2008 AMC 12B Problem 24

Question

Let

A_0 = (0, 0).

Distinct points

A_1,

A_2,

\ldots

lie on the

x

-axis, and distinct points

B_1,

B_2,

\ldots

lie on the graph of

y = \sqrt{x}.

For every positive integer

n,

A_{n-1}B_nA_n

is an equilateral triangle. What is the least

n

for which the length

A_0A_n \ge 100?

13

15

17

19

21

If

c_n = A_{n-1}A_n,

the apex

B_n

sits at height

\tfrac{\sqrt3}{2}c_n

above the midpoint; lying on

y = \sqrt{x}

gives

\tfrac34 c_n^2 = a_{n-1} + \tfrac{c_n}{2}

Subtracting consecutive relations yields

c_n = c_{n-1} + \tfrac23,

so

c_n = \tfrac{2n}{3}

and

A_0A_n = \tfrac{n(n+1)}{3}

Let

a_n = A_0A_n

and

c_n = a_n - a_{n-1}

be the base of the

n

th equilateral triangle. Its apex

B_n

lies above the midpoint at height

\tfrac{\sqrt3}{2}c_n,

and being on

y = \sqrt{x}

gives

\left(\tfrac{\sqrt3}{2}c_n\right)^2 = a_{n-1} + \tfrac{c_n}{2},

i.e.

\tfrac34 c_n^2 = a_{n-1} + \tfrac{c_n}{2}.

Writing the same relation for the previous triangle and subtracting gives

c_n = c_{n-1} + \tfrac23,

and with

c_1 = \tfrac23

we get

c_n = \tfrac{2n}{3}.

Summing,

a_n = \sum_{k=1}^n \frac{2k}{3} = \frac{n(n + 1)}{3}.

We need

\tfrac{n(n + 1)}{3} \ge 100,

i.e.

n(n + 1) \ge 300.

Since

16 \cdot 17 = 272

and

17 \cdot 18 = 306,

the least such

n

is

17.

Thus, the correct answer is C.

Problem 24 in other years