# 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?

• What is the equation of degree $4$? – Dietrich Burde Jun 12 '19 at 12:44
• @DietrichBurde solving polynomial of 4th degree ( don't know how to translate it correctly) – user560461 Jun 12 '19 at 12:46
• Yes, but what is the polynomial? Is it $x^4+x^3+x^2+x+1 = 0$? You should also show your work so that you can get the best answer possible. – Toby Mak Jun 12 '19 at 12:52
• @TobyMak I have added it to my post, thanks for advice – user560461 Jun 12 '19 at 12:56
• The polynmial has exactly one positive real root, namely $x=2.78657672265$. – Dietrich Burde Jun 12 '19 at 13:00

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 $$DE=\sqrt{1+(1-t)^2}=\sqrt{2-2t+t^2}.$$
$$\sqrt{2-2t+t^2}+\sqrt{1+t^2}=\sqrt{\frac{11}{2}}$$ 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 $$\frac{\sqrt{\frac{11}{2}}}{3}$$.