Difference between revisions of "2023 AMC 8 Problems/Problem 7"

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<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4</math>
 
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4</math>
  
==Solution==
+
==Solution 1==
 
If we extend the lines, we have
 
If we extend the lines, we have
  
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~MrThinker
 
~MrThinker
 +
 +
==Solution 2==
 +
If the analytic expression for line <math> AB </math> is <math> y=k_{1}x+b_{1} </math>, and the analytic expression for line <math> CD </math> is <math> y=k_{2}x+b_{2} </math>, we have the systems of equations:
 +
<cmath>  \begin{cases} b_{1} = 0 \\ 3k_{1} + b_{1} = 1 \end{cases}  </cmath>
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and
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<cmath>  \begin{cases} b_{2} = 10 \\ 2k_{2} + b_{2} = 9 \end{cases}  </cmath>
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Solving the systems, we have:
 +
<cmath> \begin{cases} k_{1} = \frac{1}{3} \\ b_{1} = 0 \end{cases} </cmath>
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and
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<cmath> \begin{cases} k_{2} = -\frac{1}{2} \\ b_{2} = 10 \end{cases} </cmath>
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Therefore, we can determine that the expression for line <math> AB </math> is <math> y=\frac{1}{3}x </math>. and the expression for line <math> CD </math> is <math> y=-\frac{1}{2}x + 10 </math>. When <math> x=15 </math>, the coordinates that line <math> AB </math> and line <math> CD </math> pass through are <math> (15, 5) </math> and <math> (15, \frac{5}{2}) </math>, and <math> (15, 5) </math> lies perfectly on one vertex of the rectangle while the <math> y </math> coordinate of <math> (15, \frac{5}{2}) </math> is out of the range <math> 3 \leq y \leq 5 </math> (lower than the bottom left corner of the rectangle <math> (15, 3) </math>). Considering that the <math> x </math> value of the line <math> CD </math> will only decrease, and the <math> x </math> value of the line <math> AB </math> will only increase, there will not be another point on the rectangle that lies on either of the two lines. Thus, we can conclude that the answer is <math>\boxed{\textbf{(B)}\ 1}</math>
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~Bloggish
  
 
==Video Solution by Magic Square==
 
==Video Solution by Magic Square==

Revision as of 06:15, 26 January 2023

Problem

A rectangle, with sides parallel to the $x$-axis and $y$-axis, has opposite vertices located at $(15, 3)$ and $(16, 5)$. A line drawn through points $A(0, 0)$ and $B(3, 1)$. Another line is drawn through points $C(0, 10)$ and $D(2, 9)$. How many points on the rectangle lie on at least one of the two lines?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution 1

If we extend the lines, we have

Screenshot 2023-01-25 8.26.48 AM.png

Hence, we see that the answer is $\boxed{\textbf{(B)}\ 1}$

~MrThinker

Solution 2

If the analytic expression for line $AB$ is $y=k_{1}x+b_{1}$, and the analytic expression for line $CD$ is $y=k_{2}x+b_{2}$, we have the systems of equations: \[\begin{cases} b_{1} = 0 \\ 3k_{1} + b_{1} = 1 \end{cases}\] and \[\begin{cases} b_{2} = 10 \\ 2k_{2} + b_{2} = 9 \end{cases}\] Solving the systems, we have: \[\begin{cases} k_{1} = \frac{1}{3} \\ b_{1} = 0 \end{cases}\] and \[\begin{cases} k_{2} = -\frac{1}{2} \\ b_{2} = 10 \end{cases}\] Therefore, we can determine that the expression for line $AB$ is $y=\frac{1}{3}x$. and the expression for line $CD$ is $y=-\frac{1}{2}x + 10$. When $x=15$, the coordinates that line $AB$ and line $CD$ pass through are $(15, 5)$ and $(15, \frac{5}{2})$, and $(15, 5)$ lies perfectly on one vertex of the rectangle while the $y$ coordinate of $(15, \frac{5}{2})$ is out of the range $3 \leq y \leq 5$ (lower than the bottom left corner of the rectangle $(15, 3)$). Considering that the $x$ value of the line $CD$ will only decrease, and the $x$ value of the line $AB$ will only increase, there will not be another point on the rectangle that lies on either of the two lines. Thus, we can conclude that the answer is $\boxed{\textbf{(B)}\ 1}$

~Bloggish

Video Solution by Magic Square

https://youtu.be/-N46BeEKaCQ?t=5151

See Also

2023 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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

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