Let $A$ be the point $(0, 1)$ and $B$ the point $(2, 2)$ in the plane. Consider all paths made up of the two line segments $AC$ and $CB$ as the point $C$ varies on the x-axis. Find the coordinates of $C$ for which the corresponding path has the shortest length.

My answer: If $C=(a,0)$ is the point then the lenth of the path $A \rightarrow C \rightarrow B$ is $\sqrt{a^2+1}+\sqrt{(a-2)^2+4}$. Minimising this by calculus seems a little lengthy but the answer turns out to be $C=(2/3,0)$.

Is there a less tedious and a more intuitive answer?

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    Yes there is: reflect the point (2,2) about the x-axis (so, to (2,-2)). Then any shortest path from (0,1) to this new point, a straight line, has necessarily the same length as the "reflected" version where you bounce against the x-axis. You should be able to find the actual point C by yourself. – A.Sh Jan 18 '16 at 10:50

So, just filling out the details in my comment: We want the point C=(a,0) so that ACB' is a straight line, where B'=(2,-2). Then the equation of the straight line in question is y=-3x/2 + 1. Plugging in y=0, we get x=2/3, which is our a.

The point is that finding the shortest path with one bounce against the x-axis is equivalent to finding the shortest path that goes through the x-axis but to the reflection of the goal. If you can't see this immediately, it might help to draw a picture of the situation and stare at it for a while.

Edit: I'll make the small addendum here that you don't even need coordinate geometry to finish the job. High-school Euclidean geometry (similar triangles) also do the trick.

Hint differentiating means making slope of curve $0$ so its same as $tan(0)$ now we know $$tan(\theta)=|\frac{m_1-m_2}{1+m_1m_2}|$$ so making it $0$ we get $0=-1/a+2/2-a$ $<-$ these are slopes of $AC,BC$ respectively thus $a-2+2a=0$ thus $a=2/3$ thats easy i suppose.

In mathematics, an ellipse is a curve on a plane that surrounds two focal points such that the sum of the distances to the two focal points is constant for every point on the curve.

C lays on an ellipse having as focus points A and B, and because we want the shortest distance, i.e. the smallest ellipse, we are looking for the particular ellipse which is tangent to the X axis, i.e. that intersect the X axis with 2 coincident roots.

So, you can write the equation of the ellipse having as focus points A and B based on a generic distance L, intersect with the line X = 0, impose the discriminant of the equation to be zero.

Overall, it is about the same amount of calculations.

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