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Source: iran tst 2014 third exam

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#1 h May 21, 2014, 1:22 PM • 3 Y

Y by anantmudgal09, Adventure10, Mango247

The incircle of a non-isosceles triangle $ABC$ with the center $I$ touches the sides $BC$ at $D$ .
let $X$ is a point on arc $BC$ from circumcircle of triangle $ABC$ such that if $E,F$ are feet of perpendicular from $X$ on $BI,CI$ and $M$ is midpoint of $EF$ we have $MB=MC$ .
prove that $\widehat{BAD}=\widehat{CAX}$

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#2 h May 21, 2014, 2:08 PM • 8 Y

Y by anonymouslonely, wiseman, TheBeatlesVN, enzoP14, Chokechoke, Adventure10, Mango247, and 1 other user

solution:
let $\widehat{CBA}=2a,\widehat{BCA}=2b$ .

let $Q,W,R$ are foot of perpendicular from $E,M,F$ on $BC$ .
$MB=MC$ , $\Rightarrow$ $BW=WC$ . but $M$ is mid point of $EF\Rightarrow WQ=WR \Rightarrow BQ=CR \Rightarrow BE.\cos a =CF.\cos b$ .

let $I_{a}$ is $A$ -excenter.

let $T,S$ are foot of perpendicular from $X$ to $BI_{a},CI_{a}$ , $\Rightarrow \frac{XT}{XS}=\frac{BE}{CF}=\frac{\cos b}{\cos a}=\frac{\sin \widehat{BCI_{a}}}{\sin \widehat{CBI_{a}}}=\frac{\sin \widehat{XI_{a}B}}{\sin \widehat{XI_{a}C}}\Rightarrow XI_{a}$ is symmedian of triangle $CI_{a}B$ .

let $L$ is midpoint of arc $BAC$ . $\widehat{LBC}=\widehat{LCB}=\widehat{CI_{a}}B\Rightarrow LB,LC$ are tangent to circumcircle of triangle $CI_{a}B$ .
$\frac{BX}{CX}=\frac{\sin \widehat{BLX}}{\sin \widehat{CLX}}=(\frac{\cos b}{\cos a})^{2}$ $(1)$

$\frac{\sin \widehat{BAD}}{\sin \widehat{CAD}}=\frac{AC.BD}{AB.CD}=\frac{\tan b}{\tan a}.\frac{sin 2a}{\sin 2b}=(\frac{\cos a}{\cos b})^{2}.(2)$

$(1,2)\,\Rightarrow \frac{\sin \widehat{BAD}}{\sin \widehat{CAD}}=\frac{\sin \widehat{CLX}}{\sin \widehat{XLB}},(\widehat{BAD}+\widehat{CAD}=\widehat{CLX}+\widehat{BLX})\Rightarrow \widehat{CAX}=\widehat{CLX}=\widehat{BAD}$ .

so we are done.

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#3 h Jun 3, 2014, 1:52 AM • 2 Y

Y by wiseman, Adventure10

I have a long solution.
Begin torment

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#4 h Jun 5, 2014, 3:25 AM • 2 Y

Y by wiseman, Adventure10

Note that it suffice to show $I_A, X, M_A$ are collinear, where $I_A, M_A$ are the A-excentre and midpoint of arc $BAC$ respectively.

I cant see a nicer way to show this without ratios... First, it is not too difficult to show for a point $X$ satisfying the property, $BE/CF$ is constant, which in this case is $I_AB/I_AC$ . Immediately, using $M_AB = M_AC, II_A=II_A, BP = CQ$ where $P, Q$ are the feet of $E, F$ onto $BC$ and letting $Y, Z$ be the feet of $M_A$ on $BI, IC$ , that $BE/BY = CE/CZ$ . This means that $I_A, N, M_A$ are collinear by affinity.

To see why this implies the result, we use the extraversion that the A-mixtilinear incircle point, $I$ , $M_A$ are collinear. Indeed, $X$ is just the A-mixtilinear excircle point.

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#5 h Dec 29, 2016, 7:23 AM • 7 Y

Y by anantmudgal09, Ankoganit, Aryan-23, HolyMath, rashah76, Adventure10, Mango247

The crux part of this problem is that the property of point $X$ to be proved actually holds for all points on line $I_AN$ , where $N$ is the midpoint of arc $BAC$ . This is what makes the problem hard, because there is not much special about the point $X$ which can enable one to find a neat synthetic proof without the fuss of ratio chasing things. But the above mentioned generalisation can be employed in the dynamic sense to excavate the roots of problem. Hence in my opinion, the approach below should completely demystify the problem.

First of all, as the above posts have noted, if $X$ is the intersection of $I_AN$ and $\odot ABC$ , then $\angle{BAD}=\angle{CAX}$ , which is the well known extraversion of mixtilinear touchpoint property.

In the proof, we refer to point $X$ as a variable point on line $I_AN$ , $E,F$ the perpendiculars from $X$ to $BI,CI$ and $M$ as the midpoint of $E,F$ . Thus in this context, $E,F,M$ are variable points. We shall nail the problem in two steps. First we show that the locus of $M$ must be a straight line, then we show this line is the perpendicular bisector of $BC$ and we would be through.

STEP 1

STEP 2

This post has been edited 6 times. Last edited by kapilpavase, Dec 30, 2016, 7:54 AM

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#6 h Jan 25, 2017, 5:18 AM • 3 Y

Y by SerdarBozdag, Adventure10, Mango247

It is like the IMO 2015 G4 where a point which typically lies on a line is said to lie on a circle (meaning it's their intersection) and some property is asked to hold.

Solution

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#7 h Nov 1, 2018, 4:20 PM • 2 Y

Y by top1csp2020, Adventure10

We have a lemma; X is the second intersection of A-mixtilinear excircle with (O) then AD, AX isogonal
$1$ . $IaX$ is bisector of $BXC$ : prove similarly to A-inmixtilinear (generalized by Prostasov)
$2$ . $S$ is orthocentre of $IBC$ then $S,M,Ia$ collinear and it's easy to see that $AI/AIa=SI/SD$ so $IaM//AD$
$3$ . $IAD=AIaM=RNIa(IaR^2=RM.RN)=RAX$ so $AD, AX$ isogonal
Back to the problem: From the lemma $BAD=CAX (1)$
By angle chasing $I$ is orthocentre of $EMF$ then $NEMF$ is a pallelogram so $GB=GC$
Apply $E.R.I.Q$ lemma to $(X,N,I_a)$ wrt $IB$ , $IC$ because $EK/EC=NX/NIa=FB/FJ$ so we have $H$ lies on $GM$ hence $HB=HC (2)$
Assume that there exist $X'$ distinct from $X$ satisfy $(ii)$ apply $ERIQ$ lemma again it mean $J'B/K'C=FB/EC=JB/KC$ so $J', K'$ lies on the same plane with $JK$ so $X'$ not lies on $(O) (3)$
From $(1)(2)(3)$ so Q.E.D

Attachments:

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yayups

#8 h Nov 21, 2019, 7:28 AM • 2 Y

Y by Gaussian_cyber, Adventure10

This is evidently a straightforward complex bash. Relabel $X$ in the problem to $T$ . Let $(ABC)$ be the unit circle with $a=x^2$ , $b=y^2$ , $c=z^2$ , such that the midpoints of the arcs are $p=-yz$ , $q=-zx$ , $r=-xy$ . Note that $E$ is the foot from $T$ to $BQ$ , so $e=\frac{1}{2}(t-xz+y^2+xzy^2/t).$ We similarly derive $f=\frac{1}{2}(t-xy+z^2+xyz^2/t),$ so $m=\frac{t}{2}+\frac{y^2+z^2}{4}+\frac{(y+z)x}{4}(yz/t-1).$ The problem tells us that $M$ is on the perpendicular bisector of $BC$ , which in our case means that $m/(y^2+z^2)$ is real, or that $k:=\frac{m}{y^2+z^2}-\frac{1}{4}=\frac{2t+\frac{x(y+z)}{t}(yz-t)}{y^2+z^2}$ is real. This means that $\frac{2t+\frac{x(y+z)}{t}(yz-t)}{y^2+z^2}=\frac{\frac{2}{t}+\frac{t(y+z)}{xyz}\frac{t-yz}{yzt}}{\frac{y^2+z^2}{y^2z^2}},$ or $2\left(t-\frac{y^2z^2}{t}\right)=(y+z)(t-yz)\left(\frac{1}{x}+\frac{x}{t}\right).$ Note that $t\ne yz$ , as that is not on arc $BC$ , so we may cancel the factor $t-yz$ from both sides to obtain the equation $2\frac{t+yz}{t}=\frac{y+z}{tx}(t+x^2),$ so $t=\boxed{\frac{x(2yz-xy-xz)}{y+z-2x}}$ .

It suffices to show that $T$ is the mixtilinear excircle touch point. To do this, we use the fact that $T,I_A,L$ are collinear, where $\ell=yz$ is the arc midpoint of $BAC$ . We'll solve for $t$ using this fact, and show that we get the same formula. Note that $I_A=xy+xz-yz$ , and we have $I_A$ on the chord $TL$ , so $I_A+\ell t\bar{I}_A=\ell+t,$ or $t=\frac{I_A-\ell}{1-\ell\bar{I}_A}=\frac{xy+xz-2yz}{1-\frac{1}{x}(z+y-x)}=\frac{x(2yz-xy-xz)}{y+z-2x},$ so we're done.

Remark: This complex bash actually motivates a short synthetic solution (which I think is really hard to motivate without this). Note that our proof showed that $L$ also satisfied the condition, since we factored out $t-yz=t-\ell$ . Now, the value of the midpoint of $EF$ is linear in $T$ , so since $L$ and the mixtilinear extouch point work, any point on that line must also work. From this, we could have originally just verified the problem for $I_A$ and $L$ , and finished that way, but again, I don't see how one can come up with this.

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nbdaaa

#9 h Dec 28, 2021, 5:29 PM

Y by

I just want to clarify a claim used by some of the above posts that doesn't seem to be trivial to me

Claim: Given $\triangle ABC$ with incenter $I$ , $I_a,I_b,I_c$ be the A-excenter, B-excenter, C-excenter of $\triangle ABC$ . Let $N$ be the midpoint of arc $BAC$ . Let $X$ be an arbitrary point moving on $NI_a$ . $E,F$ be the projection of $X$ on $BI,CI$ . Then prove that the locus of the midpoint $P$ of $EF$ is the perpendicular bisector of $BC$

Proof
Let $S$ be the projection of $I$ on $NI_a$
We have $\overrightarrow {PQ} = \overrightarrow {SQ} - \overrightarrow {SP} = \frac{{\overrightarrow {SB} + \overrightarrow {SC} }}{2} - \frac{{\overrightarrow {SE} + \overrightarrow {SF} }}{2} = \frac{{(\overrightarrow {SB} - \overrightarrow {SE} ) + (\overrightarrow {SC} - \overrightarrow {SF} )}}{2} = \frac{{\overrightarrow {EB} + \overrightarrow {FC} }}{2}$
So we have to prove that $\begin{array}{l} PQ \perp BC \Leftrightarrow \overrightarrow {PQ} .\overrightarrow {BC} = 0 \Leftrightarrow (\overrightarrow {BE} + \overrightarrow {CF} ).\overrightarrow {BC} = 0 \Leftrightarrow \overrightarrow {BE} .\overrightarrow {BC} - \overrightarrow {FC} .\overrightarrow {BC} = 0\\ \\ \Leftrightarrow BE.BC.\cos \frac{B}{2} - FC.BC.\cos \frac{C}{2} = 0 \Leftrightarrow \frac{{BE}}{{CF}} = \frac{{\cos \frac{C}{2}}}{{\cos \frac{B}{2}}} \end{array}$ Notice that there exists a spiral similarity with center $S$ sends $E$ to $B$ , sends $F$ to $C$ and sends $P$ to the midpoint $Q$ of $BC$ , then $\frac{{BE}}{{CF}} = \frac{{SB}}{{SC}}$ Finally by applying the law of sine, we have $\frac{{SB}}{{SC}} = \frac{{\sin SCB}}{{\sin SBC}} = \frac{{\sin N{I_a}B}}{{\sin A{I_c}B}} = \frac{{\cos \frac{C}{2}}}{{\cos \frac{B}{2}}}$

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#10 h Mar 10, 2025, 7:41 PM

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Let $J$ be $A-$ excenter, $P$ be the midpoint of arc $BC$ . We want to show that $X$ is $A-$ exmixtilinear touch point hence it sufficies to show that $X$ is $J-$ dumpty on $\triangle JBC$ .

Claim: $JBE\overset{-}{\sim} JCF$ .
Proof: Let $BF\cap CE=G$ . Since $PM$ is the newton-gauss line of quadrilateral $BCFE$ , it passes through the midpoint of $IG$ . This gives $JG\perp BC$ . DDIT on quadrilateral $IGEF$ gives $(\overline{JI},\overline{JG}),(\overline{JE},\overline{JF}),(\overline{JB},\overline{JC})$ is an involution. This must be reflection over the angle bisector of $\measuredangle BJC$ hence $\measuredangle BJE=\measuredangle FJC$ . Also $\measuredangle EBJ=90=\measuredangle JCF$ thus, the claim follows.

Claim: $X$ is $J-$ dumpty on $\triangle JBC$ .
Proof: Let $X'$ be $J-$ dumpty.
$\frac{\sin X'JC}{\sin X'JB}=\frac{JC}{JB}=\frac{CF}{BE}=\frac{\frac{CF}{JX}}{\frac{BE}{JX}}=\frac{\sin XJC}{\sin XJB}$ Since $\measuredangle BJX'+\measuredangle X'JC=\measuredangle BJX+\measuredangle XJC$ we get that $X=X'$ as desired. $\blacksquare$

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