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

(add diagram)
Line 22: Line 22:
 
label("$A$",A,N);
 
label("$A$",A,N);
 
label("$B$",B,SW);
 
label("$B$",B,SW);
label("$C$",C,S);
+
label("$C$",C,N);
 
label("$D$",D,E);
 
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>
 
</asy>
  

Revision as of 07:03, 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,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 $7+3=\boxed{10}$.

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