2003 AMC 12B Problems/Problem 22
Problem
Let be a rhombus with and . Let be a point on , and let and be the feet of the perpendiculars from to and , respectively. Which of the following is closest to the minimum possible value of ?
Solution 1
Let and intersect at . Since is a rhombus, then and are perpendicular bisectors. Thus , so is a rectangle. Since the diagonals of a rectangle are of equal length, , so we want to minimize . It follows that we want .
Finding the area in two different ways,
Solution 2 (semi-bash)
Let the intersection of and be . Since is a rhombus, we have and . Since , we have , so . Therefore, By Pythagorean Theorem, The minimum value of would give the minimum value of , so we take the derivative (or use vertex form) to find that the minimum occurs when which gives . Hence, the minimum value of is , which is closest to .
-MP8148
Solution 3
Let the intersection and be Let and Since we have We want to minimize By Cauchy, So, choice is closest to the minimum.
See also
2003 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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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