Octagon in Parallelogram
In a parallelogram ABCD, the midpoints M, N, P, Q of the sides are joined to the non-adjacent vertices. These lines form an octagon at the center of the parallelogram. Determine the area of the octagon relative to the area of the parallelogram.
What if applet does not run? |
|Activities| |Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018 Alexander Bogomolny
The applet illustrates two solutions, the second is due to Quang Tuan Bui who also came up with a third solution. A fourth solution, due to Stuart Anderson, is illustrated separately.
Both solutions utilize the fact that the lines drawn in the problem serve as medians of some triangles. For example DM is one median in ΔABD. The diagonal AC is another median in that triangle. The point of intersection X_{1} (in the applet designated by the subscript 1) divides DM in the ration
What if applet does not run? |
Solution 1
From the foregoing remarks, there are four triangles (BNX_{2}, CPX_{4}, DQX_{6}, AMX_{8}) with area 1/8 and for triangles (MBX_{1}, NCX_{3}, PDX_{5}, QAX_{7}) with area 1/12 that complement the octagon in the parallelogram. The octagon's area then is found from
1 - 4/8 - 4/12 = 1 - 1/2 - 1/3 = 1/6. |
Solution 2
It is easy to see that X_{1}X_{3}X_{5}X_{7} is a parallelogram whose area is 1/9 of the area of ABCD, which we may assume to be 1. If O is the center of ABCD (and of X_{1}X_{3}X_{5}X_{7}), then, say, the area of ΔX_{1}X_{2}X_{3} is half of the area of ΔX_{1}X_{3}O. So the areas of the four triangles surrounding X_{1}X_{3}X_{5}X_{7} add up to 4·1/4·1/9·1/2, from which the area of the octagon is
1/9 + 1/18 = 3/18 = 1/6. |
Solution 3
The midlines MP and NQ along with the diagonals AC and BD cut the parallelogram ABCD into 8 triangles each containing a triangular portion of the octagon. Consider, for example, ΔAMO and ΔX_{1}X_{2}O. Since X_{1} divides AO in the ratio
Area( ΔX_{1}X_{2}O) / Area( ΔAMO) = 1/6. |
A similar consideration applies to the remaining seven pieces of the parallelogram. Now each of the pieces measures 1/8 of the area of ABCD. Adding up all the resulting identities
(Area( ΔX_{1}X_{2}O) + Area( ΔX_{2}X_{3}O) + ... + Area( ΔX_{8}X_{1}O)) / (Area(ABCD) / 8) = 8/6. |
or,
Area(octagon) / Area(ABCD) = 1/6. |
Note
Quang Tuan Bui also observed that by redefining the points X it is possible to obtain an octagon one sixth the area of the including quadrilateral for any convex quadrilateral. To achieve that, first define O as the intersection of the midlines MP and NQ. Then let X_{2}, X_{4}, X_{6}, X_{8} be the midpoints of segments OM, ON, OP, OQ, respectively. Finally, define X_{1}, X_{3}, X_{5}, X_{7} as the intersections of QX_{2} and MX_{8}, MX_{4} and NX_{2}, NX_{6} and PX_{4}, PX_{8} and QX_{6}, respectively.
(The proof of this generalization depends on the existence of the Varignon parallelogram.)
|Activities| |Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018 Alexander Bogomolny