Hi I got a 10 cm long line, and it touches point 1,1
I need to calculate where it touches x and y.
If I think of it like an triangle i get the following information.
But how do i calculate the rest?
UPDATE Figured out that its know as the Ladder problem. http://www.mathematische-basteleien.de/ladder.htm
I also updated the image to make it more clear.
On
(if I understand correctly)
the sides of the triangle are $a,b,c$, with $c=10$. If you leave out the square, you get two small triangles which are similar. Hence $(a-1)/1=1/(b-1)$, i.e. $(a-1)(b-1)=1$, or $ab=a+b$. We also know $a^2+b^2=c^2$. From here you get $(a+b-1)^2=c^2+1$. So $(a-1)+(b-1)=-1+\sqrt{c^2+1}$, $(a-1)(b-1)=1$, i.e. $a-1$ and $b-1$ are the solutions of $x^2+(1-\sqrt{c^2+1})x+1=0$.
Looking at your figure, I do not think any of the height is $1$.
There are similar triangles in your figure: the large triangle with hypotenuse $10$ and catheti $x$ and $y$ is similar to the triangle with catheti $1$ and $y-1$ and also similar to the one with catheti $x-1$ and $1$. Using this we get that $$\frac{x}{y} = x-1.$$ Additionally, we know that $x^2 +y^2 =10^2$. Plugging in the relation $y= x/(x-1)$, we obtain $$x^2 + \left( \frac{x}{x-1}\right)^2 = 10^2$$ which is equivalent to $$x^2 + x^2 (x-1)^2 = 100 (x-1)^2$$ with the (only positive) solution (up to exchanging $x$ and $y$) $$x= \frac{1}{2} \left[\sqrt{101} +1 - \sqrt{2 (49- \sqrt{101})}\right]\approx 1.11$$ and $$y=\frac{1}{2} \left[\sqrt{101} +1 +\sqrt{2 (49- \sqrt{101})}\right] \approx 9.94.$$