## Tuesday, December 16, 2008

### Archimedes' Book of Lemmas, Proposition #11

Problem 651: Perpendicular chords and radius
Exercise your brain. Archimedes wrote the "Book of Lemmas" more than 2200 years ago. Solve the proposition #11 (high school level) and lift up your geometry skills.

Click the figure below to see the complete problem 651. gogeometry.com/ArchBooLem11.htm

1. The setting is the same as in proposition 9, and the result will help us to prove this one:

We choose the point A' on the same side of CD as A, such that arc CA'=arc AD (and consequently arc DA'=arc AC). This implies CA'=AD and DA'=AC, as congruent chords belong to congruent arcs of the same circle. Furthermore A'B=2R (as arc A'B is half a circumference).

Now we know from proposition 9, that arc A'CB is a semi-circumference, which yields a right angle: A'CB=90°. And now we can apply Pythagoras in the triangles PBC, ADP and then in A'BC:

AP^2+DP^2+BP^2+CP^2
=A'C^2+CB^2 (according to the construction of A')
=A'B^2 (Pythagoras again)
=(2R)^2=4R.
(by Thomas)

2. Draw a diameter of the circle passing through C and name the other extreme E. Now in triangles ACP & EBC, we've /_CAP=/_CEB (angles in the same sector) and /_APC=/_EBC both=90 deg. and therefore AD=BE. Now the LHS =AP^2+DP^2+BP^2+CP^2=AD^2+CB^2
=BE^2+BC^2=CE^2=4R^2. QED.
Ajit

3. Denoted the center by O.

Let M, N be the mid-points of AB, CD respectively.

AP^2 + BP^2 + CP^2 + DP^2
= (AM - PM)^2 + (AM + PM)^2 + (CN - PN)^2 + (CN + PN)^2
= 2(AM^2 + PM^2) + 2(CN^2 + PN^2)
= 2(AM^2 + PN^2) + 2(CN^2 + PM^2)
= 2 AO^2 + 2 CO^2
= 4 R^2

4. Let CP=p, BP = s, CP = p and PD = q

If O is the centre, let x = distance from O to AB and y be that from O to
CD

Then using Pythagoras we can show that

x = (q-p)/2 and y = (s-r)/2

Hence R^2 = (q-p)^2/4 + (r+s)^2/4 = (s-r)^2 + (p+q)^2/4 from which it is
easy to show that

4R^2 = p^2 + q^2 + r^2 + s^2

Sumith Peiris
Moratuwa
Sri Lanka