Dancing Rods

Geometry Level 3

Let two Rods $AB$ of length $10$ and $CD$ of length $20$ are sliding on smooth standard co-ordinate axis such that their ends are always con-cyclic. If the locus of centre of that circle which pass through all four points $A,B,C,D$ is expressed as:

$(ax-by)^2 + (bx-ay)^2 = c^2$

For positive integers $a,b,c$ . What is the minimum value of $a+b+c$

Details and Assumptions

Diagram not up to scale

Inspired from Kushal Patankar's Problem

The answer is 18.

1 solution

Deepanshu Gupta
Mar 3, 2015

$A(10\cos { \theta } ,0),\\ B(0,10\sin { \theta } ),\\ C(20\sin { \theta } ,0),\\ D(0,20\cos { \theta } )$

Let equation of con-cyclic circle : ${ x }^{ 2 }+{ y }^{ 2 }+2gx+2fy+c=0$ Now if we put x=0 , then it gives two values of 'y' which represent end's point of rod's on y-axis , which are D and B .

$x=0\quad \Rightarrow { y }^{ 2 }+2fy+c=0$ .

Using vieta's sum of root's theoram ,

$-2f={ y }_{ B }+{ y }_{ D }=10\sin { \theta } +20\cos { \theta } \\ 2h=10\sin { \theta } +20\cos { \theta } \quad .\quad .\quad .(1)$

Similarly for y=0 , we get

$2k=20\sin { \theta } +10\cos { \theta } \quad .\quad .\quad .(2)$

using eqn. (1) & (2) and eliminate variable theta and replace h and k by x and y respectively we should get locus as :

$\displaystyle{\boxed { { (2x-y) }^{ 2 }{ +(2y-x) }^{ 2 }={ 15 }^{ 2 } } }$ Ans.

Beautiful problem! I had a great time visualizing this. BTW, then I realized that the rods must move differently than the arrows shown. i.e. the ends of these 'ladders' resting on the ground must move in opposite directions. And so must the ends resting on the wall too. Might want to change that in the diagram given.

Imgur Imgur

This gives the center as midpoint or average of (x coordinates of ground points, y coordinates of wall points) $C = (\frac{10 \cos\theta+20 \sin \theta}{2}, \frac{10 \sin\theta+20 \cos \theta}{2})$ Which turned out to be an ellipse with semi major, semi minor axes 15 and 5, and rotated by 45 degrees.

Ujjwal Rane - 6 years, 3 months ago

whoah nice solution!!! can you tell me what the angle at A and the angle at B are both theta? thanks!!!

Willia Chang - 4 years, 11 months ago

Same doubt here also!! How the angles are same

ABC def - 4 years, 3 months ago

Because in a cyclic quadrilateral sum of opposite angles equal 180°

Navin Murarka - 3 years, 7 months ago

The problem says $(ax - by)^2 + (by - ax)^2 = c^2.$ It should say $(ax - by)^2 + (bx - ay)^2 = c^2.$

Jon Haussmann - 6 years, 3 months ago

Thanks Sir , I edited that ! sorry for inconvenience .

Deepanshu Gupta - 6 years, 3 months ago

A small doubt. In your diagram if O is origin ; clearly triangles OAB and ODC are similar. Therefore OA×OC=OB×OD. If I substitute values I get tan (theta) = + or - 1 which means there is only one such quadrilateral [theta=45] in first quadrant. Thus I get the circumcentre as (15/√2 ,15/√2). Then why a locus?
Is anything wrong ?please reply

Sriram G - 6 years, 3 months ago

@Sriram G – Don't confuse them as parallel lines. Figure not drawn to scale

Mayank Singh - 5 years, 10 months ago

Hats off to this one