parallel lines given, proving tangency

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parallel lines given, proving tangency

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Source: Turkey Junior National Olympiad 2021 P4

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#1 h Jan 7, 2022, 3:18 PM

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Let $X$ be a point on the segment $[BC]$ of an equilateral triangle $ABC$ and let $Y$ and $Z$ be points on the rays $[BA$ and $[CA$ such that the lines $AX, BZ, CY$ are parallel. If the intersection of $XY$ and $AC$ is $M$ and the intersection of $XZ$ and $AB$ is $N$ , prove that $MN$ is tangent to the incenter of $ABC$ .

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#2 h Jan 7, 2022, 3:29 PM

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Simple bashing

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#3 h Feb 19, 2022, 2:51 PM

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Let $O$ be the center of the incircle and let $AB$ and $BC$ be tangent to incircle at $D$ and $E$ , respectively.
$R$ is the radius of the incircle and $AB=AC=BC= 2R\sqrt3$

$\frac{BN}{BA}=\frac{BZ}{BZ+AX}=\frac{BC}{BC+CX} \Rightarrow BN = \frac{12 R^2}{CX +2R \sqrt3}$

Let $\alpha= \angle DON$ , $\beta=\angle EOM$ and $T$ and $T'$ are on incircle such that $NT$ and $MT'$ are tangent to incircle.
$NM$ is tangent to incricle if and only if $T=T' \Longleftrightarrow \angle DOT + \angle EOT' = 180 - \angle A = 120 \Longleftrightarrow \alpha + \beta = 60$

So $\tan(\alpha +\beta)$ should be equal to $\sqrt3$ .
$tan\alpha = \frac{NK}{R} = \frac{BN-BK}{R} = \frac{\frac{12R^2}{CX+2R\sqrt3}-R\sqrt3}{R}=\frac{12R}{CX+2R\sqrt3}-\sqrt3$
Similarly $tan\beta = \frac{12R}{BX+2R\sqrt3}-\sqrt3 = \frac{12R}{4R\sqrt3-CX}-\sqrt3$

Now we are ready to compute $\tan(\alpha + \beta)$
$\tan(\alpha + \beta) = \frac{tan\alpha + \tan\beta}{1- \tan\alpha\tan\beta} =\frac{ -2\sqrt3 +12R\frac{6R\sqrt3}{(CX+2R\sqrt3)(4R\sqrt3-CX)}}{1 - (3 -12R\sqrt3\frac{6R\sqrt3}{(CX+2R\sqrt3)(4R\sqrt3-CX)}+ \frac{144R^2}{(CX+2R\sqrt3)(4R\sqrt3-CX)})} = \sqrt3$
So indeed, $\alpha + \beta = 60$ and we get the desired result.

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#4 h Oct 28, 2023, 3:28 PM

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We have
$(\infty,BM\cap CY;C,Y)=-1$ which gives us that $BM$ passes through the midpoint of $[CY]\iff BM$ passes through the midpoint of $[AX]$ .
Similarily, we have
$(\infty,CN\cap BZ;B,Z)=-1$ which gives us that $CN$ passes through the midpoint of $[BZ]\iff BM$ passes through the midpoint of $[AX]$ .
So let $P$ be the midpoint of $[AX]$ .
Let's say that $|AB|=|BC|=|CA|=1$ and $|BX|=a,|XC|=1-a,|CM|=b,|MA|=1-b,|AN|=c,|NB|=1-c$
By Menelaus we get
$\frac{a}{1}.\frac{b}{1-b}.\frac{AP}{PX}=1\implies a=\frac{1-b}{b}$ $\frac{1-a}{1}.\frac{1-c}{c}.\frac{AP}{PX}=1\implies a=1-\frac{c}{1-c}$ $ab=1-b\implies b(1+a)=1\implies \boxed{b=\frac{1}{a+1}}$
$a=\frac{1-2c}{1-c}\implies a-ac=1-2c\implies 1-a=c(2-a)\implies \boxed {c=\frac{1-a}{2-a}}$
By Cosinus Theorem, we have
$|MN|^2=\frac{(1-a)^2}{(2-a)^2}+\frac{a^2}{(a+1)^2}-2.\frac{a(1-a)}{(2-a)(a+1)}.\cos60=\frac{(1-a)^2}{(2-a)^2}+\frac{a^2}{(a+1)^2}-\frac{a(1-a)}{(2-a)(a+1)}$ $=\frac{(1-a)^2(1+a)^2+a^2(2-a)^2-a(1-a)(2-a)(1+a)}{(2-a)^2(1+a)^2}$ $=\frac{a^4-2a^2+1+a^4+4a^3+4a^2-a^4+2a^3+a^2-2a}{(1+a)^2(2-a)^2}$ $=\frac{a^4-2a^3+3a^2-2a+1}{(1+a)^2(2-a)^2}$ $=\frac{(a^2-a+1)^2}{(1+a)^2(2-a)^2}$ So $|MN|=\frac{a^2-a+1}{(1+a)(2-a)}$
$|MN|=\frac{a^2-a+1}{(1+a)(2-a)}=\frac{1}{2-a}-\frac{a}{a+1}=|BN|+1-|MC|=|BN|+|MC|-|BC|$ Which gives us
$|MN|+|BC|=|BN|+|CM|$ as desired.

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