Recently I have found an old journal of Serbian Mathematical Society, and there is one neat award task for high school students.
A circle $k$ is inscribed in a square $ABCD$. For an arbitrary point $T$ of circle $k$ let $\alpha_T$ and $\beta_T$ are angles under which diagonals $AC$ and $BD$ can be seen. Prove that for each $T \in k$ it holds $\, \tan^2\alpha_T + \tan^2\beta_T =8$.
The proof that was proposed sets up a square in the coordinate system such that $A=(-a, -a), B=(a,-a), C=(a,a)$ and $D=(-a,a)$. Then for coordinates of an arbitrary point $T(x,y)$ at a circle $k$ which is inscribed in a square $ABCD$ will hold $x^2+y^2 = a^2$.
Next steps yielded the vectors $\vec{TA}=\{-a-x,-a-y\}, \vec{TB}=\{a-x,-a-y\}$, $\vec{TC}=\{a-x,a-y\}$ and $\vec{TD}=\{-a-x,a-y\}$. Now, it can easily be seen that we have: $$ \cos^2\alpha_T=\frac{a^2}{5a^2-8xy} \qquad \text{and} \qquad \cos^2\beta_T=\frac{a^2}{5a^2+8xy}. $$
Finally, we will have: $$ \tan^2\alpha_T+\tan^2\beta_T=\frac{1}{\cos^2\alpha_T}+\frac{1}{\cos^2\beta_T}-2=\frac{5a^2-8xy}{a^2}+\frac{5a^2+8xy}{a^2}-2=\boxed{8}. $$
Q: Can someone elaborate this neat award task for high school students with a geometric explanation? Thanks for reading.
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