2013 China girls' Mathematical Olympiad problem 7

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2013 China girls' Mathematical Olympiad problem 7

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#1 h Aug 13, 2013, 3:18 PM • 2 Y

Y by Adventure10, Mango247

As shown in the figure, $\odot O_1$ and $\odot O_2$ touches each other externally at a point $T$ , quadrilateral $ABCD$ is inscribed in $\odot O_1$ , and the lines $DA$ , $CB$ are tangent to $\odot O_2$ at points $E$ and $F$ respectively. Line $BN$ bisects $\angle ABF$ and meets segment $EF$ at $N$ . Line $FT$ meets the arc $\widehat{AT}$ (not passing through the point $B$ ) at another point $M$ different from $A$ . Prove that $M$ is the circumcenter of $\triangle BCN$ .

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#2 h Aug 13, 2013, 6:19 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

Let $TB,TC$ cut $\odot O_2$ again at $B',C'.$ Since $T$ is the exsimilicenter of $\odot O_1 \sim \odot O_2,$ then $B'C' \parallel BCF$ $\Longrightarrow$ $F$ is midpoint of the arc $B'TC'$ of $\odot O_2$ $\Longrightarrow$ $TF \equiv TM$ bisects $\angle B'TC' \equiv \angle BTC$ externally $\Longrightarrow$ $M$ is midpoint of the arc $BTC$ of $\odot O_1$ $\Longrightarrow$ $AM$ is external bisector of $\angle BAC$ and $MB=MC.$

Note that $\odot O_2$ is a Thebault circle of the cevian $AD$ of $\triangle ABC$ externally tangent to its circumcircle $\odot O_1.$ By Sawayama's lemma $EF$ passes through its C-excenter $\Longrightarrow$ $N$ is C-excenter of $\triangle ABC$ $\Longrightarrow$ $N \in AM.$ $\angle BNA=90^{\circ}-\frac{_1}{^2}\angle ACB=90^{\circ}-\frac{_1}{^2}\angle BMN$ $\Longrightarrow$ $\triangle BMN$ is M-isosceles, i.e. $MB=MN.$ Hence $MB=MC=MN$ $\Longrightarrow$ $M$ is circumcenter of $\triangle BCN.$

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#3 h Aug 17, 2013, 8:24 AM • 2 Y

Y by Adventure10, Mango247

s372102 wrote:

As shown in the figure, $\odot O_1$ and $\odot O_2$ touches each other externally at a point $T$ , quadrilateral $ABCD$ is inscribed in $\odot O_1$ , and the lines $DA$ , $CB$ are tangent to $\odot O_2$ at points $E$ and $F$ respectively. Line $BN$ bisects $\angle ABF$ and meets segment $EF$ at $N$ . Line $FT$ meets the arc $\widehat{AT}$ (not passing through the point $B$ ) at another point $M$ different from $A$ . Prove that $M$ is the circumcenter of $\triangle BCN$ .

As shown in the figure.

Let $N$ be the intersection of $AM$ and $EF$ .

It's fairly obvious

$\angle FET = \angle TFK =\angle TBM = \angle TCM = \angle TAM$

So we have

$\angle CBM = \angle CTM = \angle BTF = \angle BCM$
$MB^2=MB\times MF$ , and $A,E,N,T$ are concyclic.

So
$MC=MB$ , and
$\angle MNT = \angle TEA = \angle MFN$

Thus
$MN^2= MT\times MF$ ,
Consequently
$MB=MC=MN$ .

$\angle ABN = \angle ABM + \angle MBN$
$= \angle ABM + \frac{180- \angle BMN}{2}$
$=\angle ABM +90 - \frac{\angle BCA}{2}$
$= \angle ABM + 90-\frac { \angle BCM}{2} -\frac{ \angle ACM }{2}$
$=\frac{\angle CMB}{2} +\frac{\angle BCM}{2}+ \frac{\angle ACM}{2}$
$= \frac{\angle CAB}{2} +\frac{\angle BCA}{2}$
$=\frac{\angle ABF}{2}$

Namely $AN$ is the bisector of $\angle ABF$ .

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#4 h Jan 29, 2022, 3:08 PM

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Let AM meet EF at N'. we will prove N' is N.
Claim1 : M is center of BCN'.
Proof : Let BT meet $\odot O_2$ at B' and CT meet at C'. B'C' || BC so ∠C'B'F = ∠B'FB = ∠FC'B' so F is midpoint of arc C'B'T Note that FT meets $\odot O_1$ at M so M is midpoint of arc CBT so MB = MC. ∠N'ET = ∠TFB = ∠TAM so ATN'E is cyclic. ∠TN'M = ∠TEA = ∠TFN' so MN'^2 = MT.MF = MB^2 so MC = MB = MN'.

Claim2 : BN' is angle bisector of ABF.
Proof : ∠ABN' = ∠ABM + 90 - ∠BMN'/2 = ∠ABM + 90 - ∠BCA/2 = ∠ABM + 90 - ∠BCM/2 - ∠ABM/2 = ∠ABM/2 + 90 - ∠CBM/2 = ∠ABM/2 + 90 - ∠CBA/2 - ∠ABM/2 = 90 - ∠CBA/2.

we're Done.

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#5 h May 4, 2024, 6:32 PM

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As shown in the figure, $\odot O_1$ and $\odot O_2$ touches each other externally at a point $T$ , quadrilateral $ABCD$ is inscribed in $\odot O_1$ , and the lines $DA$ , $CB$ are tangent to $\odot O_2$ at points $E$ and $F$ respectively. Line $BN$ bisects $\angle ABF$ and meets segment $EF$ at $N$ . Line $FT$ meets the arc $\widehat{AT}$ (not passing through the point $B$ ) at another point $M$ different from $A$ . Prove that $M$ is the circumcenter of $\triangle BCN$ .

https://cdn.artofproblemsolving.com/attachments/9/f/932b507f4eee3e1b4c6dac51c1b3f431398322.jpg

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#6 h Apr 24, 2025, 10:34 AM

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So problem is solvable using only angle chase huh, i had no idea about that. By the way using Sawayama lemma trivialized the problem, nice solution Luis Gonzalez.

Anyway $MB=MC$ part is pure angle chase. After that problem can be viewed as following problem.

Let $ABC$ be the triangle and $I_A$ be the A-excenter. Let $T$ be arbitrary point on the $(ABC)$ . Let $\omega$ be the circle tangent to $AB$ and $(ABC)$ externally at $P$ and $T$ , respectively. Let $Q=PI_A \cap \omega$ . Then prove that $CQ$ is tangent to $\omega$ . (Since Luis Gonzalez solved this problem, i wont solve it directly. Instead let's use bc-inversion and see what happens)

Let $ABC$ be a triangle and $I$ be it's incenter. $\omega$ be the circles tangent to $AB$ and $BC$ at $P$ and $T$ , respectively. $Q=(AIP) \cap \omega$ , prove that $\omega$ and $(AQC)$ are tangent.

Proof: By an easy angle chase $CIQT$ is cyclic. Then our desired tangency is equivalent of proving $\angle ACQ+ \angle QTP=\angle AQP$ . It's clear since $\angle ACQ+\angle QTP=\angle ACI-\angle QCI+\angle QTP=\angle ACI+\angle QTP-\angle QTI=\angle ACI+\angle ITP$ . So we need to prove that $\angle ACI+\angle ITP=\angle AIP$ , which is equivalent of saying $(ACI)$ and $(PIT)$ are tangent. It's clear because of the symmetry.

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