2002 AMC 12B Problems/Problem 24
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
.
![[asy] size(200); defaultpen(0.6); pair A = (0,0), B = (40,0), C = (25.6 * 52 / 24, 19.2 * 52 / 24), D = (40 - (40-25.6)*77/32,19.2*77/32), P = (25.6,19.2), Q = (25.6, 18.5); pair E=(A+P)/2, F=(B+P)/2, G=(C+P)/2, H=(D+P)/2; draw(A--B--C--D--cycle); draw(A--P--B--P--C--P--D); label("\(A\)",A,WSW); label("\(B\)",B,ESE); label("\(C\)",C,ESE); label("\(D\)",D,NW); label("\(P\)",Q,SSW); label("24",E,WNW); label("32",F,WSW); label("28",G,ESE); label("45",H,ENE); [/asy]](http://latex.artofproblemsolving.com/9/2/c/92c3f08edb70e3f9b6b024f8f9904ab46a4c8f86.png)
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 |