2014 AMC 12A Problems/Problem 20
Problem
In , , , and . Points and lie on and respectively. What is the minimum possible value of ?
Solution 1
Let be the reflection of across , and let be the reflection of across . Then it is well-known that the quantity is minimized when it is equal to . (Proving this is a simple application of the triangle inequality; for an example of a simpler case, see Heron's Shortest Path Problem.) As lies on both and , we have . Furthermore, by the nature of the reflection, so . Therefore by the Law of Cosines
Solution 2
In , the three lines look like the Chinese character 又. Let , , and have bases , , and respectively. Then, has the same side as and the same side as . Connect all three triangles with in the center and the two triangles sharing one of its sides. Then, is formed with forming the base.
Intuitively, the pentagon's base is minimized when all three bottom sides are collinear. This is simply the original except that . (In , , and , , and the three triangles connect at to form the pentagon). Thus, ).
in this new triangle is then the minimum of . Applying law of cosines,
~bjhhar
Would prime notation be clearer?
Solution 3
(Diagram by shihan) Reflect across to . Similarly, reflect across to . Clearly, and . Thus, the sum . This value is minimized when , , and are collinear. To finish, we use the law of cosines on the triangle :
See Also
2014 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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All AMC 12 Problems and Solutions |
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