2022 AMC 10A Problems/Problem 10

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] /* Edited by MRENTHUSIASM */ size(250); real x, y; x = 6; y = 3; draw((0.5,0)--(x,0)); draw((0,0.5)--(0,y)); draw((0,y)--(x-0.5,y)); draw((x,0)--(x,y-0.5)); 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); [/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 (Coordinate Geometry)

$[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]$ 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 $D=(w,\ell),$ where $w$ is the width of the rectangle and $\ell$ is the length. Now we have vertices $E=(0,1),F=(1,0),G=(w-1,\ell),$ and $H=(w,\ell-1)$ as vertices of the irregular octagon created by cutting out the squares. Label $I=(1,1)$ and $J=(w-1, \ell-1)$ as 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

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

Solution 2 (Algebra)

Let the dimensions of the index card be $x$ and $y$ . We have $x^2 + y^2 = 64$ and $(x-2)^2 + (y-2)^2 = 32$ by Pythagoras. Subtracting, we obtain $x^2 - (x^2 - 4x + 4) + y^2 - (y^2 - 4y + 4) = 4x - 4 + 4y - 4 = 32$ . Thus, we have $x + y - 2 = 8$ , so $x + y = 10$ .