### Carpets in a Parallelogram

Points P, Q, R slide on the sides of parallelogram ABCD. Prove that, regardless of their position, the difference between the red areas and the blue area is constant.

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Copyright © 1996-2018 Alexander BogomolnyThis is problem #53 from *Which Way Did the Bicycle Go?* where the problem is solved without mentioning The Carpets Theorem. The latter, however, is obviously relevant here and makes the proof more transparent.

Forget for a moment about points P and R and subdivisions due to their presence. Let RED be the area of two triangles ADQ and BCQ. Clearly

(1)

Area(RST) + BLUE + Area(PUV) = Area(ABW).

What we have to show (see that the Hint box is unchecked) is that

[RED - Area(RST) - Area(PUV) + Area(ABW)] - BLUE

is independent of the position of the points P, Q, R. But in view of (1)

RED - Area(RST) - Area(PUV) + Area(ABW) - BLUE = RED.

Q.E.D.

### References

- J. Konhauser, D. Velleman, S. Wagon,
*Which Way Did the Bicycle Go?*, MAA, 1996, #52

### Carpets Theorem

- The Carpets Theorem
- Carpets in a Parallelogram
- Carpets in a Quadrilateral
- Carpets in a Quadrilateral II
- Square Root of 2 is Irrational
- Carpets Theorem With Parallelograms
- Two Rectangles in a Rectangle
- Bisection of Yin and Yang
- Carpets in Hexagon
- Round Carpets
- A Property of Semicircles
- Carpets in Triangle
- Carpets in Triangle, II
- Carpets in Right Triangle
- Piecewise Carpets in Parallelogram

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