[Guest]

Two circles on a plane intersect each other at two points. At one of the intersection points, A, there are two particles. One particle moves along the circumference of one circle and the other particle along that of the second circle. They start moving at the same time and reach A, at the same time, after completing one revolution.

Prove that there is a point P in the plane such that the distance of the two particles from P is always the same (the two particles are always equidistant from P).

I read this problem some time ago so may not have written it here very clearly. So in case you need clarification, just ask.

[Guest]

I think something is missing. Because if I assume of circles of the same radius, the particles moving at the same speed and at opposite directions then the answer is dead easy!

[Guest]

I assume the velocity of each particle is constant, right?

[Guest]

I assume the velocity of each particle is constant, right?

Right... Sorry forgot to mention that both particles move at constant speeds.

@ Lemeshianos - Right again. Besides, if they were of the same size, the question would probably not merit to be posted here. Consider different radii.

[Guest]

traveling both clockwise or in opposite directions?

[HoustonHokie]

If the two circles are of radius R1 and R2, then the point we're looking for is a distance R1 from the center of circle 2 and a distance R2 from the center of circle 1. Not sure how to prove it, but I can show it graphically.

The angular velocity of both particles with respect to the center of their circles is equal - they will both travel x degrees/unit of time. You can plot regular points around the circles starting from point A and traveling X degrees and then connect the pairs. A line of equidistant points for each pair is perpendicular to the midpoint of the connecting lines. All of those perpendicular lines will intersect at the common point I described above.

There are actually 2 common points such as I described in each circle. Which one gets used is dependent upon whether both are going in the same direction or opposite directions.

[Guest]

If I am visualizing this right the point is the mid point of AB where A and B are the two intersections.

[HoustonHokie]

If I am visualizing this right the point is the mid point of AB where A and B are the two intersections.

If you use the midpoint of the two intersections, that point will always be closer to the particle on the small circle than it will be to the particle on the large circle. They will only be equal when the particles are at the intersection points.

[Guest]

If the two circles are of radius R1 and R2, then the point we're looking for is a distance R1 from the center of circle 2 and a distance R2 from the center of circle 1. Not sure how to prove it, but I can show it graphically.

The angular velocity of both particles with respect to the center of their circles is equal - they will both travel x degrees/unit of time. You can plot regular points around the circles starting from point A and traveling X degrees and then connect the pairs. A line of equidistant points for each pair is perpendicular to the midpoint of the connecting lines. All of those perpendicular lines will intersect at the common point I described above.

There are actually 2 common points such as I described in each circle. Which one gets used is dependent upon whether both are going in the same direction or opposite directions.

I agree with this. I thought of the same thing but am still unable to prove it! My starting point was that any time t, the position of particles is such that the angle between lines from centre1 to particle1 and centre1 to point to intersection is the same as the angle between lines from centre2 to particle2 and centre2 to point of intersection.

Edited December 2, 2009 by DeeGee

[EventHorizon]

Nice little exercise. I'm positive there's an easier way to do this, but here's my proof.

two circles.bmp

The center of the top circle is the origin.

d is the distance between the centers of the circles.

r₁ is the radius of the top circle.

r₂ is the radius of the bottom circle.

Let t₁ be the angle from the center of the top circle to point A.

Let t₂ be the angle from the center of the bottom circle to point A.

Let t be the angle offset from the initial positions that the particles can be found in both circles.

For brevity, I'll use the following variables:

e = r₁*cos(t₁)

f = r₁*sin(t₁)

g = r₂*cos(t₂)

h = r₂*sin(t₂)

Since both particles start at the same position, these are true:

e = r₁ cos(t₁) = r₂ cos(t₂) = g

f = r₁ sin(t₁) = r₂ sin(t₂) - d = h - d

Let the point P be at (X,Y).

Let p₁ be the position of the particle on circle 1.

Let p₂ be the position of the particle on circle 2.

I'll start the particle on the top circle going counter-clockwise, and the particle on the bottom circle going clockwise.

The distance from p₁ to P is sqrt( (X - r₁*cos(t₁ + t))^2 + (Y - r₁*sin(t₁ + t))^2 )

The distance from p₂ to P is sqrt( (X - r₂*cos(t₂ - t))^2 + (Y + d - r₁*sin(t₁ - t))^2 )

Setting these equal and squaring both sides results in:

(X - r₁*cos(t₁ + t))^2 + (Y - r₁*sin(t₁ + t))^2 = (X - r₂*cos(t₂ - t))^2 + (Y + d - r₁*sin(t₁ - t))^2

Multiplying out all the squares and moving everything to one side results in:

0 = X^2 - 2Xr₁*cos(t₁+t) + (r₁*cos(t₁+1))^2 + Y^2 - 2Yr₁*sin(t₁+t) + (r₁*sin(t₁+t))^2

- X^2 + 2Xr₂*cos(t₂-t) - (r₂*cos(t₂-t))^2 - Y^2 - 2Yd + 2Yr₂*sin(t₂-t) + 2dr₂*sin(t₂-t) - d^2 - (r₂*sin(t₂-t))^2

Obviously, the X^2's and Y^2's cancel out.

Recall the following added angle identities:

sin(a +/- b) = sin(a)cos(b) +/- cos(a)sin(b)

cos(a +/- b) = cos(a)cos(b) -/+ sin(a)sin(b)

Using the added angle identities and substituting in e,f,g, & h, results in:

0 = -2X(e*cos(t) - f*sin(t)) + (e*cos(t))^2 + (f*sin(t))^2 - 2ef*cos(t)(sin(t)

- 2Y(f*cos(t) + e*sin(t)) + (f*cos(t))^2 + (e*sin(t))^2 + 2ef*cos(t)(sin(t)

+ 2X(g*cos(t) + h*sin(t)) - (g*cos(t))^2 - (h*sin(t))^2 - 2gh*cos(t)sin(t)

+ 2Y(h*cos(t) - g*sin(t)) - (h*cos(t))^2 - (g*sin(t))^2 + 2gh*cos(t)sin(t)

-2Yd - d^2 + 2d(h*cos(t) + g*sin(t))

The last terms on the first four lines cancel out:

0 = -2X(e*cos(t) - f*sin(t)) + (e*cos(t))^2 + (f*sin(t))^2

- 2Y(f*cos(t) + e*sin(t)) + (f*cos(t))^2 + (e*sin(t))^2

+ 2X(g*cos(t) + h*sin(t)) - (g*cos(t))^2 - (h*sin(t))^2

+ 2Y(h*cos(t) - g*sin(t)) - (h*cos(t))^2 - (g*sin(t))^2

-2Yd - d^2 + 2d(h*cos(t) + g*sin(t))

Using sin^2+cos^2=1, we can simplify more:

0 = -2X(e*cos(t) - f*sin(t)) + e^2 + f^2

- 2Y(f*cos(t) + e*sin(t))

+ 2X(g*cos(t) + h*sin(t)) - g^2 - h^2

+ 2Y(h*cos(t) - g*sin(t))

-2Yd - d^2 + 2d(h*cos(t) - g*sin(t))

Substituting g=e and h=f+d results in:

0 = -2X(e*cos(t) - f*sin(t)) + e^2 + f^2

- 2Y(f*cos(t) + e*sin(t))

+ 2X(e*cos(t) + (f+d)*sin(t)) - e^2 - f^2 - 2fd - d^2

+ 2Y((f+d)*cos(t) - e*sin(t))

-2Yd - d^2 + 2d((f+d)*cos(t) - e*sin(t))

Simplifying (and a little factoring) yields:

0 = 2X(2f+d)*sin(t)

+ 2Y(d*cos(t) - 2e*sin(t))

-2Yd - 2d(f+d) + 2d((f+d)*cos(t) - e*sin(t))

Moving stuff around a little (isolating X and Y) results in:

0 = X*(4f+2d)*sin(t)

+ Y*(2d*cos(t) - 2d - 4e*sin(t))

+ 2d(f+d)*cos(t) - 2d(f+d) - 2de*sin(t)

(I could used two values for t, and have two equations with 2 unknowns... but the following is easier)

Notice that cos(t) only appears twice. Let Y = -(f+d) (= -h). This will cancel out all terms without sin(t):

0 = X*(4f+2d)*sin(t)

- (f+d)*(2d*cos(t) + 2d(f+d) + (f+d)*4e*sin(t))

+ 2d(f+d)*cos(t) - 2d(f+d) - 2de*sin(t)

0 = X*(4f+2d)*sin(t) + (f+d)*4e*sin(t)) - 2de*sin(t)

0 = X*(4f+2d)*sin(t) + 2e*sin(t)*(2f+2d-d)

0 = X*(4f+2d)*sin(t) + 2e*sin(t)*(2f+d)

0 = X*(4f+2d)*sin(t) + e(4f+2d)*sin(t)

Obviously, X=-e.

So, P = (-r₁*cos(t₁), -r₂*sin(t₂))

I'll let someone else find P when both particles go counter-clockwise.

[Guest]

Nice little exercise. I'm positive there's an easier way to do this, but here's my proof.

two circles.bmp

The center of the top circle is the origin.

d is the distance between the centers of the circles.

r₁ is the radius of the top circle.

r₂ is the radius of the bottom circle.

Let t₁ be the angle from the center of the top circle to point A.

Let t₂ be the angle from the center of the bottom circle to point A.

Let t be the angle offset from the initial positions that the particles can be found in both circles.

For brevity, I'll use the following variables:

e = r₁*cos(t₁)

f = r₁*sin(t₁)

g = r₂*cos(t₂)

h = r₂*sin(t₂)

Since both particles start at the same position, these are true:

e = r₁ cos(t₁) = r₂ cos(t₂) = g

f = r₁ sin(t₁) = r₂ sin(t₂) - d = h - d

Let the point P be at (X,Y).

Let p₁ be the position of the particle on circle 1.

Let p₂ be the position of the particle on circle 2.

I'll start the particle on the top circle going counter-clockwise, and the particle on the bottom circle going clockwise.

The distance from p₁ to P is sqrt( (X - r₁*cos(t₁ + t))^2 + (Y - r₁*sin(t₁ + t))^2 )

The distance from p₂ to P is sqrt( (X - r₂*cos(t₂ - t))^2 + (Y + d - r₁*sin(t₁ - t))^2 )

Setting these equal and squaring both sides results in:

(X - r₁*cos(t₁ + t))^2 + (Y - r₁*sin(t₁ + t))^2 = (X - r₂*cos(t₂ - t))^2 + (Y + d - r₁*sin(t₁ - t))^2

Multiplying out all the squares and moving everything to one side results in:

0 = X^2 - 2Xr₁*cos(t₁+t) + (r₁*cos(t₁+1))^2 + Y^2 - 2Yr₁*sin(t₁+t) + (r₁*sin(t₁+t))^2

- X^2 + 2Xr₂*cos(t₂-t) - (r₂*cos(t₂-t))^2 - Y^2 - 2Yd + 2Yr₂*sin(t₂-t) + 2dr₂*sin(t₂-t) - d^2 - (r₂*sin(t₂-t))^2

Obviously, the X^2's and Y^2's cancel out.

Recall the following added angle identities:

sin(a +/- b) = sin(a)cos(b) +/- cos(a)sin(b)

cos(a +/- b) = cos(a)cos(b) -/+ sin(a)sin(b)

Using the added angle identities and substituting in e,f,g, & h, results in:

0 = -2X(e*cos(t) - f*sin(t)) + (e*cos(t))^2 + (f*sin(t))^2 - 2ef*cos(t)(sin(t)

- 2Y(f*cos(t) + e*sin(t)) + (f*cos(t))^2 + (e*sin(t))^2 + 2ef*cos(t)(sin(t)

+ 2X(g*cos(t) + h*sin(t)) - (g*cos(t))^2 - (h*sin(t))^2 - 2gh*cos(t)sin(t)

+ 2Y(h*cos(t) - g*sin(t)) - (h*cos(t))^2 - (g*sin(t))^2 + 2gh*cos(t)sin(t)

-2Yd - d^2 + 2d(h*cos(t) + g*sin(t))

The last terms on the first four lines cancel out:

0 = -2X(e*cos(t) - f*sin(t)) + (e*cos(t))^2 + (f*sin(t))^2

- 2Y(f*cos(t) + e*sin(t)) + (f*cos(t))^2 + (e*sin(t))^2

+ 2X(g*cos(t) + h*sin(t)) - (g*cos(t))^2 - (h*sin(t))^2

+ 2Y(h*cos(t) - g*sin(t)) - (h*cos(t))^2 - (g*sin(t))^2

-2Yd - d^2 + 2d(h*cos(t) + g*sin(t))

Using sin^2+cos^2=1, we can simplify more:

0 = -2X(e*cos(t) - f*sin(t)) + e^2 + f^2

- 2Y(f*cos(t) + e*sin(t))

+ 2X(g*cos(t) + h*sin(t)) - g^2 - h^2

+ 2Y(h*cos(t) - g*sin(t))

-2Yd - d^2 + 2d(h*cos(t) - g*sin(t))

Substituting g=e and h=f+d results in:

0 = -2X(e*cos(t) - f*sin(t)) + e^2 + f^2

- 2Y(f*cos(t) + e*sin(t))

+ 2X(e*cos(t) + (f+d)*sin(t)) - e^2 - f^2 - 2fd - d^2

+ 2Y((f+d)*cos(t) - e*sin(t))

-2Yd - d^2 + 2d((f+d)*cos(t) - e*sin(t))

Simplifying (and a little factoring) yields:

0 = 2X(2f+d)*sin(t)

+ 2Y(d*cos(t) - 2e*sin(t))

-2Yd - 2d(f+d) + 2d((f+d)*cos(t) - e*sin(t))

Moving stuff around a little (isolating X and Y) results in:

0 = X*(4f+2d)*sin(t)

+ Y*(2d*cos(t) - 2d - 4e*sin(t))

+ 2d(f+d)*cos(t) - 2d(f+d) - 2de*sin(t)

(I could used two values for t, and have two equations with 2 unknowns... but the following is easier)

Notice that cos(t) only appears twice. Let Y = -(f+d) (= -h). This will cancel out all terms without sin(t):

0 = X*(4f+2d)*sin(t)

- (f+d)*(2d*cos(t) + 2d(f+d) + (f+d)*4e*sin(t))

+ 2d(f+d)*cos(t) - 2d(f+d) - 2de*sin(t)

0 = X*(4f+2d)*sin(t) + (f+d)*4e*sin(t)) - 2de*sin(t)

0 = X*(4f+2d)*sin(t) + 2e*sin(t)*(2f+2d-d)

0 = X*(4f+2d)*sin(t) + 2e*sin(t)*(2f+d)

0 = X*(4f+2d)*sin(t) + e(4f+2d)*sin(t)

Obviously, X=-e.

So, P = (-r₁*cos(t₁), -r₂*sin(t₂))

I'll let someone else find P when both particles go counter-clockwise.

I think I did about 2/3 of that and then I gave up . Nice work.

[Guest]

Let the 1st circle (center : O1, radius : r1) & 2nd cicrcle (center : O2, radius : r2) inetersect at P1 and P2. Let the particles start from P1 in opposite direction at same angular speed.

Then the equidistant point should be some point symmetrical to both the circles. It can be P1, P2 or mid-point of O1O2 or any other similar point.

Another such similar point is the mid-point of Q1Q2 where Q1 is diametrically opposite to P1 on 1st circle and Q2 is diametrically opposite to P2 on 2nd circle. And mid point of Q1Q2 (denoted as "X") is the required point. Proof is simple. Let our particles reach R1 and R2 after some time "t". Then two triangles X O1 R1 and R2 O2 X are congruent since XO1 = R2O2 (=r2), O2X = O1R1 (=r1) and angle XO1R1= angle XO2R2 . Hence XR1 = XR2 is always true i.s. "X" is the required equidistant point. One can draw it and see it clearly.