2003 AIME I Problems/Problem 12
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 $\angle C = \angle A),
<cmath>180^2 + AD^2 - 360 \cdot AD \cos A = 180^2 + BC^2 - 360 \cdot BC \cos A </cmath> <cmath>(AD^2 - BC^2) - 360(AD - BC) \cos A = 0 </cmath> <cmath>(AD - BC)(AD + BC - 360 \cos A ) = 0 </cmath>$ (Error compiling LaTeX. Unknown error_msg)AD - BC \neq 0AD + BC - 360 \cos A = 0AD + BC = 640 - 360 = 280280 - 360 \cos A = 0 \Rightarrow \cos A = \dfrac{7}{9} = 0.777 \ldots\lfloor 1000 \cos A \rfloor = \boxed{777}$.
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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