How to find the sum of distances so that it is minimal?

Question

Question:

$A$ and $B$ are two points on the same side of a line $l$. Denote the orthogonal projections of $A$ and $B$ onto $l$ by $A^\prime$ and $B^\prime$. Suppose that the following distance are given: $d(A,A^\prime) = 5$, $d(B,B^\prime) = 4$ and $d(A^\prime,B^\prime) = 10$. Find a point on l such that the sum $d(A,C) + d(B,C)$ is minimal.

I have never seen a problem like this before, can someone help me solve this please?

So I computed the sum d(A,C) + d(B,C), took the derivative and solved for C to get 1/2, however Im unsure if this is correct, could someone solve for C and tell me if I'm right?

Answer 1

Find the symmetric point of $A$ with respect to the line $l$. Say it's $A_2$. Connect $A_2$ with $B$, let's denote the line through $A_2$ and $B$ by g. Find the intersection point of $g$ and $l$. This is the desired point $C$. Try to prove it.

Note: the given distances 5,4,10 - they don't matter; the above is generally true.

This solution is the same as method 2 here:

http://www.analyzemath.com/calculus/Problems/minimum_distance.html

Method 1 (from that page) is also nice but less elegant.

Answer 2

Try drawing a picture and make the line horizontal. It'll make the problem easier to understand/visualize.

1. Draw a horizontal line
2. Draw one vertical line going up form the line with a height of 5. With coordinate (x,y) = (0,5).
3. Draw another of height 4. With coordinate (x,y) = (10,4).
4. Draw lines from the top of the vertical lines to a point C in between them. With coordinate (x,y) = (C,0).
5. Compute the Euclidean distances, add them, and minimize w.r.t. C.

Answer 3

Consider that the locus of points $P$ for which $d(P,A)+d(P,B)=k$ is an ellipse with foci in $A$ and $B$. So the problem boils down to finding an ellipse with foci in $A$ and $B$ that is tangent in $C$ to the $l$ line. On the other hand, if we take $B''$ as the mirror image of $B$ with respect to the $l$ line, $d(A,C)+d(B,C)=d(A,C)+d(C,B'')$, hence $C$ is given by $\color{red}{AB''\cap l}$,

$$\frac{A'C}{CB'}=\frac{AA'}{B'B''}=\frac{AA'}{BB'}=\frac{5}{4}$$ and: $$ d(A,C)+d(B,C) = \sqrt{(5+4)^2+10^2} = \color{red}{\sqrt{181}}.$$

Notice that if we find the minimum of: $$ \sqrt{5^2+x^2}+\sqrt{4^2+y^2} $$ under the constraint $x+y=10$ through Lagrange multipliers, we find: $$ \frac{x}{\sqrt{5^2+x^2}}=\frac{y}{\sqrt{4^2+y^2}} $$ that is equivalent to the collinearity of $A,C,B''$.