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 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 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 $x + y = 10$ .

This means that $(x+y)^2 = x^2 + 2xy + y^2 = 100$ . Subtracting $x^2 + y^2$ from this, we get $2xy = 36$ and so $xy = 18$ .

The area of the index card is thus $\boxed{\textbf{(E) } 18}.$

~mathboy100

Video Solution 1 (Simple)

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

Video Solution 2

https://youtu.be/BIy0Koe4D4s

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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