aa r X i v : . [ m a t h . HO ] J a n Bisecting a triangle in a given direction
Robin Whitty * January 20, 2021
Abstract
Given a triangle, what is the equation of the line which bisects its area and has a givenslope? The set of all lines bisecting the area of a triangle has been elegantly determined as a certain‘deltoid’ envelope and this gives an indirect method of solution. We find that vector algebra allowsthe equation to be written down rather directly and neatly.
Keywords: triangle geometry, computational geometry, bisector, vector algebraSuppose that a triangle is specified by the coordinates of its vertices. A slope is given, and itis required to write down the equation of the line having this slope and bisecting the area ofthe triangle. In [1], Dunn and Pretty determine all area bisectors of a triangle in terms of adeltoid-shaped envelope: the line we require will be the unique tangent to this envelope havingthe given slope. This is very elegant and has inspired some attractive developments, notablyin work of Berele and Catoiu: see [2], for example, where much more is recorded aboutDunn and Pretty’s deltoid. However, it is not the most direct way to approach our problem,particularly because Dunn and Pretty reduce the area bisection question ‘a ffi nely’ to bisectinga right triangle placed at the origin.Figure 1: Tilting (anticlockwise) from a bisecting line parallel to edge b to create a bisector inthe direction of u .We take a di ff erent approach, starting with the bisecting line which lies parallel to a chosentriangle edge and tilting it to the required angle. This can be specified quite neatly in terms ofvector equations which solve to give that point on the bisecting line through which our desiredline must pass. The equation of this line is still not written down in an entirely direct mannerbecause we must choose which triangle edge to start from — e ff ectively, which Dunn andPretty deltoid edge we are tangent to — but this choice turns out to be expressed succinctly interms of solutions to pairs of simultaneous equations. * / √ − / √
2. In the figure this is depicted for edge (vector) b and we will refer to it as the b -bisect. Writing m for 1 / √
2, the b -bisect is m b as a free vector. The vector whose directionis u will correspond to a new bisector of ABC if scalars t and s are such that the triangle withsides tm b and s u has the same area as the triangle with sides (1 − t ) m b and s ′ u , both triangleshaving the same included angle α :12 | s u | × | tm b | sin α = | s ′ u | × | (1 − t ) m b | sin α, so that s ′ = t − t s . (1)Our aim is to find t since this will specify point X in figure 1, as a position vector, and hencedetermine the equation of our bisecting line.For the moment we are assuming b is the right choice of edge direction from which to tiltto give the required direction u . In fact, our approach will be to specify the collection of allvectors for which b is indeed the right choice. Since for our problem the orientation of thesevectors is immaterial we will specify that they follow the orientation of b (so, right to left infigure 1, even though the depicted vector u is orientated left to right).Now we cannot tilt away from the b -bisect clockwise beyond the median line from vertex C to the midpoint of c , which trivially bisects the triangle area. As a vector this has direction b − a . Nor can we tilt anticlockwise beyond the median line from the midpoint of a to A ; asa vector this has direction b − c . Then the collection of all free vectors between one mediandirection or the other, following the orientation of b , is given as u b = b − a + w ( a − c ) , w ∈ [0 , . We remark that w = / b − ( c + a ) = b , precisely the direction of the b -bisect.Similarly the collection of vector directions which are valid bisecting tilts from the a -bisectand c -bisect are, respectively, u a = a − c + w ( c − b ) , u c = c − b + w ( b − a ) , again with w ∈ [0 , u , exactly one of the equations in w and x , u x = u a , u x = u b , u x = u c (2)will solve for w in [0 , u is in the direction of one of the triangle medians, in whichcase there will be two solutions, corresponding to w = w =
1. The sign of x will indicatewhether the orientation of u is consistent with our ‘right-to-left’ definitions of u a , u b and u c .We return to these equations subsequently. But now we derive two vector equations fromfigure 1 which will allow us to solve for t , eliminating s . Without loss of generality we willagain assume that u b is the appropriate range of directions within which (up to orientation) liesvector u . From figure 1 we see that, for the tilted line to form triangles with the edges c and a of ABC , the values of s and t must create resultant vectors in the direction of these edges. Weexpress this as two orthogonality conditions:( − tm b − s u b ) · c ⊥ = (cid:0) (1 − t ) m b + s ′ u b (cid:1) · a ⊥ = , x ⊥ , for a vector x , denotes the perpendicular vector, whose dot product with x is zero.Substituting for u b and expanding out gives tm b · c ⊥ + s b · c ⊥ + s ( w − a · c ⊥ = − t ) m b · a ⊥ + s ′ b · a ⊥ − s ′ w c · a ⊥ = , each of which may be factored using a single dot product by virtue of the fact that a + b + c = tm + s − s ( w − b · c ⊥ = (cid:0) (1 − t ) m + s ′ + s ′ w (cid:1) b · a ⊥ = . Since no triangle edge can be orthogonal to the perpendicular of another this gives tm + s (2 − w ) = , (1 − t ) m + s ′ (1 + w ) = . We combine these two equations, using the substitution s ′ = st / (1 − t ) from equation 1, toarrive at t / (1 − t ) = (2 − w ) / (1 + w ). Solving for t gives t = (cid:16) ± √ (1 + w ) / (2 − w ) (cid:17) − , fromwhich we discard the negative square root since this gives a negative value of t at w = t = + r + w − w − . (3)The values of t , we observe, vary from 2 − √ √ −
1, as w goes from 0 to 1. Theseextreme points mark precisely the intersections of the a , b and c -bisects.It remains to determine whether the value of t identifies a point on the b -bisect, as in figure 1,or on one of the other two edge-bisects. To do this we must return to equations 2. Solving theseequations will identify an edge-bisect by giving a value of w in [0 ,
1] which can be thereafterbe substituted into equation 3. Now equations 2 are simultaneous equations in two variables, x and w , which can be written down and solved explicitly by coordinatising a , b and c . Thesolutions may be expressed tidily by defining a function of three vector variables: P ( x , y , z ) = x · y ⊥ z · y ⊥ z · x ⊥ ! , unless x · y ⊥ = , in which case the function is undefined.The value of this function, if it is defined, is a column vector whose entries are x and w . If itis not defined it is because x · y ⊥ (which is the determinant of the matrix of our simultaneousequations) is zero, i.e. x is parallel to y .We proceed thus:For u x = u a = a − c + w ( c − b ) compute P ( u , c − b , a − c ) , For u x = u b = b − a + w ( a − c ) compute P ( u , a − c , b − a ) , (4)For u x = u c = c − b + w ( b − a ) compute P ( u , b − a , c − b ) . Suppose that all three computations are defined, giving column vectors ( x a , w a ), ( x b , w b ) and( x c , w c ), say. Then there will be a unique value of { w a , w b , w c } lying in [0 , t be the valueof equation 3 when w takes this value. Then our required equation is u x + −→ C + (1 − m ) b − tm a if w a ∈ [0 , −→ A + (1 − m ) c − tm b if w b ∈ [0 , −→ B + (1 − m ) a − tm c if w c ∈ [0 , , x ∈ ( −∞ , ∞ ) . (5)3uppose, on the other hand, that, say, P ( u , a − c , b − a ) is undefined. Then u is parallel to a − c .But a − c is the median from B to the midpoint of b . Then we may place t on the a -bisect,setting w a =
0, or on the c -bisect setting w c =
1, and again specify the line equation from 5.We conclude with an example. Our triangle will be A = (4 , , B = (1 , , C = (10 , a = − B + C = (9 , − b = − C + A = ( − ,
1) and c = − A + B = ( − , b − a = ( − , c − b = (3 , a − c = (12 , − ff erent slope vectors u , as follows: u = ( − , x , w ) computed at (4) are, to two decimal places,( − . , . . , − .
14) and (39 . , . u a value of w liesin [0 , a -bisect specifiedby t (equation 3) at w = .
88, which is t = . . This is plotted in figure 2 as the dashedline meeting the vertical axis at approximately 7 . u = (3 , . , . − . , .
83) and (2 . , − . u b is selected. Again, to two decimal places, a t value of 0 .
44 is obtained andwe may plot the dashed bisecting line meeting the vertical axis at approximately 0 . u = ( − , − , , u b computation is undefined, u = ( − ,
5) is parallel to the median from B to the midpointof b . The other two computations give w = w =
1, as expected.4 eferences [1] Dunn, J.A. and Pretty, J.E., “Halving a triangle”,
Math. Gaz.
Vol. 56, No. 396, 1972, pp.105-–108.[2] Berele, A. and Catoiu, S., “Bisecting the perimeter of a triangle”,