# Generalisation of "Perimeter from Altitudes"

If $\Delta ABC$ has altitudes $AD = p$, $BE = q$, $CF = r$, then we can find the perimeter in terms of $p, q, r$.

Since the area of $\Delta ABC$ is constant, $AB \cdot r = AC \cdot q = BC \cdot p$. If we let another similar triangle $\Delta A'B'C'$ satisfy $A'B' = pq, A'C' = pr, B'C' = qr$, then we can ensure this equality holds true.

However, the length of the altitudes will not be $p, q, r$, but scaled by a constant $k$. Now let $A'F' = x$, so $B'F' = pq - x$. Computing $C'F'$ two different ways using Pythagoras, $(pr)^2 - x^2 = (qr)^2 - (pq -x)^2$, which simplifies to $(pr)^2 - x^2 = (qr)^2 - \left( (pq)^2 - 2pqx + x^2 \right)$. The $x^2$ terms cancel, so $(pr)^2 - (qr)^2 + (pq)^2 = 2pqx$ and $x = \frac{(pr)^2 - (qr)^2 + (pq)^2}{2pq}$.

We can now use the length of $x$ to find $C'F'$, which is just $\sqrt{(pr)^2 - x^2}$. The scale factor then is $\frac{r}{C'F'}$, and since all lengths in the triangle are scaled by the same factor $k$, the perimeter must also be scaled by $k$. Note by Toby M
3 months, 3 weeks ago

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