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?

  • $\begingroup$ What is the equation of degree $4$? $\endgroup$ – Dietrich Burde Jun 12 '19 at 12:44
  • $\begingroup$ @DietrichBurde solving polynomial of 4th degree ( don't know how to translate it correctly) $\endgroup$ – user560461 Jun 12 '19 at 12:46
  • $\begingroup$ 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. $\endgroup$ – Toby Mak Jun 12 '19 at 12:52
  • $\begingroup$ @TobyMak I have added it to my post, thanks for advice $\endgroup$ – user560461 Jun 12 '19 at 12:56
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    $\begingroup$ The polynmial has exactly one positive real root, namely $x=2.78657672265$. $\endgroup$ – 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 $DE=\sqrt{1+(1-t)^2}=\sqrt{2-2t+t^2}.$

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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