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IMC / 2008 / Problems / Day 2, P8

IMC 2008 · Day 2 · P8

hard

Two different ellipses are given. One focus of the first ellipse coincides with one focus of the second ellipse. Prove that the ellipses have at most two points in common.

Solution (official)

It is well known that an ellipse might be defined by a focus (a point) and a directrix (a straight line), as a locus of points such that the distance to the focus divided by the distance to the directrix is equal to a given number e<1e < 1. So, if a point XX belongs to both ellipses with the same focus FF and directrices l1l_1, l2l_2, then e1l1X=FX=e2l2Xe_1 \cdot l_1 X = FX = e_2 \cdot l_2 X (here we denote by l1Xl_1 X, l2Xl_2 X distances between the corresponding line and the point XX). The equation e1l1X=e2l2Xe_1 \cdot l_1 X = e_2 \cdot l_2 X defines two lines, whose equations are linear combinations with coefficients e1,±e2e_1, \pm e_2 of the normalized equations of lines l1,l2l_1, l_2 but of those two only one is relevant, since XX and FF should lie on the same side of each directrix. So, we have that all possible points lie on one line. The intersection of a line and an ellipse consists of at most two points.

How the field did

contestants scored
255
average (of 20)
5.78
solved (≥ 80%)
25.1%
near-0 (≤ 10%)
69.0%
discrimination
0.46

Score distribution (field cohort)

Computed on contestants with a meaningful total (field cohort); discrimination is the corrected item–total correlation.

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