[ Four points lying on one circle ]  Problem 116312 Topics: [ Four points lying on one circle ]

Points $A, B$, and $C$ lie on a circle of radius 2 with center $O$, and point $K$ lies on a line tangent to this circle at point $B$, such that $\angle A K C=46^{\circ}$, and the lengths of segments $A K, B K$, and $C K$ form an increasing geometric progression (in the given order). Find the angle $AKO$ and the distance between points $A$ and $C$. Which of the angles is larger: $ACO$ or $AKO$?
From the inequality $A K<B K<C K$ it follows that segment $C K$ intersects the circle at some point $N$, and the extension of segment $A K$ intersects the circle at some point $M$. Let $A K=x, B K=y, C K=z$. The numbers $x$, $y$, and $z$ in the given order form a geometric progression, so $y^2=xz$. On the other hand, by the tangent-secant theorem, $y^2=BK^2=x \cdot KM=z \cdot KN$, hence $x \cdot KM=xz$ and $z \cdot KN=xz$, from which we find $KM=z=KC$ and $KN=x=KA$.

Triangles $KOM$ and $KOC$ are congruent by three sides, so $KO$ is the bisector of angle $CKA$, therefore,

$$
\angle AKO=\frac{1}{2} \angle AKC=23^{\circ}
$$

From the isosceles triangle $KCM$ we find that

$$
\angle AMC=90^{\circ}-\frac{1}{2} \angle AKC=90^{\circ}-23^{\circ}=67^{\circ}
$$

Let $R=2$ be the radius of the given circle. By the sine rule,

$$
AC=2R \sin \angle AMC=2 \cdot 2 \sin 67^{\circ}=4 \sin 67^{\circ}
$$

In the quadrilateral $AOCM$, it is known that

$$
\angle AOC+\angle AKC=2 \cdot \angle AMC+\angle AKC=2 \cdot 67^{\circ}+46^{\circ}=180^{\circ}
$$

Thus, points $A, O, C$, and $K$ lie on the same circle. Therefore, $\angle ACK=\angle AOK$.

## Answer

$23^{\circ}, 4 \sin 67^{\circ} ;$ equal.