Bosnia and Herzegovina JBMO TST 2017 Problem 3

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Source: Bosnia and Herzegovina Junior Balkan Mathematical Olympiad TST 2017

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#1 h Sep 16, 2018, 12:10 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle such that $\angle ABC = 90 ^{\circ}$ . Let $I$ be an incenter of $ABC$ and let $F$ , $D$ and $E$ be points where incircle touches sides $AB$ , $BC$ and $AC$ , respectively. If lines $CI$ and $EF$ intersect at point $M$ and if $DM$ and $AB$ intersect in $N$ , prove that $AI=ND$

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#2 h Aug 22, 2019, 10:30 PM • 1 Y

Y by Adventure10

Shocked to see that no one has posted a solution yet.

.

We will use a very well known problem (or you can call it as a lemma)

Lemma: $ABC$ is a triangle and if the incircle touches $BC,CA,AB$ at $D,E,F$ and respectively and $I$ is the incenter of $\triangle ABC$ and if $CI\cap FE=M$ , then $BM\perp CM$ .

Proof of the lemma: Join $I$ to $F$ . Let $\angle BAC=\theta\implies\angle MFB=90^\circ+\frac{\theta}{2}$ also $\angle DIM=90^\circ-\frac{\theta}{2}\implies BIMF$ is a cyclic quadrilateral. So, $\angle BMC=\angle BFI=90^\circ$ .

Main Proof: Notice that $BFID, BDIM$ are cyclic quadrilaterals. Hence $FMID$ is also a cyclic quadrilateral $\implies\angle FMD=\angle FID=90^\circ$ . Let $AI\cap EF=K$ . So, $\angle FKA=90^\circ\implies ND\| AI$ .

Now it's angle chasing time...
$\angle NDI=\angle MFI=\angle FND=180^\circ-\angle DNA\implies AN\| DI$ . So, $NDIA$ is a parallelogram. Hence, $DN=IA$ . $\blacksquare$ .

This post has been edited 12 times. Last edited by amar_04, Aug 22, 2019, 10:37 PM

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#3 h Jul 29, 2020, 1:09 PM

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amar_04 wrote:

Shocked to see that no one has posted a solution yet.

.

We will use a very well known problem (or you can call it as a lemma)

Lemma: $ABC$ is a triangle and if the incircle touches $BC,CA,AB$ at $D,E,F$ and respectively and $I$ is the incenter of $\triangle ABC$ and if $CI\cap FE=M$ , then $BM\perp CM$ .

Proof of the lemma: Join $I$ to $F$ . Let $\angle BAC=\theta\implies\angle MFB=90^\circ+\frac{\theta}{2}$ also $\angle DIM=90^\circ-\frac{\theta}{2}\implies BIMF$ is a cyclic quadrilateral. So, $\angle BMC=\angle BFI=90^\circ$ .

Main Proof: Notice that $BFID, BDIM$ are cyclic quadrilaterals. Hence $FMID$ is also a cyclic quadrilateral $\implies\angle FMD=\angle FID=90^\circ$ . Let $AI\cap EF=K$ . So, $\angle FKA=90^\circ\implies ND\| AI$ .

Now it's angle chasing time...
$\angle NDI=\angle MFI=\angle FND=180^\circ-\angle DNA\implies AN\| DI$ . So, $NDIA$ is a parallelogram. Hence, $DN=IA$ . $\blacksquare$ .

How did you get $\angle DIM=90^\circ-\frac{\theta}{2}$ ?

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#4 h May 25, 2023, 6:09 PM

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We will prove that $ANDI$ is a parallelogram. First notice that $AN\perp BC, ID\perp BC$ $\implies AN\parallel ID$ .
Since $AF=AE$ and $AI$ angle bisector, $AI\perp FE$ . Let $\angle{ICB}=\alpha$ . Then $\angle{AEF}=45+\alpha$ and $\angle{AEM}$ is the external angle,so $\angle{EMC}= 45$ which implies $BFMID$ cyclic. So $\angle{FBD}=\angle{DME}=90$ , $ND\perp FE$ and we're done

This post has been edited 2 times. Last edited by rafigamath, May 25, 2023, 6:13 PM
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