Difference between revisions of "2023 AMC 8 Problems/Problem 18"
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| + | ==Problem== | ||
| + | An equilateral triangle is placed inside a larger equilateral triangle so that the region between them can be divided into three congruent trapezoids, as shown below. The side length of the inner triangle is <math>\frac23</math> the side length of the larger triangle. What is the ratio of one trapezoid to the area of the inner triangle? | ||
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| + | <need image> | ||
| + | <math>\textbf{(A) } 1 : 3 \qquad \textbf{(B) } 3 : 8 \qquad \textbf{(C) } 5 : 12 \qquad \textbf{(D) } 7 : 16 \qquad \textbf{(E) } 4 : 9</math> | ||
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==Solution== | ==Solution== | ||
We have <math>2</math> directions going <math>5</math> right or <math>3</math> left. We can assign a variable to each of these directions. We can call going right 1 direction X and we can call going 1 left Y. We can build a equation of <math>5X-3Y=2023</math>. Where we have to limit the number of moves we do. We can do this by making more of our moves the <math>5</math> move turn then the <math>3</math> move turn. The first obvious step is to go some amount of moves in the → direction then subtract off in the ← direction to land on <math>2023</math>. The least amount of <math>3</math>’s added to <math>2023</math> to make a multiple of <math>5</math> is <math>4</math> as <math>2023 + 4(3) = 2035</math>. So now we have solved the problem as we just go <math>\frac{2035}{5} = 407</math> hops right, and just do 4 more hops left. Yielding <math>407 + 4 = \boxed{\text{(D)}411}</math> as our answer. | We have <math>2</math> directions going <math>5</math> right or <math>3</math> left. We can assign a variable to each of these directions. We can call going right 1 direction X and we can call going 1 left Y. We can build a equation of <math>5X-3Y=2023</math>. Where we have to limit the number of moves we do. We can do this by making more of our moves the <math>5</math> move turn then the <math>3</math> move turn. The first obvious step is to go some amount of moves in the → direction then subtract off in the ← direction to land on <math>2023</math>. The least amount of <math>3</math>’s added to <math>2023</math> to make a multiple of <math>5</math> is <math>4</math> as <math>2023 + 4(3) = 2035</math>. So now we have solved the problem as we just go <math>\frac{2035}{5} = 407</math> hops right, and just do 4 more hops left. Yielding <math>407 + 4 = \boxed{\text{(D)}411}</math> as our answer. | ||
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==Video Solution by OmegaLearn (Using Restrictive Counting)== | ==Video Solution by OmegaLearn (Using Restrictive Counting)== | ||
https://youtu.be/gIjhiw1CUgY | https://youtu.be/gIjhiw1CUgY | ||
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| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2023|num-b=17|num-a=19}} | ||
| + | {{MAA Notice}} | ||
Revision as of 23:19, 24 January 2023
Contents
Problem
An equilateral triangle is placed inside a larger equilateral triangle so that the region between them can be divided into three congruent trapezoids, as shown below. The side length of the inner triangle is 
the side length of the larger triangle. What is the ratio of one trapezoid to the area of the inner triangle?
<need image>
Solution
We have 
directions going 
right or 
left. We can assign a variable to each of these directions. We can call going right 1 direction X and we can call going 1 left Y. We can build a equation of 
. Where we have to limit the number of moves we do. We can do this by making more of our moves the move turn then the move turn. The first obvious step is to go some amount of moves in the → direction then subtract off in the ← direction to land on 
. The least amount of ’s added to to make a multiple of is 
as 
. So now we have solved the problem as we just go 
hops right, and just do 4 more hops left. Yielding 
as our answer.
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, lpieleanu
Animated Video Solution
~Star League (https://starleague.us)
Video Solution by OmegaLearn (Using Restrictive Counting)
See Also
| 2023 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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. 
