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In the 1600s, René Descartes married algebra and geometry to create the Cartesian plane.

There exists \(2\) points on Cartesian coordinate. Point \(A\) has coordinates \((0,1)\). Point \(B\) is on the origin \((0,0)\).

Point \(A\) and \(B\) always has a constant distance of \(1\) from each other.

Point \(A\) moves with a horizontal velocity (parallel to the x-axis) of \(v\)

Point \(B\) moves with a horizontal velocity of \(2v\)

Point \(B\)'s vertical velocity (parallel to the y-axis) is \(0\) while Point \(A\) is allowed to move vertically in order to keep the constant distance of \(1\).

All this movement is happening in the first quadrant.

Given that the area made with the \(x\) and \(y\) axis and the path traveled by Point \(A\) is \(P\), find \[\left\lfloor 1000P \right\rfloor \]

After you solve this, you might want to try a continuation of this problem.

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In the diagram above, *ABCD* and *PQRS* are both rectangles. Points *P*, *Q*, *R*, and *S* lie on segments \(\overline{AB}\), \(\overline{BC}\), \(\overline{CD}\), and \(\overline{DA}\), respectively, and \(\overline{BQ} < \overline{QC}\).

If \(AB=36\) and \(BC=50\), then the maximum possible value of \(BQ\) can be written in the form \(a-\sqrt{b}\), where \(a,b \in \mathbb{N}\). What is \(a+b\)?

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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:
###### Inspired from Kushal Patankar's Problem

\[ (ax-by)^2 + (bx-ay)^2 = c^2 \]

For positive integers \(a,b,c\). What is the minimum value of \(a+b+c\)

**Details and Assumptions**

- Diagram not up to scale

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