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 rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$ . Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x$ ?

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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

Toolkit + CCSS Solution

Understand

Restated: 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.

Givens: The table is a circle of radius 4.; There are six congruent mats, arranged with six-fold symmetry.; Each mat is a rectangle of width 1 and length x.; The two endpoints of a mat's length-x side lie on the circle.; Each mat's inner corner touches an inner corner of an adjacent mat.

Unknowns: The mat length x.

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #17 Visualize Spatial Relationships, #4 Introduce a Variable

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.

Execute — Answer: C

#1 Draw a Diagram 8.G.A.5 Step 1

Set up the picture with the center

Mark the center O of the table.
Look at one mat.
Its long side of length x is a straight chord with both ends on the circle, and the mat sticks inward by its width 1, so its inner long side is another length-x segment closer to O.
Because six identical mats are spaced evenly around O, they turn into each other by 60-degree rotations, and each mat sits on its own radius that splits it in half.

$$\frac{360^\circ}{6} = 60^\circ$$

💡 Six equal mats around one center means the whole design repeats every 60 degrees, so I only need to understand one mat.

#17 Visualize Spatial Relationships 8.G.A.5 Step 2

Find the 30-degree touch direction

Draw the radius through the middle of one mat as its centerline.
The mat's two inner corners sit symmetrically, one on each side of this centerline.
Each inner corner touches the inner corner of the neighboring mat, and that meeting point must land exactly on the boundary between this mat's 60-degree wedge and the next one.
That boundary is halfway across the wedge, so each inner corner points off the centerline by half of 60 degrees, which is 30 degrees.

$$\tfrac{1}{2}\cdot 60^\circ = 30^\circ$$

💡 Neighboring corners meet right on the shared border of two wedges, and that border is 30 degrees off each mat's centerline.

#4 Introduce a Variable 8.G.B.7 Step 3

Name the outer distance and use Pythagoras

Let d be the distance from the center O to the mat's outer edge, measured along the centerline.
That centerline hits the outer edge at its midpoint and is perpendicular to it, so it splits the length-x chord into two halves of x/2.
The half-chord, the distance d, and a radius of length 4 form a right triangle with the radius as the hypotenuse.
Apply the Pythagorean theorem.

$$\left(\tfrac{x}{2}\right)^2 + d^2 = 4^2 = 16$$

💡 A radius to a chord endpoint, the perpendicular from the center, and half the chord always make a right triangle.

#17 Visualize Spatial Relationships 8.G.B.7 Step 4

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

The inner edge is 1 unit closer to O than the outer edge, so its midpoint sits at distance d - 1 from O along the centerline.
From O, look at the right triangle made by the centerline (length d - 1) and the half of the inner edge (length x/2) out to an inner corner.
From step 2 that inner corner is 30 degrees off the centerline, so this is a 30-60-90 triangle.
The side opposite 30 degrees is the short leg x/2, and the leg opposite 60 degrees is d - 1, and in a 30-60-90 triangle the 60-degree leg is the short leg times the square root of 3.

$$d - 1 = \frac{\sqrt{3}}{2}\,x \quad\Longrightarrow\quad d = 1 + \frac{\sqrt{3}}{2}\,x$$

💡 The 30-degree corner turns the inner triangle into a fixed-shape 30-60-90, locking d and x together.

#4 Introduce a Variable 8.EE.A.1 Step 5

Substitute to get one equation in x

Put the expression for d into the Pythagorean equation from step 3.
Square the sum and combine like terms.
The two x-squared pieces, one fourth and three fourths, add to a full x-squared, and the constant 1 moves to the other side of 16.

$$\left(\tfrac{x}{2}\right)^2 + \left(1 + \tfrac{\sqrt{3}}{2}x\right)^2 = 16 \;\Rightarrow\; \tfrac{x^2}{4} + 1 + \sqrt{3}\,x + \tfrac{3x^2}{4} = 16 \;\Rightarrow\; x^2 + \sqrt{3}\,x - 15 = 0$$

💡 Substituting the two facts about the same mat collapses everything into a single equation in x.

#4 Introduce a Variable 8.EE.A.2 Step 6

Solve for the positive length

Solve x squared plus root-three x minus 15 equals 0.
Completing the square (or the quadratic formula) gives x equal to negative root-three plus or minus the square root of three plus sixty, all over two.
The square root of 63 is 3 times the square root of 7.
A mat length must be positive, so keep the plus sign.
That gives x equal to (3 root 7 minus root 3) over 2, which is choice (C).

$$x = \frac{-\sqrt{3} + \sqrt{(\sqrt{3})^2 + 4\cdot 15}}{2} = \frac{-\sqrt{3} + \sqrt{63}}{2} = \frac{3\sqrt{7} - \sqrt{3}}{2}$$

💡 The equation has one positive and one negative root, and only the positive one can be a real length.

[1] #1 8.G.A.5 Mark the center O of the table. Look at one mat. Its long side of length x is a

[2] #17 8.G.A.5 Draw the radius through the middle of one mat as its centerline. The mat's two i

[3] #4 8.G.B.7 Let d be the distance from the center O to the mat's outer edge, measured along

[4] #17 8.G.B.7 The inner edge is 1 unit closer to O than the outer edge, so its midpoint sits a

[5] #4 8.EE.A.1 Put the expression for d into the Pythagorean equation from step 3. Square the s

[6] #4 8.EE.A.2 Solve x squared plus root-three x minus 15 equals 0. Completing the square (or t

Review

Reasonableness: 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).

Alternative: Use the Law of Cosines instead. The inner corners form a regular hexagon, and one can build a triangle from the center O, one inner corner, and one outer corner using the known lengths (the mat width 1, the radius 4, and the 60-degree spacing). Applying the Law of Cosines to that triangle produces the same quadratic x squared plus root-three x minus 15 equals 0, so it lands on the same answer (C).

CCSS standards used (min grade 8)

8.G.A.5 Use informal arguments to establish facts about angle sum and exterior angles (Splitting the circle into six 60-degree wedges and locating each inner corner 30 degrees off its mat's centerline.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (The radius/half-chord right triangle for the outer edge, and the 30-60-90 triangle linking d and x for the inner edge.)
8.EE.A.1 Know and apply the properties of integer exponents (Squaring and expanding the substituted expression to combine the x-squared terms into a single equation.)
8.EE.A.2 Use square root and cube root symbols to represent solutions (Solving the quadratic and simplifying the square root of 63 to express x as (3 root 7 minus root 3) over 2.)

⭐ 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.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

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

Review

CCSS standards used (min grade 8)

More like this