Kvant Math Problem 1026

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Solve time: 9m02s
Source on kvant.digital

Problem

Fig. 1

Five equal arcs $AB$, $BC$, $CD$, $DE$, $EA$ are arranged so that each is divided by its neighboring arcs into three equal parts (Fig. 1). Find the measure of each arc (in degrees).
The same question for a “rosette” consisting of $m$ equal arcs, each of which is divided by its neighboring arcs into three equal parts.

A. V. Shvetsov

Exploration

Let the common measure of each arc be $x$. Since the arcs are equal, every arc subtends the same central angle $x$.

The phrase that each arc is divided by its neighboring arcs into three equal parts means the two endpoints of the neighboring arcs lying on a given arc cut it into three arcs of length $x/3$.

For the rosette of five arcs, label the centers of the arcs by $O_A,O_B,O_C,O_D,O_E$ in cyclic order, where the arc centered at $O_A$ has endpoints $B,E$, the arc centered at $O_B$ has endpoints $C,A$, and so on. Since all arcs have equal radius, each center lies on the neighboring arcs. Because each neighboring arc trisects a given arc, the center of one neighboring arc lies at the point $x/3$ from one endpoint and the center of the other neighboring arc lies at the point $2x/3$ from that endpoint.

Hence, moving from one center to the next along the common circle of centers advances by an angle $x/3$. This suggests that the centers themselves form equally spaced points on a circle. For five arcs, after five such steps one returns to the starting center, so

$$5\cdot \frac{x}{3}=360^\circ.$$

This gives $x=216^\circ$.

Before accepting this, check consistency. If $x=216^\circ$, then $x/3=72^\circ$. Five steps of $72^\circ$ indeed complete a full turn. Each center is separated from the next by $72^\circ$, and on the arc of measure $216^\circ$ the two neighboring centers occur at distances $72^\circ$ and $144^\circ$ from one endpoint, exactly trisecting it.

The same reasoning for $m$ arcs yields

$$m\cdot \frac{x}{3}=360^\circ,$$

hence

$$x=\frac{1080^\circ}{m}.$$

The step most likely to conceal an error is the identification of the angular distance between consecutive centers with $x/3$. That must be justified carefully from the trisection condition.

Problem Understanding

We are given a rosette made from equal circular arcs. Each arc is cut by its two neighboring arcs into three equal subarcs. For five arcs, we must determine the measure of each arc. Then we must solve the same problem for a rosette consisting of $m$ equal arcs.

This is a Type C problem, since a specific numerical value must be determined.

The central difficulty is translating the geometric condition about trisection of arcs into a relation among the centers of the circles carrying those arcs.

The expected answer is

$$x=\frac{1080^\circ}{m},$$

and for $m=5$,

$$x=216^\circ.$$

The reason this should be correct is that consecutive centers turn out to be equally spaced by one third of the arc measure.

Proof Architecture

Let $x$ denote the common measure of each arc.

Lemma 1. The center of every arc lies on each neighboring arc. This follows because neighboring arcs have the same radius and pass through the endpoints of the given arc.

Lemma 2. On any arc, the centers of its two neighboring arcs are exactly the trisection points of that arc. This is a direct restatement of the hypothesis together with Lemma 1.

Lemma 3. Consecutive centers on the circle of centers are separated by an angular distance $x/3$. This follows because one neighboring center is a trisection point of the arc centered at the first center.

Lemma 4. The $m$ centers are equally spaced on a circle, and their total angular separation is $360^\circ$. Summing the $m$ equal gaps gives

m(x/3)=360^\circ.

The hardest step is Lemma 3, because it converts the trisection condition on one arc into a statement about the arrangement of all centers.

Solution

Let the common measure of every arc be $x$. Denote the centers of the circles carrying the arcs by

$$O_1,O_2,\ldots,O_m,$$

in cyclic order.

Since all arcs are equal and belong to circles of the same radius, the arc centered at $O_i$ has endpoints at the neighboring centers $O_{i-1}$ and $O_{i+1}$, with indices taken modulo $m$. Thus the center of every arc lies on each neighboring arc.

Consider the arc centered at $O_i$. By hypothesis, this arc is divided by its two neighboring arcs into three equal parts. The neighboring arcs have centers $O_{i-1}$ and $O_{i+1}$, and these centers lie on the arc centered at $O_i$. Hence the points $O_{i-1}$ and $O_{i+1}$ are precisely the two trisection points of the arc centered at $O_i$.

The arc centered at $O_i$ has measure $x$. Therefore each of its three equal parts has measure

$$\frac{x}{3}.$$

Starting from the endpoint $O_{i-1}$ of this arc and moving along it, the first trisection point is $O_{i+1}$. Consequently the central angle of the circle centered at $O_i$ subtending the arc from $O_{i-1}$ to $O_{i+1}$ equals

$$\frac{x}{3}.$$

Since $O_{i-1}$ and $O_{i+1}$ lie on the circle centered at $O_i$, this means

$$\angle O_{i-1}O_iO_{i+1}=\frac{x}{3}.$$

The same argument applies to every index $i$. Thus all consecutive centers determine equal angles $x/3$ around the polygon

$$O_1O_2\cdots O_m.$$

Traversing once around the cycle of centers, the total turn is $360^\circ$. Therefore

$$m\cdot\frac{x}{3}=360^\circ.$$

Solving,

$$x=\frac{1080^\circ}{m}.$$

For $m=5$,

$$x=\frac{1080^\circ}{5}=216^\circ.$$

Hence the measure of each arc in the five-petal rosette is

$$\boxed{216^\circ},$$

and in general the measure of each arc is

$$\boxed{\frac{1080^\circ}{m}}.$$

Verification of Key Steps

The first delicate point is proving that neighboring centers are trisection points of a given arc. The arc centered at $O_i$ has endpoints $O_{i-1}$ and $O_{i+1}$. The neighboring arcs are centered at those same points. Since the neighboring arcs divide the given arc into three equal parts and their centers lie on the given arc, the points $O_{i-1}$ and $O_{i+1}$ must be exactly the two trisection points.

The second delicate point is the identity

$$\angle O_{i-1}O_iO_{i+1}=\frac{x}{3}.$$

A careless argument might confuse arc length with angular measure. Here $x$ is already the angular measure of the entire arc. Since the neighboring centers are trisection points, the subarc from $O_{i-1}$ to $O_{i+1}$ on the circle centered at $O_i$ has measure $x/3$, and the corresponding central angle equals the same quantity.

The final step is summing the equal turns. There are exactly $m$ angles of size $x/3$, one at each center. Their total must be a full revolution, namely $360^\circ$, yielding the equation

$$m\frac{x}{3}=360^\circ.$$

No additional configurations are possible because the centers occur cyclically around the entire rosette.

Alternative Approaches

One may regard the centers $O_1,\ldots,O_m$ as vertices of a regular $m$-gon from the outset. The trisection condition implies that, on the circle centered at $O_i$, the points $O_{i-1}$ and $O_{i+1}$ are separated by one third of the full arc of measure $x$. Hence the angle at each vertex of the center polygon is $x/3$. Since the exterior angles of any cyclic equiangular polygon sum to $360^\circ$, the same equation

$$m\frac{x}{3}=360^\circ$$

follows immediately.

The main approach is preferable because it derives the equal spacing of the centers directly from the geometric meaning of the trisection condition and requires no separate discussion of polygons.