circle-trisection-by-chords

Circle Trisection by Two Chords

Summary: Problem V of chapter 22 (§535, figure 115). From a point $A$ on the unit circle, draw two chords $AB$ and $AC$ that together cut the disk into three equal-area pieces. The defining equation $s-\sin s=120^{\circ }$ reduces via $u=s-120^{\circ }$ to $u=\sin (60^{\circ }-u)$, which false-position-method solves at $u=29^{\circ }16^{\prime }27^{\prime \prime }$.

Sources: chapter22, §535. Figure 115 (figures115-118 p. 493).

Last updated: 2026-05-12.

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Setup

Generalize the bisecting-by-a-chord trick of Problem IV (equal-area-bisection-problems). Draw two chords $AB$ and $AC$ from a common point $A$ on the circumference. The picture is symmetric: by symmetry $AB=AC$, and the chord-cut region $AEB=AFC$ (segments equal). We want each segment to be one-third of the disk.

Let radius = 1, so the disk has area $\pi$ and each segment should have area $\pi /3$. With arc $AB=AC=s$,

$$\text{segment }AEB=\frac{1}{2}s-\frac{1}{2}\sin s.$$

Setting this to $\pi /3$:

$$\boxed{s-\sin s=\frac{2\pi }{3}=120^{\circ }.}$$

Reduction

Substitute $u=s-120^{\circ }$, then $\sin s=\sin (120^{\circ }+u)=\sin (60^{\circ }-(-u))=\sin (60^{\circ }-u)\cdot ?$ — Euler writes it cleanly: $\sin s=\sin (u+120^{\circ })=\sin (60^{\circ }-u)$ (using $\sin (180^{\circ }-\theta )=\sin \theta$). The equation becomes

$$u=\sin (60^{\circ }-u).$$

This is not Problem I in disguise. The unknown sits on both sides, but inside a sine on the right, with a fixed-angle offset. So a fresh false-position computation is needed.

False position

Existence: $u$ is less than $60^{\circ }$ (otherwise $\sin (60^{\circ }-u)<0$ but $u>60$ makes LHS bigger). Try $u=20^{\circ },30^{\circ },40^{\circ }$:

$$\begin{array}{rccc}u=& 20^{\circ }& 30^{\circ }& 40^{\circ }\\ 60^{\circ }-u=& 40^{\circ }& 30^{\circ }& 20^{\circ }\\ \log u=& 1.3010300& 1.4771213& 1.6020600\\ \text{(subtract 1.7581226)}& & & \\ \log u_{\text{arc}}=& 9.5429074& 9.7189987& 9.8439374\\ \log \sin (60^{\circ }-u)=& 9.8080675& 9.6989700& 9.5340517\\ \text{error}& +0.2651601& -0.0200287& -0.3098857\end{array}$$

So $u$ is between $20^{\circ }$ and $30^{\circ }$, much closer to $30^{\circ }$. Refining at $u=29^{\circ }$ gives error $+0.0075639$; combined with the $u=30^{\circ }$ error of $-0.0200287$ the proportion $275926:75639=1^{\circ }:16^{\prime },26^{\prime \prime }$, so $u\approx 29^{\circ }16^{\prime },26^{\prime \prime }$.

Minutes-level pass at $u=29^{\circ }16^{\prime }$ and $u=29^{\circ }17^{\prime }$ refines to $u=29^{\circ }16^{\prime }27^{\prime \prime }{0}^{\prime \prime \prime }$ exactly.

Final.

$$u=29^{\circ }16^{\prime }27^{\prime \prime }{0}^{\prime \prime \prime },s=AB=149^{\circ }16^{\prime }27^{\prime \prime }{0}^{\prime \prime \prime },$$

with the remaining arc $BC=360^{\circ }-2s=61^{\circ }27^{\prime }{6}^{\prime \prime }{0}^{\prime \prime \prime }$ and chord length $AB=AC=2\sin (s/2)=1.9285340$.

Geometric reading

The total circle gets cut into three regions:

1. Segment $AEB$, area $\pi /3$, subtended arc $149^{\circ }16^{\prime }$.
2. Segment $AFC$, identical to (1) by symmetry.
3. The lens $BC$-and-two-chords in the middle, area $\pi /3$, containing the arc $BC$ of $61^{\circ }27^{\prime }$ and the triangle $ABC$.

Region 3 deserves a check: its area is the whole circle minus the two segments, $\pi -2(\pi /3)=\pi /3$, as required. ✓

Why no reduction to Problem I

Problems III and IV (see equal-area-bisection-problems) reduced to $u=\cos u$ because their equations had the form ”$s$ minus a sine = a multiple of $\pi /4$ or $\pi /2$,” and a $\pi /2$ shift converts $\sin s$ into $\cos u$. Problem V has the form ”$s-\sin s=2\pi /3$,” and a $2\pi /3$ shift converts $\sin s$ into $\sin (60^{\circ }-u)$ — not a cosine of $u$. So a fresh numerical solution is required.

Figures

Figures 115–118

Related pages

• chapter-22-on-the-solution-to-several-problems-pertaining-to-the-circle — context.
• false-position-method — algorithm.
• equal-area-bisection-problems — Problem IV ($s-\sin s=\pi /2$, the two-piece analogue).