## Τρίτη, 25 Ιανουαρίου 2011

### TRIANGLE CONSTRUCTION A, h_b+a, h_c+a

To construct triangle ABC if are given A, h_b + a, h_c + a, where h_b,h_c are the altitudes from B,C, resp.

Solution 1

We have:

2Rh_b = ac
2Rh_c = ab

==>

h_b + a = (ac/2R) + a = (a/2R).(2R + c)

h_c + a = (ab/2R) + a = (a/2R).(2R + b)

We have that angle A is known ==> a/2R is known

[Geometric Proof:

Let O be the circumcenter and M the midpoint of BC.
The triangle BOC has known angles since A is known ==>

MC / OC is known ==> (a/2)/R = a/2R is known].

So the problem is equivalent to construct triangle if are given A, 2R + b, 2R + c.

Analysis:

Let ABC be the triangle in question, AOD diameter of the circumcircle and B',C' points on the extensions of AB,AC such that BB' = CC' = AD [= 2R].

A is known

AB' = AB + BB' = c + 2R, known

AC' = AC + CC' = b + 2R, known

==> The triangle AB'C' can be constructed.

We have:

DB is perpendicular to AB' (since AD is diameter) and BB' = AD [=2R]

==> the locus of D is the parabola with focus A and directrix the perpendicular to AB' at B' (see LEMMA).

Similarly:

DC is perpendicular to AC' and CC' = AD ==> the locus of D is the parabola with focus A and directric the perpendicular to AC' at C'.

Therefore D is intersection point of the two loci. B,C are the (other than A) intersections of the circle of diameter AD with the lines AB',AC' resp.

LEMMA:

In triangle ABC, let BC be fixed and D the orthogonal projection of A on BC. If BD = AC then the locus of A is a parabola.

Let A' be the orthogonal projection of A on the perpendicular to BC at B. We have AC = BD and BD = AA' ==> AC = AA' ==> the locus of A is the parabola with focus C and directrix the perpendicular to BC at B.

Solution 2

Let BB' = h_b, CC' = h_c be the altitudes from B,C, resp.

The right triangles C'AC,B'AB are similar and have known angles (since A is known)

==>

CC' / CA = BB' / BA is known ==>

h_c / b = h_b / c = (h_c - h_b) / (b - c) = [(h_c + a) - (h_b + a)] / (b - c)

==> b - c is known.

So the problem is equivalent to construct triangle if are given A, b - c, h_c + a.

Analysis:

Let ABC be the triangle in question with AC > AB.

Let D be the point on AC between A and C such that AD = AB, CE the altitude from C, and Z the intersection of the lines BD and CE.

In the triangle CDZ we have:

CD = AC - AD = b - c, known.

Angles (BDC) = (DAB) + (DBA) = A + (90 - (A/2)) = 90 + (A/2), known

(DCZ) = 90 - (CAE) = 90 - A, known.

Therefore the triangle CDZ can be constructed.

Let H be the point on the extension of CE such that EH = BC.

We have CH = CE + EH = h_c + a, known.

BE is perpendicular to CH and BC = EH ==> (according to LEMMA) the locus of B is the parabola with focus C and directrix the perpendicular to CH at H.

So B is the intersection of the line DZ and the parabola.

Solution 3

Let ABC be the triangle in question and O its circumcenter.

The perpendicular bisector of BC intersects the circumcircle at D,E (as in the figure).

Denote:

CD = DB =: m

EC = EB =: n

EA := e

We have:

b + 2R := k1, known (1)

c + 2R := k2, known (2)

m(b + c)= ad (3)(by Ptolemy Theorem in the cyclic quadrilateral ABDC)

The triangle BCD has known angles (DCB = DBC = A/2, CDB = 180 - A)

==> a/m := t is known.

==> b + c = (a/m)d = td

ae + cn = bn (4)(by Ptolemy Theorem in the cyclic quadrilateral ABCE) ==>
e = (b - c)n/a

We have:

b - c = k1 - k2, known

n/a is known since the triangle CEB has known angles (CEB = A, ECB = EBC = 90 - (A/2))

Therefore e = (b - c)n/a is known.

EA^2 = AD^2 - AD^2 (5)(by Pythagorean Theorem in the right triangle ADE)or e^2 = 4R^2 - d^2, known.

Now, from:

(1) and (2) ==> b + c = k1 + k2 - 4R (6)

(6) and (3) ==> d = (k1 + k2 - 4R) / t (7)

(7) and (5) ==> 4R^2 - ((k1 + k2 - 4R)/t)^2 = e^2

== > R is known.

So the problem is equivalent to construct triangle if are given A, b - c, R or A, b - c, a (the solution is left to the reader).

Exercises:

To construct triangle ABC if are given:

1. A, h_b - a, h_c + a

2. A, h_b - a, h_c - a