One of the Coolest Problems I've Ever Seen (though many may
- FahimFerdous
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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 $P$.
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 $P$.
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- Tahmid Hasan
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Re: One of the Coolest Problems I've Ever Seen (though many
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.
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.
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বড় ভালবাসি তোমায়,মা
Re: One of the Coolest Problems I've Ever Seen (though many
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)
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)
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- FahimFerdous
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Re: One of the Coolest Problems I've Ever Seen (though many
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.
And I used pole polar and pascal to prove the two parts.
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- FahimFerdous
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Re: One of the Coolest Problems I've Ever Seen (though many
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. :-/
Your hot head might dominate your good heart!
Re: One of the Coolest Problems I've Ever Seen (though many
Same here though I got a hint from Fahim vai about the tangent from $M$ part.FahimFerdous wrote:And I used pole polar and pascal to prove the two parts.
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.FahimFerdous wrote: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. :-/
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Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
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- Tahmid Hasan
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Re: One of the Coolest Problems I've Ever Seen (though many
Now I see what I did wrong there, I messed up the notations. Thanks Mahi for pointing that out.*Mahi* wrote: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)
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 $|XZ-ZW|$.
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- FahimFerdous
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Re: One of the Coolest Problems I've Ever Seen (though many
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.
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