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I'm trying to find a problem about right triangles with a minimalist statement that isn't too obvious. Here's what I've come up with :
ABC is an A–right triangle, H is the orthogonal projection of A onto (BC) , and P is the orthogonal projection of H onto (AB) as shown in the attached figure where the lengths of BC and HP are noted. My question is : What is the length of AH ?

I think several methods are possible but I'm looking for a short one, knowing that GeoGebra gives me 6 as an approximate answer.

So , is this really the ( only) solution ?enter image description here

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    $\begingroup$ There are two distinct solutions, thus your configuration is underspecified. Since you provide none of your own work, I will not furnish a solution. $\endgroup$ Commented Nov 24 at 18:55
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    $\begingroup$ A nice problem that you can also solve is: Given $\overline{BC} = 13$, what is the value of $\overline{PH}$ knowing that there is precisely one triangle with the given characteristics? $\endgroup$ Commented 2 days ago
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    $\begingroup$ Given the many similar triangles, you can express both $|HP|$ and $|AH|$ in terms of $|BC|$ and sines and/or cosines of, say, $\angle B$. Eliminating that angle from the two relations leaves a relation in the lengths alone. $\endgroup$ Commented 2 days ago
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    $\begingroup$ @dfnu. ..(26/9)√3. $\endgroup$ Commented 2 days ago
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    $\begingroup$ I also suggest a faster approach. Instead of using lengths, just use goniometry. Let for simplicity $\overline{BC} = 1$ and $\measuredangle ACB =\theta$. Then $\overline{PH} = \sin^2\theta\cos\theta$. $\endgroup$ Commented 2 days ago

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Expanding a comment, defining $a:=|BC|$, $h:=|HA|$, $p:=|HP|$, the similar triangles allow us to write $$\begin{align} p&=h\cos B \\ h&=a\sin B\cos B \\[8pt] \to \qquad h^2&=a^2(1-\cos^2B)\cos^2B \\[4pt] &=a^2\left(1-\frac{p^2}{h^2}\right)\frac{p^2}{h^2} \\[8pt] \to \qquad 0 &= h^6 - h^2a^2p^2 + a^2 p^4 \end{align}$$ This cubic in $h^2$ has messy exact solutions. Substituting the specific numeric values $a=13$ and $p=5$, we obtain (discarding extraneous negatives) $$h^2 \in \{36.691\ldots, 38.358\ldots\} \quad\to\quad h \in\{6.057\ldots, 6.193\ldots\} $$

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  • $\begingroup$ Yes .... Excellent.[+1] $\endgroup$ Commented yesterday
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HINT.- At first glance, and because of the Pythagorean triple $(5,12,13)$, I must confess that I thought such a configuration was not possible. But far from it, there are two possible solutions.

In fact, if $A=(0,0), C=(0,a),B=(\sqrt{169-a^2},0)$ then easily one has$H=\left(\dfrac{a^2(\sqrt{169-a^2}}{169},\dfrac{a(169-a^2}{169}\right)$ so the suitable values for the parameter are roots of the equation $x^3-169x+845=0$ whose positive roots are $a_1\approx7.33823$ and $a_2\approx 7.67164$ which yield the two configurations $$C_1=(0,7.33823),A_1=(0,0),B_1=(10.7308145295,0),H_1=(3.41923242,5)\\C_2=(0,7.67164),A_2=(0,0),B_2=(10.4950435783,0),H_2=(3,65488714,5)$$ NOTE.-The second coordinates of $H_1$ and $H_2$ are in our calculations $4.99999986$ and $4.99999998$ respectively but it is question of approximative values.

The values of $AH_1$ and $AH_2$ are now clearly determined.

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  • $\begingroup$ Yes. Good classic method. [+1] $\endgroup$ Commented yesterday
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Just to expand a little bit on my comment: for the moment, forget about the $13$ (you can scale all the results if you need), and take $\overline{BC} = 1$. If $\measuredangle ACB=\theta$ you immediately get $$\overline{PH} = f(\theta)= \sin^2\theta\cos\theta, \ \ \ \ (0\leq \theta \leq \pi/2).$$ We have $f(0) = f(\pi/2) = 0$, and $$f'(\theta) = \sin\theta (\sqrt 3\cos\theta-1)(\sqrt 3 \cos \theta+1).$$ The only zero of $f'$ in $[0,\pi/2]$ is $$\theta = \arccos\frac1{\sqrt 3}=\arctan\sqrt 2,$$ which corresponds to a maximum value of $\overline{PH}$ $$\overline{PH} = \frac{2\sqrt 3}9.$$ (Multiplying by $13$ you obtain the value you found already, of course.) For any other value of $\overline{PH}$, with $$0< \overline{PH} < \frac{2\sqrt 3}3,$$ you will find exactly two solutions to your problem.

What happens if you keep projecting vertex $A$ on $BC$ and then back on $AB$? After $n$ iterations of this double projection you get $$\overline{PH_n} = (\sin x)^{n+1}(\cos x)^n\ \ \ \ n=1,2,\dots$$ The maximum value of $\overline{PH_n}$ (which will be of course smaller and smaller) is obtained in correspondance of $$\theta = \arctan \sqrt\frac{n+1}{n} \to \frac{\pi}4 \ \ \ \mbox{as}\ \ n\to +\infty$$

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  • $\begingroup$ Very nice extension.[+1] $\endgroup$ Commented yesterday
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enter image description here

$AB = 3\frac13AP\Rightarrow PB=2\frac13 AP$

$AP\times \frac73 AP=PH^2=5^2$

$\Rightarrow AP^2=\frac{5^2\times3}7 $

$AH^2=AP^2+PH^2=5^2(1+\frac 37)$

$AH=5\sqrt {(1+\frac 37)}\approx 5.975 $

In the figure the partition of AB and BC carried out independently.This means the ratios are accurate and reliable.

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  • $\begingroup$ Can you explain why you partitioned $AB$ and $BC$ in this way? $\endgroup$ Commented yesterday
  • $\begingroup$ @sirous: I got (implicitely) the two values $6.057303$ and $6.193348$. I didn't put the downvote. Regards. $\endgroup$ Commented yesterday
  • $\begingroup$ @sirous. .By calculating the area of ∆ABC in three different ways , we can find the two solutions already mentioned. Therefore , something is wrong in your approach , but I always appreciate the effort and I never vote negatively , even if no one votes for me ! $\endgroup$ Commented yesterday
  • $\begingroup$ @JamilSanjakdar, Thanks for your respond. I knew some people will down vote my answer but it does not matter because I wanted to show a simple method. Of course the partioning must be due to some calculations. I found it by a kind of conjecture. $\endgroup$ Commented yesterday
  • $\begingroup$ I'm not the downvoter, either (I basically never downvote an answer). However I think you should clearly explain that this is not an answer but just a conjecture, and explain what this conjecture is based upon. Also, more specifically, what do you mean by: "In the figure the partition of AB and BC carried out independently. This means the ratios are accurate and reliable"? Furthermore, before posting the answer, you could have easily tested your conjecture, and thus find that the resulting value for $\overline{AH}$ did not lead to $\overline{PH} = 5$, as it was required. $\endgroup$ Commented yesterday

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