<s>[INST] Calculate the probability that the circumferecnce of triangle CDE is smaller than $1+ \sqrt \frac{11}{2}$. Let page of square ABCD have length 1. On the page AB, point E is randomly selected. Calculate the probability that the circumferecnce of triangle CDE is smaller than $1+ \sqrt \frac{11}{2}$.
I have used Pythagorean theorem for pages ED and EC, and then observed case where circumference is equal to $1+ \sqrt \frac{11}{2}$, but I got to a polynomial equation of $4$th degree ( $64x^4 - 128x^3 -112x^2-16x-175=0$, where $x$ is distance between A and E) which doesn't have "easy" readable solution. Is there any better way to solve this?
 [/INST] If you denote the distance $AE=t$ it follows that $EB=1-t$. Moreover the question can be reformulate as $CE+DE<\sqrt{\frac{11}{2}}$. $CE=\sqrt{1+t^2}$ and $DE=\sqrt{1+(1-t)^2}=\sqrt{2-2t+t^2}.$
Consider equality: 
$$\sqrt{2-2t+t^2}+\sqrt{1+t^2}=\sqrt{\frac{11}{2}}$$ 
Then, we take the square:
$$2-2t+t^2+1+t^2+2\sqrt{2-2t+t^2}\sqrt{1+t^2}=\frac{11}{2}$$
$$2\sqrt{2-2t+t^2}\sqrt{1+t^2}=-2t^2+2t+\frac{5}{2}$$
$$4(2-2t+t^2)(1+t^2)=4t^4+4t^2+\frac{25}{4}-8t^3+10t-10t^2$$
$$4t^4-8t^3+8t^2+8-8t+4t^2=4t^4+4t^2+\frac{25}{4}-8t^3+10t-10t^2$$
$$18t^2-18t+\frac{7}{4}=0$$
The equation has two symmetric solution $t_1=\frac{1}{2}-\frac{\sqrt{\frac{11}{2}}}{6}$ and $t_2=\frac{1}{2}+\frac{\sqrt{\frac{11}{2}}}{6}$. The set of point that satisfy the request has Lebesgue measure $t_2-t_1=\frac{\sqrt{\frac{11}{2}}}{3}$
If we suppose a uniform probability, the answer is $\frac{\sqrt{\frac{11}{2}}}{3}$.
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