2003 AMC 12B Problems/Problem 22
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Problem
Let be a rhombus with
and
. Let
be a point on
, and let
and
be the feet of the perpendiculars from
to
and
, respectively. Which of the following is closest to the minimum possible value of
?
![[asy] size(200); defaultpen(0.6); pair O = (15*15/17,8*15/17), C = (17,0), D = (0,0), P = (25.6,19.2), Q = (25.6, 18.5); pair A = 2*O-C, B = 2*O-D; pair P = (A+O)/2, Q=(B+O)/2, N=(A+B)/2; draw(A--B--C--D--cycle); draw(A--O--B--O--C--O--D); draw(P--N--Q); label("\(A\)",A,WNW); label("\(B\)",B,ESE); label("\(C\)",C,ESE); label("\(D\)",D,SW); label("\(P\)",P,SSW); label("\(Q\)",Q,SSE); label("\(N\)",N,NNE); [/asy]](http://latex.artofproblemsolving.com/c/2/2/c22d4d20155faaf8bd0cf51f26a5795d14b32322.png)
Solution
Let and
intersect at
. Since
is a rhombus, then
and
are perpendicular bisectors. Thus
, so
is a rectangle. Since the diagonals of a rectangle are of equal length,
, so we want to minimize
. It follows that we want
.
Finding the area in two different ways,
See also
2003 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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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