# Find the positive root of $100^{2}=x^{2}+ \left( \frac{100x}{100+x} \right)^{2}$

I was struggling with this problem:

$$100^{2}=x^{2}+ \left( \frac{100x}{100+x} \right)^{2}$$

It came up when i was developing a solution to a geometry problem. I've already checked in Mathematica and the solution is right, according to the answer. The answer needs to be a real positive number because it's a measure of a segment.

I've tried factoring, manipulating algebraically, but i couldn't solve the resulting 4th degree polynomial. I appreciate if someone could help me. Thanks!

• What's wrong with just using the solution from Mathematica? – David G. Stork Apr 9 '19 at 2:44
• It is a problem from a Olympiad contest. You can't use solvers there. – shewlong Apr 9 '19 at 3:26
• math.stackexchange.com/questions/2020139/… – lab bhattacharjee Apr 9 '19 at 5:01
• Thanks @bhattacharjee with your hint i could solve the problem. I will post my solution later! – shewlong Apr 9 '19 at 15:08

WA tell us that the positive root is $$50 (-1 + \sqrt 2 + \sqrt{2 \sqrt 2 - 1}) \approx 88.320$$ WA also tells us that the minimal polynomial of this number over $$\mathbb Q$$ is $$x^4 + 200 x^3 + 10000 x^2 - 2000000 x - 100000000$$ and so there is no simpler answer.

On the other hand, substituting $$x=50u$$ in the minimal polynomial gives $$6250000 (u^4 + 4 u^3 + 4 u^2 - 16 u - 16)$$ Now this can factored into two reasonably looking quadratics: $$u^4 + 4 u^3 + 4 u^2 - 16 u - 16 = (u^2 + (2 + 2 \sqrt 2) u + 4 \sqrt 2 + 4) (u^2 + (2 - 2 \sqrt 2) u - 4 \sqrt 2 + 4)$$ But all this is in hindsight...

• How do we know this is the minimal polynomial? What about $x - 50(-1+\sqrt{2} + \sqrt{2\sqrt{2} -1})= 0$? – David G. Stork Apr 9 '19 at 3:12
• @DavidG.Stork, see my edited answer. – lhf Apr 9 '19 at 3:13
• OK... I guess we can assume $x \in \mathbb{Q}$. – David G. Stork Apr 9 '19 at 3:52

Thanks for all the colaboration. Due to @labbhattacharjee 's comment, i was able to solve the problem analytically. Solve the equation $x^2+\frac{9x^2}{(x+3)^2}=27$ shows that, if we have an equation:

$$a^{2}+b^{2}=k$$

Then, calling $$c=\frac{ab}{a-b}$$, we can manipulate algebraically the expression above:

$$(a-b)^{2}+2c(a-b)-k=0$$

For our problem, let $$a=x$$, $$b=\frac{100x}{100+x}$$ and $$k=100^{2}$$. It turns out that $$c=100$$ and $$a-b=\frac{x^{2}}{100+x}$$. So, if we put $$u=\frac{x^{2}}{100+x}$$, we will have:

$$u^{2}+200u-100^{2}=0$$

Which the only positive root is $$100(\sqrt{2}-1)$$. Solving the equation:

$$\frac{x^{2}}{100+x}=100(\sqrt{2}-1)$$

The only positive root is $$50(\sqrt{2}-1+\sqrt{-1+2\sqrt{2}})$$.

Thanks!