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.

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

What if applet does not run?

Solution

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 X_7. 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.

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

What if applet does not run?

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.)

72552110