Try this beautiful problem from AMC 10A, 2009 based on Diamond Pattern.

## Diamond Pattern – AMC-10A, 2009- Problem 15

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 \(F_{20}\)?

- \(756\)
- \(761\)
- \(786\)

**Key Concepts**

Pattern

Sequence

Symmetry

## Check the Answer

But try the problem first…

Answer: \(761\)

AMC-10A (2009) Problem 15

Pre College Mathematics

## Try with Hints

First hint

From the above diagram we observe that in \(F_1\) the number of diamond is \(1\).in \(F_2\) the number of diamonds are \(5\).in \(F_3\) the number of diamonds are \(13\). in \(F_4\) the numbers of diamonds are \(25\).Therefore from \(F_1\) to \(F_2\) ,\((5-1)\)=\(4\) new diamonds added.from \(F_2\) to \(F_3\),\((13-5=8\) new diamonds added.from \(F_3\) to \(F_4\),\((25-13)=12\) new diamonds added.we may say that When constructing \(F_n\) from \(F_{n-1}\), we add \(4(n-1)\) new diamonds.

Can you now finish the problem ……….

Second Hint

so we may construct that Let \(S_n\) be the number of diamonds in \(F_n\). We already know that \(P_1\)=1 and for all \(n >1\) ,\(P_n=P_{n-1}+4(n-1)\).now can you find out \(P_{20}\)

can you finish the problem……..

Final Step

Now \(P_{20}\)=\(P_{19} + 4(20-1)\)

\(\Rightarrow P_{20}\)=\(P_{19} + 4.19)\)

\(\Rightarrow P_{20}\)=\(P_{18}+(4 \times 18) +( 4 \times 19)\)

\(\Rightarrow P_{20}\)= …………………………………..

\(\Rightarrow P_{20}\)=\(1+4(1+2+3+……..+18+19)\)

\(\Rightarrow P_{20}\)=\(1+ \frac{ 4 \times 19 \times 20}{2}\)

\(\Rightarrow P_{20}\)=\(761\)

## Other useful links

- https://www.cheenta.com/arithmetic-progression-amc-10b-2004-problem-21/
- https://www.youtube.com/watch?v=M_HvBNmPcfU

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