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

m (→‎Video Solution 1 (Simple))
m (→‎Solution 1 removed coordinate Geometry clutter)
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<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} \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 12\sqrt{2} \qquad \textbf{(E) } 18</math>
  
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== Solution 1 (Coordinate Geometry) ==
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== Solution 1 ==
 
<asy>
 
<asy>
 
/* Edited by MRENTHUSIASM */
 
/* Edited by MRENTHUSIASM */
Line 49: Line 49:
 
draw((x+1,0)--(x+1,y), L=L2, arrow=Arrows(),bar=Bars(15));
 
draw((x+1,0)--(x+1,y), L=L2, arrow=Arrows(),bar=Bars(15));
 
</asy>
 
</asy>
−
We will use coordinates here. Label the bottom left corner of the larger rectangle(without the square cut out) as <math>A=(0,0)</math> and the top right as <math>D=(w,\ell),</math> where <math>w</math> is the width of the rectangle and <math>\ell</math> is the length. Now we have vertices <math>E=(0,1),F=(1,0),G=(w-1,\ell),</math> and <math>H=(w,\ell-1)</math> as vertices of the irregular octagon created by cutting out the squares. Label <math>I=(1,1)</math> and <math>J=(w-1, \ell-1)</math> as the two closest vertices formed by the squares.
+
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 and </math>J<math> be the two closest vertices formed by the squares.
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The distance between the two closest vertices of the squares is thus <math>IJ=\left(4\sqrt{2}\right)^2.</math>
+
The distance between the two closest vertices of the squares is thus </math>IJ=\left(4\sqrt{2}\right)^2.<math>
 
Substituting, we get  
 
Substituting, we get  
  
 
<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>
 
<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 is <math>8,</math> we get  
+
Using the fact that the diagonal of the rectangle is </math>8,<math> we get  
 
<cmath>w^2+\ell^2 = 64.</cmath>
 
<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>
 
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>
 
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>
+
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}.$
  
 
~USAMO333
 
~USAMO333

Revision as of 10:17, 26 September 2023

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 $I and$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)^2.$Substituting, we get

<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 is$ (Error compiling LaTeX. Unknown error_msg)8,$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$2w\ell = 36,$and thus area of the original rectangle is$w\ell = \boxed{\textbf{(E) } 18}.$

~USAMO333

Edits and Diagram by ~KingRavi and ~MRENTHUSIASM

Video Solution

https://youtu.be/BIy0Koe4D4s

~Education, the Study of Everything

Video Solution 2 (Simple)

https://www.youtube.com/watch?v=joVRkVp7Qvc ~AWhiz

Video Solution 3 (Pythagorean Theorem & Square of Binomial)

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

See Also

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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