My Nineteenth Problem

Geometry Level pending

Consider two points A ≠ B A\not=B in R 2 \R^2 . Let C ∈ R 2 C\in\R^2 be such that the ratio of lengths A C : B C = a : b AC:BC=a:b for some a , b ∈ R a,b\in\R strictly positive such that a : b ≠ 1 : 1 a:b≠1:1 .

Describe geometrically the solution set for C C .

Bonus: What happens to this solution set as a : b a:b approaches 1 : 1 1:1 ? What happens as a : b a:b approaches 0 : 1 0:1 ?

Oval Parabola Circle Hyperbola Line

Daniel Ellesar by Daniel Ellesar , 24, United Kingdom

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1 solution

Let the points A , B , C A, B, C be at ( a 1 , a 2 ) (a_1,a_2) , ( b 1 , b 2 ) (b_1,b_2) and ( h , k ) (h, k) respectively. Then a 2 ( h − b 1 ) 2 + a 2 ( k − b 2 ) 2 = b 2 ( h − a 1 ) 2 + ( k − a 2 ) 2 a^2(h-b_1)^2+a^2(k-b_2)^2=b^2(h-a_1)^2+(k-a_2)^2 , or ( a 2 − b 2 ) h 2 + ( a 2 − b 2 ) k 2 − 2 ( a 2 b 1 − b 2 a 1 ) h − 2 ( a 2 b 2 − b 2 a 2 ) k + a 2 ( b 1 2 + b 2 2 ) − b 2 ( a 1 2 + a 2 2 ) = 0 (a^2-b^2)h^2+(a^2-b^2)k^2-2(a^2b_1-b^2a_1)h-2(a^2b_2-b^2a_2)k+a^2(b_1^2+b_2^2)-b^2(a_1^2+a_2^2)=0 . Hence the set of points C C represents a C i r c l e \boxed {Circle} .

Bonus

When a : b a:b approaches 1 : 1 1:1 , the set represents a straight line and when it approaches 0 : 1 0:1 , the set represents a point .

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Absolutely correct, but for the bonus, I was thinking less about 'what is the solution set at the ratio 1:1' and more about 'what happens to the solution set as the ratio approaches 1:1'. Indeed, if we approach from 1:1+h, for h>0, then the circle centre veers off to infinity along the line connecting A to B, with A closest to it. If we approach from 1+h:1, then the centre veers off to infinity with B closest to it. In both cases the radius increases, so that its points converge to the perpendicular bisector. This 'mirroring' activity depending on which side we approach seems to suggest we should view the perpendicular bisector as in fact two lines on top of each other, which preserves the fact that the solution set for C is a degree two variety.

Daniel Ellesar - 1 year, 6 months ago

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