Constant Distance of Trig Points

Geometry Level 4

The distance between the points ( 3 2 sin ⁡ ( α − c ) , 1 2 sin ⁡ ( α − c ) ) \left(\dfrac{\sqrt{3}}{2}\sin (\alpha-c), \dfrac{1}{2}\sin(\alpha-c)\right) ( 1 2 sin ⁡ ( α + c ) , 3 2 sin ⁡ ( α + c ) ) \left(\dfrac{1}{2}\sin (\alpha+c), \dfrac{\sqrt{3}}{2}\sin(\alpha+c)\right)

is constant, for some constant c c and variable α \alpha .

If the distance between the points can be expressed as p q \dfrac{p}{q} for positive coprime integers p , q p,q , then find p + q p+q .


The answer is 3.

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Daniel Liu by Daniel Liu , 21, USA

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1 solution

Ayush Garg
Aug 25, 2014

Distance between the points is given by

( 3 2 s i n ( α − c ) − 1 2 s i n ( α + c ) ) 2 + ( 1 2 s i n ( α − c ) − 3 2 s i n ( α + c ) ) 2 \sqrt { { \left( { \frac { \sqrt { 3 } }{ 2 } sin{ \left( \alpha -c \right) -\frac { 1 }{ 2 } sin{ \left( \alpha +c \right) } } } \right) }^{ 2 }+{ \left( \frac { 1 }{ 2 } sin{ \left( \alpha -c \right) -\frac { \sqrt { 3 } }{ 2 } sin{ \left( \alpha +c \right) } } \right) }^{ 2 } }

By simplifying the expression we get,

( s i n ( α − c ) ) 2 + ( s i n ( α + c ) ) 2 − 3 s i n ( α − c ) s i n ( α + c ) \sqrt { { \left( sin{ \left( \alpha -c \right) } \right) }^{ 2 }+{ \left( sin{ \left( \alpha +c \right) } \right) }^{ 2 }-\sqrt { 3 } sin{ \left( \alpha -c \right) sin{ \left( \alpha +c \right) } } }

Again using,

s i n ( α − c ) = s i n α c o s c − c o s α s i n c sin{ \left( \alpha -c \right) }=sin{ \alpha }cos{ c }-cos{ \alpha sin{ c } }

and

s i n ( α − c ) s i n ( α + c ) = ( s i n α ) 2 − ( s i n c ) 2 sin{ \left( \alpha -c \right) }sin{ \left( \alpha +c \right) }={ \left( sin{ \alpha } \right) }^{ 2 }{ -\left( sin{ c } \right) }^{ 2 }

putting these two in original equation we get

2 ( ( s i n α ) 2 ( c o s ( c ) ) 2 + ( s i n c ) 2 ( c o s α ) 2 ) − 3 ( ( s i n α ) 2 − ( s i n c ) 2 ) \sqrt { 2\left( { \left( sin{ \alpha } \right) }^{ 2 }{ \left( cos{ (c) } \right) }^{ 2 }+{ \left( sin{ c } \right) }^{ 2 }{ \left( cos{ \alpha } \right) }^{ 2 } \right) -\sqrt { 3 } \left( { \left( sin{ \alpha } \right) }^{ 2 }-{ \left( sin{ c } \right) }^{ 2 } \right) }

arranging the equation we get

( 2 ( c o s c ) 2 − 2 ( s i n c ) 2 − 3 ) ( s i n α ) 2 + 2 ( s i n c ) 2 + 3 ( s i n c ) 2 \sqrt { \left( 2{ \left( cos{ c } \right) }^{ 2 }-2{ \left( sin{ c } \right) }^{ 2 }-\sqrt { 3 } \right) { \left( sin{ \alpha } \right) }^{ 2 }+2{ \left( sin{ c } \right) }^{ 2 }+\sqrt { 3 } { \left( sin{ c } \right) }^{ 2 } }

As the distance is constant,it is independent of variable α \alpha ,so its coefficient = 0

2 ( c o s c ) 2 − 2 ( s i n c ) 2 − 3 = 0 2{ \left( cos{ c } \right) }^{ 2 }-2{ \left( sin{ c } \right) }^{ 2 }-\sqrt { 3 } =0

( c o s c ) 2 = 3 4 + 1 2 a n d ( s i n c ) 2 = 1 2 − 3 4 { \left( cos{ c } \right) }^{ 2 }=\frac { \sqrt { 3 } }{ 4 } +\frac { 1 }{ 2 } \quad and\quad { \left( sin{ c } \right) }^{ 2 }=\frac { 1 }{ 2 } -\frac { \sqrt { 3 } }{ 4 }

Putting these in original equation we get

2 ( 1 2 − 3 4 ) + 3 ( 1 2 − 3 4 ) = 1 2 \sqrt { 2\left( \frac { 1 }{ 2 } -\frac { \sqrt { 3 } }{ 4 } \right) +\sqrt { 3 } \left( \frac { 1 }{ 2 } -\frac { \sqrt { 3 } }{ 4 } \right) } =\frac { 1 }{ 2 }

p + q = 1 + 2 = 3 \boxed{p+q=1+2=3}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

This can be solved by considering the 2 points to lie on 2 circles.

Since parametric form of points lying on a circle is (asinθ , acosθ ) where a=radius and θ=angle subtended by position vector of point with x axis, Let 

                       √3÷2sin(α-c) = asinθ 
               and    1/2sin(α-c) = acosθ.............eqn 1

By comparing, a=√3/2 and θ=(α-c) putting these values in eqn 1,; 

                            1/2sin(α-c) = √3/2cos(α-c)
                             tan(α-c) = √3
                             (α-c) = 60 deg

Therefore, point 1 can be written as (√3/2sin60,1/2cos60) which is(3/4,√3/4)

Similarly for point 2 

                        1/2sin(α+c)=Asinβ and  
                     √3/2sin(α+c)=Acosβ..............eqn 2

Comparing we get A=1/2 and β=α+c Putting in eqn 2, 

                    √3/2sin(α+c)=1/2cos(α+c)
                     tan(α+c)=1/√3
                       (α+c)=30 deg.

Therefore point 2 can be written as (1/2sin30,√3/2sin30) which is(1/4,√3/4) Applying distance formula between points 1 and 2

√((3/4-1/4)²+(0)²)=1/2

p=1 and q=2 therefore p+q=3

Ashrene Roy - 6 years, 8 months ago

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