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

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<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>
 
<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>
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== Solution ==
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== Solution 1==
 
We have  
 
We have  
 
<cmath>[ABCD] = 2002 \le \frac 12 (AC \cdot BD)</cmath>
 
<cmath>[ABCD] = 2002 \le \frac 12 (AC \cdot BD)</cmath>
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The perimeter of <math>ABCD</math> is <math>AB + BC + CD + DA = 4(36 + \sqrt{113}) \Rightarrow \mathrm{(E)}</math>.
 
The perimeter of <math>ABCD</math> is <math>AB + BC + CD + DA = 4(36 + \sqrt{113}) \Rightarrow \mathrm{(E)}</math>.
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==Solution 2==
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Draw the diagram out. Notice the very peculiar sets of numbers <math>(24,32);(24,45);(28,45)</math>; these are simply multiples of the legs of well-known Pythagorean triples <math>(3,4,5);(5,12,13);(28,45,53)</math>, pointing to signs of possible right angles. We can assume that <math>\angle APB=\angle BPC=\angle CPD=\angle DPA=90^\circ</math>, so the area of the entire figure would be:
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<cmath>A=\frac{1}{2}\cdot(24+28)(45+32)=\frac{1}{2}\cdot52\cdot77=2002</cmath>
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Thus this is the correct case, so finding the side lengths of <math>ABCD</math> by the Pythagorean Theorem yields <math>AB=40</math>, <math>BC=4\sqrt{113}</math>,<math>CD=53</math>,<math>DA=51</math>, so the perimeter is <math>40+4\sqrt{130}+53+51=144+4\sqrt{130}=\boxed{\textbf{(E) }4(36+\sqrt{113})}</math>.
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~eevee9406
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 11:18, 9 November 2024

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 1

We have \[[ABCD] = 2002 \le \frac 12 (AC \cdot BD)\] (This is true for any convex quadrilateral: split the quadrilateral along $AC$ and then using the triangle area formula to evaluate $[ACB]$ and $[ACD]$), with equality only 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 $P$, as shown below.

[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); draw(rightanglemark(C,P,D,50)); [/asy]

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

Solution 2

Draw the diagram out. Notice the very peculiar sets of numbers $(24,32);(24,45);(28,45)$; these are simply multiples of the legs of well-known Pythagorean triples $(3,4,5);(5,12,13);(28,45,53)$, pointing to signs of possible right angles. We can assume that $\angle APB=\angle BPC=\angle CPD=\angle DPA=90^\circ$, so the area of the entire figure would be:

\[A=\frac{1}{2}\cdot(24+28)(45+32)=\frac{1}{2}\cdot52\cdot77=2002\]

Thus this is the correct case, so finding the side lengths of $ABCD$ by the Pythagorean Theorem yields $AB=40$, $BC=4\sqrt{113}$,$CD=53$,$DA=51$, so the perimeter is $40+4\sqrt{130}+53+51=144+4\sqrt{130}=\boxed{\textbf{(E) }4(36+\sqrt{113})}$.

~eevee9406

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

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