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
Answer: \(761\)
AMC-10A (2009) Problem 15
Pre College Mathematics
Try with Hints
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 ..........
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 \(P_{20}\)
Try this beautiful problem from Pattern based on Triangle.
Pattern - AMC-10A, 2003- Problem 23
A large equilateral triangle is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure, we have \(3\) rows of small congruent equilateral triangles, with \(5\) small triangles in the base row. How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of $2003$ small equilateral triangles?
\(1004004\)
\(1005006\)
\(1507509\)
\(3015018\)
\(6021018\)
Key Concepts
Pattern
Sequence
Symmetry
Check the Answer
Answer: \(1507509\)
AMC-10A (2003) Problem 23
Pre College Mathematics
Try with Hints
If we observe very carefully,we notice that
1st row the number of toothpicks needs a triangle is 3 i.e \(1 \times 3\)
2nd row the number of toothpicks needs a triangle is 9 i.e \( 3 \times 3\)
3rd row the number of toothpicks needs a triangle is \(18\) i.e \(6 \times 3\)
Can you now finish the problem ..........
we also observe that in the 1st row the number of triangle is 1. In the 2nd row the number of triangle is 3.In the third row the number of triangles are 5.so toothpicks is the corresponding triangular number. Since the triangle in question has \(2n-1=2003\) \(\Rightarrow n=1002\) rows.
can you finish the problem........
So number of triangle required =\(\frac{n(n+1)}{2}\)=\(\frac{(1002 )(1003)}{2}\).There are 3 toothpicks needed to form a Triangle.
Therefore required numbers of toothpicks=\( 3 \times \frac{(1002 )(1003)}{2}\)=\(1507509\)
Try this beautiful problem from Algebra based on Pattern.
Pattern - AMC-8, 2002- Problem 23
A corner of a tiled floor is shown. If the entire floor is tiled in this way and each of the four corners looks like this one, then what fraction of the tiled floor is made of darker tiles?
\(\frac{5}{9}\)
\(\frac{4}{9}\)
\(\frac{4}{7}\)
Key Concepts
Algebra
Pattern
Fraction
Check the Answer
Answer:\(\frac{4}{9}\)
AMC-8 (2002) Problem 23
Pre College Mathematics
Try with Hints
The same pattern is repeated for every \(6 \times 6 \) tile
Can you now finish the problem ..........
Looking closer, there is also symmetry of the top \(3 \times 3\) square
can you finish the problem........
If we look very carefully we must notice that, The same pattern is repeated for every \( 6 \times 6 \) tile
Looking closer, there is also symmetry of the top \( 3 \times 3\) square,
Therefore the fraction of the entire floor in dark tiles is the same as the fraction in the square Counting the tiles, there are dark tiles, and total tiles, giving a fraction of \(\frac{4}{9}\).
