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

(Replaced picture (of problem and answers) with asy)
(Problem: Condensed answers into single asy diagram)
Line 56: Line 56:
 
</asy>
 
</asy>
  
<math>\mathbf{(A)}</math>
 
 
<asy>
 
<asy>
size(3cm);
+
// Diagram by TheMathGuyd. Not sure it is worth adding the lined texture.
 +
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 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;
filldraw(sqA,mediumgrey,black);
 
</asy>
 
 
<math>\mathbf{(B)}</math>
 
<asy>
 
size(3cm);
 
 
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 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;
filldraw(sqB,mediumgrey,black);
 
</asy>
 
 
<math>\mathbf{(C)}</math><asy>
 
size(3cm);
 
path sq = (-0.5,-0.5)--(0.5,-0.5)--(0.5,0.5)--(-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 sqC = (-0.25,-0.25)--(0.25,-0.25)--(0.25,0.25)--(-0.25,0.25)--cycle;
filldraw(sq,mediumgrey,black);
+
path trD = (-0.25,0)--(0.25,0)--(0,0.25)--cycle;
filldraw(sqC,white,black);
+
path sqE = (-0.25,0)--(0,-0.25)--(0.25,0)--(0,0.25)--cycle;
</asy>
 
  
<math>\mathbf{(D)}</math><asy>
+
//ANSWERS
size(3cm);
+
real sh = 1.5;
path sq = (-0.5,-0.5)--(0.5,-0.5)--(0.5,0.5)--(-0.5,0.5)--cycle;
+
label("$\mathbf{(A)}$",(-0.5,0.5),SW);
path trD = (-0.25,0)--(0.25,0)--(0,0.25)--cycle;
+
label("$\mathbf{(B)}$",shift((sh,0))*(-0.5,0.5),SW);
filldraw(sq,mediumgrey,black);
+
label("$\mathbf{(C)}$",shift((2sh,0))*(-0.5,0.5),SW);
filldraw(trD,white,black);
+
label("$\mathbf{(D)}$",shift((0,-sh))*(-0.5,0.5),SW);
 +
label("$\mathbf{(E)}$",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>
 
</asy>
  
<math>\mathbf{(E)}</math><asy>
 
size(3cm);
 
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;
 
filldraw(sq,mediumgrey,black);
 
filldraw(sqE,white,black);
 
</asy>
 
 
==Solution 1 (Vague)==
 
==Solution 1 (Vague)==
  

Revision as of 20:39, 25 January 2023

Problem

When 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] // Diagram by TheMathGuyd. I even put the lined texture :) 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),Arrow()); //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),Arrow()); //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. Not sure it is worth adding the lined texture. 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("$\mathbf{(A)}$",(-0.5,0.5),SW); label("$\mathbf{(B)}$",shift((sh,0))*(-0.5,0.5),SW); label("$\mathbf{(C)}$",shift((2sh,0))*(-0.5,0.5),SW); label("$\mathbf{(D)}$",shift((0,-sh))*(-0.5,0.5),SW); label("$\mathbf{(E)}$",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 1 (Vague)

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.

-claregu

Solution 2 (Thorough)

Notice that when we unfold the paper from the vertical fold line, we get

Screenshot 2023-01-25 8.11.20 AM.png

Then, if we unfold the paper from the horizontal fold line, we result in

Screenshot 2023-01-25 8.14.41 AM.png

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

~MrThinker

Video Solution by Magic Square

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

See Also

2023 AMC 8 (ProblemsAnswer KeyResources)
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