TRIANGLE CONSTRUCTION A, a, h_a + h_b + h_c

To construct triangle ABC if are given A, a, h_a + h_b + h_c (sum of altitudes)

We have:

A and a known ==> R is known.

h_a + h_b + h_c = 2R(bc + ca + ab)

==> bc + ca + ab is known.

Solution 1:

bc + ca + ab = a(b+c) + bc := k^2 (known)

a^2 = b^2 + c^2 - 2bc.cosA = (b+c)^2 - 2bc(1 + cosA) = (b+c)^2 - 4bc(cos(A/2))^2

Denote b+c := X, bc := Y^2

==>

aX + Y^2 = k^2

a^2 = X^2 - 4Y^2(cos(A/2)^2

==> X^2 + 4a(cos(A/2)^2.X - 4k^2(cos(A/2)^2 - a^2 = 0

==> b+c is known and also bc is known.

==> b,c are known.

Solution 2:

We have:

2(bc + ca + ab) = (b + c)^2 - (b^2 + c^2) + 2a(b + c) (1)

Let ABC be the triangle in question. The bisector AD of A intersects the circumcircle at E. Let EF be the diameter perpendicular to BC at its midpoint M (see figure).

Denote:

AE = d, AM = m_a,

EB = EC = m, known

AF = y

EM = x, MF = 2R - x = z, known

(since the isosceles triangles EBC, FBC are known: BC = a and have known angles.)

We have:

y^2 = (2R)^2 - d^2 (from the right triangle AEF)) (2)

m(b + c) = ad (by Ptolemy Theorem in the cyclic quadril. ABEC)

==> b + c = am / d (3)

b^2 + c^2 = 2(m_a)^2 + (a^2 / 2) (Theorem of median in ABC) (4)

zd^2 + xy^2 = 2R(m_a)^2 + 2Rxz (by Stewart Theorem in AFE) (5)

(2) and (5) ==> zd^2 + x((2R)^2 - d^2) = 2R(m_a)^2 + 2Rxz (6)

(1) and (3), (4), (6) ==>

2(bc + ca + ab) = (a/m)^2d^2 - ((z-x)/R)d^2 - 4Rx + 2xz - (a^2)/2 + (2a^2/m)d

==>

[(z-x)/R - (a/m)^2]d^2 - (2a^2/m)d - 2xz + 4Rx + (a^2/2) + 2(bc + ca + ab) = 0

==> AE = d is known.

Construction:

We construct the isosceles triangle EBC with BC = a, BEC = 180 - A, BE = CE. The circle (E, d) intersects the circumcircle of EBC at A.