## 13. The Magi's Pancake

On Christmas Eve, the mother served a round "Magi's pancake" on the table. She began to cut it along the diameter but stopped when she felt the knife hit a bean. Then she made a new straight cut at a $45^{\circ}$ angle to the previous one - but, unfortunately, she hit the bean again. The children, more mathematicians than gourmands, noticed that the sum of the squares of the segments into which the bean divides the second cut is equal to twice the square of the radius of the pancake.

Do you think this is a coincidence?[^6]
13. No, not by chance. Indeed, let $O$ be the center of the pancake, and $F$ be the bean; $AB$ is the chord representing the second

![](https://cdn.mathpix.com/cropped/2024_05_21_fe999c0fe2ad81fc5164g-156.jpg?height=719&width=725&top_left_y=1548&top_left_x=674)

cut, $I$ is its midpoint. Consider triangles $OAF$ and $OBF$. By a known theorem, we have

$$
\begin{aligned}
& O A^{2}=F A^{2}+O F^{2}+2 F A \cdot F I, \\
& O B^{2}=F B^{2}+O F^{2}-2 F B \cdot F I .
\end{aligned}
$$

Adding these two equations term by term, we get

$$
2 O A^{2}=F A^{2}+F B^{2}+2 O F^{2}-4 F I^{2}
$$

But since triangle $OFI$ is a right isosceles triangle (because $\angle OFI = 45^{\circ}$), we have

$$
O F = (F I) \sqrt{2}, \text{ i.e., } 2 O F^{2} - 4 F I^{2} = 0 .
$$

Therefore,

$$
F A^{2} + F B^{2} = 2 O A^{2}
$$

- and this relationship holds for any position of point $F$.