**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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