The distance of a certain chord from the center of a circle with radius $R$ is $d$. Calculate the distance of the chord from the center, which belongs to an arc that is half as long.





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I. Solution. The distance of the chord $A A^{\prime}$ from the center $O$ of the circle is $O C=d$. The midpoint of the (smaller) arc corresponding to the chord $A A^{\prime}$ is $D$. We need to calculate the distance of the chord $A D$ from the center, $O E$. Let the projection of the midpoint $E$ of $A D$ onto $O D$ be $F$; clearly, $F$ bisects $C D$, i.e., $C F=D F$. Already in the right triangle $O E D$,

$$
\overline{O E^{2}}=\overline{O D} \cdot \overline{O F}=R \cdot \overline{O F}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_9e766d580b096d0958ffg-1.jpg?height=368&width=420&top_left_y=445&top_left_x=842)

Since

$$
O F=O C+C F \quad \text{ and } \quad O F=O D-D F
$$

$$
2 O F=O C+O D, \quad O F=\frac{1}{2}(O C+O D)=\frac{1}{2}(d+R)
$$

Thus,

$$
\overline{O E^{2}}=\frac{1}{2} R(d+R)
$$

Antal Csics (Private student, Bencés Gymnasium V. o., Bp.)

II. Solution. If $B$ is the point of the circle diametrically opposite to $D$,

$$
\left.O E=\frac{1}{2} B A . \quad \text{ Indeed, } O \text{ bisects } B D \text{ and } E \text{ bisects } A D .\right)
$$

In the right triangle $B A D$: $\overline{B A^{2}}=\overline{B D} \cdot \overline{B C}=2 R(R+d)$.

$$
\overline{O E^{2}}=\frac{1}{4} \overline{B A^{2}}=\frac{1}{2} R(R+d)
$$