AMC 10 · 2012 · #12

Grade 10 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length

coordinate-geometrycircle-equationtangent-circlesquadratic-equations convert-to-algebracoordinate-geometryidentify-subproblems ↑ Prerequisites: coordinate-geometryquadratic-equations

📏 Long solution 💡 3 insights

Problem

A square touches an inner circle on one side and has two corners on an outer circle. Find the square's side.

Pick an answer.

(A)

$\frac{\sqrt{10}+5}{10}$

(B)

$\frac{2\sqrt{5}}{5}$

(C)

$\frac{2\sqrt{2}}{3}$

(D)

$\frac{2\sqrt{19}-4}{5}$

(E)

$\frac{9-\sqrt{17}}{5}$

How to solve

Strategy Introduce a Variable

Everything in the problem is pinned to coordinates except one number, the side length, so naming it s turns the picture into equations. The tangency fact fixes the line that CD sits on, which then forces the height of AB to be 1+s. The one thing worth being careful about is the square's left-right symmetry: it is tempting to assume it from the figure, but a horizontal line meets a circle in two points that are automatically mirror images, so the symmetry can be proved instead of assumed. After that it is a single quadratic, and a final check confirms the root really produces the picture described.

1STEP 1

Fix the line that side CD lies on

The touch point fixes one side's line.

radius to (0,1) is vertical → tangent line is y=1 → CD ⊂ {y=1}, square lies in y ≥ 1

2STEP 2

Name the side, place the opposite side

The opposite side sits one side length higher.

s = side length, CD ⊂ {y=1}, AB ⊂ {y=1+s}

3STEP 3

The symmetry is forced, not assumed

The symmetry is forced, not assumed.

x²+(1+s)²=4 → x=±√(4-(1+s)²) → AB = 2√(4-(1+s)²)

4STEP 4

Match the two expressions for AB

Two expressions for that side must agree.

s = 2√(4-(1+s)²) ⟺ (s/2)²+(1+s)² = 2²

5STEP 5

Solve the quadratic

That is one plain quadratic.

5s²+8s-12=0 → s=-8±√(304)/10=(-4± 2√(19))/5

6STEP 6

Check the root really builds the square

The positive root really builds the square.

s=(2√(19)-4)/5≈ 0.9436 ＜ 1, 1+s≈ 1.9436 ＜ 2, -s/2 ＜ 0 ＜ s/2

7STEP 7

Pick the choice, carefully

A near-miss choice fails the equation, so (A) stands.

(2√(19)-4)/5≈ 0.94356 vs 2√(2)/3≈ 0.94281; 5(2√(2)/3)²+8·2√(2)/3-12≈ -0.0131 ≠ 0

Answer

(2√(19)-4)/5

Rebuild the square from s=(2√(19)-4)/5≈ 0.943560 and test every stated condition numerically. The top corners are (±s/2,1+s)≈(± 0.471780, 1.943560), and 0.471780²+1.943560²≈ 0.222576+3.777424=4.000000, so A and B really sit on x²+y²=4. The bottom side runs along y=1 from x≈-0.4718 to x≈ 0.4718 and so contains the tangency point (0,1), whose distance from the origin is exactly 1; every other point of the square has y ≥ 1 with |x| ＞ 0, hence distance ＞ 1 from the origin, so the square touches the unit circle at that single point and nowhere else. Size sanity: the square must fit in the ring between the circles along the y-axis, where only 2-1=1 unit of room exists, and s≈ 0.94 is just under 1 because the corners cut diagonally toward the outer circle. That also rules out choice (E) ≈ 0.9754 as too tall to reach, and shows (A) ≈ 0.816 and (B) ≈ 0.894 leave the top corners strictly inside the large circle.

💡Key takeaway

A horizontal line can only hit a circle at two mirror-image points, so the square's left-right symmetry is something you can prove instead of assume, and after that one Pythagorean equation finishes the job.

Fix the line that side CD lies on
Name the side, place the opposite side
The symmetry is forced, not assumed
Match the two expressions for AB
Solve the quadratic
Check the root really builds the square
Pick the choice, carefully