Τρίτη, 4 Ιανουαρίου 2011

INRADIUS 2

To construct triangle ABC if are given the radii r_1, r_2, r_3, defined
as follows: Let Da be a point on BC such that inradius of ABDa = inradius of ACDa := r_1. Similarly r_2, r_3


Solution

Let ABC be the triangle, Da, Db, Dc, three points on BC,CA,AB, resp. such that: inradius of ABDa = inradius of ACDa = r_1, and similarly Db,Dc and AHa := h_a, BH_b : = h_b, CH_c := h_c the three altitudes of ABC and r its inradius.


By this Theorem we have:

h_a = 2(r_1)^2 / (2r_1 - r) ==>

1/h_a = (2r_1 - r) / 2(r_1)^2

and similarly:

1/h_b = (2r_2 - r) / 2(r_2)^2

1/h_c = (2r_3 - r) / 2(r_3)^2

Now, since

1/r = 1/h_a + 1/h_b + 1/h_c

we have:

1/r = [(2r_1 - r) / 2(r_1)^2 ] + [(2r_2 - r) / 2(r_2)^2] + [(2r_3 - r) / 2(r_3)^2]

==>

r^2(1/(r_1)^2 + 1/(r_2)^2 + 1/(r_3)^2) - 2r(1/r_1 + 1/r_2 + 1/r_3) + 2 = 0

==> r is known ==> h_a, h_b, h_c are known (from the formulae above).

So we have to construct triangle ABC whose the three altitudes are known.
It's an easy problem.

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