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.

 

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Solution

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Copyright © 1996-2017 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 X1 (in the applet designated by the subscript 1) divides DM in the ration DX1 : MX1 = 2 : 1. Similarly we treat points X3, X5 and X7. Point X2 is the midpoint of parallelogram ABNQ and, as such, divides both AN and BQ in halves. Similar observations apply to points X4, X6, and X8.

 

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Solution 1

From the foregoing remarks, there are four triangles (BNX2, CPX4, DQX6, AMX8) with area 1/8 and for triangles (MBX1, NCX3, PDX5, QAX7) 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 X1X3X5X7 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 X1X3X5X7), then, say, the area of ΔX1X2X3 is half of the area of ΔX1X3O. So the areas of the four triangles surrounding X1X3X5X7 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 ΔX1X2O. Since X1 divides AO in the ratio 2:1, the altitude from X1 to MO measures 1/3 that from A to MO. Taking into account that X2 is the midpoint of OM we conclude that

  Area( ΔX1X2O) / 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( ΔX1X2O) + Area( ΔX2X3O) + ... + Area( ΔX8X1O)) / (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 X2, X4, X6, X8 be the midpoints of segments OM, ON, OP, OQ, respectively. Finally, define X1, X3, X5, X7 as the intersections of QX2 and MX8, MX4 and NX2, NX6 and PX4, PX8 and QX6, respectively.

(The proof of this generalization depends on the existence of the Varignon parallelogram.)

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Copyright © 1996-2017 Alexander Bogomolny

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