Bunuel
A square and a circle intersect at more than one point. Does the square have more area than the circle?
(1) there are exactly four intersection points
(2) at least two of the intersection points are on vertices of the square
Kudos for a correct solution. MAGOOSH OFFICIAL SOLUTION:Statement #1: this information, with nothing more, could mean that the circle is either smaller or larger.
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This statement, alone and by itself, is insufficient.
Statement #2: this information, with nothing more, could mean that the circle is either smaller or larger.
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This statement, alone and by itself, is insufficient.
Combined Statements: One possibility is the circle that intersects the square four times by passing through all four vertices:
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That circle is clearly bigger than the square. The circle absolutely cannot pass through exactly three vertices. If it pass through two vertices, it would have to intersect the side two more times. Possibilities include the following (point C is the center of the circle).
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Notice that, as point C approaches the top side of the square, it gets closer and closer to the circle that has this top side as a diameter, equivalent to the first circle in the statement #1 diagram. That circle is clearly has less area than the square. Well, that circle won’t work here, because it intersects at only two points, but because point C could get closer and closer to the top side without touching it, which means the area of the circle in this diagram could get closer and closer to the area of the first circle in the statement #1 diagram. This means that we could make the circle in this diagram have less area than the square has.
Thus, even with the constraints of both statements, we can construct a circle that has an area that is either greater than or less than that of the square. Even with both statements, we cannot give a definitive answer to the prompt question.
Both statements combined are insufficient.