Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m
 
(34 intermediate revisions by 13 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
−
A rectangle, with sides parallel to the <math>x</math>-axis and <math>y</math>-axis, has opposite vertices located at <math>(15, 3)</math> and <math>(16, 5)</math>. A line drawn through points <math>A(0, 0)</math> and <math>B(3, 1)</math>. Another line is drawn through points <math>C(0, 10)</math> and <math>D(2, 9)</math>. How many points on the rectangle lie on at least one of the two lines?
+
A rectangle, with sides parallel to the <math>x</math>-axis and <math>y</math>-axis, has opposite vertices located at <math>(15, 3)</math> and <math>(16, 5)</math>. A line is drawn through points <math>A(0, 0)</math> and <math>B(3, 1)</math>. Another line is drawn through points <math>C(0, 10)</math> and <math>D(2, 9)</math>. How many points on the rectangle lie on at least one of the two lines?
 +
<asy>
 +
usepackage("mathptmx");
 +
size(9cm);
 +
draw((0,-.5)--(0,11),EndArrow(size=.15cm));
 +
draw((1,0)--(1,11),mediumgray);
 +
draw((2,0)--(2,11),mediumgray);
 +
draw((3,0)--(3,11),mediumgray);
 +
draw((4,0)--(4,11),mediumgray);
 +
draw((5,0)--(5,11),mediumgray);
 +
draw((6,0)--(6,11),mediumgray);
 +
draw((7,0)--(7,11),mediumgray);
 +
draw((8,0)--(8,11),mediumgray);
 +
draw((9,0)--(9,11),mediumgray);
 +
draw((10,0)--(10,11),mediumgray);
 +
draw((11,0)--(11,11),mediumgray);
 +
draw((12,0)--(12,11),mediumgray);
 +
draw((13,0)--(13,11),mediumgray);
 +
draw((14,0)--(14,11),mediumgray);
 +
draw((15,0)--(15,11),mediumgray);
 +
draw((16,0)--(16,11),mediumgray);
  
 +
draw((-.5,0)--(17,0),EndArrow(size=.15cm));
 +
draw((0,1)--(17,1),mediumgray);
 +
draw((0,2)--(17,2),mediumgray);
 +
draw((0,3)--(17,3),mediumgray);
 +
draw((0,4)--(17,4),mediumgray);
 +
draw((0,5)--(17,5),mediumgray);
 +
draw((0,6)--(17,6),mediumgray);
 +
draw((0,7)--(17,7),mediumgray);
 +
draw((0,8)--(17,8),mediumgray);
 +
draw((0,9)--(17,9),mediumgray);
 +
draw((0,10)--(17,10),mediumgray);
 +
 +
draw((-.13,1)--(.13,1));
 +
draw((-.13,2)--(.13,2));
 +
draw((-.13,3)--(.13,3));
 +
draw((-.13,4)--(.13,4));
 +
draw((-.13,5)--(.13,5));
 +
draw((-.13,6)--(.13,6));
 +
draw((-.13,7)--(.13,7));
 +
draw((-.13,8)--(.13,8));
 +
draw((-.13,9)--(.13,9));
 +
draw((-.13,10)--(.13,10));
 +
 +
draw((1,-.13)--(1,.13));
 +
draw((2,-.13)--(2,.13));
 +
draw((3,-.13)--(3,.13));
 +
draw((4,-.13)--(4,.13));
 +
draw((5,-.13)--(5,.13));
 +
draw((6,-.13)--(6,.13));
 +
draw((7,-.13)--(7,.13));
 +
draw((8,-.13)--(8,.13));
 +
draw((9,-.13)--(9,.13));
 +
draw((10,-.13)--(10,.13));
 +
draw((11,-.13)--(11,.13));
 +
draw((12,-.13)--(12,.13));
 +
draw((13,-.13)--(13,.13));
 +
draw((14,-.13)--(14,.13));
 +
draw((15,-.13)--(15,.13));
 +
draw((16,-.13)--(16,.13));
 +
 +
label(scale(.7)*"$1$", (1,-.13), S);
 +
label(scale(.7)*"$2$", (2,-.13), S);
 +
label(scale(.7)*"$3$", (3,-.13), S);
 +
label(scale(.7)*"$4$", (4,-.13), S);
 +
label(scale(.7)*"$5$", (5,-.13), S);
 +
label(scale(.7)*"$6$", (6,-.13), S);
 +
label(scale(.7)*"$7$", (7,-.13), S);
 +
label(scale(.7)*"$8$", (8,-.13), S);
 +
label(scale(.7)*"$9$", (9,-.13), S);
 +
label(scale(.7)*"$10$", (10,-.13), S);
 +
label(scale(.7)*"$11$", (11,-.13), S);
 +
label(scale(.7)*"$12$", (12,-.13), S);
 +
label(scale(.7)*"$13$", (13,-.13), S);
 +
label(scale(.7)*"$14$", (14,-.13), S);
 +
label(scale(.7)*"$15$", (15,-.13), S);
 +
label(scale(.7)*"$16$", (16,-.13), S);
 +
 +
label(scale(.7)*"$1$", (-.13,1), W);
 +
label(scale(.7)*"$2$", (-.13,2), W);
 +
label(scale(.7)*"$3$", (-.13,3), W);
 +
label(scale(.7)*"$4$", (-.13,4), W);
 +
label(scale(.7)*"$5$", (-.13,5), W);
 +
label(scale(.7)*"$6$", (-.13,6), W);
 +
label(scale(.7)*"$7$", (-.13,7), W);
 +
label(scale(.7)*"$8$", (-.13,8), W);
 +
label(scale(.7)*"$9$", (-.13,9), W);
 +
label(scale(.7)*"$10$", (-.13,10), W);
 +
 +
 +
dot((0,0),linewidth(4));
 +
label(scale(.75)*"$A$", (0,0), NE);
 +
dot((3,1),linewidth(4));
 +
label(scale(.75)*"$B$", (3,1), NE);
 +
 +
dot((0,10),linewidth(4));
 +
label(scale(.75)*"$C$", (0,10), NE);
 +
dot((2,9),linewidth(4));
 +
label(scale(.75)*"$D$", (2,9), NE);
 +
 +
draw((15,3)--(16,3)--(16,5)--(15,5)--cycle,linewidth(1.125));
 +
dot((15,3),linewidth(4));
 +
dot((16,3),linewidth(4));
 +
dot((16,5),linewidth(4));
 +
dot((15,5),linewidth(4));
 +
</asy>
 
<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 the following diagram:
 +
<asy>
 +
usepackage("mathptmx");
 +
size(9cm);
 +
draw((0,-.5)--(0,11),EndArrow(size=.15cm));
 +
draw((1,0)--(1,11),mediumgray);
 +
draw((2,0)--(2,11),mediumgray);
 +
draw((3,0)--(3,11),mediumgray);
 +
draw((4,0)--(4,11),mediumgray);
 +
draw((5,0)--(5,11),mediumgray);
 +
draw((6,0)--(6,11),mediumgray);
 +
draw((7,0)--(7,11),mediumgray);
 +
draw((8,0)--(8,11),mediumgray);
 +
draw((9,0)--(9,11),mediumgray);
 +
draw((10,0)--(10,11),mediumgray);
 +
draw((11,0)--(11,11),mediumgray);
 +
draw((12,0)--(12,11),mediumgray);
 +
draw((13,0)--(13,11),mediumgray);
 +
draw((14,0)--(14,11),mediumgray);
 +
draw((15,0)--(15,11),mediumgray);
 +
draw((16,0)--(16,11),mediumgray);
 +
 
 +
draw((-.5,0)--(17,0),EndArrow(size=.15cm));
 +
draw((0,1)--(17,1),mediumgray);
 +
draw((0,2)--(17,2),mediumgray);
 +
draw((0,3)--(17,3),mediumgray);
 +
draw((0,4)--(17,4),mediumgray);
 +
draw((0,5)--(17,5),mediumgray);
 +
draw((0,6)--(17,6),mediumgray);
 +
draw((0,7)--(17,7),mediumgray);
 +
draw((0,8)--(17,8),mediumgray);
 +
draw((0,9)--(17,9),mediumgray);
 +
draw((0,10)--(17,10),mediumgray);
 +
 
 +
draw((-.13,1)--(.13,1));
 +
draw((-.13,2)--(.13,2));
 +
draw((-.13,3)--(.13,3));
 +
draw((-.13,4)--(.13,4));
 +
draw((-.13,5)--(.13,5));
 +
draw((-.13,6)--(.13,6));
 +
draw((-.13,7)--(.13,7));
 +
draw((-.13,8)--(.13,8));
 +
draw((-.13,9)--(.13,9));
 +
draw((-.13,10)--(.13,10));
 +
 
 +
draw((1,-.13)--(1,.13));
 +
draw((2,-.13)--(2,.13));
 +
draw((3,-.13)--(3,.13));
 +
draw((4,-.13)--(4,.13));
 +
draw((5,-.13)--(5,.13));
 +
draw((6,-.13)--(6,.13));
 +
draw((7,-.13)--(7,.13));
 +
draw((8,-.13)--(8,.13));
 +
draw((9,-.13)--(9,.13));
 +
draw((10,-.13)--(10,.13));
 +
draw((11,-.13)--(11,.13));
 +
draw((12,-.13)--(12,.13));
 +
draw((13,-.13)--(13,.13));
 +
draw((14,-.13)--(14,.13));
 +
draw((15,-.13)--(15,.13));
 +
draw((16,-.13)--(16,.13));
 +
 
 +
label(scale(.7)*"$1$", (1,-.13), S);
 +
label(scale(.7)*"$2$", (2,-.13), S);
 +
label(scale(.7)*"$3$", (3,-.13), S);
 +
label(scale(.7)*"$4$", (4,-.13), S);
 +
label(scale(.7)*"$5$", (5,-.13), S);
 +
label(scale(.7)*"$6$", (6,-.13), S);
 +
label(scale(.7)*"$7$", (7,-.13), S);
 +
label(scale(.7)*"$8$", (8,-.13), S);
 +
label(scale(.7)*"$9$", (9,-.13), S);
 +
label(scale(.7)*"$10$", (10,-.13), S);
 +
label(scale(.7)*"$11$", (11,-.13), S);
 +
label(scale(.7)*"$12$", (12,-.13), S);
 +
label(scale(.7)*"$13$", (13,-.13), S);
 +
label(scale(.7)*"$14$", (14,-.13), S);
 +
label(scale(.7)*"$15$", (15,-.13), S);
 +
label(scale(.7)*"$16$", (16,-.13), S);
 +
 
 +
label(scale(.7)*"$1$", (-.13,1), W);
 +
label(scale(.7)*"$2$", (-.13,2), W);
 +
label(scale(.7)*"$3$", (-.13,3), W);
 +
label(scale(.7)*"$4$", (-.13,4), W);
 +
label(scale(.7)*"$5$", (-.13,5), W);
 +
label(scale(.7)*"$6$", (-.13,6), W);
 +
label(scale(.7)*"$7$", (-.13,7), W);
 +
label(scale(.7)*"$8$", (-.13,8), W);
 +
label(scale(.7)*"$9$", (-.13,9), W);
 +
label(scale(.7)*"$10$", (-.13,10), W);
 +
 
 +
draw((0,10)--(17,1.5),blue);
 +
draw((0,0)--(17,17/3),blue);
 +
 
 +
dot((0,0),linewidth(4));
 +
label(scale(.75)*"$A$", (0,0), NE);
 +
dot((3,1),linewidth(4));
 +
label(scale(.75)*"$B$", (3,1), NE);
  
−
[[Image:Screenshot 2023-01-25 8.26.48 AM.png|thumb|center|400px]]
+
dot((0,10),linewidth(4));
 +
label(scale(.75)*"$C$", (0,10), NE);
 +
dot((2,9),linewidth(4));
 +
label(scale(.75)*"$D$", (2,9), NE);
  
−
Hence, we see that the answer is <math>\boxed{\textbf{(B)}\ 1}</math>
+
draw((15,3)--(16,3)--(16,5)--(15,5)--cycle,linewidth(1.125));
 +
dot((15,3),linewidth(4));
 +
dot((16,3),linewidth(4));
 +
dot((16,5),linewidth(4));
 +
dot((15,5),linewidth(4));
 +
</asy>
 +
Therefore, we see that the answer is <math>\boxed{\textbf{(B)}\ 1}.</math>
  
 
~MrThinker
 
~MrThinker
 +
 +
==Solution 2==
 +
Note that the <math> y </math>-intercepts of line <math> AB </math> and line <math> CD </math> are <math> 0 </math> and <math> 10 </math>. If the analytic expression for line <math> AB </math> is <math> y=k_{1}x </math>, and the analytic expression for line <math> CD </math> is <math> y=k_{2}x+10 </math>, we have equations: <math> 3k_{1} = 1 </math> and <math> 2k_{2} + 10 = 9 </math>. Solving these equations, we can find out that <math> k_{1} = \frac{1}{3}</math> and <math>k_{2} = -\frac{1}{2}</math>. 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> \left(15, \frac{5}{2}\right) </math>, and <math> (15, 5) </math> lies perfectly on one vertex of the rectangle while the <math> y </math> coordinate of <math> \left(15, \frac{5}{2}\right) </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> y </math> value of the line <math> CD </math> will only decrease, and the <math> y </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>
 +
 +
~[[User:Bloggish|Bloggish]]
 +
 +
==Video Solution (How to CREATIVELY THINK!!!) ==
 +
https://youtu.be/NUaes2N_4pM
 +
~Education the Study of everything
 +
 +
==Video Solution by Math-X (Smart and Simple)==
 +
https://youtu.be/Ku_c1YHnLt0?si=P3DtuhzhiVr2Jv0r&t=947 ~Math-X
 +
 +
==Video Solution by Magic Square==
 +
https://youtu.be/-N46BeEKaCQ?t=5151
 +
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=EcrktBc8zrM&ab_channel=SpreadTheMathLove (@11:08)
 +
==Video Solution by Interstigation==
 +
https://youtu.be/DBqko2xATxs&t=534
 +
 +
==Video Solution by WhyMath==
 +
https://youtu.be/jCjF9duTQxk
 +
 +
~savannahsolver
 +
 +
==Video Solution by harungurcan==
 +
https://www.youtube.com/watch?v=35BW7bsm_Cg&t=778s
 +
 +
~harungurcan
 +
 +
==Video Solution by Dr. David==
 +
https://youtu.be/LMeg3r3VFdE
  
 
==See Also==  
 
==See Also==  
−
{{AMC8 box|year=2023|num-b=4|num-a=6}}
+
{{AMC8 box|year=2023|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 01:55, 4 November 2024

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 is 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? [asy] usepackage("mathptmx"); size(9cm); draw((0,-.5)--(0,11),EndArrow(size=.15cm)); draw((1,0)--(1,11),mediumgray); draw((2,0)--(2,11),mediumgray); draw((3,0)--(3,11),mediumgray); draw((4,0)--(4,11),mediumgray); draw((5,0)--(5,11),mediumgray); draw((6,0)--(6,11),mediumgray); draw((7,0)--(7,11),mediumgray); draw((8,0)--(8,11),mediumgray); draw((9,0)--(9,11),mediumgray); draw((10,0)--(10,11),mediumgray); draw((11,0)--(11,11),mediumgray); draw((12,0)--(12,11),mediumgray); draw((13,0)--(13,11),mediumgray); draw((14,0)--(14,11),mediumgray); draw((15,0)--(15,11),mediumgray); draw((16,0)--(16,11),mediumgray); draw((-.5,0)--(17,0),EndArrow(size=.15cm)); draw((0,1)--(17,1),mediumgray); draw((0,2)--(17,2),mediumgray); draw((0,3)--(17,3),mediumgray); draw((0,4)--(17,4),mediumgray); draw((0,5)--(17,5),mediumgray); draw((0,6)--(17,6),mediumgray); draw((0,7)--(17,7),mediumgray); draw((0,8)--(17,8),mediumgray); draw((0,9)--(17,9),mediumgray); draw((0,10)--(17,10),mediumgray); draw((-.13,1)--(.13,1)); draw((-.13,2)--(.13,2)); draw((-.13,3)--(.13,3)); draw((-.13,4)--(.13,4)); draw((-.13,5)--(.13,5)); draw((-.13,6)--(.13,6)); draw((-.13,7)--(.13,7)); draw((-.13,8)--(.13,8)); draw((-.13,9)--(.13,9)); draw((-.13,10)--(.13,10)); draw((1,-.13)--(1,.13)); draw((2,-.13)--(2,.13)); draw((3,-.13)--(3,.13)); draw((4,-.13)--(4,.13)); draw((5,-.13)--(5,.13)); draw((6,-.13)--(6,.13)); draw((7,-.13)--(7,.13)); draw((8,-.13)--(8,.13)); draw((9,-.13)--(9,.13)); draw((10,-.13)--(10,.13)); draw((11,-.13)--(11,.13)); draw((12,-.13)--(12,.13)); draw((13,-.13)--(13,.13)); draw((14,-.13)--(14,.13)); draw((15,-.13)--(15,.13)); draw((16,-.13)--(16,.13)); label(scale(.7)*"$1$", (1,-.13), S); label(scale(.7)*"$2$", (2,-.13), S); label(scale(.7)*"$3$", (3,-.13), S); label(scale(.7)*"$4$", (4,-.13), S); label(scale(.7)*"$5$", (5,-.13), S); label(scale(.7)*"$6$", (6,-.13), S); label(scale(.7)*"$7$", (7,-.13), S); label(scale(.7)*"$8$", (8,-.13), S); label(scale(.7)*"$9$", (9,-.13), S); label(scale(.7)*"$10$", (10,-.13), S); label(scale(.7)*"$11$", (11,-.13), S); label(scale(.7)*"$12$", (12,-.13), S); label(scale(.7)*"$13$", (13,-.13), S); label(scale(.7)*"$14$", (14,-.13), S); label(scale(.7)*"$15$", (15,-.13), S); label(scale(.7)*"$16$", (16,-.13), S); label(scale(.7)*"$1$", (-.13,1), W); label(scale(.7)*"$2$", (-.13,2), W); label(scale(.7)*"$3$", (-.13,3), W); label(scale(.7)*"$4$", (-.13,4), W); label(scale(.7)*"$5$", (-.13,5), W); label(scale(.7)*"$6$", (-.13,6), W); label(scale(.7)*"$7$", (-.13,7), W); label(scale(.7)*"$8$", (-.13,8), W); label(scale(.7)*"$9$", (-.13,9), W); label(scale(.7)*"$10$", (-.13,10), W); dot((0,0),linewidth(4)); label(scale(.75)*"$A$", (0,0), NE); dot((3,1),linewidth(4)); label(scale(.75)*"$B$", (3,1), NE); dot((0,10),linewidth(4)); label(scale(.75)*"$C$", (0,10), NE); dot((2,9),linewidth(4)); label(scale(.75)*"$D$", (2,9), NE); draw((15,3)--(16,3)--(16,5)--(15,5)--cycle,linewidth(1.125)); dot((15,3),linewidth(4)); dot((16,3),linewidth(4)); dot((16,5),linewidth(4)); dot((15,5),linewidth(4)); [/asy] $\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution 1

If we extend the lines, we have the following diagram: [asy] usepackage("mathptmx"); size(9cm); draw((0,-.5)--(0,11),EndArrow(size=.15cm)); draw((1,0)--(1,11),mediumgray); draw((2,0)--(2,11),mediumgray); draw((3,0)--(3,11),mediumgray); draw((4,0)--(4,11),mediumgray); draw((5,0)--(5,11),mediumgray); draw((6,0)--(6,11),mediumgray); draw((7,0)--(7,11),mediumgray); draw((8,0)--(8,11),mediumgray); draw((9,0)--(9,11),mediumgray); draw((10,0)--(10,11),mediumgray); draw((11,0)--(11,11),mediumgray); draw((12,0)--(12,11),mediumgray); draw((13,0)--(13,11),mediumgray); draw((14,0)--(14,11),mediumgray); draw((15,0)--(15,11),mediumgray); draw((16,0)--(16,11),mediumgray); draw((-.5,0)--(17,0),EndArrow(size=.15cm)); draw((0,1)--(17,1),mediumgray); draw((0,2)--(17,2),mediumgray); draw((0,3)--(17,3),mediumgray); draw((0,4)--(17,4),mediumgray); draw((0,5)--(17,5),mediumgray); draw((0,6)--(17,6),mediumgray); draw((0,7)--(17,7),mediumgray); draw((0,8)--(17,8),mediumgray); draw((0,9)--(17,9),mediumgray); draw((0,10)--(17,10),mediumgray); draw((-.13,1)--(.13,1)); draw((-.13,2)--(.13,2)); draw((-.13,3)--(.13,3)); draw((-.13,4)--(.13,4)); draw((-.13,5)--(.13,5)); draw((-.13,6)--(.13,6)); draw((-.13,7)--(.13,7)); draw((-.13,8)--(.13,8)); draw((-.13,9)--(.13,9)); draw((-.13,10)--(.13,10)); draw((1,-.13)--(1,.13)); draw((2,-.13)--(2,.13)); draw((3,-.13)--(3,.13)); draw((4,-.13)--(4,.13)); draw((5,-.13)--(5,.13)); draw((6,-.13)--(6,.13)); draw((7,-.13)--(7,.13)); draw((8,-.13)--(8,.13)); draw((9,-.13)--(9,.13)); draw((10,-.13)--(10,.13)); draw((11,-.13)--(11,.13)); draw((12,-.13)--(12,.13)); draw((13,-.13)--(13,.13)); draw((14,-.13)--(14,.13)); draw((15,-.13)--(15,.13)); draw((16,-.13)--(16,.13)); label(scale(.7)*"$1$", (1,-.13), S); label(scale(.7)*"$2$", (2,-.13), S); label(scale(.7)*"$3$", (3,-.13), S); label(scale(.7)*"$4$", (4,-.13), S); label(scale(.7)*"$5$", (5,-.13), S); label(scale(.7)*"$6$", (6,-.13), S); label(scale(.7)*"$7$", (7,-.13), S); label(scale(.7)*"$8$", (8,-.13), S); label(scale(.7)*"$9$", (9,-.13), S); label(scale(.7)*"$10$", (10,-.13), S); label(scale(.7)*"$11$", (11,-.13), S); label(scale(.7)*"$12$", (12,-.13), S); label(scale(.7)*"$13$", (13,-.13), S); label(scale(.7)*"$14$", (14,-.13), S); label(scale(.7)*"$15$", (15,-.13), S); label(scale(.7)*"$16$", (16,-.13), S); label(scale(.7)*"$1$", (-.13,1), W); label(scale(.7)*"$2$", (-.13,2), W); label(scale(.7)*"$3$", (-.13,3), W); label(scale(.7)*"$4$", (-.13,4), W); label(scale(.7)*"$5$", (-.13,5), W); label(scale(.7)*"$6$", (-.13,6), W); label(scale(.7)*"$7$", (-.13,7), W); label(scale(.7)*"$8$", (-.13,8), W); label(scale(.7)*"$9$", (-.13,9), W); label(scale(.7)*"$10$", (-.13,10), W); draw((0,10)--(17,1.5),blue); draw((0,0)--(17,17/3),blue); dot((0,0),linewidth(4)); label(scale(.75)*"$A$", (0,0), NE); dot((3,1),linewidth(4)); label(scale(.75)*"$B$", (3,1), NE); dot((0,10),linewidth(4)); label(scale(.75)*"$C$", (0,10), NE); dot((2,9),linewidth(4)); label(scale(.75)*"$D$", (2,9), NE); draw((15,3)--(16,3)--(16,5)--(15,5)--cycle,linewidth(1.125)); dot((15,3),linewidth(4)); dot((16,3),linewidth(4)); dot((16,5),linewidth(4)); dot((15,5),linewidth(4)); [/asy] Therefore, we see that the answer is $\boxed{\textbf{(B)}\ 1}.$

~MrThinker

Solution 2

Note that the $y$-intercepts of line $AB$ and line $CD$ are $0$ and $10$. If the analytic expression for line $AB$ is $y=k_{1}x$, and the analytic expression for line $CD$ is $y=k_{2}x+10$, we have equations: $3k_{1} = 1$ and $2k_{2} + 10 = 9$. Solving these equations, we can find out that $k_{1} = \frac{1}{3}$ and $k_{2} = -\frac{1}{2}$. 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 $\left(15, \frac{5}{2}\right)$, and $(15, 5)$ lies perfectly on one vertex of the rectangle while the $y$ coordinate of $\left(15, \frac{5}{2}\right)$ is out of the range $3 \leq y \leq 5$ (lower than the bottom left corner of the rectangle $(15, 3)$). Considering that the $y$ value of the line $CD$ will only decrease, and the $y$ 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 (How to CREATIVELY THINK!!!)

https://youtu.be/NUaes2N_4pM ~Education the Study of everything

Video Solution by Math-X (Smart and Simple)

https://youtu.be/Ku_c1YHnLt0?si=P3DtuhzhiVr2Jv0r&t=947 ~Math-X

Video Solution by Magic Square

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

Video Solution

https://www.youtube.com/watch?v=EcrktBc8zrM&ab_channel=SpreadTheMathLove (@11:08)

Video Solution by Interstigation

https://youtu.be/DBqko2xATxs&t=534

Video Solution by WhyMath

https://youtu.be/jCjF9duTQxk

~savannahsolver

Video Solution by harungurcan

https://www.youtube.com/watch?v=35BW7bsm_Cg&t=778s

~harungurcan

Video Solution by Dr. David

https://youtu.be/LMeg3r3VFdE

See Also

2023 AMC 8 (Problems • Answer Key • Resources)
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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_8_Problems/Problem_7&oldid=231525"