Difference between revisions of "2023 AMC 8 Problems/Problem 7"
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==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 1== | ==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); | ||
− | + | 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 <math>\boxed{\textbf{(B)}\ 1}.</math> | ||
~MrThinker | ~MrThinker | ||
==Solution 2== | ==Solution 2== | ||
− | If the analytic expression for line <math> AB </math> is <math> y=k_{1}x | + | 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> |
− | < | + | |
− | and | + | ~[[User:Bloggish|Bloggish]] |
− | < | + | |
− | Solving | + | ==Video Solution (How to CREATIVELY THINK!!!) == |
− | < | + | https://youtu.be/NUaes2N_4pM |
− | and | + | ~Education the Study of everything |
− | < | + | |
− | Therefore, we can determine that the expression for line <math> AB </math> is <math> y=\frac{1}{3}x </math> | + | |
+ | ==Video Solution by Math-X (Smart and Simple)== | ||
+ | https://youtu.be/Ku_c1YHnLt0?si=P3DtuhzhiVr2Jv0r&t=947 ~Math-X | ||
− | |||
==Video Solution by Magic Square== | ==Video Solution by Magic Square== | ||
https://youtu.be/-N46BeEKaCQ?t=5151 | 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 | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2023|num-b=6|num-a=8}} | {{AMC8 box|year=2023|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:38, 23 November 2023
Contents
Problem
A rectangle, with sides parallel to the -axis and
-axis, has opposite vertices located at
and
. A line is drawn through points
and
. Another line is drawn through points
and
. How many points on the rectangle lie on at least one of the two lines?
Solution 1
If we extend the lines, we have the following diagram:
Therefore, we see that the answer is
~MrThinker
Solution 2
Note that the -intercepts of line
and line
are
and
. If the analytic expression for line
is
, and the analytic expression for line
is
, we have equations:
and
. Solving these equations, we can find out that
and
. Therefore, we can determine that the expression for line
is
and the expression for line
is
. When
, the coordinates that line
and line
pass through are
and
, and
lies perfectly on one vertex of the rectangle while the
coordinate of
is out of the range
(lower than the bottom left corner of the rectangle
). Considering that the
value of the line
will only decrease, and the
value of the line
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
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
~savannahsolver
Video Solution by harungurcan
https://www.youtube.com/watch?v=35BW7bsm_Cg&t=778s
~harungurcan
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.