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Difference between revisions of "2001 AMC 12 Problems/Problem 20"

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\text{(E) }16
 
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[[Category: Introductory Geometry Problems]]
  
 
== 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 ==
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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}}
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{{MAA Notice}}

Latest revision as of 10:21, 7 April 2022

Problem

Points $A = (3,9)$, $B = (1,1)$, $C = (5,3)$, and $D=(a,b)$ lie in the first quadrant and are the vertices of quadrilateral $ABCD$. The quadrilateral formed by joining the midpoints of $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, and $\overline{DA}$ is a square. What is the sum of the coordinates of point $D$?

$\text{(A) }7 \qquad \text{(B) }9 \qquad \text{(C) }10 \qquad \text{(D) }12 \qquad \text{(E) }16$

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: $(A+B)/2 = (2,5)$ and $(B+C)/2 = (3,2)$.

There are only two possibilities for the other vertices of the square: either they are $(6,3)$ and $(5,6)$, or they are $(0,1)$ and $(-1,4)$. The second case would give us $D$ outside the first quadrant, hence the first case is the correct one. As $(6,3)$ is the midpoint of $CD$, we can compute $D=(7,3)$, and $7+3=\boxed{10}$.

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

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