Difference between revisions of "2011 AMC 8 Problems/Problem 19"

Revision as of 21:14, 15 January 2025

Problem

How many rectangles are in this figure?

$[asy] pair A,B,C,D,E,F,G,H,I,J,K,L; A=(0,0); B=(20,0); C=(20,20); D=(0,20); draw(A--B--C--D--cycle); E=(-10,-5); F=(13,-5); G=(13,5); H=(-10,5); draw(E--F--G--H--cycle); I=(10,-20); J=(18,-20); K=(18,13); L=(10,13); draw(I--J--K--L--cycle);[/asy]$

$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$

Solution

The figure can be divided into $7$ sections. The number of rectangles with just one section is $3.$ The number of rectangles with two sections is $5.$ There are none with only three sections. The number of rectangles with four sections is $3.$

$3+5+3=\boxed{\textbf{(D)}\ 11}$

Solution 2 (Count by Reduction)

We can remove the 3 big blocks of rectangles one by one. 7 (left) + 3 (bottom) + 1 = 11 are removed in total.

~aliciawu

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=mYn6tNxrWBU Wrong video, does not contain the solution

Video Solution by WhyMath

https://youtu.be/IEeJsGh3ltk

2011 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 34: / Line 34: @@
 ==Video Solution 1 by SpreadTheMathLove==
 https://www.youtube.com/watch?v=mYn6tNxrWBU
+Wrong video, does not contain the solution
 ==Video Solution by WhyMath==

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