There are points $A$ and $B$ in the plane. Where should point $C$ be placed in the plane so that the distance $C A$ is 2.6 times the distance $C B$, and the angle $B A C$ is as large as possible?
I. solution: Point $C$ lies on that Apollonian circle for which every point $P$ satisfies $\frac{P B}{P A}=\frac{1}{2.6}=\frac{5}{13}$. The angle $B A C \varangle$ is maximal when $A P$ is tangent to the circle. This point of tangency will be the sought $C$ point. We will show that $C B A \varangle$ is a right angle.

![](https://cdn.mathpix.com/cropped/2024_05_02_569de72f43b2a4831190g-1.jpg?height=466&width=765&top_left_y=358&top_left_x=669)

Denoting the center of the Apollonian circle by $O$ and the intersection points of the line $A B$ with the circle by $M$ and $N$ (Fig. 1), the line $C M$ intersects the side $A B$ of $\triangle A B C$ in the ratio $\frac{M B}{M A}=\frac{5}{13}=\frac{C B}{C A}$, because $M$ is also a point on the circle. This implies that $C M$ is the angle bisector of $\angle A C B$,

$$
A C M \varangle + M C O \varangle = 90^{\circ},
$$

because the tangent is perpendicular to the radius of the circle.

We have seen that $A C M \varangle = M C B \varangle$. Furthermore, the triangle $C O M$ is isosceles,

$$
M C O \varangle = O M C \varangle.
$$

Substituting the angles in the equation (1) with the angles equal to them,

$$
M C B \varangle + O M C \varangle = 90^{\circ},
$$

then the third angle of $\triangle M B C$, $\angle M B C$, is also $90^{\circ}$, meaning that $\triangle A B C$ is a right triangle and thus $A B = \sqrt{A C^2 - B C^2} = B C \sqrt{2.6^2 - 1} = B C \sqrt{5.76} = 2.4 B C$, or

$$
B C = \frac{5}{12} A B
$$

Thus, the point $C$ should be placed on the perpendicular line to $A B$ at point $B$, at a distance of $\frac{5}{12} \cdot A B$ from $B$.

II. solution: Imagine the problem is solved and apply the sine theorem to $\triangle A B C$;

$$
\frac{\sin \alpha}{\sin \beta} = \frac{C B}{C A} = \frac{5}{13}
$$

from which

$$
\sin \alpha = \frac{5}{13} \cdot \sin \beta
$$

Since $\alpha$ is necessarily an acute angle, $\alpha$ is maximal when $\sin \alpha$ is maximal; $\sin \alpha$ takes its maximum value when $\sin \beta = 1$, i.e., $\beta = 90^{\circ}$.

III. solution: Consider a point $C_{1}$ such that $\frac{C_{1} B}{C_{1} A} = \frac{1}{2.6} = \frac{5}{13}$ and $\angle C_{1} B A$ is acute, and a point $C_{2}$ such that $\frac{C_{2} A}{C_{2} B}$ is also $\frac{5}{13}$, but $\angle C_{2} B A$ is obtuse. Let $C_{1} C_{1}^{\prime}$ and $C_{2} C_{2}^{\prime}$ be the distances of points $C_{1}$ and $C_{2}$ from the line $A B$ (Fig. 2) and let the angles at $A$ be denoted by $\alpha_{1}$ and $\alpha_{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_569de72f43b2a4831190g-1.jpg?height=307&width=590&top_left_y=2340&top_left_x=754)

Fig. 2

Since $C_{1} C_{1}^{\prime} < C_{1} B$ and $C_{2} C_{2}^{\prime} < C_{2} B$, we have

$$
\sin \alpha_{1} = \frac{C C_{1}}{C_{1} A} < \frac{C_{1} B}{C_{1} A} = \frac{5}{13} \text{ and } \sin \alpha_{2} = \frac{C_{2} C_{2}^{\prime}}{C_{2} B} < \frac{C_{2} B}{C_{2} A} = \frac{5}{13}
$$

Thus, the sine of the angle at $A$ is always less than $\frac{5}{13}$, except when $\angle C B A$ is a right angle, in which case $\sin \alpha = \frac{C B}{C A} = \frac{5}{13}$, meaning that $\sin \alpha$ and hence $\alpha$ are maximal.

This is how János Fehér (Győr, Révai Gymnasium) solved the problem.

IV. solution: Among all triangles $A B C$ with a common side $A B$ and satisfying the condition $A C: B C = 2.6$, the angle bisector of the angle at $C$ intersects the side $A B$ at a point $M$ such that $A M: B M = A C: B C = 2.6$. The position of point $M$ is independent of the shape of the triangle. Reflect the triangle $C M B$ over the angle bisector $C M$. The reflection of point $B$ is $B^{\prime}$, which lies on the side $A C$ and

$$
M B^{\prime} = M B
$$

is a distance independent of the position of point $C$. Therefore, all such points lie on the circle $k$ centered at $M$ and passing through $B$. (Fig. 3).

![](https://cdn.mathpix.com/cropped/2024_05_02_569de72f43b2a4831190g-2.jpg?height=282&width=658&top_left_y=862&top_left_x=728)

Conversely, if $B^{\prime}$ is any point on this circle, draw the line $A B^{\prime}$ and bisect the angle $B M B^{\prime}$. The resulting triangle $A B C$ has $M C$ as the angle bisector because the reflection of $M B^{\prime}$ over $M C$ is $M B$, and the reflection of point $C$ is itself, so the reflection of the side $C B^{\prime}$ is $C B$. This implies that

$$
A C: B C = A M: B M = 2.6
$$

so the triangle $A B C$ satisfies the conditions.

The original problem can thus be replaced by finding the point $B_{0}$ on the circle $k$ such that the angle $B_{0} A B$ is maximized. This point $B_{0}$ is clearly the point of tangency of the tangent from $A$. In this case,

$$
M B_{0} \perp A B_{0}
$$

Construct the triangle $A B C_{0}$ corresponding to this point $B_{0}$ as described above. In this triangle, $\triangle M B C_{0}$ is the reflection of $\triangle M B_{0} C_{0}$, so

$$
A B C_{0} \varangle = M B_{0} C_{0} \varangle = 90^{\circ}.
$$

V. solution: Among all triangles $A B C$ with a common side $A B$ and satisfying $A C: B C = 2.6$, find the one in which the angle $B A C \varangle$ is maximized. The angles, however, do not change if the triangles are scaled up or down. Thus, we can replace all triangles with those in which the side $A C$ is common. Then

$$
B C = \frac{A C}{2.6} = \frac{5}{13} A C
$$

is the same for all triangles, so all points $B$ lie on a circle centered at $C$ with radius $r = \frac{5}{13} \cdot A C$. The angle $B A C \varangle$ is maximized when $A B$ is tangent to this circle, so $B C \perp A B$ and by the Pythagorean theorem,

$$
A B = 2.4 B C, \text{ or } B C = \frac{5}{12} \cdot A B
$$

Many solved the problem using differentiation, which, as we have seen, is completely unnecessary.