Difference between revisions of "2023 AMC 8 Problems/Problem 18"

(Problem)
Line 3: Line 3:
  
 
<need image>
 
<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>
 
<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>
  

Revision as of 23:19, 24 January 2023

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 $\frac23$ the side length of the larger triangle. What is the ratio of one trapezoid to the area of the inner triangle?

<need image>

$\textbf{(A) } 1 : 3 \qquad \textbf{(B) } 3 : 8 \qquad \textbf{(C) } 5 : 12 \qquad \textbf{(D) } 7 : 16 \qquad \textbf{(E) } 4 : 9$

Solution

We have $2$ directions going $5$ right or $3$ 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 $5X-3Y=2023$. Where we have to limit the number of moves we do. We can do this by making more of our moves the $5$ move turn then the $3$ 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 $2023$. The least amount of $3$’s added to $2023$ to make a multiple of $5$ is $4$ as $2023 + 4(3) = 2035$. So now we have solved the problem as we just go $\frac{2035}{5} = 407$ hops right, and just do 4 more hops left. Yielding $407 + 4 = \boxed{\text{(D)}411}$ as our answer.


~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, lpieleanu

Animated Video Solution

https://youtu.be/zmRiG52jxpg

~Star League (https://starleague.us)

Video Solution by OmegaLearn (Using Restrictive Counting)

https://youtu.be/gIjhiw1CUgY


See Also

2023 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png