UGANDA INSTITUTION OF PROFESSIONAL ENGINEERS

17th National Technology Conference (NTC 2012)

Theme: Strategic Positioning for Positive Transformation of Society

Venue: Golf Course Hotel, Kampala Date: 14-15 June 2012

Innovative Approaches in Science and Engineering Chapter 1. Generalisation of Varignon and Wittenbauer Theorems Chapter 2. On the Stability of Structures Chapter 3. Extension of Integral Tables Chapter 4. Conjugate Beam Analogy. Presented by: Eng Dr. Francis Baziraake UIPE Vice President (Civil) Chairman Membership, Education and Training Committee

THE THEOREMS OF VARIGNON AND WITTENBAUER http://gogeometry.com/geometry/classical_theorems_index.html

GENERALIZATION OF VARIGNON AND WITTENBAUER

S

O A K1

K K2

D H1

J1

L H P J H2 N R J2

B I1 I I2 C M Q

Given: Quad ABCD Line Bisection point

2 sections: α=½=0.5 Trisection point 3 sections: α= ¹/3=0.33

AB H H1, H2

BC I I1,I2

CD J J1,J2

DA K K1,K2

1. Divide the sides of a quadrilateral into two equal parts. The figure formed by connecting the adjacent on either side of a polygon vertex is a parallelogram known as Varignon’s parallelogram.

2. Divide the sides of a quadrilateral into three equal parts. The figure formed by

connecting and by extending adjacent on either side of a polygon vertex is a parallelogram known as Wittenbauer’s parallelogram

3. Divide the sides of a quadrilateral into n equal parts. The figure formed by

connecting and by need extending adjacent on either side of a polygon vertex is a parallelogram known as generalised Varignon- Wittenbauer parallelograms.

Note that the points need not necessarily lie on the parallelograms! Proof A K

D

H J B C I AHK is similar to ABD due to similarity with factor 1:2 since H and K are bisectors. Since AH and AB, AK and AD are collinear, therefore HK is parallel to and has a length half that of BD which is the diagonal of the quadrilateral. Therefore area of AHK=0.52=0.25 ABD Similarly for IJ is parallel to and has a length half that of BD and area of CJI=0.52=0.25 CDB. We also note therefore that HK is equal and parallel to IJ.

Since ABD +CDB=ABCD and AHK+CJI=0.25 ABCD A

K D H J B C I By similar reasoning HI is parallel to KJ and equal in length. Area BIH +DAC=0.25 ABCD We therefore conclude HIJK is a parallelogram. Area of the triangles outside the parallelogram AHK+CJI+BIH+DAC=0.5 ABCD Case 2 General

K A h α g a l D k f j H J b α i m e B C c d

Each side of the quadrilateral is divided into n equal parts. 1/n= α Take for exampleABD. Aa/AB=Ah/AD= α. Triagles Aah and ABC are similar with a α as factor of similarity. Also Bb/AB=Dg/AD= α Therefore ah //BD (BD is diagonal of quadrilateral) or HK//BD Similarly de//BD or IJ//BD bj=(1- α)Bj jk=(1- α)jD Addition of the two equation gives: bk=(1- α)BD Therefore EH=(1- α)BD Similarly HI//AC//KJ and HI=(1- α)AC Since the opposite sides of EFGH are parallel, EFGH is a parallelogram. Area of parallelogram HK*KJ sin HKJ (1- α)2 *BD*AC sin HKJ = (1- α)2λ where λ= BD*AC sin HKJ For α=0.5 area of //gram is (1-0.5)2 λ =0.25 λ But from special case α=0.5: Area of //gram =0.5σ where σ is area of quadrilateral 0.25 λ=0.5σ λ=2 σ Area of parallelogram 0.25 λ=0.5 σ i.e half area of quad. For α=1/3=0.333 area of //gram is (1- 1/3)² *2 σ= 8/9 σ which is Wittenbauer For α=0, area of //gram is (1-0)² *2 σ =2 σ or twice area of quadrilateral Special cases Case α=0.5: H,h,g coincide