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2002 AMC 12B Problems/Problem 24

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Problem

A convex quadrilateral ABCD with area 2002 contains a point P in its interior such that PA = 24, PB = 32, PC = 28, PD = 45. Find the perimeter of ABCD.

\mathrm{(A)}\ 4\sqrt{2002} \qquad \mathrm{(B)}\ 2\sqrt{8465} \qquad \mathrm{(C)}\ 2 (48+\sqrt{2002}) \qquad \mathrm{(D)}\ 2\sqrt{8633} \qquad \mathrm{(E)}\ 4(36 + \sqrt{113})

Solution

We have [ABCD] = 2002 \le \frac 12 (AC \cdot BD) (Why is this true? Try splitting the quadrilateral along AC and then using the triangle area formula), with equality if \overline{AC} \perp \overline{BD}. By the triangle inequality,

\begin{align*}AC &\le PA + PC = 52\\BD &\le PB + PD = 77\end{align*}

with equality if P lies on \overline{AC} and \overline{BD} respectively. Thus

2002 \le \frac{1}{2} AC \cdot BD \le \frac 12 \cdot 52 \cdot 77 = 2002

Since we have the equality case, \overline{AC} \perp \overline{BD} at point P.

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...

By the Pythagorean Theorem, \begin{align*}AB = \sqrt{PA^2 + PB^2} & = \sqrt{24^2 + 32^2} = 40\\BC = \sqrt{PB^2 + PC^2} & = \sqrt{32^2 + 28^2} = 4...

The perimeter of ABCD is AB + BC + CD + DA = 4(36 + \sqrt{113}) \Rightarrow \mathrm{(E)}.

See also

2002 AMC 12B (Problems • 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
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