Golden Ratio in an Irregular Pentagon
Here is a curious problem with a surprising solution [Prasolov, #4.9, p. 77, pp. 85-86]:
| |
In a pentagon ABCDE the diagonals cut off triangles of area 1. Find the area of the pentagon.
|
(I thank Bui Quang Tuan for bringing this reference to my attention.)
References
- V. V. Prasolov, Problems in Planimetry, v 1, Nauka, Moscow, 1986 (Russian)
Solution
Copyright © 1996-2009 Alexander Bogomolny
Golden Ratio in an Irregular Pentagon
Pentagon ABCDE has the diagonals parallel to the sides. In addition, the points of intersection of the diagonals divide each in the golden ratio.
First note that, since triangles ABC and ABE have the same area, the altitudes from C and E to AB must be equal so that CE||AB. The same holds for other pairs side/diagonal.
Let M be the point of intersection of BD and CE. Then ABME is a parallelogram and ΔABE = ΔMEB. Denote Area(BCM) = x. Then
| |
Area(ABCDE) = Area(ABE) + Area(MEB) + Area(DEC) + x = 3 + x.
|
On the other hand,
| (*) |
Area(BCM) / Area(CDM) = BM / DM = Area(MEB) / Area(DEM),
|
Observe that, since BCDE is a trapezoid, Area(DEM) = Area(BCM) = x. So that, in terms of x, (*) looks like
which gives an equation for x: x² + x - 1 = 0. It follows that x = (√5 - 1)/2 implying BM/DM = φ, the golden ratio. It follows that
| |
Area(ABCDE) = 3 + (√5 - 1)/2 = (√5 + 5)/2.
|
Bui Quang Tuan gave a practical Euclidean construction of such pentagons. He also observed that the pentagon formed by the diagonals is of the same kind. We may add that, if the sides of the pentagon are extended to form a star whose vertices constitute another such pentagon.
Finally, there is an additional construction of those pentagons.
Fibonacci Numbers
- Ceva's Theorem: A Matter of Appreciation
- When the Counting Gets Tough, the Tough Count on Mathematics
- I. Sharygin's Problem of Criminal Ministers
- Single Pile Games
- Take-Away Games>
- Number 8 Is Interesting
- Curry's Paradox
- A Problem in Checker-Jumping
- Fibonacci's Quickies
Golden Ratio
- Golden Ratio in Geometry
- Golden Ratio in an Irregular Pentagon
- Golden Ratio in a Irregular Pentagon II
- Inflection Points of Fourth Degree Polynomials
- Wythoff's Nim
- Inscribing a regular pentagon in a circle – and proving it
- Cosine of 36 degrees
- Continued Fractions
Copyright © 1996-2009 Alexander Bogomolny
|