## PROBLEM 5. PROTRUDING SEMICIRCLE

Chord $A B$ of a circle with radius 1 and center at $O$ is the diameter of semicircle $A C B$, located outside the first circle (Fig. 6). It is clear that

![](https://cdn.mathpix.com/cropped/2024_05_21_2f4fa61a46fbb276c744g-13.jpg?height=680&width=865&top_left_y=1002&top_left_x=601)

Fig. 6

the point $C$ of this semicircle, which protrudes the farthest, lies on the radius $O D C$, perpendicular to $A B$. (For any other point $C^{\prime}$ of this semicircle, we have $O C^{\prime}<O D+D C^{\prime}=O D+D C=O C$.) Of course, the length of segment $O C$ depends on the choice of chord $A B$. Determine $A B$ so that segment $O C$ has the maximum length.
Solution. Let $OD = \gamma \bar{a}$; then the radius of our semicircle is

$$
AD = \sqrt{1-a} = DC
$$

Thus, $OC^2 = (OD + DC)^2 = \left(\sqrt{a} + \sqrt{1-a^2}\right) = a + 2\sqrt{a(1-a)} + 1 - a$. We get $OC^2 = 1 + 2\sqrt{a(1-a)}$. 12

This expression is maximized when the value of $a(1-a)$ is maximized. But since

$$
a(1-a) = a - a^2 = \frac{1}{4} - \left(a - \frac{1}{2}\right)^2
$$

this occurs for $a = 1/2$, in which case $OD = \sqrt{a} = \sqrt{2/2}$. For the maximum segment $OC$,

$$
AD = \sqrt{1 - OD^2} = \sqrt{1 - \frac{1}{2}} = \frac{\sqrt{2}}{2}
$$

and thus $AB = 2AD = \sqrt{2}$. Therefore, $\triangle AOB$ has sides 1, 1, and $\gamma \overline{2}$. This means that the chord $AB$ subtends a right angle with the vertex at point $O$.

The following elegant approach provides an alternative solution. Triangle $ADC$ is a right isosceles triangle, i.e., $\angle DCA = 45^\circ$ (Fig. 7). Now, if $CA$ is not tangent to the given circle, there will be a chord for which point $C$ lies further along the line $OD$. Thus, for the chord that maximizes the segment $OC$, the line $CA$ must be tangent to the circle, and in this case, $CA$ is a leg of the isosceles right triangle $OAC$. Hence, $CA = OA = 1$ and from the isosceles right triangle $DAC$ we get that

$$
AD = \frac{\sqrt{2}}{2} \text{ and } AB = \sqrt{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_2f4fa61a46fbb276c744g-14.jpg?height=418&width=1196&top_left_y=1134&top_left_x=446)

Fig. 7