[ Triangle Inequality

Point $C$ divides the chord $A B$ of a circle with radius 6 into segments $A C=4$ and $C B=5$. Find the minimum distance from point $C$ to the points on the circle.

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Let $O$ be the center of the circle. Extend the segment $OC$ beyond point $C$ to intersect the circle at point $M$, and using the triangle inequality, prove that the length of the segment $CM$ is the smallest of the distances from point $C$ to the points on the circle.

## Solution

Let $O$ be the center of the circle. Extend the segment $OC$ beyond point $C$ to intersect the circle at point $M$. We need to prove that the length of the segment $CM$ is the smallest of the distances from point $C$ to the points on the circle.

Indeed, let $X$ be any point on the circle, different from $M$. Then

$$
OC + CM = OM = OX < OC + CX
$$

(triangle inequality for triangle $OCX$). From this, it follows that $CM < CX$, which is what we needed to prove. Extend the segment $CO$ beyond point $O$ to intersect the circle at point $N$. By the theorem of the products of the segments of intersecting chords, $CM \cdot CN = AC \cdot CB$, or $CM \cdot (12 - CM) = 4 \cdot 5$. Considering the condition $CM < OM = 6$, from the obtained equation, we find that $CM = 2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_aa2960437c2efca8c53fg-34.jpg?height=681&width=652&top_left_y=2135&top_left_x=703)

## Answer

The problem