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avatar+47 

thanks to anyone who can help!

 

We can find an acute triangle with 3 altitude lengths, 1, 2, and h, if and only if  \(h^2\) belongs to interval

 \((p,q) . find (p,q)\)

 Jan 26, 2019
 #1
avatar+100028 
0

Hi Jess,

I'd really like to see someone answer this question too!   

 Jan 28, 2019
 #2
avatar+100028 
+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. 

 Jan 28, 2019
 #3
avatar+21978 
+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:

 

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

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

 

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

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

 

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

 

laugh

 Jan 28, 2019
edited by heureka  Jan 29, 2019
 #4
avatar+47 
+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.

 

smiley

 Feb 2, 2019

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