2003 AIME I Problems/Problem 12
Problem
In convex quadrilateral and The perimeter of is . 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 ),
We know that .
.
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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