Archimedean Twins
In the figure there are three semicircles of diameters A B , A C , B C and two circles S 1 and S 2 which are touching to the line C F ( C F is perpendicular to A B ). Both circles have same radius and are called Archimedean Twins. Now if A B = 1 0 , A C = 6 and B C = 4 then find the radius of circle S 1 .
The answer is 1.2.
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4 solutions
Centers are, M for pink, O for blue, P for outer.
T is the foot of the perpendicular line.
M
T
=
3
.
P
T
=
1
.
r is the radius of the blue circle, ON is the perpendicular from O
on AB, the diameter of the outer circle.
S
o
N
T
=
r
.
a
n
d
N
P
=
r
−
1
…
.
(
1
)
,
M
N
=
M
T
−
r
=
3
−
r
.
.
(
2
)
Distance between centers are
O
M
=
3
+
r
,
O
P
=
5
−
r
.
.
.
.
.
.
.
(
3
)
F
r
o
m
(
3
)
,
(
1
)
I
n
△
O
N
P
,
O
N
2
=
O
P
2
−
N
P
2
=
(
5
−
r
)
2
−
(
r
−
1
)
2
F
r
o
m
(
3
)
(
2
)
I
n
△
O
M
N
,
O
N
2
=
O
M
2
−
M
N
2
=
(
3
+
r
)
2
−
(
3
−
r
)
2
∴
(
5
−
r
)
2
−
(
r
−
1
)
2
=
(
3
+
r
)
2
−
(
3
−
r
)
2
.
S
o
l
v
i
n
g
r
=
1
.
2
Beautiful solution! Sir, I have uploaded an image with your point labels. Just append ![Imgur] on left of this and insert in your solution (https://i.imgur.com/hDMuZKb.png)
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Thanks a lot, Ujjwal. How do you do it. We can come to the same conclusion if the right side semicircle is used. That is better since we do not have N and P so near.
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My pleasure sir! I used the CAD software - MicroStation - to create that. As for the problem, out of curiosity I solved it symbolically, taking outer circle R, inner circles r1, r2 and the Archimedian circle as r. Did the derivation pretty much like you did and the result was r = R r 1 r 2
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@Ujjwal Rane – Yes. This is the formula given for Archimedean circle . You may give the proof so that others can know. That is what I have requested Rifath Rahman.
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@Niranjan Khanderia – Yes, in fact i was in the process of writing it. Only not sure if I should post it as a response to this question or a separate Note. Probably a Note could reach more users?
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@Ujjwal Rane – I agree. Note would better serve our purpose.
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@Niranjan Khanderia – Posted a note on Proof of expression for radius of Archimedes's twin circles
Sir, could you please review and give your suggestions?
Can you tell me how you know N is to the left of P? why can't it be to the right of P? thanks!!!
where r is the radius of the circle.
awesome way to solve the prob.
I too was thinking of giving alternate solution from the right side. Can you please give how you generated the file you have down loaded ? I am not able to get this size from M.S. Accessories-Paint. Thanks.
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I have used a drawing software named Edraw. Which is a very good multi designing software. I will suggest you to use it.
Let a=AC=diameter of first inner semi-circle,b=diameter of second inner semi-circle and diameter of Arcimedean circle be d,we know that d={(ab)/(a+b)}={(6 *4)/(6+4)}=24/10=2.4,so radius of Archimedean circle is 2.4/2=1.2
It is much better than mine. We can use Archimedean circle formula dirctly as you have done. But a proof would be welcome.
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Actually it came from the way you did it,you can see it in detail in wikipedia,search Archimedian twin circle
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Thanks. I did see it before I wrote my comment. I thought it would help others.
The radius r of Archimedean twin circles inscribed by three semicircles of radii a , b & a + b & having shared diameter, is given by the general formula
r = a + b a b
As per given question, setting a = 3 & b = 2 in above general formula , radius of twin circles
r = 3 + 2 3 ⋅ 2 = 5 6 = 1 . 2