+118609 I do not know what the answer is but I did find this artical.
https://brilliant.org/wiki/triangles-orthocenter/
Maybe it can help you. OR maybe you can find some other artical or you tube clip that can help you.
Let me know if you work it out. I am interested too.
+26367 We can find an acute triangle with 3 altitude lengths, 1, 2, and h,
if and only if \(h^2\) belongs to interval \((p,q)\).
Find \((p,q)\)
My attempt:
\(\mathbf{p=\ ?}\)
\(\begin{array}{|rcll|} \hline h &=& \dfrac{1\cdot 2}{\sqrt{1^2+2^2}} \\ &=& \dfrac{2}{\sqrt{5}} \\ h^2 &=& \dfrac{4}{5} \\ h^2 &=& 0.8 \\ \mathbf{p} & \mathbf{=}& \mathbf{0.8} \\ \hline \end{array}\)
\(\mathbf{q=\ ?} \)
\(\begin{array}{|rcll|} \hline 1 &=& \dfrac{2\cdot h}{\sqrt{4+h^2}} \\\\ \sqrt{4+h^2} &=& 2h \\ 4+h^2 &=& 4h^2 \\ 3h^2 &=& 4\\ h^2 &=& \dfrac{4}{3} \\ h^2 &=& 1.\overline{3} \\ \mathbf{q} & \mathbf{=}& \mathbf{1.\overline{3}} \\ \hline \end{array}\)
\((p,q) = (0.8,1.\overline{3})\)
+50 Thank you both to heureka and Melody! I read the article, and used heureka's method to find another way to solve the problem.
