Difference between revisions of "2009 AMC 12A Problems/Problem 11"
(New page: == Problem == The figures <math>F_1</math>, <math>F_2</math>, <math>F_3</math>, and <math>F_4</math> shown are the first in a sequence of figures. For <math>n\ge3</math>, <math>F_n</math> ...) |
m (→Solution 2) |
||
Line 63: | Line 63: | ||
When constructing <math>F_n</math> from <math>F_{n-1}</math>, we add <math>4(n-1)</math> new diamonds. Let <math>d_n</math> be the number of diamonds in <math>F_n</math>. We now know that <math>d_1=1</math> and <math>\forall n>1:~ d_n=d_{n-1} + 4(n-1)</math>. | When constructing <math>F_n</math> from <math>F_{n-1}</math>, we add <math>4(n-1)</math> new diamonds. Let <math>d_n</math> be the number of diamonds in <math>F_n</math>. We now know that <math>d_1=1</math> and <math>\forall n>1:~ d_n=d_{n-1} + 4(n-1)</math>. | ||
− | Hence < | + | Hence we get: |
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | d_{20} & = d_{19} + 4\cdot 19 \ | ||
+ | & = d_{18} + 4\cdot 18 + 4\cdot 19 \ | ||
+ | & = \cdots \ | ||
+ | & = 1 + 4(1+2+\cdots+18+19) \ | ||
+ | & = 1 + 4\cdot\frac{19\cdot 20}2 \ | ||
+ | & = \boxed{761} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2009|ab=A|num-b=10|num-a=12}} | {{AMC12 box|year=2009|ab=A|num-b=10|num-a=12}} |
Revision as of 13:03, 12 February 2009
Contents
[hide]Problem
The figures , , , and shown are the first in a sequence of figures. For , is constructed from by surrounding it with a square and placing one more diamond on each side of the new square than had on each side of its outside square. For example, figure has diamonds. How many diamonds are there in figure ?
Solution
Solution 1
Color the diamond layers alternately blue and red, starting from the outside. You'll get the following pattern:
In the figure , the blue diamonds form a square, and the red diamonds form a square. Hence the total number of diamonds in is .
Solution 2
When constructing from , we add new diamonds. Let be the number of diamonds in . We now know that and .
Hence we get:
See Also
2009 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |