AMC 10 · 2009 · #16

Grade 8 geometry-2d
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Synthetic Similar-Triangle Length or Area
equilateral-trianglethirty-sixty-ninety-trianglesymmetry-argument physical-representation ↑ Prerequisites: equilateral-triangle
📏 Medium solution 💡 3 insights
Problem
Points A and C lie on a circle centered at O, segments BA and BC are both tangent to that circle, and triangle ABC is equilateral. Segment BO meets the circle at D. What is the ratio BD/BO?

Pick an answer.

(A)
$\frac {\sqrt2}{3}$
(B)
$\frac {1}{2}$
(C)
$\frac {\sqrt3}{3}$
(D)
$\frac {\sqrt2}{2}$
(E)
$\frac {\sqrt3}{2}$

AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

A clear picture turns the words into two right angles and a line of symmetry. Once the radius is named r, the whole problem shrinks to one right triangle whose angles are known, and the ratio drops out of that triangle.

1STEP 1

Draw it and mark the right angles

Sketch the circle, tangents BA, BC, and triangle ABC. A tangent is square to the radius at its touch point: ∠OAB = ∠OCB = 90°, OA = OC = r.

∠ OAB = 90°, ∠ OCB = 90°, OA = OC = r
2STEP 2

Use symmetry to find the angle at B

The two tangents from B are equal, so folding across BO swaps A and C. That halves the triangle's 60° angle at B, giving ∠ABO = 30°.

∠ ABC = 60° → ∠ ABO = 1/2 · 60° = 30°
3STEP 3

Solve the right triangle for BO

Triangle OAB has 90° at A and 30° at B, a 30-60-90 triangle: the leg opposite 30° is half the hypotenuse, so BO = 2r.

OA = r (opposite 30°), BO = 2 OA = 2r
4STEP 4

Locate D and take the ratio

D lies on the circle, so OD = r, and D sits between B and O, so BD = 2r - r = r. Hence BD/BO = 1/2, choice (B).

OD = r, BD = BO - OD = 2r - r = r, BD/BO = r/2r = 1/2
Answer
1/2
Set r = 1. Then BO = 2 and OD = 1, so D is the midpoint of BO and BD = 1. The ratio BD/BO = 1/2, a clean fraction between 0 and 1, which is exactly what a point strictly between B and O should give. It matches choice (B).
💡Key takeaway

A tangent stands square to the radius; that right angle plus the triangle's symmetry makes BO two radii long, and BD is just one of them.

  • Draw it and mark the right angles
  • Use symmetry to find the angle at B
  • Solve the right triangle for BO
  • Locate D and take the ratio

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