Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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 13: Line 13:
 
Substituting, we get  
 
Substituting, we get  
  
−
  (1) <math></math>(IJ)^2<math> = </math>(w-2)^2<math> + </math>(l-2)^2<math> = </math>(4\sqrt{2})^2 = 32.<math> </math>
+
  <math></math>(IJ)^2 = (w-2)^2 + (l-2)^2 = (4\sqrt{2})^2 = 32.<math> </math>
  
 
Using the fact that the diagonal of the rectangle = <math>8</math>, we get  
 
Using the fact that the diagonal of the rectangle = <math>8</math>, we get  
  
−
(2) <cmath>w^2+l^2 = 64</cmath>.
+
<cmath>w^2+l^2 = 64</cmath>.
  
 
Subtracting <math>(2)</math> from <math>(1)</math> and simplifying, we get <math>w+l</math> = <math>10</math>.
 
Subtracting <math>(2)</math> from <math>(1)</math> and simplifying, we get <math>w+l</math> = <math>10</math>.

Revision as of 15:21, 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?

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

Solution 1(Simple coordinates and basic algebra)

(Someone please add a diagram if you can) 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 

$$ (Error compiling LaTeX. Unknown error_msg)(IJ)^2 = (w-2)^2 + (l-2)^2 = (4\sqrt{2})^2 = 32.$$ (Error compiling LaTeX. Unknown error_msg)

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

\[w^2+l^2 = 64\].

Subtracting $(2)$ from $(1)$ and simplifying, we get $w+l$ = $10$. Squaring yields $w^2 + 2wl + l^2$ = $100$ and thus area of the original rectangle = $wl$ = $18$ = $\boxed{\textbf{(E) } 18}.$

~USAMO333

Video Solution 1 (Quick and Easy)

https://youtu.be/BIy0Koe4D4s

~Education, the Study of Everything

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10A_Problems/Problem_10&oldid=181105"