AMC 10 · 2006 · #23
Grade 8 geometry-2dCircles with centers A and B have radius 3 and 8, respectively. A common internal tangent intersects the circles at C and D, respectively. Lines AB and CD intersect at E, and AE=5. What is CD?
Pick an answer.
AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Two circles have centers $A$ (radius $3$) and $B$ (radius $8$). A line that is tangent to both circles and crosses between them touches circle $A$ at $C$ and circle $B$ at $D$. This tangent line meets line $AB$ at a point $E$, with $AE = 5$. Find the length $CD$.
Givens: Circle $A$ has radius $AC = 3$; circle $B$ has radius $BD = 8$; $CD$ is a common internal tangent, touching circle $A$ at $C$ and circle $B$ at $D$; Lines $AB$ and $CD$ meet at $E$; $AE = 5$; Answer choices: (A) $13$, (B) $\frac{44}{3}$, (C) $\sqrt{221}$, (D) $\sqrt{255}$, (E) $\frac{55}{3}$
Unknowns: The length of the tangent segment $CD$
Understand
Restated: Two circles have centers $A$ (radius $3$) and $B$ (radius $8$). A line that is tangent to both circles and crosses between them touches circle $A$ at $C$ and circle $B$ at $D$. This tangent line meets line $AB$ at a point $E$, with $AE = 5$. Find the length $CD$.
Givens: Circle $A$ has radius $AC = 3$; circle $B$ has radius $BD = 8$; $CD$ is a common internal tangent, touching circle $A$ at $C$ and circle $B$ at $D$; Lines $AB$ and $CD$ meet at $E$; $AE = 5$; Answer choices: (A) $13$, (B) $\frac{44}{3}$, (C) $\sqrt{221}$, (D) $\sqrt{255}$, (E) $\frac{55}{3}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #4 Introduce a Variable, #3 Eliminate Possibilities
Tool #1 (Draw a Diagram) is the engine here: once you draw both radii to the tangent points, two right angles appear ($AC \perp CD$ and $BD \perp CD$), and the point $E$ splits the picture into two right triangles $ACE$ and $BDE$ that share the crossing at $E$. Tool #7 (Identify Subproblems) then breaks $CD$ into the two pieces $CE$ and $DE$, each living in its own right triangle. Tool #4 (Introduce a Variable / proportion) links the two triangles: they are similar, so their sides are in the fixed ratio $3:8$, which turns the known $CE$ into the unknown $DE$. Tool #3 (Eliminate Possibilities) is the safety net — the length must be a positive value a bit under $15$, and only one clean fraction matches.
Execute — Answer: B
4.G.A.1 Step 1 Draw radii, mark right angles
- Draw the radius $AC$ to the tangent point $C$ and the radius $BD$ to the tangent point $D$.
- A radius meeting a tangent at the point of contact is perpendicular to it, so $\angle ACE = 90^\circ$ and $\angle BDE = 90^\circ$.
- Record the radii: $AC = 3$ and $BD = 8$.
💡 The radius to a tangent point always makes a right angle with the tangent, so drawing it hands you two right triangles for free.
8.G.B.7 Step 2 Right triangle ACE gives CE
- Look at right triangle $ACE$.
- It has legs $AC = 3$ and $CE$, with hypotenuse $AE = 5$.
- By the Pythagorean theorem, $CE = \sqrt{AE^2 - AC^2} = \sqrt{25 - 9} = \sqrt{16} = 4$.
- (This is just the $3$–$4$–$5$ right triangle.)
💡 With the right angle at $C$, the two given lengths $3$ and $5$ fix the third side by Pythagoras.
8.G.A.5 Step 3 The two triangles are similar
- Compare triangles $ACE$ and $BDE$.
- The angles at $E$ are vertical angles, so $\angle AEC = \angle BED$.
- Both triangles also have a right angle ($\angle ACE = \angle BDE = 90^\circ$).
- Two equal angles force the third to match, so by the angle-angle criterion $\triangle ACE \sim \triangle BDE$.
💡 Vertical angles plus a shared right angle means the triangles have the same shape, just scaled.
7.RP.A.2 Step 4 Scale CE up to DE
- Similar triangles have proportional sides.
- The sides across from the equal angles pair up as $AC \leftrightarrow BD$ and $CE \leftrightarrow DE$, so $\dfrac{DE}{CE} = \dfrac{BD}{AC} = \dfrac{8}{3}$.
- Therefore $DE = CE \cdot \dfrac{8}{3} = 4 \cdot \dfrac{8}{3} = \dfrac{32}{3}$.
💡 Because the second triangle is the first blown up by the factor $\tfrac{8}{3}$, every matching length grows by $\tfrac{8}{3}$.
7.NS.A.3 Step 5 Add the two pieces
- The tangent is internal, so $E$ sits between the circles and therefore between $C$ and $D$ on the line.
- That means $CD = CE + DE = 4 + \dfrac{32}{3} = \dfrac{12}{3} + \dfrac{32}{3} = \dfrac{44}{3}$.
- This is choice (B).
💡 $E$ falls inside the gap between the circles, so the whole tangent is just its two halves added together.
4.G.A.1 Draw the radius $AC$ to the tangent point $C$ and the radius $BD$ to the tangent 8.G.B.7 Look at right triangle $ACE$. It has legs $AC = 3$ and $CE$, with hypotenuse $AE 8.G.A.5 Compare triangles $ACE$ and $BDE$. The angles at $E$ are vertical angles, so $\a 7.RP.A.2 Similar triangles have proportional sides. The sides across from the equal angle 7.NS.A.3 The tangent is internal, so $E$ sits between the circles and therefore between $ Review
Reasonableness: Cross-check with the internal-tangent-length formula. The similar-triangle ratio also gives $BE = AE \cdot \tfrac{8}{3} = \tfrac{40}{3}$, so the center distance is $AB = AE + EB = 5 + \tfrac{40}{3} = \tfrac{55}{3}$. The internal tangent length between two circles equals $\sqrt{AB^2 - (r_A + r_B)^2} = \sqrt{\left(\tfrac{55}{3}\right)^2 - 11^2} = \sqrt{\tfrac{3025}{9} - \tfrac{1089}{9}} = \sqrt{\tfrac{1936}{9}} = \tfrac{44}{3}$ — exactly the same answer, so the value is confirmed. Numerically $\tfrac{44}{3} \approx 14.7$, comfortably larger than the radii and smaller than the center distance $\tfrac{55}{3} \approx 18.3$, which is sensible. The distractors are traps: (E) $\tfrac{55}{3}$ is the center distance $AB$ itself; (A) $13$, (C) $\sqrt{221} \approx 14.9$, and (D) $\sqrt{255} \approx 16.0$ come from mixing up the internal and external tangent formulas or from an arithmetic slip.
Alternative: Skip the point $E$ and use the tangent formula from the start. Drop a line from $A$ parallel to the tangent $CD$; together with $AC$ and $BD$ it forms a right triangle whose legs are $CD$ and $r_A + r_B = 11$ (radii add because the tangent is internal) and whose hypotenuse is $AB$. First get $AB$ from the similar-triangle ratio ($AB = \tfrac{55}{3}$), then $CD = \sqrt{AB^2 - 11^2} = \tfrac{44}{3}$. Same result with one Pythagorean step.
CCSS standards used (min grade 8)
4.G.A.1Draw points, lines, line segments, rays, angles, and perpendicular and parallel lines, and identify these in two-dimensional figures (Drawing each radius to its point of tangency and recognizing the two right angles $AC \perp CD$ and $BD \perp CD$.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles in real-world and mathematical problems (Finding $CE = \sqrt{AE^2 - AC^2} = 4$ in right triangle $ACE$ (and re-deriving $CD$ in the alternative check).)8.G.A.5Use informal arguments to establish facts about angles, including the angle-angle criterion for similarity of triangles (Proving $\triangle ACE \sim \triangle BDE$ from the vertical angles at $E$ and the two right angles.)7.RP.A.2Recognize and represent proportional relationships between quantities (Using the similarity ratio $\tfrac{BD}{AC} = \tfrac{8}{3}$ to scale $CE = 4$ up to $DE = \tfrac{32}{3}$.)7.NS.A.3Solve real-world and mathematical problems involving the four operations with rational numbers (Adding $CE + DE = 4 + \tfrac{32}{3} = \tfrac{44}{3}$ over a common denominator.)
⭐ Draw the two radii to the tangent points and a pair of similar right triangles pops out: the $3$–$4$–$5$ triangle gives $CE = 4$, the $3:8$ scale gives $DE = \tfrac{32}{3}$, and since the tangent crosses between the circles you just add them to get $CD = \tfrac{44}{3}$.
⭐ Draw the two radii to the tangent points and a pair of similar right triangles pops out: the $3$–$4$–$5$ triangle gives $CE = 4$, the $3:8$ scale gives $DE = \tfrac{32}{3}$, and since the tangent crosses between the circles you just add them to get $CD = \tfrac{44}{3}$.
More like this
Same archetype — closest grade level first.
