12. It is known that $x=2026$, find $y=BC$. The answer is 90.02.

IV-A. Define $f(a)$ as the function equal to the number of distinct solutions to the equation

$$
\sin \frac{a \pi x}{x^{2}+1} + \cos \frac{\pi (x^{2} + 4 a x + 1)}{4 x^{2} + 4} = \sqrt{2 - \sqrt{2}}
$$

For example, $f(a_1) = 3$ means that when $a = a_1$, the equation has three distinct solutions $x_1, x_2, x_3$. If the equation has no solutions when $a = a_0$, then we set $f(a_0) = 0$.

Solve the inequality $f(a) \leqslant 5$. In the answer, write the total length of the resulting intervals, rounding it to two decimal places if necessary. If there are no solutions, write the number 0; if the resulting length is infinite, write the number 9999.
Solution. Let $t=\frac{a \pi x}{x^{2}+1}$. Then the equation takes the form $\sin t+\cos \left(\frac{\pi}{4}+t\right)=\sqrt{2-\sqrt{2}}$ or

\[
\begin{aligned}
& \sin t+\sin \left(\frac{\pi}{4}-t\right)=\sqrt{2-\sqrt{2}} \Longleftrightarrow \\
& 2 \sin \frac{\pi}{8} \cos \left(\frac{\pi}{8}-t\right)=\sqrt{2-\sqrt{2}} \Longleftrightarrow \\
& \cos \left(\frac{\pi}{8}-t\right)=1
\end{aligned}
\]

From the last formula, it is clear that if $t$ is replaced by $-t$, the number of solutions does not change. Therefore, the function $f(a)$ is even.

Solve the inequality $|f(a)| \leqslant 5$. The last inequality means that the initial equation should have no more than 5 roots. For this, solve the equation $\cos \left(\frac{\pi}{8}-t\right)=1$:

\[
t=\frac{\pi}{8}+2 \pi k, \quad k \in \mathbb{Z}
\]

Since $t=\frac{a \pi x}{x^{2}+1}$, then

\[
x^{2}-\frac{a}{2 k+1 / 8} x+1=0
\]

Note that for different values of $b=b_{1}$ and $b=b_{2}$, the equations $x^{2}-b_{1} x+1=0$ and $x^{2}-b_{2} x+1=0$ do not have common roots. Indeed, if these equations have a common root, then by Vieta's theorem, the other two roots will also coincide, which means $b_{1}=b_{2}$. For each fixed $k \in \mathbb{Z}$, a solution exists if $D \geqslant 0$, and when $D=0$ there is one solution, and when $D>0$ there are two. The condition $D \geqslant 0$ is equivalent to

\[
\left(\frac{a}{2 k+\frac{1}{8}}\right)^{2}-4 \geqslant 0 \Longleftrightarrow|a| \geqslant\left|2\left(2 k+\frac{1}{8}\right)\right|
\]

Therefore,

\[
\begin{aligned}
& |a|\frac{17}{4} \Longrightarrow \text { more than } 5 \text { solutions. }
\end{aligned}
\]

Answer: $a \in\left[-\frac{17}{4} ; \frac{17}{4}\right]$, the length of the interval is 8.5.