We read the following fragment on a torn-out page of a book: "If two vertices of a regular pentagon are given: $P, R$, then by drawing the following 8 circles using only a compass, we can mark the missing vertices of the pentagon. The center of $k_{1}$ is $P$, its radius is $P R$; we will abbreviate this as: $k_{1}=P(P R)$, and we will use similar abbreviations. $k_{2}=R(P R)$, this intersects $k_{1}$ at points $X$ and $Y$. $k_{3}=X(X Y)$ intersects $k_{1}$ at $Y$ and $Z$, $k_{4}=P(X Y)$ intersects $k_{3}$ at $V$ and $W$. $k_{5}=V(R Z)$ and $k_{6}=W(R Z)$ intersect at points $U$ and $T$." - What could be the center and radius of the remaining two circles?
In a regular pentagon, any two vertices are connected either by a side or a diagonal, and all five diagonals are equal; thus, the segment $P R$ is either a side or a diagonal of the pentagon. In the first case, $P$ and $R$ are adjacent vertices, the other vertex adjacent to $P$ will be on $k_{1}$, from which we need to intersect with a suitable circle - because we can only use a compass -, and the other vertex adjacent to $R$ will be on $k_{2}$. In the second case, $P$ and $R$ are opposite vertices, so the other vertex opposite to $P$ will be on $k_{1}$, and the other vertex opposite to $R$ will be on $k_{2}$.

In the first case, it is natural to think that the intersection of the vertex $A$ adjacent to $P$ from $k_{1}$ is completed with another circle around $R$. Then the radius of $k_{7}$ is the diagonal of a regular pentagon with side $P R$. A circle of the same radius around $P$ can intersect $k_{2}$ at the vertex $C$ adjacent to $R$, and the intersection of these two circles can give the vertex $B$ of the pentagon opposite to the side $P R$, thus completing the construction in the remaining two steps. Thus - if our assumption is correct - the further two steps only serve to use the diagonal, so the length of the diagonal must already be on the diagram, namely as the distance between a pair of the named points, - since the first six circles provide numerous additional intersections, and the book certainly only labeled the points to be used further. Continuing, the pair of points giving the length of the diagonal is one of the points obtained by $k_{6}$, $U$ or $T$, otherwise their construction and labeling would be unnecessary, and the construction could be completed with fewer circles.
![](https://cdn.mathpix.com/cropped/2024_05_02_fbf8ed20f90a931863c2g-1.jpg?height=1292&width=832&top_left_y=786&top_left_x=641)

The ratio of the diagonal to the side of a regular pentagon, cut by a diagonal in an isosceles triangle, is $2 \cos 36^{\circ} = (\sqrt{5}+1) / 2$. We will show that, taking the distance $P R$ as the unit, the measure of the distance $P T$ is exactly this ratio.

$P R X$ and $P R Y$ are equilateral triangles (Fig. 1, for better overview, we have omitted the circles and drawn the relevant distances as straight lines), so $X Y = \sqrt{3}$, and $\angle X P Y = 120^{\circ}$. The quadrilateral $X V P W$ is a rhombus, so $V W$ perpendicularly bisects $P X$, and $V W^2 = 4 X Y^2 - P X^2 = 11$, since in any parallelogram the sum of the squares of the diagonals is equal to the sum of the squares of the sides. The triangles $X P Y$ and $X P Z$ are congruent, as their sides are pairwise equal, so $\angle X P Z = 120^{\circ}$, hence $Z$ lies on the line determined by the points $P$ and $R$, thus $R Z = 2$. Similarly, the quadrilateral $U V T W$ is a rhombus, and $U T$ perpendicularly bisects $V W$, so $U$ and $T$ lie on the line determined by $P$ and $X$, and $U T^2 = 4 R Z^2 - V W^2 = 5$. Now, denoting the common midpoint of $P X$, $V W$, and $U T$ by $F$,

$$
P T = P F + F T = \frac{P X + U T}{2} = \frac{1 + \sqrt{5}}{2},
$$
this is what we wanted to prove, and based on this - as we have seen - the circles $k_{7} = P(P T)$ and $k_{8} = R(P T)$ intersect at the missing vertices $A, B, C$, and $A, B, C, R, P$ are the vertices of a regular pentagon.

We can similarly reach the goal if we consider the given $P$ and $R$ as two non-adjacent vertices of the sought regular pentagon, and look for the length of the side of the pentagon in the phase after the drawing of $k_{6}$. The side is $\frac{1}{2 \cos 36^{\circ}} = \frac{2}{\sqrt{5}+1} = \frac{\sqrt{5}-1}{2}$ of the diagonal, which, according to the above, is given by the distance $P U$, so the circles $k_{7}' = P(P U)$, $k_{8}' = R(P U)$ intersect $k_{2}$, $k_{1}$, and each other at the missing vertices of the pentagon.

In both cases, we get two pentagons, which are mirror images of each other with respect to the line through $P$ and $R$.

![](https://cdn.mathpix.com/cropped/2024_05_02_fbf8ed20f90a931863c2g-2.jpg?height=282&width=637&top_left_y=545&top_left_x=733)

Remark. The ratio of the diagonal to the side of a regular pentagon can also be determined without using trigonometric functions. Let $A B C$ be a triangle with $A B = B C = 1$, $A C = d$, and $\angle A B C = 108^{\circ}$. Measure $A B$ on $A C$ from $A$, and let the endpoint be $D$. Then $\angle C A B = 36^{\circ}$; $\angle A B D = \angle A D B = 72^{\circ}$; $\angle D B C = 36^{\circ} = \angle D C B$; so $\triangle C D B \sim \triangle A B C$; $C D : C B = A B : A C$, that is, $(d-1) : 1 = 1 : d$; $d^2 - d - 1 = 0$, and the positive root of this is $d = \frac{\sqrt{5}+1}{2}$.


[^0]:    ${ }^{1}$ We use the value of $\cos 36^{\circ}$ cited in problem 1213, see K. M. L. 27 (1963/9. 19.).