One of the Coolest Problems I've Ever Seen (though many may

[FahimFerdous]

Let $ABC$ be a triangle with incentre $I$ and $D, E, F$ be the touchpoints of the incircle with sides $BC, CA, AB$. $FX$ is the perpendicular to $DF$ at $F$. Now let $DI \cap AB=S$, $DE \cap FX=T$, $ST \cap EF=R$. Let the circle with diametre $IR$ intersect the incircle at $P$ where $A$ and $P$ lies on opposite sides of $IR$. Let's call this point $P_{ABC}$.

Now let ${XYZ}$ be an isosceles triangle with $XZ=YZ$ and $XY \le YZ$. Let $W$ be a point on $YZ$ such that $YW \le YX$. Let $K=P_{ZXW}$ and $L=P_{YXW}$. Prove that $XY \geq 2KL$.

Small Hint: Find TWO properties of $P$.

[Tahmid Hasan]

Let $D'$ be the diametrically opposite point point of $D$. Hence $FX$ goes through $D'$.
Let $D'D' \cap EF=R'$. Applying Pascal's theorem on cyclic hexagon $FFEDD'D'$ we get $S,R',T$ are collinear.
So $ST,D'D',EF$ are concurrent at $R'$ i.e. $R'=ST \cap EF$ which implies $R=R'$.
So $\angle RD'I=90^{\circ} \Rightarrow D' \in$ the circle with diameter $IR$ and $RD'$ is tangent to $\odot DEF$.
Let $RP$ be the other tangent from $R$ to $\odot DEF$ with $P' \in \odot DEF$.
Then $RP'I=90^{\circ} \Rightarrow P' \in$ the circle with diameter $IR$. Hence $P=P'$.
So we conclude $PD'$ is the polar of $R$ wrt $\odot DEF$. Since $R$ lies on $EF$, the polar of $A$ by La Hire's theorem
we get $A$ lies on $PD'$. So we can define $P_{ABC}=AD' \cap \odot DEF \neq D'$.
Let $AD' \cap BC=G$. By lemma-2 Yufei Zhao we get $D,G$ are symmetric wrt $M$, the midpoint of $BC$.
Now $\angle DPG=90^{\circ} \Rightarrow M$ is the circumcentre of $\triangle DPG$.
Hence $MD=MP \Rightarrow \angle DPM=\angle PDM=\angle PD'D \Rightarrow MP$ is tangent to $\odot DEF$.
And $MP=\frac{1}{2}|b-c|$.
Now I have a problem with my proof.
It is obvious that $\triangle XYZ$ is acute. So we can write $\angle ZXY=\angle XYZ=60^{\circ}+\alpha, \angle XZY=60^{\circ}-2\alpha$
where $0^{\circ} \le \alpha <30^{\circ}$.
Now $\angle XWZ=\angle 60^{\circ}+\alpha+\angle YXW \ge 60^{\circ}-2\alpha=\angle XZW \Rightarrow XZ \ge XW$.
Let $M_1,M_2$ be the midpoints of $WZ,WY$ respectively.
Lemma: Let $ABCD$ be a non-concave non-cross(!) quadrilateral then $AB+BC+CD \geq DA$.
Proof: Triangle inequality two times.
Applying the lemma on $KLM_2M_1$ we get $2KL+XZ-XW+|XW-XY| \ge YZ$.
If $XW \ge XY$ then clearly $2KL \ge XY$.
But if $XW<XY$ the problem gets messed up which I can't complete.
Here's a counterexample picture(for the modulus thingy).
Draw a circle with centre $X$ and radius $XY$ which intersects $YZ$ at $K$.
If $W$ lies on segment $KY$ then $XW<XY$ and so $2KL \ge 2XW-XY$ and I can't get the desired inequality from there.
And in the picture clearly $W$ would satisfy $YW \le YX$, help me out here.

[*Mahi*]

Let me just finish Tahmid's proof from the part $MP = \frac 1 2 |b-c| $.
Let $M'$ be the midpoint of $XW$.
Then $2KL \leq 2(M'K + M'L) = (|YX - YW| + |XZ- ZW|) = XY + XZ - YZ = XY$.
So $XY \geq 2KL$
(also, I just noticed that the question was posted wrong edited now, sorry tahmid)

[FahimFerdous]

Very very sorry for the typo. When I typed it, I didn't have the question, I typed it from memory, so the mistake occured.

And I used pole polar and pascal to prove the two parts.

[FahimFerdous]

Wait a min. I believe that the ques we had in the exams had a typo. I mean that I PROVED that XY is LESS/EQUAL to 2KL, not GREATER/EQUAL. And my proof has no flaw as long as I'm concerned. I'll post it this friday as I can't post now. :-/

[*Mahi*]

And I used pole polar and pascal to prove the two parts.

Same here though I got a hint from Fahim vai about the tangent from $M$ part.

Wait a min. I believe that the ques we had in the exams had a typo. I mean that I PROVED that XY is LESS/EQUAL to 2KL, not GREATER/EQUAL. And my proof has no flaw as long as I'm concerned. I'll post it this friday as I can't post now. :-/

I think you had some of the assumptions of this question in reverse as you have seen, the proof of $XY \geq 2 KL$ was proved, so please post your proof as soon as possible. I don't think there is a typo in the question, as I just tried geogebra and it showed the same results for the lengths of the segments.

[Tahmid Hasan]

Let me just finish Tahmid's proof from the part $MP = \frac 1 2 |b-c| $.
Let $M'$ be the midpoint of $XW$.
Then $2KL \leq 2(M'K + M'L) = (|YX - YW| + |XZ- XW|) = XY + XZ - YZ = XY$.
So $XY \geq 2KL$
(also, I just noticed that the question was posted wrong edited now, sorry tahmid)

Now I see what I did wrong there, I messed up the notations. Thanks Mahi for pointing that out.
Let $M$ be the midpoint of $XW$
$P_{ABC}$ denotes the touch point of the second tangent from the midpoint of the opposite side of $A$.[Notice that $A$ is written first in the notation $P_{ABC}$.]
So $P_{ZXW}$ denotes the touch point of second tangent from the mipoint of the opposite side of $Z$ i.e. $XW$ to the incircle of $\triangle ZXW$.
Similarly $P_{YXW}$ denotes the touch point of second tangent from the mipoint of the opposite side of $Z$ i.e. $XW$ to the incircle of $\triangle YXW$.
Hence $MK=\frac 12|XZ-ZW|=\frac 12(XZ-ZW)$ and $ML=\frac 12|XY-YW|=\frac 12(XY-YW)$.
So applying triangle inequality on $\triangle MKL$ we get $2KL \le XZ-ZW+XY-YW=XY$. Done!
And Mahi, there's a little typo in your solution you wrote $|XZ-XW|$, it should be $|XZ-ZW|$.

[FahimFerdous]

Oh, maybe there's some mistake in my calculation, I'll check it. However, calculative mistakes ACTUALLY don't count as mistakes. The main part was the property of P.