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 (Coordinate Geometry)
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
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
Solution 2 (Algebra)
Let the dimensions of the index card be and . We have and by Pythagoras. Subtracting, we obtain . Thus, we have , so .
This means that . Subtracting from this, we get and so .
The area of the index card is thus
~mathboy100
Video Solution 1 (Simple)
https://www.youtube.com/watch?v=joVRkVp7Qvc ~AWhiz
Video Solution 2
Video Solution 3 (Pythagorean Theorem & Square of Binomial)
https://www.youtube.com/watch?v=rJ61GqU6NWM&list=PLmpPPbOoDfgj5BlPtEAGcB7BR_UA5FgFj&index=5
See Also
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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All AMC 10 Problems and Solutions |
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