AMC 10 · 2008 · #22

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

pythagorean-theoremthirty-sixty-ninety-trianglechord-perpendicular-from-center spatial-visualizationconvert-to-algebra ↑ Prerequisites: pythagorean-theorem

📏 Long solution 💡 3 insights 📊 Diagram

Problem

Six identical mats are laid around a round table, each with its outer edge a chord of the rim. Each mat's inner corner touches the next mat's inner corner. Find the mat's length.

Pick an answer.

(A)

$2\sqrt{5}-\sqrt{3}$

(B)

(C)

$\frac{3\sqrt{7}-\sqrt{3}}{2}$

(D)

$2\sqrt{3}$

(E)

$\frac{5+2\sqrt{3}}{2}$

How to solve

Strategy Draw a Diagram

The whole puzzle is about where corners land on a circle, so a clean picture with the center marked is the anchor. From the picture, the six-fold symmetry pins down the angles, and two right triangles sharing the mat's centerline turn the geometry into an equation. Naming the distance from the center to the outer edge lets the Pythagorean theorem and a 30-60-90 triangle do the rest.

1STEP 1

Set up the picture with the center

Six mats divide the circle into sixty degree steps.

360°/6 = 60°

2STEP 2

Find the 30-degree touch direction

The touch direction sits halfway between two mats.

1/2 · 60° = 30°

3STEP 3

Name the outer distance and use Pythagoras

A right triangle links the chord to the radius.

(x/2)² + d² = 4² = 16

4STEP 4

Relate d to x with a 30-60-90 triangle

The touch angle relates the same distance to the length.

d - 1 = √(3)/2 x ⟹ d = 1 + √(3)/2 x

5STEP 5

Substitute to get one equation in x

Substituting gives one quadratic.

(x/2)² + (1 + √(3)/2x)² = 16 → x²/4 + 1 + √(3) x + 3x²/4 = 16 → x² + √(3) x - 15 = 0

6STEP 6

Solve for the positive length

Taking the positive root gives (3√(7)-√(3))/2, choice (A).

x = (-√(3) + √((√(3))² + 4 · 15))/2 = (-√(3) + √(63))/2 = (3√(7) - √(3))/2

Answer

(3√(7)-√(3))/2

Numerically, (3 root 7 minus root 3) over 2 is about (7.94 - 1.73) / 2, near 3.10. That is between the choices 3 and 2 root 3 (about 3.46) and is safely less than the diameter 8, so a mat of that length fits. Checking the inner corner distance from O: with x about 3.10, d = 1 + (root 3 / 2) x is about 3.69, and the outer corner distance is the square root of (1.55 squared plus 3.69 squared), which is exactly 4, matching the table's radius. The geometry is consistent, confirming (C).

💡Key takeaway

Mark the center, use the even spacing to fix the angles, and let two right triangles turn the picture into one equation you can solve.

Set up the picture with the center
Find the 30-degree touch direction
Name the outer distance and use Pythagoras
Relate d to x with a 30-60-90 triangle
Substitute to get one equation in x
Solve for the positive length