AMC 10 · 2006 · #16

Grade 8 geometry-2d
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Synthetic Similar-Triangle Length or Area
similar-trianglespythagorean-theoremtangent-circles convert-to-algebra ↑ Prerequisites: similar-trianglespythagorean-theorem
📏 Long solution 💡 3 insights 📊 Diagram
Problem
A line tangent to two circles passes between them, meeting the line of centres at a point 5 from the smaller centre. The radii are 3 and 8. Find the distance between the two tangent points.

Pick an answer.

(A)
13
(B)
$\frac{44}{3}$
(C)
$\sqrt{221}$
(D)
$\sqrt{255}$
(E)
$\frac{55}{3}$
How to solve
Strategy Draw a Diagram

Tool #1 (Draw a Diagram) is the engine here: once you draw both radii to the tangent points, two right angles appear (AC ⊥ CD and BD ⊥ 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.

1STEP 1

Draw radii, mark right angles

Each radius meets the tangent at a right angle.

AC = 3, BD = 8, ∠ ACE = ∠ BDE = 90°
2STEP 2

Right triangle ACE gives CE

The first right triangle gives one piece as 4.

CE = √(5² - 3²) = √(16) = 4
3STEP 3

The two triangles are similar

Vertical angles and the right angles make the triangles similar.

∠ AEC = ∠ BED, ∠ ACE = ∠ BDE = 90° → △ ACE ∼ △ BDE
4STEP 4

Scale CE up to DE

The ratio of radii scales that piece up to 32/3.

DE/CE = BD/AC = 8/3 → DE = 4 · 8/3 = 32/3
5STEP 5

Add the two pieces

The crossing point lies between them, so adding gives 44/3, choice (B).

CD = CE + DE = 4 + 32/3 = 44/3 → (B)
Answer
44/3
Cross-check with the internal-tangent-length formula. The similar-triangle ratio also gives BE = AE · 8/3 = 40/3, so the center distance is AB = AE + EB = 5 + 40/3 = 55/3. The internal tangent length between two circles equals √(AB² - (r_A + r_B)²) = √((55/3)² - 11²) = √(3025/9 - 1089/9) = √(1936/9) = 44/3 — exactly the same answer, so the value is confirmed. Numerically 44/3 ≈ 14.7, comfortably larger than the radii and smaller than the center distance 55/3 ≈ 18.3, which is sensible. The distractors are traps: (E) 55/3 is the center distance AB itself; (A) 13, (C) √(221) ≈ 14.9, and (D) √(255) ≈ 16.0 come from mixing up the internal and external tangent formulas or from an arithmetic slip.
💡Key takeaway

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 = 32/3, and since the tangent crosses between the circles you just add them to get CD = 44/3.

  • Draw radii, mark right angles
  • Right triangle ACE gives CE
  • The two triangles are similar
  • Scale CE up to DE
  • Add the two pieces

More like this

Same archetype — closest grade level first.