Difference between revisions of "2004 AMC 10A Problems/Problem 16"

Latest revision as of 00:37, 14 August 2024

Problem

The $5\times 5$ grid shown contains a collection of squares with sizes from $1\times 1$ to $5\times 5$ . How many of these squares contain the black center square?

$[asy] for (int i=0; i<5; ++i) { for (int j=0; j<5; ++j) { draw((-2.5+i, -2.5+j)--(-1.5+i, -2.5+j) -- (-1.5+i, -1.5+j) -- (-2.5+i, -1.5+j)--cycle); } fill((-0.5,-0.5)--(-0.5, 0.5)--(0.5,0.5) -- (0.5,-0.5)--cycle, black); } [/asy]$

$\mathrm{(A) \ } 12 \qquad \mathrm{(B) \ } 15 \qquad \mathrm{(C) \ } 17 \qquad \mathrm{(D) \ } 19\qquad \mathrm{(E) \ } 20$

Solution

Solution 1

Since there are five types of squares: $1 \times 1, 2 \times 2, 3 \times 3, 4 \times 4,$ and $5 \times 5.$ We must find how many of each square contain the black shaded square in the center.

If we list them, we get that

There is $1$ of all $1\times 1$ squares, containing the black square
There are $4$ of all $2\times 2$ squares, containing the black square
There are $9$ of all $3\times 3$ squares, containing the black square
There are $4$ of all $4\times 4$ squares, containing the black square
There is $1$ of all $5\times 5$ squares, containing the black square

Thus, the answer is $1+4+9+4+1=19\Rightarrow\boxed{\mathrm{(D)}\ 19}$ .

Solution 2

We use complementary counting. There are only $2\times2$ and $1\times1$ squares that do not contain the black square. Counting, there are $12$ - $2\times2$ squares, and $25-1 = 24$ $1\times1$ squares that do not contain the black square. That gives $12+24=36$ squares that don't contain it. There are a total of $25+16+9+4+1 = 55$ squares possible $(25$ - $1\times1$ squares $16$ - $2\times2$ squares $9$ - $3\times3$ squares $4$ - $4\times4$ squares and $1$ - $5\times5$ square), therefore there are $55-36 = 19$ squares that contain the black square, which is $\boxed{\mathrm{(D)}\ 19}$ .

Video Solution by OmegaLearn

https://youtu.be/HhdpuJt78Hg?t=168

~ pi_is_3.14 ~VictorZhang

@@ Line 1: / Line 1: @@
-The answer is 19 or (D) since the number of ways of arranging squares 1x1 through 3x3 are squares( as in power of degree) of their sides. As for the 4x4 and 5x5, it's easy to find the few ways visually.
+==Problem==
+The <math>5\times 5</math> grid shown contains a collection of squares with sizes from <math>1\times 1</math> to <math>5\times 5</math>. How many of these squares contain the black center square?
+<asy>
+for (int i=0; i<5; ++i) {
+ for (int j=0; j<5; ++j) {
+   draw((-2.5+i, -2.5+j)--(-1.5+i, -2.5+j) -- (-1.5+i, -1.5+j) -- (-2.5+i, -1.5+j)--cycle);
+ }
+ fill((-0.5,-0.5)--(-0.5, 0.5)--(0.5,0.5) -- (0.5,-0.5)--cycle, black);
+}
+</asy>
+<math> \mathrm{(A) \ } 12 \qquad \mathrm{(B) \ } 15 \qquad \mathrm{(C) \ } 17 \qquad \mathrm{(D) \ }  19\qquad \mathrm{(E) \ } 20  </math>
+==Solution==
+===Solution 1===
+Since there are five types of squares: <math>1 \times 1, 2 \times 2, 3 \times 3, 4 \times 4,</math> and <math>5 \times 5.</math> We must find how many of each square contain the black shaded square in the center.
+If we list them, we get that
+*There is <math>1</math> of all <math>1\times 1</math> squares, containing the black square
+*There are <math>4</math> of all <math>2\times 2</math> squares, containing the black square
+*There are <math>9</math> of all <math>3\times 3</math> squares, containing the black square
+*There are <math>4</math> of all <math>4\times 4</math> squares, containing the black square
+*There is <math>1</math> of all <math>5\times 5</math> squares, containing the black square
+Thus, the answer is <math>1+4+9+4+1=19\Rightarrow\boxed{\mathrm{(D)}\ 19}</math>.
+===Solution 2===
+We use complementary counting. There are only <math>2\times2</math> and <math>1\times1</math> squares that do not contain the black square. Counting, there are <math>12</math>-<math>2\times2</math> squares, and <math>25-1 = 24</math> <math>1\times1</math> squares that do not contain the black square. That gives <math>12+24=36</math> squares that don't contain it. There are a total of <math>25+16+9+4+1 = 55</math> squares possible <math>(25</math> - <math>1\times1</math> squares <math>16</math> - <math>2\times2</math> squares <math>9</math> - <math>3\times3</math> squares <math>4</math> - <math>4\times4</math> squares and <math>1</math> - <math>5\times5</math> square), therefore there are <math>55-36 = 19</math> squares that contain the black square, which is <math>\boxed{\mathrm{(D)}\ 19}</math>.
+== Video Solution by OmegaLearn ==
+https://youtu.be/HhdpuJt78Hg?t=168
+~ pi_is_3.14
+~VictorZhang
+== Video Solutions ==
+*https://youtu.be/0W3VmFp55cM?t=4697
+*https://youtu.be/aMmF6jz6xA4
+==See also==
+{{AMC10 box|year=2004|ab=A|num-b=15|num-a=17}}
+[[Category:Introductory Combinatorics Problems]]
+{{MAA Notice}}

2004 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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