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June 18, 2020

Circle Problem | AMC 10A, 2006 | Problem 23

Try this beautiful problem from Geometry: Circle

Circle Problem - AMC-10A, 2006- Problem 23

Circles with centers $A$ and $B$ have radii 3 and 8 , respectively. A common internal tangent intersects the circles at $C$ and $D$, respectively. Lines $A B$ and $C D$ intersect at $E,$ and $A E=5 .$ What is $C D ?$



  • $13$
  • $\frac{44}{3} $
  • $\sqrt{221}$
  • $\sqrt{255}$
  • \(\frac{55}{3}\)

Key Concepts




Check the Answer

Answer: $ \frac{44}{3}$

AMC-10 (2006) Problem 23

Pre College Mathematics

Try with Hints

Circle Problem

Given that Circles with centers $A$ and $B$ have radii 3 and 8 and $A E=5 .$.we have to find out \(CD\).So join \(BC\) and \(AD\).then clearly \(\triangle BCE\) and \(\triangle ADE\) are Right-Triangle(as \(CD\) is the common tangent ).Now \(\triangle BCE\) and \(\triangle ADE\) are similar.Can you proof \(\triangle BCE\) and \(\triangle ADE\)?

Can you now finish the problem ..........

Circle Problem

$\angle A E D$ and $\angle B E C$ are vertical angles so they are congruent, as are angles $\angle A D E$ and $\angle B C E$ (both are right angles because the radius and tangent line at a point on a circle are always perpendicular). Thus, $\triangle A C E \sim \triangle B D E$.

By the Pythagorean Theorem, line segment \(DE=4\)

Therefore from the similarity we can say that \(\frac{D E}{A D}=\frac{C E}{B C} \Rightarrow \frac{4}{3}=\frac{C E}{8}\) .

Therefore \(C E=\frac{32}{3}\)

can you finish the problem........

Therefore \(CD=CE+DE=4+\frac{32}{3}=\frac{44}{3}\)

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One comment on “Circle Problem | AMC 10A, 2006 | Problem 23”

  1. Interesting problem from lativia.2015 ones are written in a row..It is allowed to delete any two written numbers a and b and is replaced by a+b/4 the process is continued as long as one number remains or the number is less than .0001.Prove that the last number is greater than .0001

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