+0

# deleted

+3
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deleted, deleted

Jan 26, 2019
edited by jess.shen2024  Apr 27, 2019

#1
+104436
0

Hi Jess,

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

Jan 28, 2019
#2
+104436
+2

I do not know what the answer is but I did find this artical.

https://brilliant.org/wiki/triangles-orthocenter/

Let me know if you work it out. I am interested too.

Jan 28, 2019
#3
+23145
+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:

$$\mathbf{p=\ ?}$$

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

$$\mathbf{q=\ ?}$$

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

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

Jan 28, 2019
edited by heureka  Jan 29, 2019
#4
+48
+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.

Feb 2, 2019