Difference between revisions of "2001 AMC 12 Problems/Problem 20"
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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>. | ||
| + | |||
| + | == Video Solution == | ||
| + | https://www.youtube.com/watch?v=wR0mphUnhLc | ||
== See Also == | == See Also == | ||
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
![[asy] pair A=(3,9), B=(1,1), C=(5,3), D=(7,3); draw(A--B--C--D--cycle); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,N); label("$D$",D,E); pair AB = (A + B)/2, BC = (B + C)/2, CD = (C + D)/2, DA = (D + A)/2; draw(AB--BC--CD--DA--cycle); [/asy]](http://latex.artofproblemsolving.com/8/c/f/8cfb9d6da99381de67ea5f718469d0ecfdf8c71c.png)
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. 
