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AMC 8 Math Olympiad USA Math Olympiad

Diamond Pattern | AMC-10A, 2009 | Problem 15

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 \(F_{20}\)?

Diamond Pattern
  • \(756\)
  • \(761\)
  • \(786\)

Key Concepts


Pattern

Sequence

Symmetry

Check the Answer


Answer: \(761\)

AMC-10A (2009) Problem 15

Pre College Mathematics

Try with Hints


Diamond Pattern

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 \(P_{20}\)

can you finish the problem……..

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

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

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

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

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

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

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

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AMC 8 Math Olympiad USA Math Olympiad

Pattern Problem | AMC-10A, 2003 | Problem 23

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?

Pattern Problem
  • \(1004004\)
  • \(1005006\)
  • \(1507509\)
  • \(3015018\)
  • \(6021018\)

Key Concepts


Pattern

Sequence

Symmetry

Check the Answer


Answer: \(1507509\)

AMC-10A (2003) Problem 23

Pre College Mathematics

Try with Hints


Pattern in a triangle

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=\( 3 \times \frac{(1002 )(1003)}{2}\)=\(1507509\)

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AMC 8 Math Olympiad USA Math Olympiad

Pattern Problem| AMC 8, 2002| Problem 23

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?

Pattern
  • \(\frac{5}{9}\)
  • \(\frac{4}{9}\)
  • \(\frac{4}{7}\)

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……..

Pattern

If we look very carefully we must notice that,
The same pattern is repeated for every \( 6 \times 6 \) tile

pattern 6/6


Looking closer, there is also symmetry of the top \( 3 \times 3\) square,

Pattern 3/3


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 \(\frac{4}{9}\).

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