ordinates at every abscissa, every triple forming an isosceles-perimeter constant-area triangle.

Pattern (§379–380 closing)

The chapter’s punchline: any condition on $n$-tuples of ordinates that is symmetric in the tuple — power sums, root sums, triangle area, more general symmetric functions — leads to a polynomial constraint on the elementary symmetric coefficients $P,Q,R,\dots$ by Newton’s identities. The constraint pins one coefficient in terms of the others; the rest remain free, generating curve families. Four-or-more ordinates follow the same template with no extra difficulty.

Related pages

• chapter-16-on-finding-curves-from-properties-of-the-ordinate — chapter summary.
• sum-of-ordinate-powers-curves — the two-ordinate analogue.
• chord-product-cubic — chapter 10 cubic chord-product theorem, the forward-direction counterpart to constant $R=pqr$ here.
• diameter-and-center-of-cubic — chapter 10 lo+mo=no diameter, the forward-direction counterpart to constant $P=p+q+r$ here.