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 $ABC$.
Problem
Source: Turkey Junior National Olympiad 2021 P4
Tags: geometry, tangent
BarisKoyuncu
07.01.2022 18:29
See that this question is equivalent to proving $BCMN$ is a tangential quadrilateral.
Let $|BX|=a$ and $|CX|=b$. We have
$$\frac{|AN|}{|NB|}=\frac{|AX|}{|XB|}=\frac{|CX|}{|CB|}=\frac b{a+b}\text{ and }|AN|+|NB|=a+b\Rightarrow |AN|=\frac{b(a+b)}{2b+a}\text{ and }|NB|=\frac{(a+b)^2}{2b+a}$$Similarly, $|AM|=\frac{a(a+b)}{2a+b}\text{ and }|MC|=\frac{(a+b)^2}{2a+b}$.
Since $\angle NAM=60^\circ$, we have
$$|MN|^2=|AN|^2+|AM|^2-|AN|\cdot |AM|=\frac{(a+b)^2(a^2+ab+b^2)^2}{(2a+b)^2(2b+a)^2}\Rightarrow |MN|=\frac{(a+b)(a^2+ab+b^2)}{(2a+b)(2b+a)}$$We need to prove that
$$|NB|+|MC|=|BC|+|MN|\Leftrightarrow \frac{(a+b)^2}{2b+a}+\frac{(a+b)^2}{2a+b}=a+b+\frac{(a+b)(a^2+ab+b^2)}{(2a+b)(2b+a)}$$, which is correct. Hence completes the proof.
Infinityfun
19.02.2022 17:51
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.
bin_sherlo
28.10.2023 18:28
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.