Is there an acute triangle of integral side lengths, integral area and nonintegral altitudes?

Edit: if you were wondering, Heron's formula states that $$S_{ABC}=\frac{1}{4}\sqrt{2b^2c^2+2c^2a^2+2a^2b^2-a^4-b^4-c^4}$$ where $a$, $b$ and $c$ are the side lengths of the triangle. The condition for acuteness is $b^2+c^2>a^2$ if $a$ is the longest side.

Context: Without the condition of acuteness, I have made this question: for which primes $p$ is there a triangle with sides $p$, $x$, $y$ satisfying the same conditions above? With the transformation into the variables $p$, $x$, $x+k$, one obtains a Pell-like equation which one may show admits a solution only when $4|p-1$, and further show that one can jump from there to a solution with nonintegral altitudes (I will have to check my proof of this again, but it does lead to 5-29-30 and 13-150-157). However, this means that the other two sides are quite big, and so the triangle ceases to be acute, leading directly to my question.

  • $\begingroup$ Write a system of equations. So not clear what to look for. Ambiguous wording. $\endgroup$ – individ Mar 16 '17 at 16:45
  • $\begingroup$ Sorry, last time I checked, $30^2-29^2=59>5^2$ though, and $157^2-150^2=2149>13^2$ $\endgroup$ – tehjh Mar 16 '17 at 16:47
  • $\begingroup$ There's only one equation really, which is Heron's formula, but I will write it in anyway if you feel that would aid your understanding. $\endgroup$ – tehjh Mar 16 '17 at 16:50
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    $\begingroup$ Heron's formula comes down to this decision... math.stackexchange.com/questions/944814/… $\endgroup$ – individ Mar 16 '17 at 17:01
  • $\begingroup$ Thank you. I seem to have failed in forgetting the key phrase primitive heronian triangles. $\endgroup$ – tehjh Mar 16 '17 at 17:12

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