Difference between revisions of "2001 AMC 12 Problems/Problem 20"
(New page: == Problem == Points <math>A = (3,9)</math>, <math>B = (1,1)</math>, <math>C = (5,3)</math>, and <math>D=(a,b)</math> lie in the first quadrant and are the vertices of quadrilateral <math...) |
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\text{(E) }16 | \text{(E) }16 | ||
</math> | </math> | ||
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+ | [[Category: Introductory Geometry Problems]] | ||
== Solution == | == 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> | ||
We already know two vertices of the square: <math>(A+B)/2 = (2,5)</math> and <math>(B+C)/2 = (3,2)</math>. | We already know two vertices of the square: <math>(A+B)/2 = (2,5)</math> and <math>(B+C)/2 = (3,2)</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>. | 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 == | ||
{{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 10:21, 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.