POLYA006:  Construction of triangle ABC with given A, a, and sum of altitudes h_a+h_b+h_c  (proposed by Antreas P. Hatzipolakis, 11/9/01).

Give a ruler-and-compass construction of triangle ABC,  given the magnitude of angle A,  the length a of side BC, A, and the sum h of the heights h_a, h_b, and h_c.

Discussion. [APH, 11/14/01]: We know A and a. What length X can we get from these?

[PY, 11/14/01]: By the law of sines, a/sin A = 2R, the diameter of the circumcircle. We can actually draw the circumcircle, or the part of the circle on which the vertex A lies.

[APH, 11/14/01]: We know h_a + h_b + h_c : = k. Can we replace this expression with another one in terms of triangle sides [and the length X] ?

[PY, 11/14/01]: If T is the area of the triangle, the altitudes are (2T)/a, (2T)/b, and (2T)/c. Their sum k = (2T)(1/a + 1/b + 1/c) = 2T(ab+bc+ca)/(abc).

[APH, 11/14/01]: Hmmm... "in terms of triangle sides [and the <known> length X]".  Directly:  h_a = b sin C = bc/(2R),  or 2R.h_a = bc. Similarly,   2R.h_b = ca, and  2R.h_c = ab.  ==>  ab + bc + ca = 2Rk = l^2 [known].  Indirectly from your formula and the formula: abc = 4RT.

Now, since we know the circumcircle and the side BC = a, we draw them! In that configuration, what fixed points / constant (known) lengths
do you see?

[PY, 11/14/01]: I see the fixed points B, C and the circumcenter O. The point A moves on the arc of the circle such that angle BOC = 2.angle A.

[APH]: Well... the more we look at a picture, the more we observe things on it!

            K
A

B-------M-------C

            E

Fixed point: M = midpoint of BC. If we draw the perp. bisector of BC, we get the antipodal  points K, E (fixed) and the fixed lengths: KM = f, ME = g, KE = KM + ME = 2R, BK, CK, BE = CE := m. Now, we have a number of fixed points, and a number of fixed lengths. How could we determine A, once we know that it lies on a known circle?

[PY,11/14/01]: The position of A depends on the lengths of b and c. If I know b+c, then I know how to find A, since the construction of ABC from A, a, and b+c is quite easy.

[APH, 11/14/01]: Good idea! But how could you find it?

[PY, 11/14/01]: From ab+bc+ca = 2Rk, [R = circumradius and k = given sum of heights] and the law of cosines, we have

a(b+c) + bc = 2Rk, and
(b+c)^2 - 2bc(1+cos A) = a^2.

Eliminating bc we obtain a quadratic in b+c, namely,

(b+c)^2 + 2a(1+cos A)(b+c) = a^2 + 2Rk(1+cos A).

[APH,11/14/01]: Very good! But keep in mind that usually trigonometry is not allowed!

[PY, 11/14/01]: Well, R(1+cos A) is exactly  KM in your picture, and it is not difficult to construct a segment with length a(1+cos A).

[LZ, 11/18/01]: There was an easier problem in The AMATYC Review, whose solution already appeared in the Fall/2001 issue.
Problem AL-2. (proposed by Bill Leonard, California State U., Fullerton, CA) Construct triangle ABC given a=BC, angle A, and the inradius r.
Of course, it's easier to find b+c in this case. Are there still other interesting ways of posing the similar question?

[PY, 11/18/01]: This one I know. With a and A we construct the circumcircle as above. The perpendicular bisector of BC intersects the circumcircle at E, on the arc that does NOT contain A. The incenter I is a point on the circle, center E, passing through B (and C). It is also at a given distance r from BC.  Join EI to intersect the circumcircle again at A. This gives triangle ABC. There are two possible positions for I. But they give congruent triangles ABC.

[LL, 11/20/01]: Here is a construction of ABC, given  A, a, b+c . This is item 17, page 22 in Court. I quote him.
" Let ABC be the required triangle. Produce BA and lay off AD = AC. In the isosceles triangle ACD we have angle D = ACD = BAC/2 = A/2.
Thus in the triangle BCD we know the base BC = a, the side BD = b+c, and the angle D = A/2; hence this triangle may be constructed. The vertices B, C
belong also to the required triangle, and the third vertex A is the intersection of BD with the mediator of the segment CD. Discussion. The problem is not possible, unless a < (b+c). Assuming that this condition is satisfied, we have given in the auxiliary triangle BCD the angle opposite the smaller side; hence we may have two such triangles, one, or none. From each auxiliary triangle we obtain one and only one required triangle; hence the problem may have two, one, or no solutions."
 

Bibliography. [1] N.A.Court, College Geometry, 2nd edition, 1952, Barnes and Noble.

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