This is another problem from the Math15 Forum; this mathematics forum is quite well-done. ^_^
Given five fixed points A, B, C, D, E, construct a pentagon FGHIJ such that the midpoint of each side corresponds to these five fixed points.
BoJone habitually analyzes these types of problems using coordinates. As shown in the figure below, we place the five known vertices into a coordinate system. Let the side DE lie on the x-axis, with E as the origin. We define the five points as: A(a,b); B(c,d); C(e,f); D(g,0); E(0,0).
If we can determine any single vertex of the pentagon FGHIJ, then the entire pentagon can be easily constructed. Let us attempt to find the coordinates of point J. Let the coordinates of J be (x, y). Since J, A, F are collinear and JA = AF (as A is the midpoint of JF), we can determine the coordinates of F as (2a-x, 2b-y). Similarly:
The coordinates of point G are (2c-2a+x, 2d-2b+y);
The coordinates of point H are (2e-2c+2a-x, 2f-2d+2b-y);
The coordinates of point I are (2g-2e+2c-2a+x, -2f+2d-2b+y);
The coordinates of point J are (-2g+2e-2c+2a-x, 2f-2d+2b-y).
Recall that we initially set J = (x, y), so we must have: \begin{gathered} \textcolor{red}{-2g+2e-2c+2a-x = x} \\ \textcolor{red}{2f-2d+2b-y = y} \end{gathered}
Solving these, we find x = -g+e-c+a and y = f-d+b. Heavens! Constructing these quantities is incredibly simple!! It only requires adding and subtracting the known values.
The construction method is briefly described below:
1. Select ED as the x-axis, with E as the origin, and construct the y-axis.
2. Mark the projections of A, B, C onto the y-axis (relative to E).
3. On the y-axis, construct a segment equal to “the sum of the projections of A and C minus the projection of B”.
4. Mark the projections of A, B, C onto the x-axis (relative to E).
5. On the x-axis, construct a segment equal to “the sum of the projections of A and C minus the sum of the projection of B and the length ED”.
6. Using the coordinates determined in steps 3 and 5, construct point J.
7. Draw the line JA and mark point F such that AF = JA; follow similar steps for the remaining vertices until the pentagon is complete!
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