1996 AMC 12 Problem 28

Attempt Problem 28 of the 1996 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 1996 AMC 12 solutions, or check the answer key.

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28.

On a 4×4×34\times4\times3 rectangular parallelepiped, vertices A,A, B,B, and CC are adjacent to vertex D.D. The perpendicular distance from DD to the plane containing A,A, B,B, and CC is closest to
1.61.6
1.91.9
2.12.1
2.72.7
2.92.9
Answer: C
Concepts:planespoint-to-plane distance
Small hint:
Place DD at the origin with the three adjacent edges along coordinate axes
Big hint:
Write the intercept form of the plane through (4,0,0),(0,4,0),(0,0,3)(4,0,0),(0,4,0),(0,0,3)
Solution:
Put D=(0,0,0)D=(0,0,0) and take the adjacent vertices as (4,0,0),(0,4,0),(0,0,3).(4,0,0),(0,4,0),(0,0,3). Their plane is x4+y4+z3=1. \frac{x}{4}+\frac{y}{4}+\frac{z}{3}=1. Its distance from the origin is 1(14)2+(14)2+(13)2=1234≈2.06, \begin{aligned} \frac1{\sqrt{\begin{gathered} (\frac{1}{4})^2+(\frac{1}{4})^2\\ {}+(\frac{1}{3})^2 \end{gathered}}} &=\frac{12}{\sqrt{34}}\\ &\approx2.06, \end{aligned} closest to 2.1.2.1. Thus the correct answer is C.
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