#2**+2 **

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.

Melody Jan 28, 2019

#3**+8 **

**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:

\(\mathbf{p=\ ?}\)

\(\begin{array}{|rcll|} \hline h &=& \dfrac{1\cdot 2}{\sqrt{1^2+2^2}} \\ &=& \dfrac{2}{\sqrt{5}} \\ h^2 &=& \dfrac{4}{5} \\ h^2 &=& 0.8 \\ \mathbf{p} & \mathbf{=}& \mathbf{0.8} \\ \hline \end{array}\)

\(\mathbf{q=\ ?} \)

\(\begin{array}{|rcll|} \hline 1 &=& \dfrac{2\cdot h}{\sqrt{4+h^2}} \\\\ \sqrt{4+h^2} &=& 2h \\ 4+h^2 &=& 4h^2 \\ 3h^2 &=& 4\\ h^2 &=& \dfrac{4}{3} \\ h^2 &=& 1.\overline{3} \\ \mathbf{q} & \mathbf{=}& \mathbf{1.\overline{3}} \\ \hline \end{array}\)

\((p,q) = (0.8,1.\overline{3})\)

heureka Jan 28, 2019

#4**+3 **

Thank you both to heureka and Melody! I read the article, and used heureka's method to find another way to solve the problem.

jess.shen2024 Feb 2, 2019