Difference between revisions of "2003 AMC 12B Problems/Problem 22"
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\qquad\mathrm{(E)}\ 7.5</math> | \qquad\mathrm{(E)}\ 7.5</math> | ||
== Solution 1 == | == Solution 1 == | ||
− | Let <math>\overline{AC}</math> and <math>\overline{BD}</math> intersect at <math>O</math>. Since <math>ABCD</math> is a rhombus, then <math>\overline{AC}</math> and <math>\overline{BD}</math> are [[perpendicular | + | Let <math>\overline{AC}</math> and <math>\overline{BD}</math> intersect at <math>O</math>. Since <math>ABCD</math> is a rhombus, then <math>\overline{AC}</math> and <math>\overline{BD}</math> are [[perpendicular bisector]]s. Thus <math>\angle POQ = 90^{\circ}</math>, so <math>OPNQ</math> is a [[rectangle]]. Since the diagonals of a rectangle are of equal length, <math>PQ = ON</math>, so we want to minimize <math>ON</math>. It follows that we want <math>ON \perp AB</math>. |
Finding the area in two different ways, | Finding the area in two different ways, |
Revision as of 17:51, 17 January 2021
Contents
[hide]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 1
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,
Solution 2 (semi-bash)
Let the intersection of and
be
. Since
is a rhombus, we have
and
. Since
, we have
, so
. Therefore,
By Pythagorean Theorem,
The minimum value of
would give the minimum value of
, so we take the derivative (or use vertex form) to find that the minimum occurs when
which gives
. Hence, the minimum value of
is
, which is closest to
.
-MP8148
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.