Department of Mathematics Education
Problem 1. If a triangle ABC is given and a random point P on the triangle is selected, construct two lines through P to divide the triangle into three regions with equal area.
Problem 2. Given a triangle ABC, find a point D such that line segments AD, BD, and CD trisect the area of the triangle into three regions with equal areas. Define D and prove that the triangle is divided into three regions of equal area. Show a construction for finding D.
Problem 3. Given a triangle ABC, and given a point E. such that line segments AE, BE, and CE trisect the area of the triangle into three regions with equal areas. Show a construction and prove that it divides the triangle into three regions of equal area.
Problem 4. Given a triangle ABC. Construct two line segments parallel to the base BC to divide the triangle into three regions with equal areas. Prove that the construction divides the triangle into three regions of equal area.
If there is given a point P on the side of the triangle, then
there are seven ways to construct three equal areas with lines
drawn from P. The sequence below shows the seven patterns with
P moving from right to left.
Click here to see a GSP animation
of this problem.
In the first case, P is the base vertex on the right hand side
of the triangle, and in the seventh case, P is the base vertex
on the left hand side of the triangle. The construction in each
case involves trisecting the segment on the opposite side of the
triangle and connecting the vertex by segments to the two trisection
points. Since we know at least five ways to construct the trisection
of a line, the construction task for these two cases can be assumed.
Much needs to be discovered in the other cases.
For the third and the fifth cases, P is located such that a segment
drawn to the opposite vertex determines a triangle that is one-third
the area of the original triangle. This means P is located at
a trisection point on the base of the triangle. But then the remaining
two-thirds of the original triangle must be divided into two equal
areas by a line from P through the opposite side, making two triangles.
These triangles have the same height, so the point must be the
midpoint of that side. That is, if P is a trisection point of
the base, then the original triangle is divided into three equal
areas by lines from P to the vertex and from P to the midpoint
of the side along the two-thirds section.
This leaves three constructions to be determined, the second,
fourth, and sixth in the sequence above. These correspond to the
cases where P is neither a vertex nor a trisection point on the
base. In the second case, P is in the right third of the base;
in the fourth case, P is in the center third of the base; and
in the sixth case, P is in the left third of the base. Clearly,
the constructions for the second and sixth cases will be virtually
the same.
Each case could be constructed by using the result of a slightly
different construction. That is,
Given a triangle ABC with a given point on the base that is neither a vertex nor a trisection point, construct a line through P that cuts off a triangle one third the area of ABC.
Note that if we know this construction, we could use it to construct
the shaded triangles in cases two, four, and six. Case four would
be finished. Cases two and six could be completed by deteriming
another triangle with the same length base opposite of P.
For triangle ABC, construct the altitude and find its trisection
points. Again we know at least five different constructions for
trisecting a line segment. The point P is given on base AC and
is neither a vertex nor a trisection point of AC.
Determine which end of the base if nearest to point P. Construct
a perpendiclar to AC at that end. Construct a parallel line to
AC through the nearest trisection point on the altitude and construct
a perpendicular segment from P to this line at Q.
The areas of triangle AQC is one third the area of triangle
ABC.
Extend line AQ to intersect with the perpendicular to C at
point D.
Triangle APD has the same area as triangle AQC because AP/AC
= PQ/ CD and therefore (AP)(CD) = (PQ)(AC). Each side of this
latter equation is twice the area of the respective triangles.
Now any point on a line parallel to AC through D could be the
vertex of a triangle with base AP that has area one-third the
area triangle ABC. The intersection point with side AB will determine
a vertex E of the desired triangle. Draw line PE.
