$\mathbb{\text { Franklyn } 5 . P .}$

Let \$A_1A_2A_3\$ be an acute triangle, the radius of the circumscribed circle is $\$ 1 \$$, \$O\$ is the center of the circumscribed circle. From the vertices \$A_i\$ the cevians are drawn through \$O\$ to their intersections with the opposite sides at points \$B_i\$ respectively $\$(\mathrm{i}=1,2,3) \$$.

(a) Among the three segments \$B_iO\$, choose the longest one. What is its smallest possible length?

(b) Among the three segments \$B_iO\$, choose the shortest one. What is its largest possible length?
First, we show that of the two segments \(B_iO\), the one directed towards the smaller side is longer (it is clear that when the sides are equal, the segments are equal). Let, for example, \(A_1A_3 < A_2A_3\). Since \(\angle OA_1A_2 = \angle OA_2A_1\), we have \(\angle OA_2B_1 < \angle OA_1B_2\). In triangles \(A_1OB_2\) and \(A_2OB_1\), we have \(A_1O = A_2O\) and \(\angle A_1OB_2 = \angle A_2OB_1\). Therefore, \(B_1O < B_2O\), which is what we needed to show.

(a) Let in an acute triangle \(A_1A_2\), the smallest side is \(A_1A_2\). Then the segment \(B_3O\) is the longest among \(B_iO\). Since \(\angle A_3 \le 60^\circ\),

![](https://cdn.mathpix.com/cropped/2024_05_06_c317f3f1f50d1d30304bg-37.jpg?height=55&width=1995&top_left_y=2528&top_left_x=14)
\(A_1A_2\). Then \(\frac{1}{2} \le OP \le B_3O\). Equality is achieved in an equilateral triangle.

(b) Let in an acute triangle \(\mathrm{A}_1 \mathrm{A}_2 \mathrm{A}_3\), inscribed in a circle of radius \(1\), the side \(A_1A_2\) is the smallest, and the side \(A_1A_3\) is the largest. We will move the point \(A_1\) along the circumscribed circle in the direction of point \(A_2\). During this movement, the segment \(B_1O\) is equal to \(180^\circ - 2\angle A_2A_1A_3\), resulting in an isosceles triangle where \(A_1A_3 = A_2A_3 \ge A_1A_2\).

In triangle \(A_1B_1A_3\), the segment \(A_3O\) is the angle bisector, so

\(\frac{B_1O}{A_1O} = \frac{B_1A_3}{A_1A_3} = \frac{B_1A_3}{A_2A_3}\). It is not difficult to see that the last ratio is no more than \(\frac{1}{2}\) when \(A_1A_2 \le A_1A_3\). Therefore, \(B_1O \le \frac{1}{2}\). Equality is achieved in an equilateral triangle.

## Answer

(a), (b) \(\frac{1}{2}\).