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Geometry Level pending

Consider two fixed points F ′ F' and F F so that F ′ F = 4 F' F = 4 .

Let M M be a variable point. Denote K K and L L to be the respective orthogonal projections of F F and F ′ F' on the bisector of the angle F ′ M ^ F F' \hat M F .

Assume that F ′ M ^ L = F M ^ K = α F' \hat M L = F \hat M K = \alpha , and F K × F ′ L = 3 FK \times F' L = 3 . Find ( M F − M F ′ ) 2 (MF - MF')^2 .


The answer is 4.

Mohammad Hamdar by Mohammad Hamdar , 22, Lebanon

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1 solution

Mohammad Hamdar
May 6, 2016

U s i n g t h e r i g h t t r i a n g l e s M F K a n d M F ′ L , Using\quad the\quad right\quad triangles\quad MFK\quad and\quad MF'L,\quad sin ⁡ α = F ′ L M F ′ a n d sin ⁡ α = F K M F \\ \sin { \alpha } =\frac { F'L }{ MF' } \quad and\quad \sin { \alpha } =\frac { FK }{ MF } \quad s o , sin ⁡ 2 α = F K × F ′ L M F × M F ′ = 3 M F × M F ′ . \\ so,\quad \sin ^{ 2 }{ \alpha } =\frac { FK\times F'L }{ MF\times MF' } =\frac { 3 }{ MF\times MF' } . U s i n g t h e c o s i n e l a w i n t r i a n g l e F ′ M F , \\ Using\quad the\quad cosine\quad law\quad in\quad triangle\quad F'MF, F F ′ 2 = M F 2 + M F ′ 2 − 2 M F × M F ′ cos ⁡ 2 α \\ { FF' }^{ 2 }={ MF }^{ 2 }+{ MF' }^{ 2 }-2MF\times MF'\cos { 2\alpha } 16 = M F 2 + M F ′ 2 − 6 sin ⁡ 2 α ( cos ⁡ 2 α ) 16={ MF }^{ 2 }+{ MF' }^{ 2 }-\frac { 6 }{ \sin ^{ 2 }{ \alpha } } (\cos { 2\alpha } )\\ N o w u s i n g t h e i d e n t i t y a 2 + b 2 = ( a − b ) 2 + 2 a b a n d 1 − cos ⁡ 2 α = 2 sin ⁡ 2 α ; Now\quad using\quad the\quad identity\quad { a }^{ 2 }+{ b }^{ 2 }={ (a-b) }^{ 2 }+2ab\\ and\quad 1-\cos { 2\alpha } =2\sin ^{ 2 }{ \alpha } ; 16 = ( M F − M F ′ ) 2 + 2 M F × M F ′ − 6 sin ⁡ 2 α ( 1 − 2 sin ⁡ 2 α ) \\ 16={ (MF-MF') }^{ 2 }+2MF\times MF'-\frac { 6 }{ \sin ^{ 2 }{ \alpha } } (1-2\sin ^{ 2 }{ \alpha } ) 4 + 6 sin ⁡ 2 α = ( M F − M F ′ ) 2 + 2 ( 3 sin ⁡ 2 α ) \\ 4+\frac { 6 }{ \sin ^{ 2 }{ \alpha } } ={ (MF-MF') }^{ 2 }+2(\frac { 3 }{ \sin ^{ 2 }{ \alpha } } ) ⟹ ( M F − M F ′ ) 2 = 4 \\ \Longrightarrow { (MF-MF') }^{ 2 }=4 .

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