Difference between revisions of "2003 AIME I Problems/Problem 12"
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− | By the [[Law of Cosines]] on <math>\triangle ABD</math> at angle <math>A</math> and on <math>\triangle BCD</math> at angle <math>C</math> (note <math>\angle C = \angle A), | + | By the [[Law of Cosines]] on <math>\triangle ABD</math> at angle <math>A</math> and on <math>\triangle BCD</math> at angle <math>C</math> (note <math>\angle C = \angle A</math>), |
<cmath>180^2 + AD^2 - 360 \cdot AD \cos A = 180^2 + BC^2 - 360 \cdot BC \cos A </cmath> | <cmath>180^2 + AD^2 - 360 \cdot AD \cos A = 180^2 + BC^2 - 360 \cdot BC \cos A </cmath> | ||
<cmath>(AD^2 - BC^2) - 360(AD - BC) \cos A = 0 </cmath> | <cmath>(AD^2 - BC^2) - 360(AD - BC) \cos A = 0 </cmath> | ||
<cmath>(AD - BC)(AD + BC - 360 \cos A ) = 0 </cmath> | <cmath>(AD - BC)(AD + BC - 360 \cos A ) = 0 </cmath> | ||
− | < | + | <math>AD - BC \neq 0</math>, so <math>AD + BC - 360 \cos A = 0</math>. We know that <math>AD + BC = 640 - 360 = 280</math>, so <math>280 - 360 \cos A = 0 \Rightarrow \cos A = \dfrac{7}{9} = 0.777 \ldots</math>, and <math>\lfloor 1000 \cos A \rfloor = \boxed{777}</math>. |
===Solution 2=== | ===Solution 2=== |
Revision as of 15:08, 10 August 2013
Contents
[hide]Problem
In convex quadrilateral and The perimeter of is 640. Find (The notation means the greatest integer that is less than or equal to )
Solution
Solution 1
By the Law of Cosines on at angle and on at angle (note ),
, so . We know that , so , and .
Solution 2
Notice that , and , and , so we have side-side-angle matching on triangles and . Since the problem does not allow , we know that is not a right angle, and there is a unique other triangle with the matching side-side-angle.
Extend to so that is isosceles with . Then notice that has matching side-side-angle, and yet because is not right. Therefore is the unique triangle mentioned above, so is congruent, in some order of vertices, to . Since would imply , making quadrilateral degenerate, we must have .
Since the perimeter of is , . Hence . Drop the altitude of from and call the foot . Then right triangle trigonometry on shows that , so .
See also
2003 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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