Difference between revisions of "2022 AMC 10A Problems/Problem 10"
(→Solution 1(Simple coordinates and basic algebra)) |
(→Solution 1(Simple coordinates and basic algebra)) |
||
| Line 8: | Line 8: | ||
<math>\textbf{(A) }14 \qquad \textbf{(B) }10 \sqrt{2}</math> <math> \qquad \textbf{(C) }16 \qquad \textbf{(D) }12 \sqrt{2}</math> <math> \qquad \textbf{(E) }18</math> | <math>\textbf{(A) }14 \qquad \textbf{(B) }10 \sqrt{2}</math> <math> \qquad \textbf{(C) }16 \qquad \textbf{(D) }12 \sqrt{2}</math> <math> \qquad \textbf{(E) }18</math> | ||
== Solution 1(Simple coordinates and basic algebra) == | == Solution 1(Simple coordinates and basic algebra) == | ||
| − | + | ||
We will use coordinates here. Label the bottom left corner of the larger rectangle(without the square cut out) as <math>A(0,0)</math> and the top right as <math>D(w,l)</math>, where <math>w</math> is the width of the rectangle and <math>l</math> is the length. Now we have vertices <math>E(0,1)</math> , <math>F(1,0)</math> , <math>G(w-1,l)</math>, and <math>H(w,l-1)</math> as vertices of the irregular octagon created by cutting out the squares. Label <math>I(1,1)</math> and <math>J(w-1, l-1)</math> as the two closest vertices formed by the squares. | We will use coordinates here. Label the bottom left corner of the larger rectangle(without the square cut out) as <math>A(0,0)</math> and the top right as <math>D(w,l)</math>, where <math>w</math> is the width of the rectangle and <math>l</math> is the length. Now we have vertices <math>E(0,1)</math> , <math>F(1,0)</math> , <math>G(w-1,l)</math>, and <math>H(w,l-1)</math> as vertices of the irregular octagon created by cutting out the squares. Label <math>I(1,1)</math> and <math>J(w-1, l-1)</math> as the two closest vertices formed by the squares. | ||
The distance between the two closest vertices of the squares is thus <math>IJ</math> = (<math>4</math> <math>\sqrt{2})^2.</math> | The distance between the two closest vertices of the squares is thus <math>IJ</math> = (<math>4</math> <math>\sqrt{2})^2.</math> | ||
| Line 23: | Line 23: | ||
~USAMO333 | ~USAMO333 | ||
| + | Edits by ~KingRavi | ||
==Video Solution 1 (Quick and Easy)== | ==Video Solution 1 (Quick and Easy)== | ||
Revision as of 14:22, 12 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?
Solution 1(Simple coordinates and basic algebra)
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)
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 
from 
and simplifying, we get 
= 
.
Squaring yields 
= 
and thus area of the original rectangle = 
= 
= 
~USAMO333 Edits by ~KingRavi
Video Solution 1 (Quick and Easy)
~Education, the Study of Everything
