Excircle in Equilateral Triangle

Choose the circle $x^2+y^2=1.\,$ Since that's the $A\text{-excircle}\,$ of $\Delta ABC,\,$ we'll choose $BC:\,x\sqrt{3}+y=2,\,$ $AC:\,x\sqrt{3}-y+2,\,$ $AB:\,y=-1,\,$ and $P=(\cos t,\sin t),\,$ with $\displaystyle -\frac{\pi}{2}\lt t\lt\frac{5\pi}{6}.$

We get $MP=1+\sin t=1+\cos \left(\frac{\pi}{2}-t\right)=2\cos^2\left(\frac{\pi}{4}-\frac{t}{2}\right),\,$ implying $\sqrt{MP}=\sqrt{2}\displaystyle\cos\left(\frac{\pi}{4}-\frac{t}{2}\right).\,$ Similarly,

$\displaystyle NP=\left|\frac{\sqrt{3}}{2}\cos t-\frac{1}{2}\sin t+1\right|=1+\cos\left(\frac{\pi}{6}+t\right)=2\cos^2\left(\frac{\pi}{12}+\frac{t}{2}\right),$

implying $\displaystyle\sqrt{NP}=\sqrt{2}\cos\left(\frac{\pi}{12}+\frac{t}{2}\right).\,$ Further,

$\displaystyle LP=\left|\frac{\sqrt{3}}{2}\cos t+\frac{1}{2}\sin t+1\right|=1+\cos\left(\frac{\pi}{6}-t\right)=2\cos^2\left(\frac{\pi}{12}-\frac{t}{2}\right),$

implying $\displaystyle\sqrt{LP}=\sqrt{2}\cos\left(\frac{\pi}{12}-\frac{t}{2}\right).\,$ From here,

$\displaystyle\frac{\sqrt{MP}+\sqrt{NP}}{\sqrt{LP}}=\frac{\displaystyle\cos\left(\frac{\pi}{12}+\frac{t}{2}\right)+\cos\left(\frac{\pi}{4}-\frac{t}{2}\right)}{\displaystyle\cos\left(\frac{\pi}{12}-\frac{t}{2}\right)}=\frac{\sqrt{3}\displaystyle\cos\left(\frac{\pi}{12}-\frac{t}{2}\right)}{\displaystyle\cos\left(\frac{\pi}{12}-\frac{t}{2}\right)}=\sqrt{3}.$

In what follows we'll invoke repeatedly the following result from the diagram below.

Assuming $R\,$ is the radius of the excircle, we apply the above lemma to three triangles below:

$PS=\sqrt{MP\cdot(2R)},\,$ $PT=\sqrt{NP\cdot(2R)},\,$ $PQ=\sqrt{LP\cdot(2R)}.\,$

$\overparen{SQ}=\overparen{QT}=60^{\circ},\,$ hence, $\overparen{ST}=120^{\circ},\,$ so that $QS=QT=R\,$ and, subsequently, $ST=R\sqrt{3}.$

By Ptolemy's theorem in quadrilateral $QSPT,\,$ $PQ\cdot ST=PT\cdot QS+PS\cdot QT.\,$ A substitution then gives

$\sqrt{LP\cdot (2R)}\cdot R\sqrt{3}=\sqrt{NP\cdot (2R)}\cdot R+\sqrt{MP\cdot ((2R)}\cdot R,$

i.e., $\sqrt{LP}\sqrt{3}=\sqrt{MP}+\sqrt{NP}.$

Leo Giugiuc has kindly communicated to me another problem from the peru geometrico facebook group. The problem has been posted by Edson Curahua Ortega, with credits to Edson Curahua. Solution 1 is by Leo Giugiuc; Solution 2 is by John Ascona Briceño.