LAST EDITED ON Jul-18-03 AT 08:23 PM (EST)
 
Hi, Bractals

Your conjecture is not entirely correct:
Any point inside of the deltoid (on a median or not) has 3 different area bisectors. In addition, the deltoid boundaries are not circular arcs, but segments of hyperbolas, the extended triangle sides being their asymptotes.

Draw a line from a point Q inside the DABC such that it cuts the triangle area in half.

Put the origin of the xy coordinate system into the triangle vertex C and the y-axis along the side a (i.e., through the vertex B). If the triangle "DABC is labeled counterclockwise, the y-coordinate of the vertex B is positive. If the triangle DABC is labeled clockwise, the y-coordinate of the vertex B is negative. In that case, reflect both the DABC and the point Q in the x-axis, reducing the problem to the former case. Select the unit length a = CB = 1.

To simplify the problem, we transform the general DABC to an isosceles right angle DA'B'C' by the following linear transformation (see the attached drawing):

u = x/xA
v = y - yA/xA x

http://www.cut-the-knot.org/htdocs/dcforum/User_files/3f185050569e7500.gif

Subsequently, we refrain from using the dashes for DA'B'C' and re-label it simply as DABC.

http://www.cut-the-knot.org/htdocs/dcforum/User_files/3f18506256b0d683.gif

Consider the area bisectors intersecting the line segments BMa at point P and AMb at point S as they slide on the triangle sides. Their equations are

u/uS + v/vP = 1

Since the lines are area bisectors, we have

uSvP = 1/2 or uS = 1/(2vP)

Substituting this to the bisector line equation

2vPu + v/vP = 1 or 2vP2u + v - vP = 0

This is a family of lines dependent on the parameter vP. The equation of their envelope (if it exists) is given by

F(u, v, vP) = 2vP2u + v - vP = 0
¶F(u, v, vP)/¶vP = 4vPu - 1 = 0

Eliminating the parameter vP we get the envelope equation

v = 1/(8u)

So the envelope is a segment of hyperbola with asymptotes along the sides a (v = 0) and b (u = 0) of the DABC. The equations of the medians ma, mb and their midpoints Da, Db are:

ma: 2v + u = 1, Da = (1/2, 1/4)
mb: v + 2u = 1, Db = (1/4, 1/2)

The hyperbola intersects (actually, touches - after all it is an envelope) the medians at their midpoints: Combining v = 1/(8u) and 2v + u = 1 (ma) and solving for u, v, we get u = 1/2, v = 1/4, or Da. Combining v = 1/(8u) and v + 2u = 1 (mb) and solving for u, v, we get u = 1/4, v = 1/2, or Db.

If we select a point (u0, v0) on the hyperbolic segment, then v0 = 1/(8u0). Substituting this to the general equation of the bisector

2vPu0 + v0/vP = 1
2vPu0 + 1/(8u0vP) = 1
16vP2u02 - 8u0vP + 1 = 0
(4vPu0 - 1)2 = 0
vP = 1/4u0

We have a double root - the only solution. Therefore, there is only one area bisector through the point (u0, v0) intersecting the DABC on the line segments BMa and AMb. If we select a point (u1, v1) above the hyperbolic segment, then v1 > 1/(8u1). Substituting this to the general equation of the bisector just like for the point (u0, v0) we get

(4vPu1 - 1)2 < 0

Since this is impossible, there is no area bisector through the point (u1, v1) intersecting the DABC on the line segments BMa and AMb. If we select a point (u2, v2) below the hyperbolic segment, then v2 < 1/(8u2). Denote w2 = 1 - 8u2v2 > 0. Substituting this to the general equation of the bisector just like for the point (u0, v0) we get

(4vPu2 - 1)2 = w2 > 0
vP = (1 ± w) / (4u2)

Taking into account the limits 1/2 £ vP £ 1 separately for each root

1/2 £ (1 ± w) / (4u2) £ 1
2u2 - 1 £ ± w £ 4u2 - 1

In order to square an inequality, we have to make sure that the smaller side is positive (or zero). Therefore, we can square the left inequality for the - sign root (after multiplcation by -1) and the right inequality for the + sign root. For the - sign root we get

w £ 2u2 - 1
w2 = 1 - 8u2v2 £ (2u2 - 1)2
u2 + 2v2 ³ 1

For the + sign root we get

w £ 4u2 - 1
w2 = 1 - 8u2v2 £ (4u2 - 1)2
2u2 + v2 ³ 1

Therefore, the point (u2, v2) must be above the median ma for the - sign root to be acceptable and above the median mb for the + sign root to be acceptable. Any point inside the DAGMb and any point inside the DBGMa is below the hyperbolic envelope, above one median, and below the other. Therefore, only one root is acceptable and there is only one area bisector through the point (u2, v2) intersecting the DABC on the line segments BMa and AMb. For any point below the hyperbolic segment DaDb and above both medians ma, mb, there are two such area bisectors.

By cyclic permutation of the vertices A, B, and C, we can find the hyperbolic segments DbDc and DcDa and all other results obtained for the vertex C at the origin. For any point inside the triangle pair DBGMc and DCGMb, there is only one area bisector through this point intersecting the DABC on the line segments CMb and BMc. For any point bounded by the hyperbolic segment DbDc and the medians mb, mc there are two such area bisectors. Likewise, for any point inside the triangle pair DCGMa and DAGMc, there is only one area bisector through this point intersecting the DABC on the line segments AMc and CMa. For any point bounded by the hyperbolic segment DcDa and the medians mc, ma there are two such area bisectors. Adding up the 3 families of area bisectors, for any point inside the DABC and outside of the hyperbolic "deltoid" DaDbDc, there is only one area bisector through this point. For any point inside the hyperbolic deltoid, there are 3 different area bisectors through this point. For any point on the hyperbolic deltoid boundary and not coincident with one of the deltoid vertices Da, Db, Dc, there are 2 different area bisectors through this point. At the vertices, 2 different hyperbolic segments touch the median (and therefore touch each other). The two area bisectors on the deltoid boundary, each from a different family, fuse into the single median. Therefore, there is only one area bisector through each vertex of the deltoid.