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I'm trying to solve this geometry problem. All my efforts so far have resulted in 4th grade equations which I can only solve numerically.

Two squares aligned as shown in the image. What is the length of x?

enter image description here

Dan Byström
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  • What equation did you get? Fourth-degree looks correct to me. – Martin R Feb 05 '24 at 12:48
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    The obvious one is: y^2+x^2=10^2 where y = 10x/(10+x) – Dan Byström Feb 05 '24 at 12:51
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    That is what I get as well, equivalently $x^4+20 x^3+100 x^2-2000 x-10000 = 0$. Quartic equations can be solved exactly, but it is not nice. – Martin R Feb 05 '24 at 12:57
  • I think Dan has the right idea. If you set $y$ to be the height of the smaller triangle (on the right), you get $$ \left{ \begin{split} x^2 + y^2 = 10^2 \ \frac{y}{x} = \frac{10}{10+x} \end{split} \right. $$ – Matti P. Feb 05 '24 at 12:58
  • I saw this problem in an Instagram post a couple of months or so ago, I opened up the comments and saw a solution (which involved trig-functions, that's all I remember). I quickly closed it again not to spoil the fun solving it. But now I fail both at solving the problem AND finding that post again. :-( – Dan Byström Feb 05 '24 at 13:03
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    Welcome to Math.SE! ... I concur about your system, and about the 4th degree equation. As it happens, the quartic factors into a pair of nice quadratics ... provided you get clever with $\sqrt{2}$. (This is non-obvious. I only realized it by calculating the exact roots in Mathematica, seeing a bunch of $\sqrt{2}$s, and then re-factoring the over the corresponding "field extension".) Defining $s:=10$ (to help track how this parameter affects the outcome), we have $$x^4+ 2 s x^3+ s^2 x^2- 2 s^3 x -s^4 = ( (\sqrt2-1) x^2+sx+s^2) ( (\sqrt2+1) x^2 - sx - s^2)$$ – Blue Feb 05 '24 at 13:09
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    Given Blue's comment, one could then hope for a direct path to $(\sqrt2+1)x^2-sx-s^2=0$. I have no idea if that's feasible – Milten Feb 07 '24 at 14:59

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COMMENT.-Taking as unknown $t$ the upper vertical segment you have the equation $$(\sqrt{100+t^2}+10)^2=10^2+\left(10+\sqrt{10^2-(10-t)^2}\right)^2$$ which gives the solution $$t=5\left(1+\sqrt2-\sqrt{2\sqrt2-1}\right)$$ This way you have $$\boxed{x=\sqrt{10^2-(10-t)^2}}$$

Piquito
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enter image description here

In picture $CE=x$, $AF=y$ and $CF=z$

$\triangle EFC\sim\triangle EAC \Rightarrow \frac x{x+10}=\frac {10}{y+10}$

$\Rightarrow xy+10x=10x+100\Rightarrow xy =100$

$y^2=10^2+(10-z)^2=200-20z+z^2$

$x^2=10^2-z^2$

Multiplying both sides of these relations we get:

$(xy)^2=100=(10-z^2)(z^2-20z+200)$

Which is an equation of 4 degree in terms of z. Wolfram gives positive root as:

$z=5(1-\sqrt 2+\sqrt{2\sqrt 2 -1})\approx 4.7$

Hence the measure of x is:

$x=\sqrt{10^2-4.7^2}\approx 8.825$

sirous
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