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...)
 
Line 16: Line 16:
  
 
== Solution ==
 
== Solution ==
 +
 +
<asy>
 +
pair A=(3,9), B=(1,1), C=(5,3), D=(7,3);
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draw(A--B--C--D--cycle);
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label("$A$",A,N);
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label("$B$",B,SW);
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label("$C$",C,S);
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label("$D$",D,E);
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</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>.  

Revision as of 07:00, 9 May 2011

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,S); label("$D$",D,E); [/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 $7+3=\boxed{10}$.

See Also

2001 AMC 12 (ProblemsAnswer KeyResources)
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