Difference between revisions of "2002 AMC 12B Problems/Problem 24"

Revision as of 11:37, 19 January 2008

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 $ABCD$ .

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

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,

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

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 $P$ .

File:2002 12B AMC-24.png

By the Pythagorean Theorem, $\begin{align*} AB = \sqrt{PA^2 + PB^2} & = \sqrt{24^2 + 32^2} = 40\\ BC = \sqrt{PB^2 + PC^2} & = \sqrt{32^2 + 28^2} = 4\sqrt{113}\\ CD = \sqrt{PC^2 + PD^2} & = \sqrt{28^2 + 45^2} = 53\\ DA = \sqrt{PD^2 + PA^2} & = \sqrt{45^2 + 24^2} = 51 \end{align*}$

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

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
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Difference between revisions of "2002 AMC 12B Problems/Problem 24"

Revision as of 11:37, 19 January 2008

Problem

Solution

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