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Difference between revisions of "2022 AMC 10A Problems/Problem 10"

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== Problem ==
 
== Problem ==
  
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Daniel finds a rectangular index card and measures its diagonal to be 8 centimeters.
+
Daniel finds a rectangular index card and measures its diagonal to be <math>8</math> centimeters.
−
Daniel then cuts out equal squares of side 1 cm at two opposite corners of the index
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Daniel then cuts out equal squares of side <math>1</math> cm at two opposite corners of the index card and measures the distance between the two closest vertices of these squares to be <math>4\sqrt{2}</math> centimeters, as shown below. What is the area of the original index card?
−
card and measures the distance between the two closest vertices of these squares to
+
<asy>
−
be centimeters, as shown below. What is the area of the original index card?
+
// Diagram by MRENTHUSIASM, edited by Djmathman
 +
size(200);
 +
defaultpen(linewidth(0.6));
 +
draw((489.5,-213) -- (225.5,-213) -- (225.5,-185) -- (199.5,-185) -- (198.5,-62) -- (457.5,-62) -- (457.5,-93) -- (489.5,-93) -- cycle);
 +
draw((206.29,-70.89) -- (480.21,-207.11), linetype ("6 6"),Arrows(size=4,arrowhead=HookHead));
 +
draw((237.85,-182.24) -- (448.65,-95.76),linetype ("6 6"),Arrows(size=4,arrowhead=HookHead));
 +
label("$1$",(450,-80));
 +
label("$1$",(475,-106));
 +
label("$8$",(300,-103));
 +
label("$4\sqrt 2$",(300,-173));
 +
</asy>
 +
<math>\textbf{(A) } 14 \qquad \textbf{(B) } 10\sqrt{2} \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 12\sqrt{2} \qquad \textbf{(E) } 18</math>
  
−
<math>\textbf{(A) }14 \qquad \textbf{(B) }10 \sqrt{2}</math> <math> \qquad \textbf{(C) }16 \qquad \textbf{(D) }12 \sqrt{2}</math> <math> \qquad \textbf{(E) }18</math>
 
 
== Solution 1 ==
 
== Solution 1 ==
−
(Someone please add a diagram if you can)
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<asy>
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We will use coordinates here. Label the bottom left corner of the larger rectangle(without the square cut out) as A(0,0) and the top right as D(w,l), where w is the width of the rectangle and l is the length. Now we have vertices E(0,1) , F(1,0) , G(w-1,l), and H(w,l-1) as vertices of the irregular octagon created by cutting out the squares. Label I(1,1) and J(w-1, l-1) as the two closest vertices formed by the squares.
+
/* Edited by MRENTHUSIASM */
−
The distance between the two closest vertices of the squares is thus IJ = (<math>4</math> <math> \sqrt{2}</math>).
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size(250);
 +
real x, y;
 +
x = 6;
 +
y = 3;
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draw((0,0)--(x,0));
 +
draw((0,0)--(0,y));
 +
draw((0,y)--(x,y));
 +
draw((x,0)--(x,y));
 +
draw((0.5,0)--(0.5,0.5)--(0,0.5));
 +
draw((x-0.5,y)--(x-0.5,y-0.5)--(x,y-0.5));
 +
draw((0.5,0.5)--(x-0.5,y-0.5),dashed,Arrows());
 +
draw((x,0)--(0,y),dashed,Arrows());
 +
label("$1$",(x-0.5,y-0.25),W);
 +
label("$1$",(x-0.25,y-0.5),S);
 +
label("$8$",midpoint((0.5,y-0.5)--(x/2,y/2)),(0,2.5));
 +
label("$4\sqrt{2}$",midpoint((0.5,0.5)--(x/2,y/2)),S);
 +
label("$A$",(0,0),SW);
 +
label("$E$",(0,0.5),W);
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label("$F$",(0.5,0),S);
 +
label("$I$",(0.5,0.5),N);
 +
label("$D$",(x,y),NE);
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label("$G$",(x-0.5,y),N);
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label("$H$",(x,y-0.5),E);
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label("$J$",(x-0.5,y-0.5),S);
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Label L1 = Label("$w$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white));
 +
Label L2 = Label("$\ell$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white));
 +
draw((0,-1)--(x,-1), L=L1, arrow=Arrows(),bar=Bars(15));
 +
draw((x+1,0)--(x+1,y), L=L2, arrow=Arrows(),bar=Bars(15));
 +
</asy>
 +
Label the bottom left corner of the larger rectangle (without the square cut out) as <math>A</math> and the top right as <math>D</math>. <math>w</math> is the width of the rectangle and <math>\ell</math> is the length. Now we have vertices <math>E, F, G, H</math> as vertices of the irregular octagon created by cutting out the squares. Let <math>I, J</math> be the two closest vertices formed by the squares.
 +
The distance between the two closest vertices of the squares is thus <math>IJ=\left(4\sqrt{2}\right).</math>
 
Substituting, we get  
 
Substituting, we get  
  
−
(1) IJ^2 = (w-2)^2 + (l-2)^2 = (4sqrt(2))^2 = 32.
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<cmath>(IJ)^2 = (w-2)^2 + (\ell-2)^2 = \left(4\sqrt{2}\right)^2 = 32 \implies w^2+\ell^2-4w-4\ell = 24.</cmath>
−
Using the fact that the diagonal of the rectangle = 8, we get  
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Using the fact that the diagonal of the rectangle is <math>8,</math> we get
 +
<cmath>w^2+\ell^2 = 64.</cmath>
 +
Subtracting the first equation from the second equation, we get <cmath>4w+4\ell=40 \implies w+\ell = 10.</cmath>
 +
Squaring yields <cmath>w^2 + 2w\ell + \ell^2 = 100.</cmath>
 +
Subtracting the second equation from this, we get <math>2w\ell = 36,</math> and thus area of the original rectangle is <math>w\ell = \boxed{\textbf{(E) } 18}.</math>
  
−
(2) w^2 + l^2 = 64.
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~USAMO333
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Subtracting (2) from (1) and simplifying, we get w+l = 10.
+
 
−
Squaring yields w^2 + 2wl + l^2 = 100 and thus wl = 18.
+
Edits and Diagram by ~KingRavi and ~MRENTHUSIASM
 +
 
 +
Minor edit by yanes04
 +
 
 +
== Video Solution==
 +
https://youtu.be/BIy0Koe4D4s
 +
 
 +
~Education, the Study of Everything
 +
 
 +
== Video Solution 2 (Pythagorean Theorem & Square of Binomial)==
 +
https://www.youtube.com/watch?v=rJ61GqU6NWM&list=PLmpPPbOoDfgj5BlPtEAGcB7BR_UA5FgFj&index=5
 +
 
 +
==Solution 2==
 +
https://youtu.be/SRV4e74qsM8
 +
 
 +
Let x be the width of the original rectangle and y be the height. Through observation and logic, we can then conclude that (x-2)^2 + (y-2)^2 = 32. After expanding and simplifying the expressions, you end up with x+y=10. If you then solve for x in terms of y, you end up with x=10=y. Since 8 is the diagonal of the original rectangle, we can write that (10-y)^2+y^2=64. Use the quadratic formula to solve for y, and then solve for x. Using either root, yields a product of the difference of squares, 25-7=18. Therefore, 18 (E) is our solution!
 +
 
 +
~Namya
 +
{{AMC10 box|year=2022|ab=A|num-b=9|num-a=11}}

Latest revision as of 15:18, 31 October 2024

Problem

Daniel finds a rectangular index card and measures its diagonal to be $8$ centimeters. Daniel then cuts out equal squares of side $1$ cm at two opposite corners of the index card and measures the distance between the two closest vertices of these squares to be $4\sqrt{2}$ centimeters, as shown below. What is the area of the original index card? [asy] // Diagram by MRENTHUSIASM, edited by Djmathman size(200); defaultpen(linewidth(0.6)); draw((489.5,-213) -- (225.5,-213) -- (225.5,-185) -- (199.5,-185) -- (198.5,-62) -- (457.5,-62) -- (457.5,-93) -- (489.5,-93) -- cycle); draw((206.29,-70.89) -- (480.21,-207.11), linetype ("6 6"),Arrows(size=4,arrowhead=HookHead)); draw((237.85,-182.24) -- (448.65,-95.76),linetype ("6 6"),Arrows(size=4,arrowhead=HookHead)); label("$1$",(450,-80)); label("$1$",(475,-106)); label("$8$",(300,-103)); label("$4\sqrt 2$",(300,-173)); [/asy] $\textbf{(A) } 14 \qquad \textbf{(B) } 10\sqrt{2} \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 12\sqrt{2} \qquad \textbf{(E) } 18$

Solution 1

[asy] /* Edited by MRENTHUSIASM */ size(250); real x, y; x = 6; y = 3; draw((0,0)--(x,0)); draw((0,0)--(0,y)); draw((0,y)--(x,y)); draw((x,0)--(x,y)); draw((0.5,0)--(0.5,0.5)--(0,0.5)); draw((x-0.5,y)--(x-0.5,y-0.5)--(x,y-0.5)); draw((0.5,0.5)--(x-0.5,y-0.5),dashed,Arrows()); draw((x,0)--(0,y),dashed,Arrows()); label("$1$",(x-0.5,y-0.25),W); label("$1$",(x-0.25,y-0.5),S); label("$8$",midpoint((0.5,y-0.5)--(x/2,y/2)),(0,2.5)); label("$4\sqrt{2}$",midpoint((0.5,0.5)--(x/2,y/2)),S); label("$A$",(0,0),SW); label("$E$",(0,0.5),W); label("$F$",(0.5,0),S); label("$I$",(0.5,0.5),N); label("$D$",(x,y),NE); label("$G$",(x-0.5,y),N); label("$H$",(x,y-0.5),E); label("$J$",(x-0.5,y-0.5),S); Label L1 = Label("$w$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label("$\ell$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); draw((0,-1)--(x,-1), L=L1, arrow=Arrows(),bar=Bars(15)); draw((x+1,0)--(x+1,y), L=L2, arrow=Arrows(),bar=Bars(15)); [/asy] Label the bottom left corner of the larger rectangle (without the square cut out) as $A$ and the top right as $D$. $w$ is the width of the rectangle and $\ell$ is the length. Now we have vertices $E, F, G, H$ as vertices of the irregular octagon created by cutting out the squares. Let $I, J$ be the two closest vertices formed by the squares. The distance between the two closest vertices of the squares is thus $IJ=\left(4\sqrt{2}\right).$ Substituting, we get

\[(IJ)^2 = (w-2)^2 + (\ell-2)^2 = \left(4\sqrt{2}\right)^2 = 32 \implies w^2+\ell^2-4w-4\ell = 24.\] Using the fact that the diagonal of the rectangle is $8,$ we get \[w^2+\ell^2 = 64.\] Subtracting the first equation from the second equation, we get \[4w+4\ell=40 \implies w+\ell = 10.\] Squaring yields \[w^2 + 2w\ell + \ell^2 = 100.\] Subtracting the second equation from this, we get $2w\ell = 36,$ and thus area of the original rectangle is $w\ell = \boxed{\textbf{(E) } 18}.$

~USAMO333

Edits and Diagram by ~KingRavi and ~MRENTHUSIASM

Minor edit by yanes04

Video Solution

https://youtu.be/BIy0Koe4D4s

~Education, the Study of Everything

Video Solution 2 (Pythagorean Theorem & Square of Binomial)

https://www.youtube.com/watch?v=rJ61GqU6NWM&list=PLmpPPbOoDfgj5BlPtEAGcB7BR_UA5FgFj&index=5

Solution 2

https://youtu.be/SRV4e74qsM8

Let x be the width of the original rectangle and y be the height. Through observation and logic, we can then conclude that (x-2)^2 + (y-2)^2 = 32. After expanding and simplifying the expressions, you end up with x+y=10. If you then solve for x in terms of y, you end up with x=10=y. Since 8 is the diagonal of the original rectangle, we can write that (10-y)^2+y^2=64. Use the quadratic formula to solve for y, and then solve for x. Using either root, yields a product of the difference of squares, 25-7=18. Therefore, 18 (E) is our solution!

~Namya

2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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 AMC 10 Problems and Solutions
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