Given a circle and a point $P$ outside the circle. Construct a secant through $P$ such that the chord $A B$ is equal to $B P$.
I. Solution: Let $P B=x$, then $P A=2 x$. The geometric mean of the segments of the secant from $P$ is the length of the tangent segment $P E=t$ (Figure 1).

Thus, $x \cdot 2 x=t^{2}$, from which $x=\frac{t}{\sqrt{2}}=\frac{t}{2} \sqrt{2}$. Therefore, $x$ is the hypotenuse of an isosceles right triangle whose legs are $\frac{t}{2}$ in length. The circle drawn with radius $x$ around $P$ intersects the given circle at points $B_{1}$ and $B_{2}$. The solvability condition is: Denoting the center of the given circle by $O$ and its radius by $r$, it must be that $O P>r$ and $O P > O P - r$. Since both sides are positive,

$$
t^{2}>2(O P-r)^{2}
$$

But $t^{2}=O P^{2}-r^{2}$, and thus $O P^{2}-r^{2}>2 \cdot O P^{2}-4 \cdot O P \cdot r+2 r^{2}$

or

$$
O P^{2}-4 \cdot O P \cdot r+3 r^{2}=(O P-r)(O P-3 r)<0
$$

From this, it follows that there are two distinct solutions when $r<O P<3 r$. (Allowing $O P=r$ or $O P=3 r$ leads to a degenerate or trivial solution.)

Schmidt Eligius (Bp. I., Fürst S. g. II. o. t.)

![](https://cdn.mathpix.com/cropped/2024_05_02_e91bb2a661d684f75ad2g-1.jpg?height=588&width=658&top_left_y=877&top_left_x=712)

Figure 1

II. Solution: Imagine the problem is solved (Figure 1). The line through $B_{2}$ parallel to $A_{2} O$ intersects and bisects the distance $O P$ at point $F$, and $F B_{2}=\frac{O A_{2}}{2}=\frac{r}{2}$. Therefore, the circle drawn with radius $\frac{r}{2}$ around the midpoint of the distance $O P$ intersects the given circle at points $B_{1}$ and $B_{2}$.

The problem is solvable if the triangle $O F B_{2}$ is constructible, i.e.,

$$
r-\frac{r}{2}<O F<r+\frac{r}{2}, \quad \text { i.e., (since } \quad 2 O F=O P) \quad r<O P<3 r
$$

Huszár Károly (Bp. II., Rákóczi g. II. o. t.)

Most solutions based on similarity transformations essentially coincide with this solution.

III. Solution: Imagine the problem is solved (Figure 2). Denote the midpoint of the chord $A_{1} B_{1}$ by $M$. The line through $B_{1}$ parallel to $O M$ intersects $O P$ at point $Q$.

![](https://cdn.mathpix.com/cropped/2024_05_02_e91bb2a661d684f75ad2g-1.jpg?height=645&width=667&top_left_y=2060&top_left_x=707)

Figure 2

According to the problem, $P B_{1}: P M=2: 3=P Q: P O$, so $P Q=\frac{2}{3} P O$.

Since $O M \perp P M$, $Q B_{1}$ is also perpendicular to $P B_{1}$, meaning point $B_{1}$ lies on the Thales circle drawn over $P Q$ as the diameter. The construction process is thus: construct point $Q$ and the Thales circle drawn over $P Q$ intersects the given circle at the desired points $B_{1}$ and $B_{2}$.

The construction is feasible if, in addition to $P$ being outside the given circle (a condition of the problem), $Q$ is inside the given circle, i.e., $O Q=\frac{O P}{3}<r<O P$, or $r<O P<3 r$.

Rácz Márton (Bp. II., Rákóczi g. II. o. t.)

IV. Solution: Imagine the problem is solved. Reflect the given circle over point $B$. The central reflection of point $A$ is clearly point $P$ according to the problem, and the entire circle's reflection is a circle congruent to the given circle, passing through $P$ and tangent to the given circle at $B$. The center of the reflection is thus the intersection of the circles with radii $2 r$ and $r$ centered at $O$ and $P$, respectively, at points $O_{1}$ and $O_{2}$ (Figure 2). The central lines $O O_{1}$ and $O O_{2}$ intersect the given circle at the desired points $B_{1}$ and $B_{2}$. The solvability condition is that a triangle can be constructed with sides $O P$, $r$, and $2 r$, i.e., $2 r-r<O P<2 r+r$, or $r<O P<3 r$.

Szendrei István (Kunszentmiklós, Damjanich g. I. o. t.)

![](https://cdn.mathpix.com/cropped/2024_05_02_e91bb2a661d684f75ad2g-2.jpg?height=494&width=634&top_left_y=962&top_left_x=716)

Figure 3

V. Solution: If we imagine the problem is solved, then (Figure 3) the perpendicular from $B_{1}$ to the secant $P A_{1}$—by Thales' theorem—intersects the circle at the antipodal point of $A_{1}$, denoted as $A_{1}^{*}$, and $P A_{1}^{*}=A_{1} A_{1}^{*}=2 r$. Therefore, the circle drawn with radius $2 r$ around $P$ intersects the given circle at points $A_{1}^{*}$ and $A_{2}^{*}$, whose antipodal points are the desired points $A_{1}$ and $A_{2}$. The solvability condition is the same as in Solution IV, word for word.