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Difference between revisions of "2022 AMC 12A Problems/Problem 3"

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=Solution 1 (List)=
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<asy>
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fill((3,2.5)--(3,4.5)--(5.3,4.5)--(5.3,2.5)--cycle,mediumgray);
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draw((0,0)--(7,0)--(7,7)--(0,7)--(0,0));
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draw((3,0)--(3,4.5));
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draw((0,4.5)--(5.3,4.5));
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draw((5.3,7)--(5.3,2.5));
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draw((7,2.5)--(3,2.5));
  
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label("A",(0,0),S);
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label("B",(3,0),S);
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label("C",(7,0),S);
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label("D",(7.5,3),S);
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label("E",(7.5,7.8),S);
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label("F",(5.5,7.8),S);
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label("G",(-.5,7.8),S);
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label("H",(-.5,5),S);
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</asy>
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 +
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By finding the dimensions of the middle rectangle, we need to find the dimensions of the other 4 rectangles. By doing this, there is going to be a rule.
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Rule: <math>AB + BC = CD + DE = EF + FG = GH + AH</math>
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Let's make a list of all the dimensions of the rectangles from the diagram. We have to fill in the dimensions from up above, but still apply to the rule.
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<math>AH\times AB</math>
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<math>BC\times CD</math>
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<math>DE\times EF</math>
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<math>FG\times GH</math>
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By applying the rule, we get <math>AB=6, BC=2, CD=7, DE=1, EF=6, FG=2</math>, and <math>GH=3</math>
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By substitution, we get this list
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<math>5\times 6</math>
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<math>2\times 7</math>
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<math>1\times 6</math>
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<math>2\times 3</math>
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Notice how the only dimension not used in the list was <math>2\times 4</math> and that corresponds with B so the answer is, <math>\textbf{(B) }B.</math>
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~ghfhgvghj10

Revision as of 16:52, 19 November 2022

Solution 1 (List)

[asy] fill((3,2.5)--(3,4.5)--(5.3,4.5)--(5.3,2.5)--cycle,mediumgray); draw((0,0)--(7,0)--(7,7)--(0,7)--(0,0)); draw((3,0)--(3,4.5)); draw((0,4.5)--(5.3,4.5)); draw((5.3,7)--(5.3,2.5)); draw((7,2.5)--(3,2.5)); label("A",(0,0),S); label("B",(3,0),S); label("C",(7,0),S); label("D",(7.5,3),S); label("E",(7.5,7.8),S); label("F",(5.5,7.8),S); label("G",(-.5,7.8),S); label("H",(-.5,5),S); [/asy]


By finding the dimensions of the middle rectangle, we need to find the dimensions of the other 4 rectangles. By doing this, there is going to be a rule.

Rule: $AB + BC = CD + DE = EF + FG = GH + AH$

Let's make a list of all the dimensions of the rectangles from the diagram. We have to fill in the dimensions from up above, but still apply to the rule.

$AH\times AB$

$BC\times CD$

$DE\times EF$

$FG\times GH$

By applying the rule, we get $AB=6, BC=2, CD=7, DE=1, EF=6, FG=2$, and $GH=3$

By substitution, we get this list

$5\times 6$

$2\times 7$

$1\times 6$

$2\times 3$

Notice how the only dimension not used in the list was $2\times 4$ and that corresponds with B so the answer is, $\textbf{(B) }B.$

~ghfhgvghj10

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