Difference between revisions of "2002 AMC 12B Problems/Problem 24"
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A [[convex polygon|convex]] [[quadrilateral]] <math>ABCD</math> with area <math>2002</math> contains a point <math>P</math> in its interior such that <math>PA = 24, PB = 32, PC = 28, PD = 45</math>. Find the perimeter of <math>ABCD</math>. | A [[convex polygon|convex]] [[quadrilateral]] <math>ABCD</math> with area <math>2002</math> contains a point <math>P</math> in its interior such that <math>PA = 24, PB = 32, PC = 28, PD = 45</math>. Find the perimeter of <math>ABCD</math>. | ||
− | <math>\mathrm{(A)}\ 4\sqrt{2002} \qquad \mathrm{(B)}\ 2\sqrt{8465} \qquad \mathrm{(C)}\ 2<math> < | + | <math>\mathrm{(A)}\ 4\sqrt{2002} \qquad \mathrm{(B)}\ 2\sqrt{8465} \qquad \mathrm{(C)}\ 2</math> <math>(48+\sqrt{2002}) \qquad \mathrm{(D)}\ 2\sqrt{8633} \qquad \mathrm{(E)}\ 4(36 + \sqrt{113})</math> |
== Solution == | == Solution == | ||
We have | We have |
Revision as of 12:38, 19 January 2008
Problem
A convex quadrilateral with area
contains a point
in its interior such that
. Find the perimeter of
.
Solution
We have
(Why is this true? Try splitting the quadrilateral along
and then using the triangle area formula), with equality if
. By the triangle inequality,
with equality if lies on
and
respectively. Thus
Since we have the equality case, at point
.
By the Pythagorean Theorem,
The perimeter of is
.
See also
2002 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 |