1979 AMC 12 Problem 17

Attempt Problem 17 of the 1979 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1979 AMC 12 solutions, or check the answer key.

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17.

Points A,A, B,B, C,C, and DD are distinct and lie, in the given order, on a straight line. Line segments AB,AB, AC,AC, and ADAD have lengths x,x, y,y, and z,z, respectively. If line segments ABAB and CDCD may be rotated about points BB and C,C, respectively, so that points AA and DD coincide, to form a triangle with positive area, then which of the following three inequalities must be satisfied? I. x<z2II. y<x+z2III. y<z2 \begin{aligned} &\text{I. }x\lt\frac z2\\ &\text{II. }y\lt x+\frac z2\\ &\text{III. }y\lt\frac z2 \end{aligned}
I\mathrm{I} only
II\mathrm{II} only
I\mathrm{I} and II\mathrm{II} only
II\mathrm{II} and III\mathrm{III} only
I,\mathrm{I}, II\mathrm{II} and III\mathrm{III}
Answer: C
Concepts:triangle inequalitycaseworklogical deduction
Small hint:
After the rotations, identify the three side lengths of the new triangle
Big hint:
Apply all three strict triangle inequalities to x,x, y−x,y-x, and z−yz-y
Solution:
The new triangle has side lengths x,x, y−x,y-x, and z−y.z-y. Its triangle inequalities give x<(y−x)+(z−y),⟹x<z2;y−x<x+(z−y),⟹y<x+z2;z−y<x+(y−x),⟹y>z2. \begin{aligned} x&\lt(y-x)+(z-y),\\ &\Longrightarrow x\lt \frac{z}{2};\\ y-x&\lt x+(z-y),\\ &\Longrightarrow y\lt x+\frac{z}{2};\\ z-y&\lt x+(y-x),\\ &\Longrightarrow y\gt \frac{z}{2}. \end{aligned} Thus I and II must hold, while III must fail.
Therefore, the correct answer is C.
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