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Difference between revisions of "2023 AMC 8 Problems/Problem 2"

(→‎Solution 2 (Similar to Solution 3 but without pictures))
(→‎*Easy Video Explanation by MathTalks_Now*)
 
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==Problem==
 
==Problem==
 
A square piece of paper is folded twice into four equal quarters, as shown below, then cut along the dashed line. When unfolded, the paper will match which of the following figures?
 
A square piece of paper is folded twice into four equal quarters, as shown below, then cut along the dashed line. When unfolded, the paper will match which of the following figures?
−
[[Image:2023 AMC 8-2.png|thumb|center|400px]]
+
<asy>
 +
//Restored original diagram. Alter it if you would like, but it was made by TheMathGuyd,
 +
// Diagram by TheMathGuyd. I even put the lined texture :)
 +
// Thank you Kante314 for inspiring thicker arrows. They do look much better
 +
size(0,3cm);
 +
path sq = (-0.5,-0.5)--(0.5,-0.5)--(0.5,0.5)--(-0.5,0.5)--cycle;
 +
path rh = (-0.125,-0.125)--(0.5,-0.5)--(0.5,0.5)--(-0.125,0.875)--cycle;
 +
path sqA = (-0.5,-0.5)--(-0.25,-0.5)--(0,-0.25)--(0.25,-0.5)--(0.5,-0.5)--(0.5,-0.25)--(0.25,0)--(0.5,0.25)--(0.5,0.5)--(0.25,0.5)--(0,0.25)--(-0.25,0.5)--(-0.5,0.5)--(-0.5,0.25)--(-0.25,0)--(-0.5,-0.25)--cycle;
 +
path sqB = (-0.5,-0.5)--(-0.25,-0.5)--(0,-0.25)--(0.25,-0.5)--(0.5,-0.5)--(0.5,0.5)--(0.25,0.5)--(0,0.25)--(-0.25,0.5)--(-0.5,0.5)--cycle;
 +
path sqC = (-0.25,-0.25)--(0.25,-0.25)--(0.25,0.25)--(-0.25,0.25)--cycle;
 +
path trD = (-0.25,0)--(0.25,0)--(0,0.25)--cycle;
 +
path sqE = (-0.25,0)--(0,-0.25)--(0.25,0)--(0,0.25)--cycle;
 +
filldraw(sq,mediumgrey,black);
 +
draw((0.75,0)--(1.25,0),currentpen+1,Arrow(size=6));
 +
//folding
 +
path sqside = (-0.5,-0.5)--(0.5,-0.5);
 +
path rhside = (-0.125,-0.125)--(0.5,-0.5);
 +
transform fld = shift((1.75,0))*scale(0.5);
 +
draw(fld*sq,black);
 +
int i;
 +
for(i=0; i<10; i=i+1)
 +
{
 +
  draw(shift(0,0.05*i)*fld*sqside,deepblue);
 +
}
 +
path rhedge = (-0.125,-0.125)--(-0.125,0.8)--(-0.2,0.85)--cycle;
 +
filldraw(fld*rhedge,grey);
 +
path sqedge = (-0.5,-0.5)--(-0.5,0.4475)--(-0.575,0.45)--cycle;
 +
filldraw(fld*sqedge,grey);
 +
filldraw(fld*rh,white,black);
 +
int i;
 +
for(i=0; i<10; i=i+1)
 +
{
 +
  draw(shift(0,0.05*i)*fld*rhside,deepblue);
 +
}
 +
draw((2.25,0)--(2.75,0),currentpen+1,Arrow(size=6));
 +
//cutting
 +
transform cut = shift((3.25,0))*scale(0.5);
 +
draw(shift((-0.01,+0.01))*cut*sq);
 +
draw(cut*sq);
 +
filldraw(shift((0.01,-0.01))*cut*sq,white,black);
 +
int j;
 +
for(j=0; j<10; j=j+1)
 +
{
 +
draw(shift(0,0.05*j)*cut*sqside,deepblue);
 +
}
 +
draw(shift((0.01,-0.01))*cut*(0,-0.5)--shift((0.01,-0.01))*cut*(0.5,0),dashed);
 +
//Answers Below, but already Separated
 +
//filldraw(sqA,grey,black);
 +
//filldraw(sqB,grey,black);
 +
//filldraw(sq,grey,black);
 +
//filldraw(sqC,white,black);
 +
//filldraw(sq,grey,black);
 +
//filldraw(trD,white,black);
 +
//filldraw(sq,grey,black);
 +
//filldraw(sqE,white,black);
 +
</asy>
  
−
==Solution 1==
+
<asy>
−
Take a paper and fold it using the given conditions to see the resulting answer
+
// Diagram by TheMathGuyd.
−
-apex304
+
size(0,7.5cm);
 +
path sq = (-0.5,-0.5)--(0.5,-0.5)--(0.5,0.5)--(-0.5,0.5)--cycle;
 +
path rh = (-0.125,-0.125)--(0.5,-0.5)--(0.5,0.5)--(-0.125,0.875)--cycle;
 +
path sqA = (-0.5,-0.5)--(-0.25,-0.5)--(0,-0.25)--(0.25,-0.5)--(0.5,-0.5)--(0.5,-0.25)--(0.25,0)--(0.5,0.25)--(0.5,0.5)--(0.25,0.5)--(0,0.25)--(-0.25,0.5)--(-0.5,0.5)--(-0.5,0.25)--(-0.25,0)--(-0.5,-0.25)--cycle;
 +
path sqB = (-0.5,-0.5)--(-0.25,-0.5)--(0,-0.25)--(0.25,-0.5)--(0.5,-0.5)--(0.5,0.5)--(0.25,0.5)--(0,0.25)--(-0.25,0.5)--(-0.5,0.5)--cycle;
 +
path sqC = (-0.25,-0.25)--(0.25,-0.25)--(0.25,0.25)--(-0.25,0.25)--cycle;
 +
path trD = (-0.25,0)--(0.25,0)--(0,0.25)--cycle;
 +
path sqE = (-0.25,0)--(0,-0.25)--(0.25,0)--(0,0.25)--cycle;
  
−
==Solution 2 (Vague)==
+
//ANSWERS
 +
real sh = 1.5;
 +
label("$\textbf{(A)}$",(-0.5,0.5),SW);
 +
label("$\textbf{(B)}$",shift((sh,0))*(-0.5,0.5),SW);
 +
label("$\textbf{(C)}$",shift((2sh,0))*(-0.5,0.5),SW);
 +
label("$\textbf{(D)}$",shift((0,-sh))*(-0.5,0.5),SW);
 +
label("$\textbf{(E)}$",shift((sh,-sh))*(-0.5,0.5),SW);
 +
filldraw(sqA,mediumgrey,black);
 +
filldraw(shift((sh,0))*sqB,mediumgrey,black);
 +
filldraw(shift((2*sh,0))*sq,mediumgrey,black);
 +
filldraw(shift((2*sh,0))*sqC,white,black);
 +
filldraw(shift((0,-sh))*sq,mediumgrey,black);
 +
filldraw(shift((0,-sh))*trD,white,black);
 +
filldraw(shift((sh,-sh))*sq,mediumgrey,black);
 +
filldraw(shift((sh,-sh))*sqE,white,black);
 +
</asy>
  
−
Notice how the paper is folded. The bottom right corner of the twice-folded paper has to be the middle of the unfolded paper. So if you cut it in the way that it is shown in the problem, you find (it has to be symmetrical) that the cuts make an equilateral rhombus [tilted square] centered in the middle of the paper.
+
==Solution==
  
−
-claregu
+
Notice that when we unfold the paper along the vertical fold line, we get the following shape:
  
−
==Solution 3 (Thorough)==
 
  
−
Notice that when we unfold the paper from the vertical fold line, we get
+
<asy>
  
−
[[Image:Screenshot_2023-01-25_8.11.20_AM.png|thumb|center|200px]]
+
size(90);
 +
path sq = (-0.5,0)--(0.5,0)--(0.5,0.5)--(-0.5,0.5)--cycle;
 +
path trE = (-0.25,0)--(0.25,0)--(0,0.25)--cycle;
  
−
Then, if we unfold the paper from the horizontal fold line, we result in
+
real sh = 1.5;
 +
filldraw(shift((sh,-sh))*sq,mediumgrey,black);
 +
filldraw(shift((sh,-sh))*trE,white,black);
  
−
[[Image:Screenshot_2023-01-25_8.14.41_AM.png|thumb|center|200px]]
+
size(90);
 +
path sq = (-0.5,-0.5)--(0.5,-0.5)--(0.5,0.5)--(-0.5,0.5)--cycle;
 +
path sqE = (-0.25,0)--(0,-0.25)--(0.25,0)--(0,0.25)--cycle;
  
−
It is clear that the answer is <math>\boxed{\textbf{(E)}}</math>
+
real sh = 1.5;
 +
filldraw(shift((sh,-sh))*sq,mediumgrey,black);
 +
filldraw(shift((sh,-sh))*sqE,white,black);
 +
</asy>
 +
 
 +
It is clear that the answer is <math>\boxed{\textbf{(E)}}</math>.
  
 
~MrThinker
 
~MrThinker
 +
 +
==Solution 2==
 +
 +
When you fold the paper in that specific way, it shows the top left square. That means that any cut you make on that folded part will look like that on that top left square. Since it is folded into four parts, the cut will reflect on all of the other 3 parts. Since there is a cut diagonally on the bottom right of that square, that will reflect on the other squares, making the shape of <math>\boxed{\textbf{(E)}}</math> <asy>
 +
 +
size(90);
 +
path sq = (-0.5,0)--(0.5,0)--(0.5,0.5)--(-0.5,0.5)--cycle;
 +
path trE = (-0.25,0)--(0.25,0)--(0,0.25)--cycle;
 +
 +
real sh = 1.5;
 +
filldraw(shift((sh,-sh))*sq,mediumgrey,black);
 +
filldraw(shift((sh,-sh))*trE,white,black);
 +
 +
size(90);
 +
path sq = (-0.5,-0.5)--(0.5,-0.5)--(0.5,0.5)--(-0.5,0.5)--cycle;
 +
path sqE = (-0.25,0)--(0,-0.25)--(0.25,0)--(0,0.25)--cycle;
 +
 +
real sh = 1.5;
 +
filldraw(shift((sh,-sh))*sq,mediumgrey,black);
 +
filldraw(shift((sh,-sh))*sqE,white,black);
 +
</asy>
 +
 +
~AliceDubbleYou
 +
 +
==*Easy Video Explanation by MathTalks_Now*==
 +
https://studio.youtube.com/video/PMOeiGLkDH0/edit
 +
 +
==Video Solution by Math-X (Smart and Simple)==
 +
https://youtu.be/Ku_c1YHnLt0?si=ZucTBcN42MKGX2Ty&t=115 ~Math-X
 +
 +
==Video Solution (How to Creatively THINK!!!)==
 +
https://youtu.be/suFxwnH-ak8
 +
~Education the Study of everything
 +
 +
==Video Solution by Magic Square==
 +
https://youtu.be/-N46BeEKaCQ?t=5658
 +
 +
==Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=EcrktBc8zrM
 +
==Video Solution by Interstigation==
 +
https://youtu.be/DBqko2xATxs&t=67
 +
 +
==Video Solution by WhyMath==
 +
 +
https://youtu.be/z6SxQkQACjo?si=WJAMIdKzUO7oGLGc
 +
 +
==Video Solution by harungurcan==
 +
https://www.youtube.com/watch?v=35BW7bsm_Cg&t=97s
 +
 +
~harungurcan
 +
 +
==Video Solution by Dr. David==
 +
https://youtu.be/octW02FH-iU
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2023|num-b=1|num-a=3}}
 
{{AMC8 box|year=2023|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 01:53, 4 November 2024

Problem

A square piece of paper is folded twice into four equal quarters, as shown below, then cut along the dashed line. When unfolded, the paper will match which of the following figures? [asy] //Restored original diagram. Alter it if you would like, but it was made by TheMathGuyd, // Diagram by TheMathGuyd. I even put the lined texture :) // Thank you Kante314 for inspiring thicker arrows. They do look much better size(0,3cm); path sq = (-0.5,-0.5)--(0.5,-0.5)--(0.5,0.5)--(-0.5,0.5)--cycle; path rh = (-0.125,-0.125)--(0.5,-0.5)--(0.5,0.5)--(-0.125,0.875)--cycle; path sqA = (-0.5,-0.5)--(-0.25,-0.5)--(0,-0.25)--(0.25,-0.5)--(0.5,-0.5)--(0.5,-0.25)--(0.25,0)--(0.5,0.25)--(0.5,0.5)--(0.25,0.5)--(0,0.25)--(-0.25,0.5)--(-0.5,0.5)--(-0.5,0.25)--(-0.25,0)--(-0.5,-0.25)--cycle; path sqB = (-0.5,-0.5)--(-0.25,-0.5)--(0,-0.25)--(0.25,-0.5)--(0.5,-0.5)--(0.5,0.5)--(0.25,0.5)--(0,0.25)--(-0.25,0.5)--(-0.5,0.5)--cycle; path sqC = (-0.25,-0.25)--(0.25,-0.25)--(0.25,0.25)--(-0.25,0.25)--cycle; path trD = (-0.25,0)--(0.25,0)--(0,0.25)--cycle; path sqE = (-0.25,0)--(0,-0.25)--(0.25,0)--(0,0.25)--cycle; filldraw(sq,mediumgrey,black); draw((0.75,0)--(1.25,0),currentpen+1,Arrow(size=6)); //folding path sqside = (-0.5,-0.5)--(0.5,-0.5); path rhside = (-0.125,-0.125)--(0.5,-0.5); transform fld = shift((1.75,0))*scale(0.5); draw(fld*sq,black); int i; for(i=0; i<10; i=i+1) { draw(shift(0,0.05*i)*fld*sqside,deepblue); } path rhedge = (-0.125,-0.125)--(-0.125,0.8)--(-0.2,0.85)--cycle; filldraw(fld*rhedge,grey); path sqedge = (-0.5,-0.5)--(-0.5,0.4475)--(-0.575,0.45)--cycle; filldraw(fld*sqedge,grey); filldraw(fld*rh,white,black); int i; for(i=0; i<10; i=i+1) { draw(shift(0,0.05*i)*fld*rhside,deepblue); } draw((2.25,0)--(2.75,0),currentpen+1,Arrow(size=6)); //cutting transform cut = shift((3.25,0))*scale(0.5); draw(shift((-0.01,+0.01))*cut*sq); draw(cut*sq); filldraw(shift((0.01,-0.01))*cut*sq,white,black); int j; for(j=0; j<10; j=j+1) { draw(shift(0,0.05*j)*cut*sqside,deepblue); } draw(shift((0.01,-0.01))*cut*(0,-0.5)--shift((0.01,-0.01))*cut*(0.5,0),dashed); //Answers Below, but already Separated //filldraw(sqA,grey,black); //filldraw(sqB,grey,black); //filldraw(sq,grey,black); //filldraw(sqC,white,black); //filldraw(sq,grey,black); //filldraw(trD,white,black); //filldraw(sq,grey,black); //filldraw(sqE,white,black); [/asy]

[asy] // Diagram by TheMathGuyd. size(0,7.5cm); path sq = (-0.5,-0.5)--(0.5,-0.5)--(0.5,0.5)--(-0.5,0.5)--cycle; path rh = (-0.125,-0.125)--(0.5,-0.5)--(0.5,0.5)--(-0.125,0.875)--cycle; path sqA = (-0.5,-0.5)--(-0.25,-0.5)--(0,-0.25)--(0.25,-0.5)--(0.5,-0.5)--(0.5,-0.25)--(0.25,0)--(0.5,0.25)--(0.5,0.5)--(0.25,0.5)--(0,0.25)--(-0.25,0.5)--(-0.5,0.5)--(-0.5,0.25)--(-0.25,0)--(-0.5,-0.25)--cycle; path sqB = (-0.5,-0.5)--(-0.25,-0.5)--(0,-0.25)--(0.25,-0.5)--(0.5,-0.5)--(0.5,0.5)--(0.25,0.5)--(0,0.25)--(-0.25,0.5)--(-0.5,0.5)--cycle; path sqC = (-0.25,-0.25)--(0.25,-0.25)--(0.25,0.25)--(-0.25,0.25)--cycle; path trD = (-0.25,0)--(0.25,0)--(0,0.25)--cycle; path sqE = (-0.25,0)--(0,-0.25)--(0.25,0)--(0,0.25)--cycle; //ANSWERS real sh = 1.5; label("$\textbf{(A)}$",(-0.5,0.5),SW); label("$\textbf{(B)}$",shift((sh,0))*(-0.5,0.5),SW); label("$\textbf{(C)}$",shift((2sh,0))*(-0.5,0.5),SW); label("$\textbf{(D)}$",shift((0,-sh))*(-0.5,0.5),SW); label("$\textbf{(E)}$",shift((sh,-sh))*(-0.5,0.5),SW); filldraw(sqA,mediumgrey,black); filldraw(shift((sh,0))*sqB,mediumgrey,black); filldraw(shift((2*sh,0))*sq,mediumgrey,black); filldraw(shift((2*sh,0))*sqC,white,black); filldraw(shift((0,-sh))*sq,mediumgrey,black); filldraw(shift((0,-sh))*trD,white,black); filldraw(shift((sh,-sh))*sq,mediumgrey,black); filldraw(shift((sh,-sh))*sqE,white,black); [/asy]

Solution

Notice that when we unfold the paper along the vertical fold line, we get the following shape:


[asy] size(90); path sq = (-0.5,0)--(0.5,0)--(0.5,0.5)--(-0.5,0.5)--cycle; path trE = (-0.25,0)--(0.25,0)--(0,0.25)--cycle; real sh = 1.5; filldraw(shift((sh,-sh))*sq,mediumgrey,black); filldraw(shift((sh,-sh))*trE,white,black); size(90); path sq = (-0.5,-0.5)--(0.5,-0.5)--(0.5,0.5)--(-0.5,0.5)--cycle; path sqE = (-0.25,0)--(0,-0.25)--(0.25,0)--(0,0.25)--cycle; real sh = 1.5; filldraw(shift((sh,-sh))*sq,mediumgrey,black); filldraw(shift((sh,-sh))*sqE,white,black); [/asy]

It is clear that the answer is $\boxed{\textbf{(E)}}$.

~MrThinker

Solution 2

When you fold the paper in that specific way, it shows the top left square. That means that any cut you make on that folded part will look like that on that top left square. Since it is folded into four parts, the cut will reflect on all of the other 3 parts. Since there is a cut diagonally on the bottom right of that square, that will reflect on the other squares, making the shape of $\boxed{\textbf{(E)}}$ [asy] size(90); path sq = (-0.5,0)--(0.5,0)--(0.5,0.5)--(-0.5,0.5)--cycle; path trE = (-0.25,0)--(0.25,0)--(0,0.25)--cycle; real sh = 1.5; filldraw(shift((sh,-sh))*sq,mediumgrey,black); filldraw(shift((sh,-sh))*trE,white,black); size(90); path sq = (-0.5,-0.5)--(0.5,-0.5)--(0.5,0.5)--(-0.5,0.5)--cycle; path sqE = (-0.25,0)--(0,-0.25)--(0.25,0)--(0,0.25)--cycle; real sh = 1.5; filldraw(shift((sh,-sh))*sq,mediumgrey,black); filldraw(shift((sh,-sh))*sqE,white,black); [/asy]

~AliceDubbleYou

*Easy Video Explanation by MathTalks_Now*

https://studio.youtube.com/video/PMOeiGLkDH0/edit

Video Solution by Math-X (Smart and Simple)

https://youtu.be/Ku_c1YHnLt0?si=ZucTBcN42MKGX2Ty&t=115 ~Math-X

Video Solution (How to Creatively THINK!!!)

https://youtu.be/suFxwnH-ak8 ~Education the Study of everything

Video Solution by Magic Square

https://youtu.be/-N46BeEKaCQ?t=5658

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=EcrktBc8zrM

Video Solution by Interstigation

https://youtu.be/DBqko2xATxs&t=67

Video Solution by WhyMath

https://youtu.be/z6SxQkQACjo?si=WJAMIdKzUO7oGLGc

Video Solution by harungurcan

https://www.youtube.com/watch?v=35BW7bsm_Cg&t=97s

~harungurcan

Video Solution by Dr. David

https://youtu.be/octW02FH-iU

See Also

2023 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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

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