Difference between revisions of "2021 Fall AMC 12B Problems/Problem 2"
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==Video Solution by TheBeautyofMath== | ==Video Solution by TheBeautyofMath== | ||
For AMC 10: https://youtu.be/lC7naDZ1Eu4?t=512 | For AMC 10: https://youtu.be/lC7naDZ1Eu4?t=512 | ||
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For AMC 12: https://youtu.be/yaE5aAmeesc?t=512 | For AMC 12: https://youtu.be/yaE5aAmeesc?t=512 | ||
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{{AMC10 box|year=2021 Fall|ab=B|num-a=3|num-b=1}} | {{AMC10 box|year=2021 Fall|ab=B|num-a=3|num-b=1}} | ||
{{AMC12 box|year=2021 Fall|ab=B|num-a=3|num-b=1}} | {{AMC12 box|year=2021 Fall|ab=B|num-a=3|num-b=1}} | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 23:13, 11 November 2023
- The following problem is from both the 2021 Fall AMC 10B #2 and 2021 Fall AMC 12B #2, so both problems redirect to this page.
Contents
Problem
What is the area of the shaded figure shown below?
![[asy] size(200); defaultpen(linewidth(0.4)+fontsize(12)); pen s = linewidth(0.8)+fontsize(8); pair O,X,Y; O = origin; X = (6,0); Y = (0,5); fill((1,0)--(3,5)--(5,0)--(3,2)--cycle, palegray+opacity(0.2)); for (int i=1; i<7; ++i) { draw((i,0)--(i,5), gray+dashed); label("${"+string(i)+"}$", (i,0), 2*S); if (i<6) { draw((0,i)--(6,i), gray+dashed); label("${"+string(i)+"}$", (0,i), 2*W); } } label("$0$", O, 2*SW); draw(O--X+(0.35,0), black+1.5, EndArrow(10)); draw(O--Y+(0,0.35), black+1.5, EndArrow(10)); draw((1,0)--(3,5)--(5,0)--(3,2)--(1,0), black+1.5); [/asy]](http://latex.artofproblemsolving.com/9/5/0/950ff8f5847f0f8acb13a489295e0c08395cb448.png)
Solution 1 (Area Addition)
The line of symmetry divides the shaded figure into two congruent triangles, each with base 
and height 
Therefore, the area of the shaded figure is ![\[2\cdot\left(\frac12\cdot3\cdot2\right)=2\cdot3=\boxed{\textbf{(B)} \: 6}.\]](http://latex.artofproblemsolving.com/e/2/d/e2d4bdf0660f7ecc1c8db029d77a71a94d5925fc.png)
~MRENTHUSIASM ~Wilhelm Z
Solution 2 (Area Subtraction)
To find the area of the shaded figure, we subtract the area of the smaller triangle (base 
and height 
) from the area of the larger triangle (base and height 
): ![\[\frac12\cdot4\cdot5-\frac12\cdot4\cdot2=10-4=\boxed{\textbf{(B)} \: 6}.\]](http://latex.artofproblemsolving.com/f/4/5/f4549488e5575113e4bc3c0ded44f9db3c88a1f1.png)
~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com)
Solution 3 (Shoelace Theorem)
The consecutive vertices of the shaded figure are 
and 
By the Shoelace Theorem, the area is ![\[\frac12\cdot|(1\cdot2+3\cdot0+5\cdot5+3\cdot0)-(0\cdot3+2\cdot5+0\cdot3+5\cdot1)|=\frac12\cdot12=\boxed{\textbf{(B)} \: 6}.\]](http://latex.artofproblemsolving.com/6/2/9/62964fd1a93ce6cac7270ea14e2167c9365cd3aa.png)
~Taco12 ~I-AM-DA-KING
Solution 4 (Pick's Theorem)
We have lattice points in the interior and 
lattice points on the boundary. By Pick's Theorem, the area of the shaded figure is ![\[4+\frac{6}{2}-1 = 4+3-1 = \boxed{\textbf{(B)} \: 6}.\]](http://latex.artofproblemsolving.com/8/1/4/81411734430f77e31c966a110c816c2ef4d4b3fc.png)
~danprathab
Video Solution by Interstigation
https://youtu.be/p9_RH4s-kBA?t=110
~Interstigation
Video Solution
~Education, the Study of Everything
Video Solution by WhyMath
~savannahsolver
Video Solution by TheBeautyofMath
For AMC 10: https://youtu.be/lC7naDZ1Eu4?t=512
For AMC 12: https://youtu.be/yaE5aAmeesc?t=512
~IceMatrix
See Also
| 2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 10 Problems and Solutions | ||
| 2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 1 |
Followed by Problem 3 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
