2009 AMC 10A Problem 15

P11-15~4 minPrint

Sequences & Series Compute small cases, spot the pattern, generalize

The figures $F_1$ , $F_2$ , $F_3$ , and $F_4$ shown are the first in a sequence of figures. For $n\ge3$ , $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13$ diamonds. How many diamonds are there in figure $F_{20}$ ?

Show solution without answering

Key insight

Each new ring with n diamonds per side holds 4(n - 1) diamonds, so F_20 = 1 + 4(1 + 2 + ... + 19) = 761.

Solution

1. Count the first figures: $F_1 = 1$ , $F_2 = 5$ , $F_3 = 13$ , $F_4 = 25$ . The jumps are $4, 8, 12$ .
2. See where the jumps come from. The new outside square of $F_n$ has $n$ diamonds on each side. Four sides of $n$ would be $4n$ , but the $4$ corner diamonds are each counted twice, so the ring holds $4n - 4 = 4(n - 1)$ diamonds. For $n = 3$ that is $8$ , for $n = 4$ it is $12$ . This matches.
3. So each figure is the one before plus $4(n-1)$ , and
$F_{20} = 1 + 4(1) + 4(2) + 4(3) + \cdots + 4(19) = 1 + 4(1 + 2 + \cdots + 19).$
4. The sum $1 + 2 + \cdots + 19$ is $\dfrac{19 \times 20}{2} = 190$ .
5. So $F_{20} = 1 + 4 \times 190 = 761$ .
6. Check the method on a figure we know: $F_4 = 1 + 4(1 + 2 + 3) = 1 + 24 = 25$ . Correct.

The answer is $\boxed{\textbf{(E)}\ 761}$ .

Why this works

When a figure grows by adding a layer each time, count what one layer adds, then add up the layers. Here the layers add $4, 8, 12, \ldots$ , so the total uses the sum $1 + 2 + \cdots + n$ . Always test your method on the figures the problem shows you before using it for $n = 20$ .

Alternative approach

Look at the diamonds in rows along the diagonals. $F_n$ is an $n$ by $n$ grid of diamonds with an $(n-1)$ by $(n-1)$ grid fitted into the gaps: $F_3 = 9 + 4 = 13$ and $F_4 = 16 + 9 = 25$ . So $F_{20} = 20^2 + 19^2 = 400 + 361 = 761$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Count the diamonds in the first figures: $F_1 = 1$ , $F_2 = 5$ , $F_3 = 13$ , $F_4 = 25$ . The increases are $4, 8, 12$ .

Why: the outer square of $F_n$ has $n$ diamonds on each side, and a square with $n$ diamonds per side contains $4n - 4 = 4(n-1)$ diamonds (four corners are shared). So
$F_n = F_{n-1} + 4(n-1).$

Summing the increments from $F_1 = 1$ :
$F_n = 1 + 4\big(1 + 2 + \cdots + (n-1)\big) = 1 + 4 \cdot \frac{(n-1)n}{2} = 2n(n-1) + 1.$

Check: $F_3 = 2 \cdot 3 \cdot 2 + 1 = 13$ and $F_4 = 2 \cdot 4 \cdot 3 + 1 = 25$ , matching the figure.

Then $F_{20} = 2 \cdot 20 \cdot 19 + 1 = 760 + 1 = 761$ .

The answer is $\boxed{\textbf{(E)}\ 761}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.1)
Intermediate Algebra, Chapter 17: More Sequences and Series (§17.1, §17.2)
Introduction to Number Theory, Chapter 12: Introduction to Modular Arithmetic (§12.6)