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Printjmc
algebra senior
Problem
Three noncollinear points and a line are given in the plane. Suppose no two of the points lie on a line parallel to (or itself). There are exactly lines perpendicular to with the following property: the three circles with centers at the given points and tangent to line all concur at some point. Find all possible values of .
Enter all possible values of separated by commas.
Enter all possible values of separated by commas.
Solution
The condition for line is that each of the three points lies at an equal distance from the line as from some fixed point; in other words, the line is the directrix of a parabola containing the three points, and the fixed point is the focus.
Three noncollinear points in the coordinate plane determine a quadratic polynomial in unless two of the points have the same -coordinate. Therefore, given the direction of the directrix, three noncollinear points determine a parabola, unless two of the points lie on a line perpendicular to the directrix. This case is ruled out by the given condition, so the answer is .
Three noncollinear points in the coordinate plane determine a quadratic polynomial in unless two of the points have the same -coordinate. Therefore, given the direction of the directrix, three noncollinear points determine a parabola, unless two of the points lie on a line perpendicular to the directrix. This case is ruled out by the given condition, so the answer is .
Final answer
1