IRL_ABooklet_2023

We connect $C$ to $A$ and $E$. Because triangles $ABC$ and $CDE$ are isosceles with an angle of $120^{\circ }$, we have $\angle BAC=\angle BCA=\angle DCE=\angle DEC=30^{\circ }$.



Hence, $\angle ACE=40^{\circ }$. Let $x=|AE|$ and $d=|CA|=|CE|$. The Cosine rule for $△CDE$ gives $d^2=2-2\cos (120^{\circ })=3$. The Cosine Rule for $△ACE$ then says $x^2=2d^2-2d^2\cos (40^{\circ })=6(1-\cos (40^{\circ }))$, and so $x>1$ iff $\cos (40^{\circ })<\frac{5}{6}$. To show this, we consider a regular pentagon with side length 1 and draw two diagonals that start at the same vertex. Because the internal angles of a regular pentagon have measure $108^{\circ }$, the angles are as indicated in the diagram below. Our aim is to prove $\cos (36^{\circ })<5/6$.



Letting $y=|CE|=|CA|$ and considering the perpendicular bisector of $EA$ in triangle $ACE$, we obtain $\cos (72^{\circ })=\cos (\angle AEC)=\frac{1}{2y}$. The Cosine Rule for triangle $ACE$ gives $1=2y^2-2y^2\cos (36^{\circ })$, hence $\cos (36^{\circ })=1-\frac{1}{2y^2}$. On the other hand, the Cosine Rule for triangle $ABC$ tells us $y^2=2-2\cos (108^{\circ })$. Using $\cos (72^{\circ })=-\cos (108^{\circ })$, this implies $y^2=2+2\cos (72^{\circ })=2+\frac{1}{y}$. Because $y$ is opposite the largest angle in triangle $ABC$, it must be its largest side, i.e. $y>1$. Therefore, $y^2=2+\frac{1}{y}<3$. This then implies $$\cos (36^{\circ })=1-\frac{1}{2y^2}<1-\frac{1}{6}=\frac{5}{6}.$$ Finally, we obtain $\cos (40^{\circ })<\cos (36^{\circ })<\frac{5}{6}$ which shows that $|EA|=x>1$ in the originally given pentagon.

Remark. From triangle $ABC$ we obtain $1=1+y^2-2y\cos (36^{\circ })$ (Cosine Rule) and so $\cos (36^{\circ })=\frac{y}{2}$. Using our previous value, we get $\frac{y}{2}=1-\frac{1}{2y^2}$, hence, after multiplying by $2y$, $y^2=2y-\frac{1}{y}$. Comparing this with an earlier calculation gives $2y-\frac{1}{y}=2+\frac{1}{y}$, i.e. $y^2-y-1=0$. The positive root of this quadratic is $y=\frac{1+\sqrt{5}}{2}$, the golden ratio. This gives us the precise value $$\cos (36^{\circ })=\frac{y}{2}=\frac{1+\sqrt{5}}{4}$$ which can easily be shown to be smaller than $5/6$.