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

Revision as of 18:02, 25 November 2022

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 centimeters, as shown below. What is the area of the original index card?

$\textbf{(A) }14 \qquad \textbf{(B) }10 \sqrt{2}$ $\qquad \textbf{(C) }16 \qquad \textbf{(D) }12 \sqrt{2}$ $\qquad \textbf{(E) }18$

Solution (Simple coordinates and basic algebra)

$[asy] size(8cm); draw((0,0)--(6,0)); draw((0,0)--(0,3)); draw((0,3)--(6,3)); draw((6,0)--(6,3)); draw((0,1)--(1,1)); draw((1,1)--(1,0)); draw((5,2)--(6,2)); draw((5,2)--(5,3)); draw((1,1)--(5,2),dashed); draw((0,3)--(6,0),dashed); label("$A$",(0,0),W); label("$D$",(6,3),E); label("$E$",(0,1),W); label("$F$",(1,0),S); label("$4\sqrt{2}$",(2,1.25),S); label("$8$",(2,2),N); label("$H$",(6,2),E); label("$G$",(5,3),N); label("$I$",(1,1),N); label("$J$",(5,2),S); label("$1$",(0.5,0),S); label("$1$",(0,0.5),W); label("$1$",(6,2.5),E); label("$1$",(5.5,3),N); label("$w$",(3,-0.5),S); label("$l$",(7,1.5),E); [/asy]$

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. The distance between the two closest vertices of the squares is thus $IJ$ = ( $4$ $\sqrt{2})^2.$ Substituting, we get

$IJ^2 = (w-2)^2 + (l-2)^2 = (4\sqrt{2})^2 = 32.$ $\implies w^2+l^2-4w-4l = 24$

Using the fact that the diagonal of the rectangle = $8$ , we get

$w^2+l^2 = 64$ .

Subtracting the first equation from the second equation, we get $4w+4l=40 \implies w+l = 10$ Squaring yields $w^2 + 2wl + l^2 = 100$ Subtracting the second equation from this, we get $2wl = 36$ , and thus area of the original rectangle is $wl = \boxed{\textbf{(E) } 18}.$

~USAMO333

Edits and Diagram by ~KingRavi

Video Solution 1 (Simple)

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

Video Solution 2

https://youtu.be/BIy0Koe4D4s

@@ Line 43: / Line 43: @@
 Substituting, we get
-<cmath>(IJ)^2 = (w-2)^2 + (l-2)^2 = (4\sqrt{2})^2 = 32. </cmath>
+<cmath>IJ^2 = (w-2)^2 + (l-2)^2 = (4\sqrt{2})^2 = 32. </cmath>
 <cmath> \implies w^2+l^2-4w-4l = 24</cmath>
@@ Line 52: / Line 52: @@
 Subtracting the first equation from the second equation, we get <cmath>4w+4l=40 \implies w+l = 10</cmath>
 Squaring yields <cmath>w^2 + 2wl + l^2 = 100</cmath>
-Subtracting the second equation from this, we get <math>2wl = 36</math>, and thus area of the original rectangle = <math>wl</math> = <math>18</math> = <math>\boxed{\textbf{(E) } 18}.</math>
+Subtracting the second equation from this, we get <math>2wl = 36</math>, and thus area of the original rectangle is <math>wl = \boxed{\textbf{(E) } 18}.</math>
 ~USAMO333

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