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Source: 2011 China Western MO

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#1 h Mar 11, 2012, 12:41 PM • 2 Y

Y by Adventure10, Mango247

In triangle $ABC$ with $AB>AC$ and incenter $I$ , the incircle touches $BC,CA,AB$ at $D,E,F$ respectively. $M$ is the midpoint of $BC$ , and the altitude at $A$ meets $BC$ at $H$ . Ray $AI$ meets lines $DE$ and $DF$ at $K$ and $L$ , respectively. Prove that the points $M,L,H,K$ are concyclic.

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#2 h Mar 11, 2012, 2:36 PM • 2 Y

Y by Adventure10, Mango247

i use two lemma to proof problem and proof it soon

first lemma:
$BK$ is perpendicular $AI$ and $CL$ is perpendicular to $AI$
second lemma:
$LM$ is parallel to $AB$
proof of problem:
by lemma 1 $A,K,H,B$ are concyclic and $A,H,C,L$ are concyclic so $\angle KHM = \angle BAI$ and by lemma 2 we know that $\angle MLA = \angle BAL$ so $\angle MLK = \angle MHK$
i write the lemma proof soon

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9775 posts

jayme

#3 h Mar 11, 2012, 2:51 PM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
always nice... based
1. on a result of Altshiller-Court that you can see in my article : An unlikely concurrence... vol. 4 p. 5
2. on the Lieutenant Calabre...: Symétrique de (OI)... vol. 4 p. 6
in
http://perso.orange.fr/jl.ayme

Sincerely
Jean-Louis

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1347 posts

WakeUp

#4 h May 22, 2012, 12:10 PM • 1 Y

Y by Adventure10

Of course, $\angle LFE=\angle DFE=90^{\circ}-C$ so $AI\perp EF$ we have $\angle FLI=\angle\frac{C}{2}$ . So $\angle FLI=\angle ICD$ . This implies $ICDL$ is cyclic, and clearly $E$ also lies on this circle, which has $IC$ as diameter. Thus $\angle ALC=90^{\circ}$ . Let $CL$ meet $AB$ at $C'$ . Then $AI\perp CC'$ and since $\angle C'AL=\angle CAL$ , it is clear that $L$ is the midpoint of $CC'$ . Then since $M$ is the midpoint of $BC$ , it follows that $BC'||LM$ i.e. $AB||LM$ . By a symmetric argument, using the point $B'$ etc, we can prove that $MK||AC$ . Now $ALHC$ is cyclic since $\angle ALC=\angle AHC=90^{\circ}$ . Thus $\angle LHM=\angle LAC=\frac{1}{2}A$ . But since $MK||AC$ , we also have $\angle MKL=\angle LAC=\frac{A}{2}$ . So $\angle MHL=\angle MKL$ and the conclusion follows.

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154 posts

vslmat

#5 h Sep 19, 2012, 3:52 PM • 2 Y

Y by Adventure10, Mango247

Another solution:
Let $AI$ cut the circumcircle of $ABC$ at $G$ , the middle point of the lower arc $BC$ . It is obvious that $O, M, G$ are collinear.
$\angle AEK = 90^{\circ} + \angle \frac{ACB}{2}$ , thus $\angle AKE = \angle \frac{ABC}{2}$ , hence $BIDK$ is cyclic, from which we obtain a very well known fact that $BK\perp AK$ . Similarly $CL\perp AL$ .
$ALHC$ is cyclic, hence $\angle KLH = \angle HCA = \angle BGA$ .
But $BMKG$ is also cyclic, hence $\angle BGA = \angle KMH$ , thus $\angle KLH = \angle KMH, MKHL$ is cyclic.

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sayantanchakraborty

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sayantanchakraborty

#6 h Sep 29, 2014, 12:38 PM • 1 Y

Y by Adventure10

Before giving a solution I shall present two prerequisite lemmas.

Lemma 1: $CK \perp AK$ .

Solution: $\angle{CIA}=90+\frac{B}{2}$ and $\angle{KDB}=\angle{EDB}=90-\frac{B}{2}$ so ${CIKD}$ concyclic.So $\angle{CKA}=\angle{CKI}=\angle{IDC}=90^{\circ}$ .

Lemma 2: $BL \perp AL$

Solution: $\angle{BIL}=90-\frac{C}{2}$ and $\angle{BDL}=\angle{FDC}=90-\frac{C}{2}$ .Thus $\angle{ALB}=\angle{ILB}=\angle{IDB}=90^{\circ}$ .

Now I come back to the main problem.Let $N$ be the midpoint of $AB$ .Then clearly $N$ is the circumcenter of $ALB$ so $\angle{LNB}=2\angle{LAB}=A$ .In other words $NL \parallel AC$ or $N,M,L$ are collinear.Thus $\angle{HML}=\angle{CML}=C$ .Also note that $\angle{AHC}=\angle{AKC}=90^{\circ} \implies ACHK$ cyclic.So $\angle{HKL}=\angle{ACH}=C$ .Combining these we have $\angle{HKL}=\angle{HML}=C$ or $MKHL$ is cyclic,as desired.

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#7 h Jan 9, 2021, 12:37 AM

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Nice one

$\color{black}\rule{25cm}{1pt}$
Let $G$ be the intersection of $AI$ with $BC$ , let $T$ be the intersection of $AI$ with the circumcircle of $ABC$ and let $N$ be the midpoint of $AC$ .
We have that:
$\angle GLD = \angle FLA = 180-\left( \angle LAF + \angle LFA\right) = 180-\left( \frac{1}{2} \angle A + 90 + \frac{1}{2}\angle B\right) = \frac{1}{2} \angle C$ this implies that $CDLIE$ is a cyclic quad.
Similarly we have that $\angle DKI = \frac{1}{2} \angle B$ , which implies that $BKDIF$ is cyclic as well.
The statements above imply that $BK \perp AI$ and $CK \perp AI$ .
This implies that $BKHA$ and $CHLA$ are both cyclic.

Now we have the following lemma, which ends the problem.
$\color{red}\rule{25cm}{0.5pt}$
Lemma: $LM \parallel AB$ .
Proof (Sketch):
Throw this onto the complex plane with the circumcircle being the unit circle and let $c=\frac{1}{b}$ .
By this way we have that $t=-1$
Since we have that $CL \perp AI$ and $L \in AI$ we easily get that:
$l=\frac{2b^4-2ab^4+(1+ab^3)(b^2+1)}{2b(b^3+1}$ but this implies that:
$\frac{m-l}{\overline{m-l}}=\frac{m-n}{\overline{m-n}}$ thus we have that $M,L$ and $N$ are colinear.
$\color{red}\rule{25cm}{0.5pt}$

Now back to our problem.

We have that:
$\angle LMH = \angle LMG = \angle GBA = \angle HBA = \angle HKA = \angle HKL$ thus we have that $KMLH$ is a cyclic quadrilateral.

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