Generalisation of "Perimeter from Altitudes"

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

Since the area of ΔABC\Delta ABC is constant, AB⋅r=AC⋅q=BC⋅pAB \cdot r = AC \cdot q = BC \cdot p. If we let another similar triangle ΔA′B′C′\Delta A'B'C' satisfy A′B′=pq,A′C′=pr,B′C′=qrA'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,rp, q, r, but scaled by a constant kk. Now let A′F′=xA'F' = x, so B′F′=pq−xB'F' = pq - x. Computing C′F′C'F' two different ways using Pythagoras, (pr)2−x2=(qr)2−(pq−x)2(pr)^2 - x^2 = (qr)^2 - (pq -x)^2 , which simplifies to (pr)2−x2=(qr)2−((pq)2−2pqx+x2)(pr)^2 - x^2 = (qr)^2 - \left( (pq)^2 - 2pqx + x^2 \right) . The x2x^2 terms cancel, so (pr)2−(qr)2+(pq)2=2pqx (pr)^2 - (qr)^2 + (pq)^2 = 2pqx and x=(pr)2−(qr)2+(pq)22pqx = \frac{(pr)^2 - (qr)^2 + (pq)^2}{2pq} .

We can now use the length of xx to find C′F′C'F', which is just (pr)2−x2\sqrt{(pr)^2 - x^2}. The scale factor then is rC′F′\frac{r}{C'F'}, and since all lengths in the triangle are scaled by the same factor kk, the perimeter must also be scaled by kk.

#Geometry

Note by Toby M
11 months, 2 weeks ago

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