AMC 10 · 2008 · #25

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

A round table has radius 4. Six identical rectangular mats, each 1 wide and x long, are laid around the table. Each mat's long side of length x is a chord of the table's edge (both of its outer corners sit on the circle). The mats point inward, and each inner corner just touches an inner corner of the next mat. Find the length x.

Pick an answer.

(A)

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

(B)

$\ 3$

(C)

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

(D)

$\ 2\sqrt{3}$

(E)

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

AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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

Mark the center O. A mat's outer long side is a chord of length x, and the six identical mats repeat every 60 degrees around O.

360°/6 = 60°

2STEP 2

Find the 30-degree touch direction

A mat's centerline radius halves its wedge, so a touching inner corner sits 30 degrees off that centerline.

1/2 · 60° = 30°

3STEP 3

Name the outer distance and use Pythagoras

Let d be the centerline distance from O to the outer edge. The half-chord x/2, d, and the radius 4 form a right triangle.

(x/2)² + d² = 4² = 16

4STEP 4

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

The inner edge lies 1 closer, so O to its midpoint is d minus 1, and the 30-60-90 triangle makes that root-3 times x/2.

d - 1 = √(3)/2 x ⟹ d = 1 + √(3)/2 x

5STEP 5

Substitute to get one equation in x

Substitute d into the Pythagorean equation; the x-squared terms merge, leaving x squared plus root-3 x minus 15 equals 0.

(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

The quadratic formula turns root-63 into 3 root 7, and a length must be positive, so x is (3 root 7 minus root 3) over 2 — choice (C).

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