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 $\widehat{BAD}=\widehat{CAX}$
Problem
Source: iran tst 2014 third exam
Tags: geometry, circumcircle, trigonometry, ratio, symmetry, geometry proposed
nima1376
21.05.2014 17:08
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.
leader
03.06.2014 04:52
I have a long solution.
Let $P,Q,R$ be feet of $E,F,M$ on $BC$. Now let $I_a$ be the $A$-excenter of $ABC$ and let the perpendicular from $I_a$ to $AI_a$ cut $AB,AC$ at $T,S$. Now let $N$ be the midpoint of arc $BAC$. Let $K,L$ be feet of $N$ on $BI,CI$.
Now by the condition $RB=RC$ and since $EM=MF$ we have $PR=RQ$ now $BP=CQ$. Sicne $\angle ICI_a=\angle IBI_a=90$ then $ICI_aB$ is cyclic and $RS$ touches the circumcircle of that quadrilateral. Now $\angle EBP=\angle II_aC$ and $\angle ICI_a=90=\angle EPB$ giving $EBP\sim II_aC$ so $II_a*BP=BE*CI_a$. Similarly $EC*BI_a=CQ*II_a=BP*II_a=BE*CI_a$ so $BE/CF=BI_a/CI_a$(1) but we also have $\angle NBK=\angle ICB=\angle II_aB$ giving $IBI_a\sim NKB$ so $BK/BI_a=BN/II_a=CN/II_a=CL/CI_a$ so $BK/CL=BI_a/CI_a$ now because of (1) we have that $BE/BK=CF/CL$ so that $EX,FX$ intersect $NI_a$ at the same ratio giving $N,X,I_a$ collinear. Now $\angle NXB=\angle BXC/2=(180-\angle BAC)/2=\angle I_aTB$ so $XI_aTB$ is cyclic. and so is $XI_aSC$ similarly. Now now that $\angle CXS=\angle CI_aS=\angle I_aBC=\angle XBC+\angle XTS$ so circles $ABC$ and $XTS$ touch. But \[\angle TXS=\angle TBI_a+\angle SCI_a=\angle CBI_a+\angle SCI_a=\angle CI_aS+\angle SCI_a=180-\angle AST=180-\angle ATS\] so circle $XTS$ touches $AB,AC$ Now note that the composition of inversion with respect to cecnter $A$ and circle with square radius $AB*AC$ and the symmetry wrt $\angle BAC$ pictures the incircle into a circle touching $AB,AC$ and circle $ABC$ externaly. So the incircle goes to circle $XTS$ and now $D$ goes to a point on circle $XTS$ touching circle $ABC$(point $X$) so $D$ goes to $X$ giving FINALLY $\angle BAD=\angle CAX$.
IDMasterz
05.06.2014 06:25
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.
kapilpavase
29.12.2016 10:23
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.
[asy][asy]
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[/asy][/asy]
This is not tough. Consider $X$ moving with a constant velocity. Then points $E,F$ move with constant velocity too, and hence their centre of mass $M$ must move along a straight line. (There is no net force on the system)
call this locus line as $\ell$. We show that for two special locations of $X$ the two corresponding locations of $M$ are on the perp bisector of $BC$. Hence $\ell$ would be the perpendicular bisector as required.
[asy][asy]
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[/asy][/asy]
First let $X=I_A$,. Then clearly, $M=P$ , where $P$ is the midpoint of $BC$. Hence $P$ lies on $\ell$
Next, let $X=N$, where $N$ is the midpoint of arc $BAC$. So if $NR\perp BI, NQ\perp CI$ and $Y$ is the midpoint of $QR$(as in figure), We need to show that $YP\perp BC$.
Since $\angle{PQC}=\angle{PNC}=A/2$. So $QP\perp BI$. Similarly, $RP\perp CI$ .Thus, $NQPR$ is a parallelogram, and hence $N,Y,P $ are collinear, i.e. $Y $ lies on peep bisector of $BC $. So, $\ell=YP$, the perpendicular bisector of $BC$, which was to be proved
anantmudgal09
25.01.2017 08:18
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.
Let $\triangle J_AJ_BJ_C$ be the excentral triangle of $\triangle ABC$; $L$ be the midpoint of $\overline{BC}$ and $N$ be the midpoint of $\overline{J_BJ_C}$ (also the midpoint of arc $BAC$), $J'$ be the point symmetric to $J_A$ in $N$ and $S$ be the point on $\overline{J_AN}$ such that $IS \perp J_AN$.
For any point $P$ in the plane; define $P_B, P_C$ as the feet of perpendiculars from $P$ to $\overline{BI}, \overline{CI}$ respectively and $P^*$ as the midpoint of $\overline{P_BP_C}$. Note that for any point $R$ there exists a unique point $T$ such that $R=T^*$.
Let $P$ be any point on $\overline{J_AJ'}$. Spiral similarity about $S$ shows that $P^*$ moves uniformly on a line as $P$ moves uniformly on line $\overline{J_AJ'}$. For $P=J$ we get $P^*=L$ and for $P=J'$ we get $P^*=N$. It follows that the locus of $P^*$ is the perpendicular bisector of $\overline{BC}$.
In conclusion, $P^*$ lies on line $\overline{LN}$ (i.e., $BP^*=CP^*$) if and only if $P$ lies on $\overline{J_AN}$. It follows that $X \ne N$ is the intersection of this line with the circumcircle of triangle $ABC$. Applying $\sqrt{bc}$ inversion, $J_A \rightarrow I$ and $N \rightarrow K$ ($K$ is the feet of the $A$ external bisector) so $X \rightarrow D \Longrightarrow \angle BAX=\angle CAD$ as desired. $\, \square$
nguyenhaan2209
01.11.2018 19:20
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:
yayups
21.11.2019 10:28
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.
nbdaaa
28.12.2021 20:29
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 \[\frac{{SB}}{{SC}} = \frac{{\sin SCB}}{{\sin SBC}} = \frac{{\sin N{I_a}B}}{{\sin A{I_c}B}} = \frac{{\cos \frac{C}{2}}}{{\cos \frac{B}{2}}}\]