AMC 10 · 2006 · #17

Grade 10 geometry-2d
tangent-circlespower-of-a-pointcoordinate-geometry convert-to-algebra ↑ Prerequisites: pythagorean-theoremcoordinate-geometry
📏 Long solution 💡 3 insights 📊 Diagram
Problem
A square and a circle share a rational side and radius, the circle passes through one corner, and a tangent from another corner has a known length. Find the ratio of the radius to the side.

Pick an answer.

(A)
$\frac{1}{2}$
(B)
$\frac{5}{9}$
(C)
$\frac{3}{5}$
(D)
$\frac{5}{3}$
(E)
$\frac{9}{5}$
How to solve
Strategy Convert to Algebra

The picture hands you exactly one measurement, AF, and asks for a ratio of two unknowns. One equation, two unknowns — so the finish cannot be ordinary algebra alone. Tool #13 (Convert to Algebra) turns the geometry into that one equation: tools #17 and #1 pin down where E actually sits and put the square on axes, tool #7 splits off the tangent-length subproblem AF² = AE² - r², and tool #4 names the coordinates. What finishes the problem is tool #15 (Organize Information in More Ways): the equation lands in the shape s² + √(2) sr = 9 + 5√(2), and re-reading it as "rational part" plus "√(2) part" — which is legal only because r and s are rational — splits one equation into two. That split is the whole problem; everything before it is bookkeeping.

1STEP 1

Pin down where E sits

Two clues force the centre onto the diagonal, past the corner.

ED = r, B, D, E collinear with D between them, BE = s√(2) + r
2STEP 2

Put the square on axes

Coordinates write the centre with one expression.

A=(0,s), B=(0,0), C=(s,0), D=(s,s), E = (s + r/√(2), s + r/√(2))
3STEP 3

Trade the tangent for a right angle

The tangent becomes a right angle, so the touch point never needs locating.

AF² + EF² = AE² → 9 + 5√(2) = AE² - r²
4STEP 4

Compute AE squared

Computing the long distance makes the squared radius cancel.

AE² = (s+r/√(2))² + (r/√(2))² = s² + √(2) sr + r²/2 + r²/2 = s² + √(2) sr + r²
5STEP 5

Split the equation at the radical

Rationality splits the equation at the radical into two equations.

s² + √(2) sr = 9 + 5√(2) ⟹ s² - 9 = (5 - sr)√(2) ⟹ sr = 5 and s² = 9
6STEP 6

Solve and take the ratio

Solving gives the ratio 5/9, choice (B).

s = 3, r = 5/3, r/s = 5/3/3 = 5/9 → (B)
Answer
5/9
Plug the values back into the original picture. With s=3 and r=5/3: E = (3+5/3√(2), 3+5/3√(2)) ≈ (4.1785, 4.1785) and A=(0,3), so AE² ≈ 18.849 and AF² = AE² - r² ≈ 18.849 - 2.778 = 16.071. Meanwhile 9+5√(2) ≈ 16.071 — an exact match, so the configuration is consistent, not just algebraically forced. Also AE ≈ 4.34 > r, confirming A is outside the circle so a tangent from A exists. The ratio 5/9 ≈ 0.56 passes the eyeball test: in the figure the radius is a bit more than half the side. The distractors are exactly the wrong things to report: (D) 5/3 is r itself, (C) 3/5 is 1/r, (E) 9/5 is the flipped ratio s/r, and (A) 1/2 is what you get by eyeballing the figure and rounding. Notice the answer choices carry a warning: (D) and (E) exceed 1, and r > s would put A inside or near the circle, so they were never plausible.
💡Key takeaway

The tangent turns into a right triangle and the whole picture collapses to s² + √(2) sr = 9 + 5√(2); because r and s are rational, the plain parts and the √(2) parts must match separately, so s²=9 and sr=5 and r/s = 5/9.

  • Pin down where E sits
  • Put the square on axes
  • Trade the tangent for a right angle
  • Compute AE squared
  • Split the equation at the radical
  • Solve and take the ratio