Problem 644: Arbelos, Archimedean Twins

Exercise your brain. Archimedes wrote the "Book of Lemmas" more than 2200 years ago. Solve the proposition #5 (high school level) and lift up your geometry skills.

Continue reading at:

gogeometry.com/ArchBooLem05.htm

## Tuesday, December 16, 2008

### Archimedes' Book of Lemmas, Proposition #5

Labels:
arbelo,
Archimedean Twins,
Archimedes,
book of lemmas,
circle,
diameter,
perpendicular

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The proof is not very elegant, but in my opinion, it is correct:

ReplyDeleteFirst, we introduce some definitions: Let O be the midpoint of AB, O_1 the midpoint of AD and O_2 the midpoint of DB. Let r_1 be the radius of the left twin and r_2 the radius of the right one. Now the feet of the perpendiculars through E and F on AB will be called H_1 and H_2, respectively. Then we will refer to EH_1 as h_1 and to EH_2 as h_2. Furthermore, we introduce q :=AD.

We can assume that CD=1 (otherwise there would be a similar figure with CD=1). We will also assume, that the left semicircle is smaller than the right one, which can always be achieved by reflection.

Then, as the triangle ABC would be right, we have the formula for the altitude AD*DB=CD=1. As we introduced q:=AD, it follows, that DB=1/q. So, the radii of the semicircles are q/2 (left) and 1/2q (right, where 1/2q always means 1/(2q)), and the big semicircle has radius q/2+1/2q. As the left semicircle and the left twin are tangent, it follows, that EO_1=q/2+r_1 and similarly FO_2=1/2q+r_2. As the points E, H_1, D and the tangency point of the left twin on CD form a rectangle, we have O_1H_1=q/2-r_1, and similarly O_2H_2=1/2q-r_2.

Then we have right triangles O_1H_1E and H_2O_2F, and Pythagoras yields

(q/2+r_1)^2=h_1^2+(q/2-r_1)^2

and (1/2q+r_2)=h_2^2+(1/2q-r_2)^2.

By expanding and simplifying, we get

h_1^2=2r_1q (1)

and h_2^2=2r_2/q (2)

We consider the right triangles OEH_1 and OFH_2 in the same way. As the twins are tangent to the big semicircle, we have the equations OE=AB/2-r_1=q/2+1/2q-r_1. Similarly, OF=q/2+1/2q-r_2. We also need the lengths OH_1=AB/2-AD+AH_1=q/2+1/2q-q+r_1=1/2q-q/2+r_1 and OH_2=AB/2-AD-DO_2=q/2+1/2q-q-r_2=1/2q-q/2-r_2. Now we apply Pythagoras to the triangles OEH_1 and OFH_2:

(q/2+1/2q-r_1)^2=h_1^2+(1/2q-q/2+r_1)^2

and (q/2+1/2q-r_2)^2=h_2^2+(1/2q-q/2-r_2)^2

in which we can replace h_1^2 and h_2^2 by the results of (1) and (2):

(q/2+1/2q-r_1)^2=2r_1q+(1/2q-q/2+r_1)^2 (3)

and (q/2+1/2q-r_2)^2=2r_2/q +(1/2q-q/2-r_2)^2 (4)

By expanding and simplifying the terms in (3) and (4) we get r_1=1/(2/q+2q) and r_2=1/(2/q+2q), which completes the proof.

As I said initially, the proof is not a beautiful one. I would like to see a better one, so post it if you find one. (by Thomas)

Let AC=2a,CB=2b, O be the midpoint of AB

ReplyDeleteLet X,O,Y, in that order, be the centers of the semicircles[denoted as (X),(O),(Y)]on AC,AB, CB.Note the radii of these are a,a + b,b resp.

Let r be the radius of cicle (E).

Look at the triangle EXO.

Let EM, EM be drawn perpendicular to AB,CD resp.

We calculate EN(=MC) in two different ways.

First note that

XE=a+t since (X),(E) touch;

NC=EM=t since (E) touches CD;

XN=XC-NC=a-t;CO=AO-AC=(a+b)-2a=b-a;

NO=NC+CO=t+b-a=b-(a-t);

Also EO=(a+b)-t=b+(a-t) since (E)touches(O)internally

Now

EN^2=EX^2-XN^2=(a+t)^2-(a-t)^2=4at;

Also

EN^2=EO^2-NO^2=[b+(a-t)]^2-[b-(a-t)]^2=4b(a-t)

Hence

at=b(a-t),t(a+b)=ab,

Radius of(E)=t=ab/(a+b)which is symmetric in a,b

By a similar argument

Radius of(F)=ab/(a+b) also

--Pravin--

(Pravin)Typo error: 6th line should read as:

ReplyDelete"Let EN, EM be drawn perpendicular to AB,CD resp."

This problem is the same as problem 295!

ReplyDeleteSee: http://gogeometry.blogspot.com/search?q=295

Let O be the centre of the semi circle of diameter AB. O1 of the semi circle of diameter AD, O2 of DB

ReplyDeleteLet the respective radii be a+b, a and b

Let the radii of circle E be e and of F be f

Consider Tr. EOO1

Writing 2 expressions using Pythagoras for the altitude from E we have

(a+e)^2 - (a-e)^2 = (a+b-e)^2 - (a+b-a (a-e))^2

Simplifying using the difference of 2 squares,

4ae = 2b(2a-2e)

Do ae + be = ab and so e = ab/(a+b)

Similarly we can prove that f = ab/(a+b)

Hence e = f

Sumith Peiris

Moratuwa

Sri Lanka

Let AB= 2R , AD= 2r

ReplyDeletePer the result of problem 635 , radius x of circle E is

X=r(R-r)/R….. (1)

Radius x1 of circle F is calculated by replacing r by R- r in (1)

X1=(R-r)(R-R+r)/R= X

So x=x1