Tuesday, December 16, 2008

Archimedes' Book of Lemmas, Proposition #12

Problem 652: Diameter, chords, perpendicular, tangents
Exercise your brain. Archimedes wrote the "Book of Lemmas" more than 2200 years ago. Solve the proposition #12 (high school level) and lift up your geometry skills.

Click the figure below to see the complete problem 652

gogeometry.com/ArchBooLem12.htm

1. http://img193.imageshack.us/img193/4671/problem652.png

Connect OC , DE
AD cut BF at L ( see picture)
Quadrilateral LDFE is cyclic with LF as diameter ( Angles D=E = 90)
Note that angle(DAE)= angle (CDE)
So angle (DLE)= angle (DCO) ( angles complement to congruence angles)
OC is perpendicular bisector of DE . On OC only center of circle LDFE have property (DLE)= ½ (DCE)
So C is the center of circle LDFE and diameter LF pass through center C
In Triangle ABL , F is the orthocenter so line LCF will perpendicular to AB
Peter Tran

2. Nice solution!

3. Peter Tran's solution suggests the following (perhaps easier) problem:

"Let D, E be any two points on the semicircle on AB as diameter. Let AE, BD intersect at F; AD, BE intersect at L. Prove LF is perpendicular to AB."

Pravin

4. Peter Tran's solution gives me this idea:
To construct the tangents we draw the circle with diameter CO
This cercle intersects the given semi circle at D and E
he passes trough O midpoint of AB and the altitude feet E,D, he is the nine point circle of triangle ABL
let be M the second intersection with AB
M is an altitude foot
CM perpendicular to AB (CO diameter)
hence CM is an altitude and passes trough orthocenter F and M=G
FC perpendicular to AB

5. Let AD, BE meet at M. MF is obviously perpendicular to AB

Let DC meet MF at S and EC meet MF at T

< SDM = < SMD = 90 - A so DS = SM ....(1)
Similarly MT = TE .....(2)
DC = CE......(3)

(1) +(2) -(3) yields DS + MT - DC = MS + ET - CE which in turn yields SC+CT+TS = 0 which can only happen if C,T,S coincide

So C must lie on MF which completes our proof

Sumith Peiris
Moratuwa
Sri Lanka