The chord of circle #x^2 + y^2 = 25# connecting #A(0,5)# and #B(4,3)# is the base of an equilateral triangle with the third vertex #C# in the first quadrant. What is the smallest possible #x#-coordinate of #C#?

5 Answers
CW
Apr 2, 2018

Smallest possible x-coordinate of #C=(2-sqrt3)#

Explanation:

enter image source here
given a circle #x^2+y^2=25, => "center " O=(0,0), and "radius " r=5#,
#AB=sqrt((4-0)^2+(3-5)^2))=sqrt(16+4)=sqrt20=2sqrt5# units
#=>AC=CB=AB=2sqrt5# units
slope of #AB=m_(AB)=(3-5)/(4-0)=-1/2#
let #D# be the midpoint of #AB#,
#=># co-ordinates of #D=(x_d,y_d)=(2,4)#,
As #AC=BC, and D# is the midpoint of #AB#,
#=> angleCDA=angleCDB=90^@#
#=> CD=CAsin60=2sqrt5*sqrt3/2=sqrt15# units
as #CD# is perpendicular to #AB#
slope of #CD=m_(CD)=2#
#=> tanalpha=2, => sinalpha=2/sqrt5, cosalpha=1/sqrt5#
Let co-ordinates of #C=(x,y) #
#=> C(x,y)=(x_d-CDcosalpha," " y_d-CDsinalpha)#
#=(2-sqrt15*1/sqrt5, " " 4-sqrt15*2/sqrt5)#
#=(2-sqrt3, " "4-2sqrt3)#

CW
Apr 2, 2018

Smallest possible x-co-ordinate of #C=(2-sqrt3)#

Explanation:

solution 2 :
enter image source here
Given a circle #x^2+y^2=25, => " center " O=(0,0), and "radius " r=5# units
let #D# be the midpoint of chord #AB#,
#=> OD# is perpendicular to #AB#,
as #AC=BC# and #D# is midpoint of #AB, C# lies on line #OD#
#AB=sqrt(4^2+2^2)=sqrt20=2sqrt5# units
#OD=sqrt(2^2+4^2)=sqrt20=2sqrt5# units
#CD=ACsin60=ABsin60=2sqrt5*sqrt3/2=sqrt15#
#=> OC=OD-CD=sqrt20-sqrt15# units
slope of #OD=m_(OD)=4/2=2#
#tanalpha=2, =>sinalpha=2/sqrt5, cosalpha=1/sqrt5#
Let co-ordinates of #C=(x,y)#
#=> C(x,y)=(OCcosalpha, OCsinalpha)#
#= ((sqrt20-sqrt15)*1/sqrt5, " "(sqrt20-sqrt15)*2/sqrt5)#
#=(sqrt4-sqrt3, " " 2(sqrt4-sqrt3)#
#=(2-sqrt3, " " 4-2sqrt3)#

Hence, smallest possible x-co-ordinate of #C=(2-sqrt3)#

Ratnaker Mehta
Apr 2, 2018

#C=C(2-sqrt3,4-2sqrt3)#.

Explanation:

Let the third vertex of the equilateral #DeltaABC# be

#C(x,y)#, where, #A(0,5) and B(4,3)#.

We have, #AB^2=BC^2=CA^2#.

Using the distance formula,

# CA^2=BC^2 rArr (x-0)^2+(y-5)^2=(x-4)^2+(y-3)^2#.

#:. x^2+y^2-10y+25=x^2+y^2-8x-6y+16+9#.

#:. -4y=-8x, or, y=2x...............(diamond_1)#.

Also #CA^2=AB^2 rArr (x-0)^2+(y-5)^2=(0-4)^2+(5-3)^2.#

#:. x^2+(2x-5)^2=16+4, i.e., 5x^2-20x=20-25, or, #

# x^2-4x=-1#, and completing the square on the L.H.S.,

# x^2-4x+4=-1+4 :. (x-2)^2=3 rArr x=2+-sqrt3#.

Clearly, #x_(min)=2-sqrt3#, and, the corresponding

#y=2x=2(2-sqrt3)=4-2sqrt3#.

Thus, #C=C(2-sqrt3,4-2sqrt3)# is the desired third vertex, as

Respected CW has readily obtained!

Dean R.
May 20, 2018

#C = (2-sqrt{3},4-2sqrt{3})#

Explanation:

We're oddly given a circle in this problem and awkward wording asking for the smallest possible x coordinate.

I'm not sure what the circle has to do with anything. We could list equations endlessly for circles containing these two points. Forget about the circle.

Given one side, there will be two equilateral triangles with that as a side. Let's rewrite the problem:

Find all possible third vertices #C(x,y)# which make an equilateral triangle with #A(a,b)# and #B(c,d)#.

There are a few different ways to do this. I'd lean toward complex numbers, but I probably should try to keep this simpler than that.

Let's review. The altitude #h# of an equilateral triangle bisects a side #s# so #(s/2)^2 + h^2 = s^2# or # h= \sqrt{3}/2 s#

We just need to go #pm h# along the perpendicular bisector of AB and we'll get to our two possible Cs.

The midpoint D of AB is #D({a+c}/2, {b+d}/2)#

The direction vector from A to B is #B-A=(c-a,d-b)#.

For the perpendicular direction vector we swap and negate one:

#P=(d-b,a-c)#

Now schematically what we're doing is

# C = D \pm h P/|P| = D \pm h/|P| \ P #

We have #s=|AB|=|P|# so # h/|P|= sqrt3/2 #

#C = D pm sqrt{3}/2 P #

# C = ({a+c}/2, {b+d}/2) pm sqrt3/2 (d-b,a-c) #

That's the general solution; let's apply it to

# (a,b)=(0,5), (c,d)=(4,3)#

# C= ( (0+4)/2, (5+3)/2) pm sqrt3/2 (-2, -4) #

# C = (2,4) pm sqrt(3)(1,2) #

# C=(2+sqrt{3},4+2sqrt{3}) or (2-sqrt{3},4-2sqrt{3})#

Uh, the one with the least x coordinate is

#C = (2-sqrt{3},4-2sqrt{3})#

Check:

We check the squared distance to each point

# |C-A|^2 = (2-sqrt{3})^2 + (4-2sqrt{3}-5)^2 = 7 -4sqrt{3} + 13 +4 sqrt{3} = 20#

# |C-B|^2 = (2-sqrt{3}- 4 )^2 + (4-2sqrt{3} - 3 )^2 = 20 quad sqrt#

Dean R.
May 20, 2018

#x=2 - \sqrt{3}#

Explanation:

I did a geometric construction turned into algebra via the Cartesian plane in the last answer. Let's try something a bit different here.

Given #A(0,5)# and #B(4,3)# for an equilateral triangle with #C(x,y)#

#|AB|^2 = |AC|^2 = |BC|^2#

#|AB|^2=(0-4)^2+(3-5)^2=20#

#x^2+(y-5)^2=(x-4)^2+(y-3)^2=20#

Two equations, two unknowns. First one first.

# x^2 + y^2 - 10 y + 25 = x^2 - 8x + 16 + y^2 - 6y + 9 #

#-10y = -8x -6y#

#8x = 4y#

#y=2x#

That must be the equation for the perpendicular bisector of AB.

# x^2 + (2x-5)^2 = 5x^2 - 20 x + 25 = 20 #

#x^2-4x +1 = 0#

The Shakespeare Quadratic Formula (#2b# or #-2b#) says #x^2-2bx+c# has zeros #x=b pm sqrt{b^2=c}#.

#x = 2 pm \sqrt{3} #

The smallest root is

#x=2 - \sqrt{3}#

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