2022 AMC 10A Problems/Problem 10
Contents
Problem
Daniel finds a rectangular index card and measures its diagonal to be centimeters. Daniel then cuts out equal squares of side 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
Label the bottom left corner of the larger rectangle (without the square cut out) as and the top right as . is the width of the rectangle and is the length. Now we have vertices as vertices of the irregular octagon created by cutting out the squares. Let be the two closest vertices formed by the squares. The distance between the two closest vertices of the squares is thus Substituting, we get
Using the fact that the diagonal of the rectangle is we get Subtracting the first equation from the second equation, we get Squaring yields Subtracting the second equation from this, we get and thus area of the original rectangle is
~USAMO333
Edits and Diagram by ~KingRavi and ~MRENTHUSIASM
Minor edit by yanes04
Video Solution
~Education, the Study of Everything
Video Solution 2 (Simple)
https://www.youtube.com/watch?v=joVRkVp7Qvc ~AWhiz
Video Solution 3 (Pythagorean Theorem & Square of Binomial)
https://www.youtube.com/watch?v=rJ61GqU6NWM&list=PLmpPPbOoDfgj5BlPtEAGcB7BR_UA5FgFj&index=5
Solution 2
Let x be the width of the original rectangle and y be the height. Through observation and logic, we can then conclude that (x-2)^2 + (y-2)^2 = 32. After expanding and simplifying the expressions, you end up with x+y=10. If you then solve for x in terms of y, you end up with x=10=y. Since 8 is the diagonal of the original rectangle, we can write that (10-y)^2+y^2=64. Use the quadratic formula to solve for y, and then solve for x. Using either root, yields a product of the difference of squares, 25-7=18. Therefore, 18 (E) is our solution!
~Namya
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 |