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2008 AMC 10A Problem 25

2008 AMC 10A Problem 25AMC 10 Step by Step

Name: Date: September 13, 2026

Circles · difficulty 5 of 5 · about 8 min · position P21-25

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 $x$ . Furthermore, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x$ ?

A)
$2\sqrt{5}-\sqrt{3}$
B)
$3$
C)
$\frac{3\sqrt{7}-\sqrt{3}}{2}$
D)
$2\sqrt{3}$
E)
$\frac{5+2\sqrt{3}}{2}$

Answer

Key insight

The six inner corners form a regular hexagon with side x, so the inner edge is x sqrt(3)/2 from the center; add 1 and apply Pythagoras.

Solution

Let $O$ be the table's center. Take one mat with outer corners $E, F$ on the circle and inner corners $G, H$ , so $EF = GH = x$ and $EG = FH = 1$ .

By the six-fold symmetry, the six inner corners (each shared by two adjacent mats) form a regular hexagon centered at $O$ , and $GH$ is one of its sides. A regular hexagon's circumradius equals its side, so $OG = OH = x$ and $\triangle OGH$ is equilateral. Its height is the distance from $O$ to the inner edge $GH$ :
$d(O, GH) = \frac{\sqrt3}{2}x.$

The outer edge $EF$ is parallel to $GH$ and $1$ unit farther from $O$ , so $d(O, EF) = \frac{\sqrt3}{2}x + 1$ . Let $M$ be the midpoint of $EF$ ; then $ME = \frac{x}{2}$ and $OE = 4$ . Pythagoras in $\triangle OME$ :
$\left(\frac{\sqrt3}{2}x + 1\right)^2 + \left(\frac{x}{2}\right)^2 = 16.$
Expanding, $\frac34 x^2 + \sqrt3 x + 1 + \frac14 x^2 = 16$ , so
$x^2 + \sqrt3\, x - 15 = 0 \quad\Longrightarrow\quad x = \frac{-\sqrt3 + \sqrt{3 + 60}}{2} = \frac{3\sqrt7 - \sqrt3}{2}.$

The answer is $\boxed{\textbf{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}}$ .

Why this works

Rotational symmetry converts "inner corners touch" into a regular hexagon, and the hexagon's side-equals-radius property turns the unknown into a distance from the center. Once one mat is described by a distance-to-chord picture, everything reduces to one right triangle. The lesson: in symmetric arrangements, identify the regular polygon hiding in the contact points, then work with a single representative piece.

Alternative approach

Coordinates: put $G = (x/2, \frac{\sqrt3}{2}x)$ and $H = (-x/2, \frac{\sqrt3}{2}x)$ as adjacent hexagon vertices, so the outer corner is $E = (x/2, \frac{\sqrt3}{2}x + 1)$ . Setting $|OE|^2 = 16$ gives the same quadratic. Numerically $x \approx 3.10$ , which rules out (B) $3$ and (D) $2\sqrt3 \approx 3.46$ once you trust the equation.

The trap

Assuming the inner corners lie on a circle of radius 3 (the table radius minus the mat width), rather than at distance x from the center.

Common mistakes

Assuming the inner corners lie on a circle of radius 3 (the table radius minus the mat width), rather than at distance x from the center.
Sign error when solving the quadratic, e.g. taking $\frac{\sqrt3 + 3\sqrt7}{2}$ , or dropping the cross term $\sqrt3 x$ from the expansion.

Techniques

Add construction lines/points (drop altitudes, extend segments, connect centers) · Exploit symmetry to reduce work or pair up objects