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

The circle \(O\) has a center at the point of origin with point \(P = (3 , 4)\) on the circle. The line (red) \(l : 3x + 4y - 7 = 0\) intersects the circle at points \(A\) and \(B\) as shown above.

What is the area of the quadrilateral \(AOBP\)?

\[ \large {\left\{\begin{matrix}A_{n+1}=\alpha A_n+(1-\alpha)B_n \\ B_{n+1}=\alpha B_n+(1-\alpha)C_n \\ C_{n+1}=\alpha C_n+(1-\alpha)A_n\end{matrix}\right.} \]

Let \(A_1,B_1,C_1\) be three distinct non-collinear points on the coordinate plane.

Also \(A_n,B_n,C_n\) satisfy the recurrence relation above (\(0<\alpha<1\)).

Then \(\displaystyle\lim_{n\to\infty}A_n\) is the \(\text{__________} \) of the triangle \(A_1B_1C_1\).

Did you know that we can have coordinate systems where the coordinate axes are not perpendicular to each other? Such coordinate systems are known as *oblique coordinate systems*.

The above figure shows an oblique coordinate system. In such systems, the \(x\)-coordinate of a point is found by measuring the distance from the point to the \(y\)-axis parallel to the \( x \)-axis. Similarly, the \(y\)-coordinate of a point is found by measuring the distance from the point to the \(x \)-axis parallel to the \(y\)-axis. The red point in the above figure has the coordinates \((a, b)\).

**Question:** We are given two points: \(A \ (1,6)\) and \(B \ (5,2) \) in an oblique coordinate system where the angle between the positive axes is \( 60^{\circ} \). What is the distance between the two points, \(AB\)?

The \(\triangle AOB\) consists of point \(A = (0 , 1)\), point \(O = (0 , 0)\), and point \(B\) lying somewhere on the \(x\)-axis.

Let \(P\) be the point in the first quadrant such that \(AP = PB\) and \(AO \parallel PB\), as shown above.

If the length of \(OP\) is \(\sqrt{194}\), what is the area of the quadrilateral \( AOBP? \)

The answer would come in the form square root of \(x\), submit your answer as \(x\).

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