AMC 10 · 2009 · #15

Grade 6 geometry-2d

Archetype Lattice / Grid Pattern from Small Cases └ Sub-archetype Figure-N Growth Sequence

pattern-recognitionsequences-arithmeticsystematic-enumeration easier-related-problempattern-recognition ↑ Prerequisites: sequences-arithmetic

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

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 $F_{20}$ ?

Pick an answer.

(A)

401

(B)

485

(C)

585

(D)

626

(E)

761

AMC 10 2009 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.

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Toolkit + CCSS Solution

Understand

Restated: A sequence of figures is built from diamonds. F_1 is a single diamond. For each n at least 3, F_n is made from F_{n-1} by wrapping it in a new square frame that has one more diamond on each side than the outer square of F_{n-1} had. The count grows as F_1 = 1, F_2 = 5, F_3 = 13. Find how many diamonds are in F_20.

Givens: F_1 has 1 diamond, F_2 has 5 diamonds, and F_3 has 13 diamonds.; For n at least 3, F_n is F_{n-1} wrapped in a new square frame.; The new frame has one more diamond on each side than F_{n-1}'s outer square had on each side.; The figure shows F_1 through F_4 as nested diamond squares.

Unknowns: The number of diamonds in figure F_20.

Understand

Plan

Primary tool: #5 Look for a Pattern

Secondary: #1 Draw a Diagram, #4 Introduce a Variable

Reaching F_20 by drawing all twenty figures is hopeless, so the move is to see how the count climbs from one figure to the next. The figure and the building rule tell you exactly how big each new ring is. Once that ring size follows a clean pattern, add the rings up into a single formula and evaluate it at n = 20 instead of drawing anything.

Execute — Answer: E

#1 Draw a Diagram 4.OA.C.5 Step 1

Read the counts from the figure

Look at the first figures.
F_1 is one diamond.
F_2 is a small square of diamonds totaling 5.
F_3, given in the problem, has 13.
Notice how each figure is the previous one with a fresh square ring of diamonds wrapped around the outside; the inside never changes.
So the whole job is to measure how big each new ring is.

$$F_1 = 1,\quad F_2 = 5,\quad F_3 = 13$$

💡 Each figure is just the one before it plus a border, so track the borders.

#5 Look for a Pattern 4.OA.C.5 Step 2

Size of each new ring

The rule says the new frame in F_n has one more diamond per side than the frame before.
Counting per side, F_2's outer square has 2 diamonds on each side, F_3's has 3, F_4's has 4, so F_n's new frame has n diamonds on each side.
A square frame with n diamonds on a side holds 4n minus 4 diamonds, because the 4 corner diamonds would otherwise be counted twice.
So building F_n adds 4n minus 4 diamonds.
Check: F_3 adds 4 times 3 minus 4 = 8, and 5 + 8 = 13, which matches.

$$\text{ring}(n) = 4n - 4;\qquad F_3 = 5 + (4\cdot 3 - 4) = 5 + 8 = 13$$

💡 A square border with n per side has 4n spots, minus the 4 corners shared by two sides.

#4 Introduce a Variable 6.EE.B.6 Step 3

Add up all the rings into a formula

Start from F_1 = 1 diamond, then add one ring for each step up to n.
The rings are 4(2) - 4, 4(3) - 4, and so on up to 4n - 4, which are 4, 8, 12, up to 4(n-1).
Pull out the 4: the total added is 4 times (1 + 2 + ...
+ (n-1)).
The sum 1 + 2 + ...
+ (n-1) equals (n-1)n over 2, so the added part is 4 times (n-1)n over 2 = 2n(n-1).
Adding the starting diamond gives a formula for any figure.

$$F_n = 1 + 4\bigl(1 + 2 + \cdots + (n-1)\bigr) = 1 + 4\cdot\frac{(n-1)n}{2} = 2n(n-1) + 1$$

💡 Adding a steady staircase of rings collapses into one triangular-number sum.

#5 Look for a Pattern 6.EE.A.1 Step 4

Evaluate at n = 20

Plug n = 20 into the formula.
That gives 2 times 20 times 19, plus 1.
First 20 times 19 is 380, then times 2 is 760, and adding 1 gives 761.
As a sanity anchor the formula also reads 2 times 20 squared minus 2 times 20 plus 1 = 800 - 40 + 1 = 761.
So F_20 has 761 diamonds, which is answer (E).

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

💡 With the formula in hand, the twentieth figure is one substitution, not twenty drawings.

[1] #1 4.OA.C.5 Look at the first figures. F_1 is one diamond. F_2 is a small square of diamonds

[2] #5 4.OA.C.5 The rule says the new frame in F_n has one more diamond per side than the frame

[3] #4 6.EE.B.6 Start from F_1 = 1 diamond, then add one ring for each step up to n. The rings a

[4] #5 6.EE.A.1 Plug n = 20 into the formula. That gives 2 times 20 times 19, plus 1. First 20 t

Review

Reasonableness: The count grows like 2n squared, so near n = 20 it should be a bit under 2 times 400 = 800; 761 sits right there, while 401 and 485 are far too small for a figure this deep and 626 does not match the 2n(n-1)+1 pattern that the earlier figures obey. The formula reproduces every known value: n = 1 gives 1, n = 2 gives 5, n = 3 gives 13, so trusting it at n = 20 giving 761 is sound.

Alternative: Color the diamond rings alternately, outermost first. In F_20 one color fills a full 20 by 20 square of diamonds and the other fills a 19 by 19 square, so the total is 20 squared + 19 squared = 400 + 361 = 761. This centered-square view gives the same answer without summing rings.

CCSS standards used (min grade 6)

4.OA.C.5 Generate a number or shape pattern following a given rule (Reading the counts 1, 5, 13 off the figures and using the building rule to see each figure adds an outer square ring.)
6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Summing the rings 4(1 + 2 + ... + (n-1)) and folding them into the general expression 2n(n-1) + 1.)
6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents (Substituting n = 20 into the formula and computing 2(20)(19) + 1 = 761.)

⭐ When a shape grows by a bigger border each step, add the borders up into one formula instead of drawing every stage.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Read the counts from the figure

Size of each new ring

Add up all the rings into a formula

Evaluate at n = 20

Review

CCSS standards used (min grade 6)

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