Let ABCD be a trapezoid; find point M on AB, such that $DM+MC$ is minimal possible

Question

Let ABCD be a trapezoid with $A=B=90$, $AD=a$ and $AB=BC=2a$.

Find point M on AB, such that $DM+MC$ is minimal possible.

I have been trying to do this question, but without success. I have succeeded in proving that $DA+AC<DB+BC$ something which I believe would help in finding this minimal value, however I have not succeeded in finishing it off. Moreover, I have also found that from Pythagoras we have that $DA+AC=a+2a\sqrt{2}$ and $DB+BC=2a+a\sqrt{5}$. Can someone please help me finish the question off?

Answer 1

Another way to solve -

Choose a point $M$ on $AB$ such that $AM = x, \,$ then BM = $(2a-x)$

$DM = \sqrt{AD^2 + AM^2} = \sqrt{a^2+x^2}$

Similarly $MC = \sqrt{BC^2 + BM^2} = \sqrt{4a^2+(2a-x)^2}$

$L = DM + MC = \sqrt{a^2+x^2} + \sqrt{8a^2+ x^2 - 4ax}$

To find min value of $L$,

$ \displaystyle \frac{dL}{dx} = \frac {x} {\sqrt{a^2+x^2}} + \frac{x-2a}{\sqrt{8a^2+ x^2 - 4ax}} = 0$

$\displaystyle x \sqrt{8a^2+ x^2 - 4ax} + (x-2a) \sqrt{a^2+x^2} = 0$

$\displaystyle x \sqrt{8a^2+ x^2 - 4ax} = (2a-x) \sqrt{a^2+x^2}$

$\displaystyle x^2 (8a^2+ x^2 - 4ax) = (2a-x)^2 (a^2+x^2)$

Solving we get,

$ \displaystyle (x + \frac{2a}{3})^2 = \frac{16a^2}{9}$

So, $x = \frac{2a}{3}$ or $x = -2a$

Discarding $x = -2a$, $AM = x = \frac{2a}{3}$ gives you min.

EDIT: Just added a diagram for what Calvin Lin suggested which uses reflection -

$DM + MC$ will minimize when it is a straight line.

Answer 2

Hint: Reflect $C$ about $AB$ to $C'$.