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Geometry Difficulty 2.7 Junior Find the answer

Given that α\alpha, β\beta, and γ\gamma are three distinct planes, consider the following propositions:
- Proposition p: If αβ\alpha \perp \beta and βγ\beta \perp \gamma, then αγ\alpha \parallel \gamma.
- Proposition q: If three non-collinear points on plane α\alpha are equidistant from plane β\beta, then αβ\alpha \parallel \beta.

Among the following conclusions, the correct one is

Pick one

Solution

Since when αβ\alpha \perp \beta and βγ\beta \perp \gamma, it is possible for α\alpha to be parallel or perpendicular to γ\gamma, the proposition p is a false proposition. Furthermore, if three non-collinear points on plane α\alpha are equidistant from plane β\beta, it is possible for α\alpha and β\beta to be parallel or intersecting. Therefore, proposition q is also a false proposition. Consequently, the proposition "p and q" is false, the proposition "p or not q" is true, the proposition "p or q" is false, and the proposition "not p and not q" is true. Therefore, the correct choice is C\boxed{\text{C}}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.