No Calculus Allowed!

Consider a pentagon $OABCD$ such that

$OA=7$

$OB=a$

$OC=b$

$OD=5\sqrt2$

$\angle AOB=\angle BOC=\angle COD=45^\circ$

NoCalculusAllowed

By Cosine Theorem ,

$AB=\sqrt{\strut 49+a^2-7\sqrt2a}$

$BC=\sqrt{\strut a^2+b^2-\sqrt2ab}$

$CD=\sqrt{\strut 50+b^2-10b}$

By Triangle Inequality ,

$AB+BC+CD$

$\geq AD$

$\geq\sqrt{\strut 7^2+(5\sqrt2)^2-2(7)(5\sqrt2)\cos135^\circ}$

$\geq\sqrt{\strut 49+50+2(7)(5\sqrt2)(\frac{\sqrt2}{2})}$

$\geq13$

Hence, the minimum value is $13$ .

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This is also the suggested solution from the paper.

Consider this figure: Title of picture By the Law of Cosines:

$AB=\sqrt{49+a^2-7\sqrt{2}a}$

$BC=\sqrt{a^2+b^2-\sqrt{2}ab}$

By the Pythagorean Theorem:

$CD=\sqrt{50+b^2-10b}$

The sum $AB+BC+CD$ can be minimized to $AD$ when $a$ and $b$ are such that $B$ and $C$ lie on $\overline{AD}$ . By the Pythagorean Theorem, $AD=13$ .

Lets try to solve this by basic reasoning approach

As the expression is [ $\sqrt {49+a^{2}-7\sqrt {2} a}$ + $\sqrt {a^{2}+b^{2}-\sqrt{2}ab}$ + $\sqrt {50+b^{2}-10b}$ ] .

Now $\sqrt {50+b^{2}-10b}$ = $\sqrt{[b-5]^{2}+25}$ .

So $\sqrt {50+b^{2}-10b}$ is minimized, when b = 5 .

Now, putting b = 5 in second term, we get, [ $\sqrt {25+a^{2}-5\sqrt {2} a}$ ]

Now, the first two terms can be written as,
[ $\sqrt{{[a-\frac{7}{\sqrt{2}}]}^{2}+\frac{49}{2}}$ + $\sqrt{{[a-\frac{5}{\sqrt{2}}]}^{2}+\frac{25}{2}}$ ]

Now, just a small work, find such value of a, that minimizes the expression. As [ $a=\frac{7}{\sqrt{2} }$ ] , minimizes first term and [ $a=\frac{5}{\sqrt{2} }$ ] minimizes second term, So, obviously to minimize both the terms [ $a=\frac{\frac{7}{ \sqrt{2}}+\frac{5}{\sqrt{2}}}{2}=\frac{6}{ \sqrt{2}}$ ]

Thus the minimum values for each term can be given by,
[ $\sqrt {49+a^{2}-7\sqrt {2} a}$ ] has minimum value = 5
[ $\sqrt {50+a^{2}-5\sqrt {2} a}$ ] has minimum value = $\sqrt{13}$
[ $\sqrt {50+b^{2}-10b}$ ] has minimum value = 5

So, 10 + $\sqrt{13}$ is the minimum value. i.e., 13.60555127546 , So integer value is 13 .

Why not try Minkowski inequality as $\sqrt {\left( a-\frac {7}{\sqrt 2}\right)^2 +\left(\frac {7}{\sqrt 2}\right) ^2}+\sqrt {\left( -a+\frac {b}{\sqrt 2}\right)^2 +\left(\frac {b}{\sqrt 2}\right) ^2}+\sqrt {\left( 5\sqrt 2 -\frac {b}{\sqrt 2}\right)^2 +\left(\frac {-b}{\sqrt 2}\right) ^2}\ge \sqrt {\left(\frac {17}{\sqrt 2}\right) ^2+\left(\frac {7}{\sqrt 2}\right) ^2}=13$

Firstly i interpreted that these are magnitude of sum of two vectors having magnitude '7' and 'a' with angle between them 135 degree and similarly 'a' and 'b' with 135 degree angle , $\sqrt{50}$ . and 'b' with 135 degree angle...!! now drawn these vectors on Cartesian plane with coordinates and express these in vectors. now use Cauchy inequality by using concept of vectors,and by multiplying(dot product) these 3 vectors by appropriate vectors so that terms a,b are cancel out on summing them. this yield directly our required answer...!! { i am here only post key lines of my solution)