## 15. The Great Mass

The main stained glass window of a modern cathedral is a circle with a diameter of 2 m, intersected by a cross formed by two perpendicular lines that intersect at a point 50 cm away from the center of the window. During the great mass, one slightly absent-minded Parisian decided to calculate the sum of the squares of the lengths of the two segments of this cross.

"Look at this! - he exclaimed. - The number of square meters is equal to the number of mortal sins ${ }^{2}$".

Why did this happen?
15. Let the cross form segments $A B$ and $C D$, intersecting at point $M$; further, $O$ - the center of the stained glass, $H$ - its projection on $A B$, and finally, $\alpha$ - the measure of $O M H$. Consider the right triangle $O H B$:

$$
H B^{2}=O B^{2}-O H^{2}=O B^{2}-O M^{2} \cdot \sin ^{2} \alpha
$$

But $A B=2 H B$. Therefore, $A B^{2}=4 O B^{2}-4 O M^{2} \cdot \sin ^{2} \alpha=\left(4-\sin ^{2} \alpha\right) \mathrm{M}^{2}$. $C D^{2}$ can be calculated similarly, but instead of $\alpha$, we now take $90^{\circ}-\alpha$. In the end, we get

$C D^{2}=4 O C^{2}-4 O M^{2} \cdot \sin ^{2}\left(90^{\circ}-\alpha\right)=\left(4-\cos ^{2} \alpha\right) \mathrm{M}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_fe999c0fe2ad81fc5164g-158.jpg?height=943&width=932&top_left_y=568&top_left_x=582)

$$
A B^{2}+C D^{2} \text { - the number of deadly sins }
$$

Thus,

$$
A B^{2}+C D^{2}=7 \mathrm{~m}^{2}
$$

but 7 is the number of deadly sins.