Difference between revisions of "2003 AMC 12B Problems/Problem 22"
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+ | == Solution 3 == | ||
+ | Let the intersection <math>\overline{AC}</math> and <math>\overline{BD}</math> be <math>E.</math> Let <math>\overline{PE}=y \implies \overline{AP}=8-y</math> and <math>\overline{EQ}=x \implies \overline {QB}=15-x.</math> Since <math>\triangle NQB \sim \triangle APN,</math> we have <math>\frac{8-y}{x}=\frac{y}{15-x} \implies 120=15y+8x.</math> We want to minimize <math>\overline{PQ}=\sqrt{x^2+y^2}.</math> By Cauchy, <math>17^2(x^2+y^2)=(x^2+y^2)(8^2+15^2) \ge (8x+15y)^2=120^2 \implies \sqrt{x^2+y^2} \ge \frac{120}{17} \approx 7.</math> So, choice <math>\boxed{C}</math> is closest to the minimum. | ||
== See also == | == See also == |
Latest revision as of 14:17, 23 June 2022
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
Solution 3
Let the intersection and
be
Let
and
Since
we have
We want to minimize
By Cauchy,
So, choice
is closest to the minimum.
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.