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 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 | |
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 |
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