Difference between revisions of "2022 AMC 10A Problems/Problem 10"
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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. | 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) | + | 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 | ||
| − | < | + | <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 < | + | 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 < | + | 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> |
~USAMO333 | ~USAMO333 | ||
Edits and Diagram by ~KingRavi and ~MRENTHUSIASM | Edits and Diagram by ~KingRavi and ~MRENTHUSIASM | ||
| + | |||
| + | Minor edit by yanes04 | ||
== Video Solution== | == Video Solution== | ||
| Line 69: | Line 71: | ||
~Education, the Study of Everything | ~Education, the Study of Everything | ||
| − | == Video Solution 2 ( | + | == Video Solution 2 (Pythagorean Theorem & Square of Binomial)== |
| − | https://www.youtube.com/watch?v= | + | https://www.youtube.com/watch?v=rJ61GqU6NWM&list=PLmpPPbOoDfgj5BlPtEAGcB7BR_UA5FgFj&index=5 |
| − | |||
| − | == | + | ==Solution 2== |
| − | https:// | + | 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}} | {{AMC10 box|year=2022|ab=A|num-b=9|num-a=11}} | ||
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Latest revision as of 15:18, 31 October 2024
Contents
Problem
Daniel finds a rectangular index card and measures its diagonal to be 
centimeters.
Daniel then cuts out equal squares of side 
cm at two opposite corners of the index card and measures the distance between the two closest vertices of these squares to be 
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]](http://latex.artofproblemsolving.com/2/2/f/22f9f7a96f6ad58ba5a6b8396b925cd53152cf24.png)
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]](http://latex.artofproblemsolving.com/4/6/c/46c58f38312c56694cd1890f9c251f64ec6f7dd2.png)
Label the bottom left corner of the larger rectangle (without the square cut out) as 
and the top right as 
. 
is the width of the rectangle and 
is the length. Now we have vertices 
as vertices of the irregular octagon created by cutting out the squares. Let 
be the two closest vertices formed by the squares.
The distance between the two closest vertices of the squares is thus 
Substituting, we get
![\[(IJ)^2 = (w-2)^2 + (\ell-2)^2 = \left(4\sqrt{2}\right)^2 = 32 \implies w^2+\ell^2-4w-4\ell = 24.\]](http://latex.artofproblemsolving.com/a/e/3/ae31eb53986921a8e5b41f6caa23b2f3797d15ff.png)
Using the fact that the diagonal of the rectangle is 
we get
![\[w^2+\ell^2 = 64.\]](http://latex.artofproblemsolving.com/4/8/e/48e7b02cdfb36277be0b58a3c913b8018e90ad5b.png)
Subtracting the first equation from the second equation, we get ![\[4w+4\ell=40 \implies w+\ell = 10.\]](http://latex.artofproblemsolving.com/a/3/b/a3bf4cdf8bfc438cd8f50589b76aa8b4ab67d736.png)
Squaring yields ![\[w^2 + 2w\ell + \ell^2 = 100.\]](http://latex.artofproblemsolving.com/9/6/4/96447fbf7ed654c45257ac25f40fda25ca5a46f3.png)
Subtracting the second equation from this, we get 
and thus area of the original rectangle is 
~USAMO333
Edits and Diagram by ~KingRavi and ~MRENTHUSIASM
Minor edit by yanes04
Video Solution
~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
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 | ||
