An isosceles $\triangle ABC$ has $\angle BAC =\angle ABC =72^{o}$. The angle bisector $AL$ meets the line through $C$ parallel to $AB$ at $D$. $a)$ Prove that the circumcenter of $\triangle ADC$ lies on $BD$. $b)$ Prove that $\frac {BE} {BL}$ is irrational.
Problem
Source: Bulgarian Spring Tournament 2023 10.2
Tags: geometry
StarLex1
26.03.2023 05:24
WLOG $A(0,0) , B(2,0) ,C(1,n)$
notice CAB = 72 then by angle inclination $Grad_{AC}=\tan(72)$ , $Grad_{AD}=\tan(36)$
thus $C(1,n) , C(1,\frac{\sqrt{10+2\sqrt{5}}}{\sqrt{5}-1})$
equation line AD
$ \frac{\sqrt{10+2\sqrt{5}}}{\sqrt{5}-1} = \frac{\sqrt{10-2\sqrt{5}}}{\sqrt{5}+1}*x$
$D = (2+\sqrt{5},\frac{\sqrt{10+2\sqrt{5}}}{\sqrt{5}-1})$
Equation of circle formed by 3 point
$0^2+0^2+a*0+b*0+c=0$
$c=0$
$1+\frac{(5+\sqrt{5})}{3-\sqrt{5}}+a+b*\frac{\sqrt{10+2\sqrt{5}}}{\sqrt{5}-1}=0$
$(2+\sqrt{5})^2+\frac{(5+\sqrt{5})}{3-\sqrt{5}}+a*(2+\sqrt{5})+b\frac{(\sqrt{10+2\sqrt{5}}}{\sqrt{5}-1}=0$
$-\frac{a}{2}=\frac{ -8-4\sqrt{5}}{1+\sqrt{5}}=\frac{3+\sqrt{5}}{2}$
$-\frac{b}{2} = \frac{3+\sqrt{5}}{2\sqrt{5+2\sqrt{5}}}$
Equation BD
$\frac{y }{\frac{\sqrt{10+2\sqrt{5}}}{\sqrt{5}-1}}= \frac{(x-2)}{\sqrt{5}}$
sub $x = \frac{3+\sqrt{5}}{2}$
which will give u $y = \frac{3+\sqrt{5}}{2\sqrt{5+2\sqrt{5}}}$
SevroMyName
26.03.2023 13:04
For this solution you would have gotten 5/6 cause you haven't explicitly proven y is irrational.