Three segments don't simply become altitudes of a triangle!

The reciprocals of the altitudes of any triangle can themselves form a triangle, but $\frac{1}{7} > \frac{1}{47} + \frac{1}{51}$ , so 7, 47, and 51 cannot be altitudes of the same triangle.

The solution is quite elementary. I am posting because no one else is posting.

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• Let $a, b$ and $c$ be the sides of the triangle with altitudes $p, q$ and $r$ ; with $p$ is the altitude from $a$ , $q$ is the altitude from $b$ and $r$ is the altitude from $c$ .

• So, the area of the triangle, $\Delta = \frac{1}{2} ap = \frac{1}{2} bq = \frac{1}{2} cr$ . Additionally, if $p < q< r$ , then $a>b>c$ .

• We know, the necessary and sufficient condition for three segments to be the lengths of the sides of a triangle is: Sum of two shorter segments is larger than the longest one .

• That is, $a < b + c \Leftrightarrow \frac{2\Delta}{p} < \frac{2\Delta}{q} + \frac{2\Delta}{r} \Leftrightarrow \frac{1}{p} < \frac{1}{q} + \frac{1}{r}$ .

• So, we have found the necessary and sufficient condition for three segments to be the lengths of the three altitudes of a triangle─

The sum of the reciprocals of the two longer segments is greater than the reciprocal of the shortest segment.

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Only $(p,q,r) = \boxed{(7, 47, 51)}$ ,among the $4$ alternatives, breaks the condition.