Difference between revisions of "2022 AMC 10A Problems/Problem 10"
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~Education, the Study of Everything | ~Education, the Study of Everything | ||
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| + | {{AMC10 box|year=2022|ab=A|num-b=9|num-a=11}} | ||
Revision as of 14:47, 12 November 2022
Contents
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?
Solution 1(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]](http://latex.artofproblemsolving.com/f/a/f/faf8ada9f98a94526b6e69122423cbfd7f73c37d.png)
We will use coordinates here. Label the bottom left corner of the larger rectangle(without the square cut out) as 
and the top right as 
, where 
is the width of the rectangle and 
is the length. Now we have vertices 
, 
, 
, and 
as vertices of the irregular octagon created by cutting out the squares. Label 
and 
as the two closest vertices formed by the squares.
The distance between the two closest vertices of the squares is thus 
= (
Substituting, we get
![\[(IJ)^2 = (w-2)^2 + (l-2)^2 = (4\sqrt{2})^2 = 32.\]](http://latex.artofproblemsolving.com/6/7/2/6726745f425e077bd1c9e2cd621040ebb19bc08a.png)
![\[\implies w^2+l^2-4w-4l = 24\]](http://latex.artofproblemsolving.com/7/1/2/71256843dde8938f2a562f52e589a03910377da7.png)
Using the fact that the diagonal of the rectangle = 
, we get
![\[w^2+l^2 = 64\]](http://latex.artofproblemsolving.com/e/7/a/e7a153fc2d18b8b1ca8bd8882cf8a267b3e1c4de.png)
.
Subtracting the first equation from the second equation, we get ![\[4w+4l=40 \implies w+l = 10\]](http://latex.artofproblemsolving.com/f/d/a/fdaa8743e7b8a2ddf9d47f6a144044ed5df11701.png)
Squaring yields ![\[w^2 + 2wl + l^2 = 100\]](http://latex.artofproblemsolving.com/c/9/a/c9a2cad334a489d0814683ac04bea06efacd45cc.png)
Subtracting the second equation from this, we get 
, and thus area of the original rectangle = 
= 
= 
~USAMO333
Edits and Diagram by ~KingRavi
Video Solution 1 (Quick and Easy)
~Education, the Study of Everything
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 | ||
