# Square inside a triangle problem [see desc.]

I Have a problem solving this problem only using high school math(Geometry) without using digital help (computer). The solution should be $\cfrac{11\sqrt{77}}{2}$.

How can I get this solution by hand? I get only a fourthgrade equation that I can not solve: $(x+1)^2(x^2+1)= 100x^2$ using similar triangles. There most be an easier way?

It should be $$\frac{x}{x+1}=\frac{\sqrt{x^2+1}}{10},$$ which indeed gives $$100x^2=(x+1)^2(x^2+1).$$

Now, let $x+\frac{1}{x}=t$.

Thus, $$100=(t+2)t$$ or $$(t+1)^2=101,$$ which gives $$t=\sqrt{101}-1$$ and $$x+\frac{1}{x}=\sqrt{101}-1.$$

• Thank you so much. But how do you choose the substitution t ? And can you show me how you get 100 = (t+2)t by substituting, your calculations ? – Eirik Feb 2 '18 at 13:33
• I divided both sides by $x^2$ and used $(x+1)^2=x^2+2x+1$. – Michael Rozenberg Feb 2 '18 at 13:33
• Can you explain how you thought when you chose your substituion? Or If you know somewhere I can find videoes or anyting explaining this? – Eirik Feb 3 '18 at 15:50
• I just solved some number of problems during my life. I just see this substitution. – Michael Rozenberg Feb 3 '18 at 17:04

Call the upper part of the vertical leg $y$. Then $$(1+x)^2+(1+y)^2=100$$ and ${y\over1}={1\over x}$, hence $xy=1$. Letting $x+y=:s$ we therefore have $$2+2s+(s^2-2xy)=100\ ,$$ or $(s+1)^2=101$. This gives $s=\sqrt{101}-1$, so that by Vieta's theorem $x$ and $y$ are the solutions of the quadratic $$z^2-(\sqrt{101}-1)z+1=0\ .$$