Summer Tuggle

Assignment 6.1

1.      Construct a triangle and its medians. Construct a second triangle with the three sides having the lengths of the three medians from your first triangle. Find some relationship between the two triangles. (E.g., are they congruent? similar? have same area? same perimeter? ratio of areas? ratio or perimeters?) Prove whatever you find.

Triangle ABC is the original triangle.  Line segments RB, SC, and TA are the medians of triangle ABC.

Let AT act as one side of our new triangle whose side lengths are equal to the medians of the triangle ABC.  From center T create a circle with radius SC.  Then from point T create a line that is parallel to CS.  This radius (TG) will serve as the second side of our new triangle.  Connect point A and the new point G to get the third side of the triangle.

Half of triangle AGT lies inside the original triangle ABC.  Since the medians of a triangle divide the original triangle into 6 equal parts, the yellow area below is 1/6 of triangle ABC.

A new triangle, ETB is formed.

The medians of this triangle can be constructed.  Since TK intersects the other two constructed medians at the centroid, TK is also a median.  Therefore, TK bisects EB.

The purple area, triangle EKT is half of the triangle ETB.  Triangle ETB is 1/6 of the original triangle ABC so triangle EKT is 1/12 of the original triangle ABC.

Point K and point L are midpoints of their respective sides thus dividing triangle ESB into a triangle and a trapezoid.  The triangle BLK is similar to the original triangle ESB and scaled by 1/2.  This means that the area of KLB is 1/4 of the triangle ESB and it follows that trapezoid EKLS is 3/4 of the triangle ESB (which is 1/6 of the triangle ABC).  Therefore EKLS is 1/8 of the triangle ABC.

The sum of the three colored areas (triangle ATL)

(1/6)x + (1/12)x + (1/8)x is (3/8)x where x is the area of the original triangle ABC.  Since AL is the median of triangle ATG, triangle ALG is equal in area to triangle ATL.

Therefore the area of triangle ATG is ]2(3/8)x[ which results in (3/4)x.  It can be concluded that the area of the triangle formed by the medians of another triangle is 3/4 of the area of the original triangle.

(Dr. Mitch Rothstein assisted in solving this problem)