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Found answer to 3 inscribed tangent circles inside another circle, but solution used trig. This question is about 2 inscribed tangent circles ON THE DIAMETER OF THE LARGEST CIRCLE, and this GRE question cannot use trig as a solution. The answer seems intuitive, but I can't explain it.

"Three circles with centers on line segment PQ are tangent at points P, Q, R, where point R lies on line segment PQ." (PQ is diameter of largest circle.)

Which is greater (or equal)? Quantity A: Circumference of largest circle? Quantity B: Sum of circumference of 2 smaller inscribed circles?

Set D=10 for larger circle. Makes r=5 (becomes diameter of smaller circles). Qty A: $C=\pi D$ $C=10 \pi = 31.42$ (Circumference of larg circle) Qty B: Sum of $\pi (d)$ (small diameter) = 2[(Pd)(d)] = 31.42 Answer is: QtyA = QtyB

  1. IS THERE A RULE IN GEOMETRY about tangent circles on the Diameter of circle? I haven't found one.
  2. Am I missing something? Thanks. Charlie
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  • $\begingroup$ The circumference of each circle is proportional to the diameters. No need to calculate circumferences. Just work with the diameters. PR + RQ = PQ $\endgroup$ – Doug M Sep 21 '17 at 16:05
  • $\begingroup$ Don't try to dig where there is no earth ... No underlying "big theory"... $\endgroup$ – Jean Marie Sep 21 '17 at 16:14
  • $\begingroup$ But why do you assume that the the small circles have equal radii ? The quantities will be equal even if these circles have unequal radii. $\endgroup$ – Jean Marie Sep 21 '17 at 16:17
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"Inscribed" is the key word. You necessarily have despite $d_1$ and $d_2$ can vary a lot $$D=d_1+d_2$$ where $D,d_1,d_2$ are the diameters. Thus the equivalent equality $$D\pi=d_1\pi+d_2\pi$$ Consequently the sums are equal.

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The main thing here is not that the two smaller circles are on the diameter of the larger circle and tangent with each other and the larger circle. That's just to let you know you have two circles, the sum of whose diameters equals the diameter of a larger circle. Since you're asked to compare the circumference of the larger with the sum of the circumferences of the two smaller circles, and since a circle's circumference is proportional to its diameter, you have$$\frac{C}{c_1}=\frac{D}{d_1}$$from which, by separation of ratios $$\frac{C}{C-c_1}=\frac{D}{D-d_1}$$ But $$D-d_1=d_2$$ Therefore $$C-c_1=c_2$$ Hence the circumference of the larger circle equals the sum of the circumferences of the two smaller circles

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  • $\begingroup$ As I just said to the OP : the text doesn't say that the two small circles have equal radii ! $\endgroup$ – Jean Marie Sep 21 '17 at 16:19
  • $\begingroup$ I realized that and have tried to correct. $\endgroup$ – Edward Porcella Sep 21 '17 at 16:33
  • $\begingroup$ Thanks. It just takes a thought process and there's no stated axiom, although both comments come close to doing that. Thanks again. $\endgroup$ – CBowles Sep 21 '17 at 16:42
  • $\begingroup$ True about the statement, but there's a diagram I didn't know how to upload and it shows the 2 small circles tangent at point Q, which is the center point of large diameter. $\endgroup$ – CBowles Sep 21 '17 at 16:51
  • $\begingroup$ @CBowles No, the small circles do not need to be tangent at the midpoint of the diameter. Once again, they do not need to be with a same size. $\endgroup$ – Jean Marie Sep 21 '17 at 16:54

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