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Hi I got a 10 cm long line, and it touches point 1,1

I need to calculate where it touches x and y.

enter image description here

If I think of it like an triangle i get the following information.

  • One side is 10 cm.
  • You get an angle of 90
  • and an Height of 1 cm.

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.

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    What are you given exactly? When you talk about angle, height, etc., **what** angle, **what** height are you talking about? Your question is unclear and the diagram does not provide enough information to solve the problem.2011-04-10
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    Along your sloping line, is the distance from the y-intercept to (1,1) 10 cm, or is it the distance from the y-intercept to the x-axis?2011-04-10
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    Question updated2011-04-10
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    you updated your question with a link containing a detailed solution, so I erased my answer as unnecessary2011-04-10
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    @user8268: Your solution was clearer and in better English; also the link might get broken later on; so I think it would be good to retain your answer.2011-04-10
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    @Also the link doesn't contain a general solution, only a specific solution for length $5$. It would be good to be able to compare your answer with Fabian's.2011-04-10

2 Answers 2

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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 $$y=\frac{1}{2} \left[\sqrt{101} +1 +\sqrt{2 (49- \sqrt{101})}\right] \approx 9.94.$$

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    This can't be right, since there must be two solutions.2011-04-10
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    The second you obtain via symmetry ... But let me check again.2011-04-10
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    Also you don't have $x^2+y^2=100$.2011-04-10
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    There can't be another solution from symmetry if all your steps are implications. There would have to be an arbitrary choice somewhere along the way for the second solution to get lost.2011-04-10
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    @joriki: thank you, I got some signs mixed up...2011-04-10
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    @Fabian: Equation $x^2+x^2(x-1)^2=100(x-1)^2$ is equivalent to $x^4-2x^3-98x^2+200x-100=0$. How did you find the exact solution?2011-04-10
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    @Américo Tavares: Wolfram Alpha helps. If you want to do it yourself. Substituting $x=y+1/2$ leads to the depressed quartic $y^4- 199y^2/2 +101 y-395/6=0$ for which Ferrari found a solution (http://en.wikipedia.org/wiki/Quartic_function#Ferrari.27s_solution).2011-04-10
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    @Fabian: Thanks! +1.2011-04-10
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    Wow, I've never seen the word cathetus (plural catheti) used for a side of a right triangle other than the hypotenuse. I need to get out more. http://en.wikipedia.org/wiki/Cathetus2011-04-11
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(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 $x^2+(1-\sqrt{c^2+1})x+1=0$.

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    Note that in the figure $b$ denotes the side length $b=1$ of the square.2011-04-10
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    undeleted at yoriki's request :) (sorry for the $b$ notation - I had it before it appeared on the picture)2011-04-10