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Another easy geometry from Taiwan TST

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Source: 2021 Taiwan TST Round 2 Independent Study 2-G

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42 posts

Li4

#1 h Apr 8, 2021, 8:54 AM • 5 Y

Y by FaThEr-SqUiRrEl, p_square, SINNDT---loveMATHforever, tiendung2006, Rounak_iitr

Let $ABC$ be a triangle with circumcircle $\Gamma$ , and points $E$ and $F$ are chosen from sides $CA$ , $AB$ , respectively. Let the circumcircle of triangle $AEF$ and $\Gamma$ intersect again at point $X$ . Let the circumcircles of triangle $ABE$ and $ACF$ intersect again at point $K$ . Line $AK$ intersect with $\Gamma$ again at point $M$ other than $A$ , and $N$ be the reflection point of $M$ with respect to line $BC$ . Let $XN$ intersect with $\Gamma$ again at point $S$ other that $X$ .

Prove that $SM$ is parallel to $BC$ .

Proposed by Ming Hsiao

This post has been edited 1 time. Last edited by Li4, Apr 8, 2021, 10:40 AM
Reason: typo

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#2 h Apr 8, 2021, 9:08 AM • 1 Y

Y by FaThEr-SqUiRrEl

By inversion wrt A, we see that XMBC are harmonic, and then with perspective at S, the problem is equivalent to show that N, X, the midpoint of BC(M') are colinear

However, its well known that BC bisects XM'M, the colinearity is obvious, the result follows

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#3 h Apr 8, 2021, 9:56 AM • 1 Y

Y by FaThEr-SqUiRrEl

I want to kill myself after I heard my friends' solution.
My solution given in the test is just some barycentric coordinate stuff and finding out the fact that the midpoint of $BC$ lies on $S'N$ , where $S'$ is the point on $\Gamma$ such that $MS'||BC$ .

This post has been edited 1 time. Last edited by j20210301, Apr 8, 2021, 9:57 AM

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#4 h Apr 8, 2021, 10:05 AM • 1 Y

Y by FaThEr-SqUiRrEl

In fact, after finding out that MS' pass through the midpoint of BC, then bashing it is not as complicated as how it seem.

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2016 posts

PROF65

#5 h Apr 8, 2021, 7:28 PM

Y by

lemma: let $AP$ cuts the circle $(ABC)$ at $Q$ ;if $R$ the reflection of $Q$ in $BC$ and $Q'$ is the second intersection of the parallel to $BC$ through $Q$ and $(ABC)$ . then $A,R$ and $Q'$ are colinear iff $AP$ is the $A-$ symmedian line .
it s easy to prove it
back to the problem
from the lemma it remain to prove that $XM$ is the $X-$ symmedian of $XBC$ ;
let $L=(AEF)\cap AM$
since $X$ is the Miquel point or the similicenter of $BF\to CE$ it suffices to show that $XL$ is the $X-$ symmedian of $XEF$ i.e. $\frac{XE}{XF}=\frac{EL}{FL}$ :
$\frac{EL}{LF}=\frac{\sin \angle EAL}{\sin\angle LAF}=\frac{\sin \angle EAK}{\sin\angle KAC}=\frac{EK}{KC }=\frac{EB}{FC }$ (because $K$ is the Miquel point of $CE\to FB)$
but $X$ is the similicenter of $BE\to CF$ then $\frac{EB}{FC }=\frac{XE}{XF }$ hence $\frac{EL}{LF}=\frac{XE}{XF}$ which ends the proof.
RH HAS

This post has been edited 1 time. Last edited by PROF65, Apr 8, 2021, 9:00 PM
Reason: typo

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#6 h Apr 11, 2021, 1:50 AM

Y by

j20210301 wrote:

In fact, after finding out that MS' pass through the midpoint of BC, then bashing it is not as complicated as how it seem.

I considered complex bashing the colinearity, but I forgot the formula for reflecting a point through a line... sad

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#7 h Apr 12, 2021, 1:21 PM

Y by

wayneyam wrote:

j20210301 wrote:

In fact, after finding out that MS' pass through the midpoint of BC, then bashing it is not as complicated as how it seem.

I considered complex bashing the colinearity, but I forgot the formula for reflecting a point through a line... sad

I thought that it can be quickly derived if you have remembered some high school complex number stuff??

btw, you are so strong that you have high possibility to be in Taiwan team.

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222 posts

#8 h Apr 13, 2021, 7:16 PM

Y by

Solved with BOBTHEGR8, Pluto1708, Arwen713

Observe that $(XM; BC)=-1$ . Now, let the tangent from $X,M$ to $(ABC)$ intersect $BC$ at $Y$ , then $Y$ is the circumcenter of $(XMN)$ . Let $(XMN) \cap BC=D$ internally, i.e. $D$ lies on segment $BC$ . Now, $DM=DN$ thus, $XD$ is angle bisector of $\angle MXN$ . So, if we show $XD$ is angle bisector of $\angle BXC$ , we will get that $XM,XN$ are isogonal in $\angle BXC$ which implies the result.

Now, we have $\angle XCB+\angle BXD=\angle YXB+\angle BXD=\angle YXD=\angle XDY=\angle XCB+\angle DXC\implies \angle BXD=\angle DXC$ and we are done.

This post has been edited 2 times. Last edited by Rg230403, Apr 14, 2021, 6:01 PM
Reason: Forgot to credit

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#9 h Apr 17, 2021, 2:49 PM • 1 Y

Y by hsiangshen

Here comes my whole solution.

Let $S'$ be the point on $\Gamma$ such that $MS'\parallel BC$ . Notice that $NS'$ passes through the midpoint of $BC$ , since quadrilateral $BMS'C$ is a trapezoid. Our goal now is to prove that $S', \text{the midpoint of }BC, X$ are collinear.(By the uniqueness of the point $S$ .)

Let $P=BE \cap CF$ , and take $\triangle ABC$ as the reference triangle of barycentric coordinate. Denote $a=\overline{BC},b=\overline{CA},c=\overline{AB}$ .
Suppose that point $P$ has the coordinate $[u:v:w]$ , then we have $E=[u:0:w],F=[u:v:0]$ . By some easy computation, we can write down following equation.

$(AEF):a^2yz+b^2zx+c^2xy=(x+y+z)(\frac{c^2u}{u+v}y+\frac{b^2u}{u+w}z).$
$(ABE):a^2yz+b^2zx+c^2xy=(x+y+z)(\frac{b^2u}{u+w}z).$
$(ACF):a^2yz+b^2zx+c^2xy=(x+y+z)(\frac{c^2u}{u+v}y).$ By using the fact that subtracting the equation of a circle with another could get the equation of their radical axis. More precisely, for two circles $\omega_{1},\omega_{2}$ with equations as following:
$\begin{cases}\omega_{1}:a^2yz+b^2zx+c^2xy=(x+y+z)(p_{1}x+q_{1}y+r_{1}z)\\ \omega_{2}:a^2yz+b^2zx+c^2xy=(x+y+z)(p_{2}x+q_{2}y+r_{2}z) \end{cases}.$ then the equation of their radical axis can be written as as $(p_{1}-p_{2})x+(q_{1}-q_{2})y+(r_{1}-r_{2})z=0.$ Now we can write down the equations of line $AX$ and line $AM$ as the immediate result of above discussion.
$AK:\frac{c^2}{u+v}y=\frac{b^2}{u+w}z.$
$AX:\frac{c^2}{u+v}y+\frac{b^2}{u+w}z=0.$ Hence, the coordinate of $X$ has the form: $[t:(b^2)/(u+w):-(c^2)/(u+v)]$ , plug this into the equation of $(ABC)$ , and get:
$\frac{-(abc)^2}{(u+w)(u+v)}+\frac{-b^2c^2}{u+v}t+\frac{b^2c^2}{u+w}=0\implies (v-w)t=a^2.$ Now, we'll begin to deal with the coordinate of $S'$ . Notice that line $AS'$ is isogonal to line $AM$ with respect to $\triangle ABC$ . Therefore, $S'$ has the form: $[s:u+w:u+v]$ . Plug this into the equation of $(ABC)$ and simplify it, and get:
$(b^2(u+v)+c^2(u+w))s=-a^2(u+w)(u+v).$
Back to the original problem, $S', \text{the midpoint of }BC, X$ collinear is equivalent to
$\begin{aligned} \left|\begin{array}{cccc} s & u+w & u+v \\ t & (b^2)/(u+w) &-(c^2)/(u+v)\\ 0 & 1 & 1 \end{array}\right|=0 &\iff s(\dfrac{b^2}{u+w}+\dfrac{c^2}{u+v})=t(w-v)\\ &\iff -a^2=-a^2. \end{aligned}$ Then the proof is complete.

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#10 h Sep 26, 2024, 3:48 PM

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Invert around $A$ . Then we get the default cevian harmonic configuration, so $(X',M';B',C')=-1$ . Therefore,
$(X,M;B,C)=(AX,AM;AB,AC)=(AX',AM';AB',AC')=-1.$ Let $T$ be the point on $\Gamma$ such that $XT\parallel BC$ . Then projecting through $T$ , we get that $T,M,M_{BC}$ are collinear, where $M_{BC}$ is the midpoint of $BC$ . Therefore, $T$ , $M_{BC}$ , and $2M_{BC}-M$ are collinear, so reflecting over the perpendicular bisector of $BC$ , $X,N,M_{BC}$ are collinear. Thus $S$ is indeed just $2M_{BC}-N$ , so we win. $\blacksquare$

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