LAST EDITED ON Jul-23-03 AT 05:02 PM (EST)
 
Hi, Bractals

I can follow your construction now. Your analysis is an excellent post. I am also presenting my (independent) analysis. It melts down to the same thing - solving a quadratic equation.

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

The extended medians of a triangle divide the plane outside the triangle into 6 segments (see the attached drawing). If the point Q is on one of the extended medians, the problem is solved. Suppose the point Q is between the extended medians ma and mb. 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.

http://www.cut-the-knot.org/htdocs/dcforum/User_files/3f171aa13433dd89.gif

The line cutting the triangle area in half intersect the triangle in 2 points: P (on the triangle side a = CB) and S (on the triangle side b = CA). The equation of the line QP is

y - yP = (yQ - yP) / (xQ - xP) * (x - xP) = (yQ - yP) / xQ * x

because xP = 0. Since the points Q, P, S are colinear (on a line), we have the following equation for their coordinates:

(1) yS - yP = (yQ - yP) / xQ * xS

Since the point S and the triangle vertex A are on a line through the coordinate origin,

(2) tan(g) = yA/xA = yS/xS

Finally, since the area of DSPC is half of the area of DABC

1/2 CS * CP sin(g) = 1/2 (1/2 CA * CB sin(g))

CS * CP = 1/2 CA * CB

Writing this using the xy coordinates of the points P, S, and A and using the unit length CB = 1, we get

(3) yP Ö(xS2 + yS2) = 1/2 Ö(xA2 + yA2)

We have 3 equations for 3 unknowns yP, yS, and xS. Simplifying the 3rd equation using the 2nd equation:

yP Ö(xS2 + yS2) = yP.xS Ö(1 + (yS/xS)2) = yP xS Ö(1 + (yA/xA)2) = yP xS/xA Ö(xA2 + yA2) = 1/2 Ö(xA2 + yA2)

yP xS/xA = 1/2

Combining this with the 2nd equation we get

xS = xA/(2yP)
yS = xS yA/xA = xA/(2yP) yA/xA = yA/(2yP)

Substituting to the 1st equation we get the following quadratic equation

yP2 - xA/(2xQ) yP - xA/2 (yA/xA - yQ/xQ) = 0

The equations of medians ma = AGa and mb = BGb are

ma: y = (yA - 1/2) * x/xA + 1/2
mb: y = (yA - 1) * x/xA + 1

Since the point Q is between these 2 medians,

(yA - 1/2) * xQ/xA + 1/2 < yQ < (yA - 1) * xQ/xA + 1

Since xQ < 0

-(yA - 1/2)/xA - 1/(2xQ) < -yQ/xQ < -(yA - 1)/xA - 1/xQ

1/(2xA) - 1/(2xQ) < yA/xA - yQ/xQ < 1/xA - 1/xQ

1/4 - xA/(4xQ) < xA/2 (yQ/xQ - yA/xA) < 1/2 - xA/(2xQ)

Again, since xQ < 0 and xA > 0, both the left and right side of this inequality are positive and we can denote

u = -xA/(2xQ) > 0

v2 = xA/2 (yQ/xQ - yA/xA) > 0

and write our quadratic equation as

yP2 + uyP - v2 = 0

The solution is yP = -u/2 ± Ö(u2/4 + v2). The root with minus sign before the square root is negative and cannot be on the side a = CB of our DABC.

To perform the construction, we have to construct the lengths u > 0 and v > 0. The square root in the formula for yP will be a hypothenuse of a right triangle with sides u/2 and v. Finaly, we will subtract u/2 from the length of this hypothenuse to get the length of yP (see the attached drawing).

http://www.cut-the-knot.org/htdocs/dcforum/User_files/3f171ae93521d184.gif

1. Draw a unit circle (radius a = CB = 1) with the center at the origin C. The circle intersects the x-axis in points D and E.

2. Erect normals to the x-axis at the points D and E. These normals intersect the lines CA and CQ at points D' and E', respectively. This gives us DD' = yA/xA and EE' = -yQ/xQ (remember, xQ < 0).

3. Add these two lengths: Drop a normal to the y-axis from the point E'. Denote the heel of the normal H. Connect the points D and H. Draw a line through the point D' parallel to DH. The line intersects the y-axis at point H'. This gives us CH' = yA/xA - yQ/xQ.

4. Multiply this length by xA: Drop a normal from the point A to the x-axis, denote A' the heel of this normal, CA' = xA. Connect points D and H'. Draw a line parallel to DH' through the point A'. The line intersects the y-axis at point K. This gives as CK = xA (yA/xA - yQ/xQ).

5. Divide the line segment CK in half. This gives as CL = xA/2 (yA/xA - yQ/xQ).

6. Find the square root of this distance: The unit circle intersects the negative portion of the y-axis opposite to the point L at point F. Find the center of the line segment LF - point M. Construct a circle with center M and radius ML = MF. The circle intrsects the y-axis at point N. This gives us v = CN = Ö{xA/2 (yA/xA - yQ/xQ)}.

7. Divide the length xA by the length xQ: Copy the length of the line segment xA = CA' to the negative portion of the y-axis (the 1st quadrant is already too messy), i.e., CR = CA'. Connect points Q' and R. Draw a line through the point E parallel to Q'R. The line intersects the y-axis at the point R'. This gives us CR' = -xA/xQ (remember, xQ < 0).

8. Divide the line segment CR' in four. This gives as u/2 = CT = -xA/(4xQ).

9. The distance TN is the hypotenuse of the right DNTC with sides u/2 = CT and v = CN, i.e., TN = Ö(u2/4 + v2).

10. Find the roots of our quadratic equation by substracting the length u/2 from the hypotenuse or adding the length u/2 to the hypotenuse: Draw a circle with center at the point T and radius TN. The circle intersects the y-axis at points P and P'. CP = TN - CT = Ö(u2/4 + v2) - u/2 and -CP' = -(TN + CT) = -{Ö(u2/4 + v2) + u/2} are the roots of our quadratic equation.

11. Only the first root is acceptable, yP = CP. Connect the points Q and P and extend the line through the DABC. QP is the desired line.

If the point Q is inside the triangle, the analysis and construction are the same as if the point is outside, with one exeption. We cannot trivially determine which two sides of the triangle the area bisector intersects, so that we have to try all 3 pairs. Aparently, when the point Q is close to one side, there is not much difference from the case when the point Q is outside close to the same side. However, when the point Q is deep inside the triangle, there might be 3 different area bisectors through a given point. I will analyze this in the next post.