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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.

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. – Alex Becker Apr 10 '11 at 20:09
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? – Matthew Conroy Apr 10 '11 at 20:19
Question updated – gulbaek Apr 10 '11 at 20:37
you updated your question with a link containing a detailed solution, so I erased my answer as unnecessary – user8268 Apr 10 '11 at 20:50
@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. – joriki Apr 10 '11 at 20:56
up vote 2 down vote accepted

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. – joriki Apr 10 '11 at 21:01
The second you obtain via symmetry ... But let me check again. – Fabian Apr 10 '11 at 21:05
Also you don't have $x^2+y^2=100$. – joriki Apr 10 '11 at 21:07
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. – joriki Apr 10 '11 at 21:08
@joriki: thank you, I got some signs mixed up... – Fabian Apr 10 '11 at 21:09

(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. – joriki Apr 10 '11 at 20:50
undeleted at yoriki's request :) (sorry for the $b$ notation - I had it before it appeared on the picture) – user8268 Apr 10 '11 at 21:38

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