Find the smallest value for sum of square roots? (2008 AMC Senior)

Question

I change the expression into the square roots of sum of two squares.

The result should be the length of the tree blue lines, but I still can not find the smallest value.

Answer 1

Let $\measuredangle AOD=135^{\circ},$ rays $OB$ and $OC$ are placed between rays $OA$ and $OD$ so that $$\measuredangle AOB=\measuredangle BOC=\measuredangle COD=45^{\circ}.$$ Also, let $OA=7$, $OB=a$, $OC=b$ and $OD=5\sqrt2.$

Thus, by the triangle inequality: $$AB+BC+CD\geq AD$$ or $$\sqrt{7^2+a^2-2\cdot7\cdot a\cdot\cos45^{\circ}}+\sqrt{a^2+b^2-2ab\cos45^{\circ}}+$$ $$+\sqrt{b^2+(5\sqrt2)^2-2b\cdot5\sqrt2\cos45^{\circ}}\geq\sqrt{7^2+(5\sqrt2)^2-2\cdot7\cdot5\sqrt2\cos45^{\circ}}$$ or $$\sqrt{a^2-7\sqrt2a+49}+\sqrt{a^2-\sqrt2ab+b^2}+\sqrt{b^2-10b+50}\geq13.$$ The equality occurs when $A$, $B$, $C$ and $D$ are placed on the segment $AD,$ which says that we got a minimal value.

Also, we can get that the minimum occurs for $$(a,b)=\left(\frac{35\sqrt2}{17},\frac{35}{12}\right).$$

Answer 2

The distance between $(b,5)$ and $(5,0)$ is the same as the distance between $(b,0)$ and $(5,5)$. Ponder upon the following visual:

Now, $(\frac{a}{\sqrt2},\frac{a}{\sqrt2})$ lies on the line $y=x$. Plotting the other points and lines, we have the following diagram:

We have to minimize $DA + AB+BC$

Recall that the path taken by light between two points is the shortest. Imagine that a light ray emerges from D, reflects at the x-axis, then reflects again on $y=x$ and reaches point C.

See my answer on determining the path of light ray:

Calculate if a line will pass through a given point?

Hence, the following construction gives us the path of light ray (Minimum length)

The length $DA + AB+BC$ is the same as $DC''$. (I am leaving the proof to the reader)

So the minimum length is $13$

Answer 3

Your work can be modified to consider instead the points $$A = \left(\frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}}\right), \\ B = (0,b), \\ C = (7,0), \\ D = (-5,5).$$ Then you can note that $$CA = \sqrt{49 + a^2 - 7\sqrt{2} a}, \\ AB = \sqrt{a^2 + b^2 - \sqrt{2} ab}, \\ BD = \sqrt{50 + b^2 - 10b}.$$ Consequently the minimum value of $CA + AB + BD$ is attained when $C, A, B, D$ are collinear; i.e., $$CA + AB + BD \ge CD = \sqrt{(7+5)^2 + (0-5)^2} = 13.$$