Shortest Distance From Circle to Parabola

Cheaty solution

Sketch the situation, and you'll see that in the $xy$ -plane, the parabola and circle are really close: $(\Delta x)^2 + (\Delta y)^2 << 1$ . Therefore the shortest distance will be approximately $d^2 = \sqrt{\left[(\Delta x)^2 + (\Delta y)^2\right] + (\Delta z)^2} \approx \sqrt{0 + 1} = 1.$ With the low precision Brilliant requires, "1" is an accepted solution!

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Goody solution

Since the circle and parabola lie in parallel planes at distance $\Delta z = 1$ , the problem may be reduced to the shortest distance in the $xy$ -plane.

If the parabola crosses the circle, the shortest distance is obviously zero, or 1 if we include $\Delta z$ . A sketch suggests that this is not the case: the parabola lies entirely outside of the circle.

Outside a circle, the shortest distance to the circle is obtained at the shortest distance to its center. Therefore we find the shortest distance between the parabola $y = \tfrac12 + (x - \tfrac32)^2$ and the origin.

At the shortest distance between a point and a curve, the curve has a tangent perpendicular to the connecting line segment. Thus we solve $x\:dx + y\:dy = 0.$ Now for the parabola $dy = 2(x - \tfrac32)\:dx$ , so we must solve $x + (x - \tfrac32)(\tfrac12 + (x - \tfrac32)^2) = 0.$ This cubic equation may be reduced to the standard form $8x^3 - 36 x^2 + 62 x - 33 = 0,$ but there are no obvious ways to simplify this. Solving numerically, I found $x \approx 0.93264; \\ y \approx \tfrac12 + (0.93264 - \tfrac32)^2 \approx 0.8219; \\ \sqrt{x^2 + y^2} \approx 1.24311\ \ \text{distance to origin in}\ xy\text{-plane}; \\ d_{xy} \approx 0.24311\ \ \text{distance to circle in}\ xy\text{-plane}; \\ d \approx \sqrt{0.024311^2 + 1^2} \approx \boxed{1.02913}.$

(The exact form for $x$ is $x = 1\frac12 - 2\sqrt[3]{\frac 1{3(\sqrt{921} - 27)}} + \frac16\sqrt[3]{3(\sqrt{921} - 27)},$ which is not at all helpful in solving this problem.)