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> | ||
− | == Solution == | + | == 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> | ||
Line 45: | Line 45: | ||
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>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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: | ||
+ | |||
+ | <cmath>A=\frac{1}{2}\cdot(24+28)(45+32)=\frac{1}{2}\cdot52\cdot77=2002</cmath> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ~eevee9406 | ||
== See also == | == See also == |
Latest revision as of 11:18, 9 November 2024
Contents
[hide]Problem
A convex quadrilateral with area contains a point in its interior such that . Find the perimeter of .
Solution 1
We have (This is true for any convex quadrilateral: split the quadrilateral along and then using the triangle area formula to evaluate and ), with equality only if . By the triangle inequality,
with equality if lies on and respectively. Thus
Since we have the equality case, at point , as shown below.
By the Pythagorean Theorem,
The perimeter of is .
Solution 2
Draw the diagram out. Notice the very peculiar sets of numbers ; these are simply multiples of the legs of well-known Pythagorean triples , pointing to signs of possible right angles. We can assume that , so the area of the entire figure would be:
Thus this is the correct case, so finding the side lengths of by the Pythagorean Theorem yields , ,,, so the perimeter is .
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.