Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.3 AIME, harder Find the answer

5. In the Cartesian coordinate system, with the point $(1,0)$ as the center and $r$ as the radius, a circle intersects the parabola $y^{2}=x$ at four points $A$ , $B, C, D$ . If the intersection point $F$ of $A C$ and $B D$ is exactly the focus of the parabola, then $r=$ $\qquad$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

5. $\frac{\sqrt{15}}{4}$ .

Combining the equations of the circle and the parabola, we get
$\left\{\begin{array}{l} y^{2}=x \\ (x-1)^{2}+y^{2}=r^{2} \end{array}\right.$

Eliminating $y$ , we get $x^{2}-x+1-r^{2}=0$ . From $\Delta=1-4\left(1-r^{2}\right)>0$ , we solve to get $r>\frac{\sqrt{3}}{2}$ .
According to the problem, quadrilateral $A B C D$ is an isosceles trapezoid, and the intersection point of $A C$ and $B D$ is the focus $F\left(\frac{1}{4}, 0\right)$ of the parabola, so points $A, F, C$ are collinear. Let $A\left(x_{1}, \sqrt{x_{1}}\right), C\left(x_{2},-\sqrt{x_{2}}\right)\left(0\frac{\sqrt{3}}{2}$ , which meets the requirements of the problem.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.