Difference between revisions of "2001 AMC 12 Problems/Problem 20"
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== Solution == | == Solution == | ||
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There are only two possibilities for the other vertices of the square: either they are <math>(6,3)</math> and <math>(5,6)</math>, or they are <math>(0,1)</math> and <math>(-1,4)</math>. The second case would give us <math>D</math> outside the first quadrant, hence the first case is the correct one. As <math>(6,3)</math> is the midpoint of <math>CD</math>, we can compute <math>D=(7,3)</math>, and <math>7+3=\boxed{10}</math>. | There are only two possibilities for the other vertices of the square: either they are <math>(6,3)</math> and <math>(5,6)</math>, or they are <math>(0,1)</math> and <math>(-1,4)</math>. The second case would give us <math>D</math> outside the first quadrant, hence the first case is the correct one. As <math>(6,3)</math> is the midpoint of <math>CD</math>, we can compute <math>D=(7,3)</math>, and <math>7+3=\boxed{10}</math>. | ||
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+ | == Video Solution == | ||
+ | https://www.youtube.com/watch?v=wR0mphUnhLc | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2001|num-b=19|num-a=21}} | {{AMC12 box|year=2001|num-b=19|num-a=21}} | ||
+ | {{MAA Notice}} |
Latest revision as of 11:21, 7 April 2022
Contents
Problem
Points ,
,
, and
lie in the first quadrant and are the vertices of quadrilateral
. The quadrilateral formed by joining the midpoints of
,
,
, and
is a square. What is the sum of the coordinates of point
?
Solution
We already know two vertices of the square: and
.
There are only two possibilities for the other vertices of the square: either they are and
, or they are
and
. The second case would give us
outside the first quadrant, hence the first case is the correct one. As
is the midpoint of
, we can compute
, and
.
Video Solution
https://www.youtube.com/watch?v=wR0mphUnhLc
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.