I Need Help in a Challenge

Question

My teacher challenged me with the question below:

$$\sqrt{\frac{\sqrt{41}+\sqrt{29}+\sqrt{10}}{2}\ast \left ( \frac{\sqrt{41}+\sqrt{29}+\sqrt{10}}{2} - \frac{\sqrt{41}}{1} \right )\ast \left ( \frac{\sqrt{41}+\sqrt{29}+\sqrt{10}}{2} - \frac{\sqrt{29}}{1} \right )\ast \left ( \frac{\sqrt{41}+\sqrt{29}+\sqrt{10}}{2} - \frac{\sqrt{10}}{1} \right )}$$

And I tried for a lot of time, but I stuck after doing things like 25 + 16 and 25+4, I don't even know if this is a way to solve this, if someone can do a step-by-step I would love, but do what you can and thanks already.

Sorry for the bad english

Answer 1

We can simplify it bit by bit that is the easiest way!

First by combining and expanding using surd rules:

$$\frac{\sqrt{41}+\sqrt{29}+\sqrt{10}}{2}\ast \left ( \frac{\sqrt{41}+\sqrt{29}+\sqrt{10}}{2} - \frac{\sqrt{41}}{1} \right )=\frac{\sqrt{290}-1}{2}$$

Then by combining the root $29$'s and $10$'s and expanding (FOIL):

$$ \left ( \frac{\sqrt{41}+\sqrt{29}+\sqrt{10}}{2} - \frac{\sqrt{29}}{1} \right )\ast \left ( \frac{\sqrt{41}+\sqrt{29}+\sqrt{10}}{2} - \frac{\sqrt{10}}{1} \right )=\sqrt{290}+\frac{1-\sqrt{290}}{2}$$

Then finally:

$$\frac{\sqrt{290}-1}{2} \ast \left (\sqrt{290}+\frac{1-\sqrt{290}}{2}\right)= \frac{289}{4} $$

And square root leaving you with $\frac{17}{2}$

Answer 2

It seems the whole point of the challenge is to recognize the horrible expression at hand is the area of a triangle computed using Heron's formula.

Reall for any triangle with sides $a, b, c$, its area $\Delta$ is given by the formula:

$$\Delta = \frac14 \sqrt{(a+b+c)(b+c-a)(a-b+c)(a+b-c)}$$

Compare that to the horrible expression, one find it equals to the area of a triangle with sides $a = \sqrt{41},b = \sqrt{29},c = \sqrt{10}$. Notice

$$41 = 4^2 + 5^2,\quad 29 = 5^2 + 2^2,\quad 10 = 1 + 3^2 = (4-5)^2 + (5-2)^2$$

We can realize such a triangle by placing the three vertices at

$$A : (5,2),\quad B : (4,5),\quad C : (0,0)$$

The value we seek is simply

$$\frac12\left|\begin{matrix}5 & 2 & 1\\4 & 5 & 1\\ 0 & 0 & 1\end{matrix}\right| = \frac12 (5\cdot 5 - 2\cdot 4) = \frac{17}{2}$$