Passes through tangent point

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Passes through tangent point

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#1 h Jan 1, 2016, 7:14 AM • 14 Y

Y by buzzychaoz, dragonx111, anantmudgal09, huynguyen, mihaith, MihneaD, Scorpion.k48, solver6, teddy8732, Mz_T, Siddharth03, Adventure10, Mango247, nguyenducmanh2705

Let $ABC$ be a triangle inscribed in circle $(O)$ with $A$ -excircle $(J)$ . Circle passing through $A,B$ touches $(J)$ at $M$ . Circle passing through $A,C$ touches $(J)$ at $N$ . $BM$ cuts $CN$ at $P$ . Prove that $AP$ passes through tangent point of $A$ -mixtilinear incircle with $(O)$ .

Happy new year 2016!

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#2 h Jan 2, 2016, 8:55 AM • 2 Y

Y by Adventure10, Mango247

Up.I hope to see a solution to this nice problem.

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#5 h Jan 2, 2016, 8:03 PM • 5 Y

Y by huynguyen, mihaith, Scorpion.k48, Adventure10, Mango247

Using variation problem for incircle as following

Let $ABC$ be a triangle with incircle $(I)$ touches $BC$ at $D$ . Circle passing through $A,B$ touches $(I)$ at $M$ . Circle passing through $A,C$ touches $(I)$ at $N$ . $BM$ cuts $CN$ at $P$ . Prove that $\angle PAB=\angle DAC$ .

Lemma. Let $ABC$ be a triangle and incircle $(I,r)$ touches $BC,CA,AB$ at $D,E,F$ , resp. Circle passes through $B,C$ and touches $(I)$ at $X$ then

i) $\dfrac{XE.XF}{XD^2}=\dfrac{r^2}{IB.IC}.$

ii) $\dfrac{XE}{XF}=\dfrac{IB.DE^2}{IC.DF^2}$

Consequence. $\dfrac{XE^2}{XD^2}=\dfrac{r^2.DE^2}{IB^2.DF^2}$ and $\dfrac{XF^2}{XD^2}=\dfrac{r^2.DF^2}{IC^2.DE^2}$ .

Proof of problem. Let $(I)$ touches $CA,AB$ at $E,F$ and $X,Y,Z$ is cevian triangle of $P$ . Let $BY$ cuts $DF$ at $K$ . We have

$\dfrac{YC}{YA}=\dfrac{[YBC]}{[YBA]}=\dfrac{[YBC]}{[KBD]}.\dfrac{[KBD]}{[KBF]}.\dfrac{[KBF]}{[YBA]}=\dfrac{BC.BY}{BD.BK}.\dfrac{KD}{KF}.\dfrac{BF.BK}{BY.BA}=\dfrac{BC}{BA}\dfrac{MD^2}{MF^2}=\dfrac{BC}{BA}.\dfrac{r^2.DF^2}{EF^2.IA^2}$ .

Similarly, $\dfrac{ZB}{ZA}=\dfrac{BC}{CA}.\dfrac{r^2.DE^2}{EF^2.IA^2}$ .

Now by Ceva's theorem then $\dfrac{XB}{XC}=\dfrac{YA}{YC}.\dfrac{ZB}{ZA}=\dfrac{AB.DE^2}{AC.DF^2}$ .

So $\dfrac{DB}{DC}.\dfrac{XB}{XC}=\dfrac{p-b}{p-c}.\dfrac{AB.DE^2}{AC.DF^2}=\dfrac{DF.IB}{DE.IC}.\dfrac{AB.DE^2}{AC.DF^2}=\dfrac{AB^2}{AC^2}$ . Thus $AD,AX$ are isogonal.

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#6 h Jan 3, 2016, 2:33 AM • 4 Y

Y by huynguyen, mihaith, Adventure10, Mango247

Proof of lemma. Follow topic IMO ShortList 2002, geometry problem 7, let $DK$ is diameter of $(I)$ then $XK,EF,BC$ are concurrent at $G$ . $AD$ cuts $EF$ at $H$ .

We have $\dfrac{XE}{XF}.\dfrac{KE}{KF}=\dfrac{GE}{GF}=\dfrac{HE}{HF}=\dfrac{[AED]}{[AFD]}=\dfrac{DE}{DF}.\dfrac{DB}{IC}.\dfrac{IB}{DC}$ and note that $KE.IC=2r^2=KF.IB$ . So $\dfrac{XE}{XF}=\dfrac{IB.DE^2}{IC.DF^2}.$

And $\dfrac{FX}{FK}.\dfrac{EX}{EK}=\dfrac{GX}{GK}=\dfrac{GD^2}{GK^2}=\dfrac{XD^2}{4r^2}$ and note that $KE.IC=2r^2=KF.IB$ . So $\dfrac{XE.XF}{XD^2}=\dfrac{r^2}{IB.IC}.$

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#7 h Jan 3, 2016, 2:42 AM • 5 Y

Y by huynguyen, dragonx111, mihaith, Adventure10, Mango247

Solution of original problem by inversion. Using $\sqrt{bc}$ inversion we need to prove the problem

Let $ABC$ be a triangle inscriebd in circle $(O)$ . Circle $(K)$ touches $CA,AB$ and $(O)$ internally. Draw tangent line $CM,BN$ to $(K)$ with $M,N$ lie on $(K)$ . Circles $(ABN),(ACM)$ intersect again at $P$ . Prove that $AP$ passes through Nagel point.

Proof. Let $(ABN)$ cuts $CA$ again at $Y$ and $(ACM)$ cuts $AB$ again at $Z$ . $AP$ cuts $(O)$ again at $X$ . We easily seen $\dfrac{XB}{XC}=\dfrac{ZB}{YC}$ . But by inversion, we can prove $\dfrac{ZB}{YC}=\dfrac{DE^2}{DF^2}$ . So $\dfrac{XB}{XC}=\dfrac{DE^2}{DF^2}$ . From this, we can prove that $AP$ passes through Nagel point.

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#8 h Jan 3, 2016, 2:46 AM • 2 Y

Y by Adventure10, Mango247

Thank you, mr.buratinogiggle, for the lemma and the solution.

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#9 h Jan 3, 2016, 7:16 AM • 3 Y

Y by buratinogigle, Adventure10, Mango247

And how we get those ratios in the inversion proof, dear buratinogigle ?

This post has been edited 5 times. Last edited by mihaith, Jan 26, 2016, 5:34 PM

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#10 h Jan 3, 2016, 8:06 AM • 3 Y

Y by dragonx111, mihaith, Adventure10

Dear mihaith, actually I get it from the above lemmas before I use inversion.

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#11 h Jan 4, 2016, 1:27 PM • 2 Y

Y by Adventure10, Mango247

buratinogigle wrote:

Dear mihaith, actually I get it from the above lemmas before I use inversion.

I understand, thank you.

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#12 h Jan 5, 2016, 4:45 PM • 3 Y

Y by Adventure10, Mango247, nguyenducmanh2705

Generalization of the inversion version:
Given triangle $ABC$ . Let $(O)$ be an arbitrary circle passing through $B,C$ and cuts $AB,AC$ at $E$ and $F$ , respectively. Let $\omega$ be circle internally tangent to $(O)$ and tangent to rays $AB$ and $AC.$ Construct tangents $BM$ and $CN$ to $\omega$ . The $A$ -excircle of triangle $ABC$ touches $BC$ at $S$ . Then $AS$ is the radical axis of $(EBM)$ and $(FCN).$

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#16 h Jan 6, 2016, 1:08 PM • 2 Y

Y by Adventure10, Mango247

Another way to generalize the version created by inversion

Let $ABC$ be a triangle inscribed in circle $(O)$ . $K$ is a point on bisector of angle $A$ and circle $(K)$ touches $(O)$ internally at $L$ . $E,F$ lie on $(K)$ such that $CE,DF$ are tangents of $(K)$ with $E,F$ and $A$ are not the same side with respect to $BC$ . Circumcircle of triangle $ACE$ and $ABF$ intersect again at $P$ . Prove that $\angle PAC=\angle LAB$ .

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#17 h Jan 19, 2016, 11:43 AM • 4 Y

Y by Re1gnover, buratinogigle, enhanced, Adventure10

buratinogigle wrote:

Using variation problem for incircle as following

Let $ABC$ be a triangle with incircle $(I)$ touches $BC$ at $D$ . Circle passing through $A,B$ touches $(I)$ at $M$ . Circle passing through $A,C$ touches $(I)$ at $N$ . $BM$ cuts $CN$ at $P$ . Prove that $\angle PAB=\angle DAC$ .

Let $\odot (I)$ touches $CA,$ $AB$ at $E,$ $F$ , respectively and let $R$ $\equiv$ $EF$ $\cap$ $BC$ . Let $\odot (BTC)$ be the circle passing through $B,$ $C$ and tangent to $\odot (I)$ at $T$ . Let $J$ $\equiv$ $DT$ $\cap$ $EN$ $\cap$ $FM$ be the perspector of the intouch triangle of $\triangle ABC$ and the excentral triangle of $\triangle ABC$ ( $X_{57}$ of $\triangle ABC$ (well-known)). From Steinbart's theorem we get $AT,$ $BN,$ $CM$ are concurrent at $K$ , so from the dual of Desargue involution theorem (for $BMCN$ ) $\Longrightarrow$ there is an involution $\Phi$ that swaps $(AB,AC),$ $(AK,AP),$ $(AM,AN)$ . On the other hand, it's easy to see that the pole $L$ of $MN$ WRT $\odot (I)$ lies on $AJ$ , so from the dual of Desargue involution theorem (for $LMLN$ ) $\Longrightarrow$ $AL$ is fixed under $\Phi$ .

Let $AL,$ $AP,$ $AK$ cuts $\odot (ABC)$ again at $X,$ $Y,$ $Z$ , respectively. From the discussion above $\Longrightarrow$ $XS$ is tangent to $\odot (ABC)$ at $X$ where $S$ $\equiv$ $BC$ $\cap$ $YZ$ , so $S$ is the midpoint of $DR$ (well-known property of $X_{57}$ ). From $(B,C;D,R)$ $=$ $-1$ we get ${SD}^2$ $=$ $SB$ $\cdot$ $SC$ , so $S$ lies on the radical axis of $\odot (I)$ and $\odot (BTC)$ $\Longrightarrow$ $ST$ is tangent to $\odot (BTC)$ at $T$ .

Let $V$ $\equiv$ $AT$ $\cap$ $BC$ . Since the isogonal conjugate of $TV$ WRT $\angle BTC$ passes through the tangency point of the A-excircle of $\triangle ABC$ with $BC$ which is the isotomic conjugate of $D$ WRT $B,$ $C$ (well-known), so $BV$ $:$ $CV$ $=$ ${BD}^3$ $:$ ${CD}^3$ $\Longrightarrow$ $Y( \infty, S; B, C) = \frac{CS}{BS} =\left( \frac{CT}{BT} \right)^2 = \left( \frac{CD}{BD} \right)^2 = \frac{BD}{CD} \cdot \left( \frac{CD}{BD} \right)^3 = \frac{BD}{CD} \cdot \frac{CV}{BV} = (D,V;B,C),$ hence if $AD$ cuts $\odot (ABC)$ again at $U$ then $YU$ $\parallel$ $BC$ $\Longrightarrow$ $\angle PAB$ $=$ $\angle DAC$ .

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#18 h Jan 20, 2016, 4:08 AM • 5 Y

Y by buratinogigle, enhanced, Adventure10, Mango247, nguyenducmanh2705

buratinogigle wrote:

Using variation problem for incircle as following

Let $ABC$ be a triangle with incircle $(I)$ touches $BC$ at $D$ . Circle passing through $A,B$ touches $(I)$ at $M$ . Circle passing through $A,C$ touches $(I)$ at $N$ . $BM$ cuts $CN$ at $P$ . Prove that $\angle PAB=\angle DAC$ .

Another way to finish the proof (without calculation) :

Let $\odot (I)$ touches $CA,$ $AB$ at $E,$ $F$ , respectively and let $R$ $\equiv$ $EF$ $\cap$ $BC$ . Let $\odot (BTC)$ be the circle passing through $B,$ $C$ and tangent to $\odot (I)$ at $T$ . Let $J$ $\equiv$ $DT$ $\cap$ $EN$ $\cap$ $FM$ be the perspector of the intouch triangle of $\triangle ABC$ and the excentral triangle of $\triangle ABC$ ( $X_{57}$ of $\triangle ABC$ (well-known)). From Steinbart's theorem we get $AT,$ $BN,$ $CM$ are concurrent at $K$ , so from the dual of Desargue involution theorem (for $BMCN$ ) $\Longrightarrow$ there is an involution $\Phi$ that swaps $(AB,AC),$ $(AK,AP),$ $(AM,AN)$ . On the other hand, it's easy to see that the pole $L$ of $MN$ WRT $\odot (I)$ lies on $AJ$ , so from the dual of Desargue involution theorem (for $LMLN$ ) $\Longrightarrow$ $AL$ is fixed under $\Phi$ .

Let $AL,$ $AP,$ $AK$ cuts $\odot (ABC)$ again at $X,$ $Y,$ $Z$ , respectively. From the discussion above $\Longrightarrow$ $XS$ is tangent to $\odot (ABC)$ at $X$ where $S$ $\equiv$ $BC$ $\cap$ $YZ$ , so $S$ is the midpoint of $DR$ (well-known property of $X_{57}$ ). From $(B,C;D,R)$ $=$ $-1$ we get ${SD}^2$ $=$ $SB$ $\cdot$ $SC$ , so $S$ lies on the radical axis of $\odot (I)$ and $\odot (BTC)$ $\Longrightarrow$ $ST$ is tangent to $\odot (BTC)$ at $T$ .

Let $Q$ $\equiv$ $AT$ $\cap$ $\odot (BTC),$ $V$ $\equiv$ $AT$ $\cap$ $BC$ . It's well-known that there is a circle tangent to $AB,$ $AC$ and tangent to $\odot (BTC)$ at $Q$ , so $QI$ is the bisector of $\angle CQB$ (see incenter of triangle) $\Longrightarrow$ $QI$ passes through the midpoint of arc $BC$ in $\odot (BTC)$ , hence notice $TD$ is the bisector of $\angle BTC$ we get $D,$ $I,$ $Q,$ $S,$ $T$ are concyclic (Reim theorem). Since $VA$ $\cdot$ $VZ$ $=$ $VB$ $\cdot$ $VC$ $=$ $VT$ $\cdot$ $VQ$ $=$ $VD$ $\cdot$ $VS$ , so $A,$ $D,$ $S,$ $Z$ are concyclic, hence we conclude that $\measuredangle BAP = \measuredangle BZS = \measuredangle AZS - \measuredangle AZB = \measuredangle ADB - \measuredangle ACB = \measuredangle DAC.$

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#19 h Jan 24, 2016, 3:49 AM • 7 Y

Y by buratinogigle, quangMavis1999, Omez, AlastorMoody, Adventure10, Mango247, nguyenducmanh2705

Generalization: The incircle $(I)$ of $\triangle ABC$ touches $BC,CA,AB$ at $D,E,F$ and let $\mathcal{H}_A$ be the conic through $A,D,E,F$ and the isogonal $T \equiv X_{57}$ of the Mittenpunkt of $\triangle ABC.$ $J$ is an arbitrary point on $\mathcal{H}_A$ and $JF,JE$ cut $(I)$ again at $N,M.$ If $P \equiv BM \cap CN,$ then $AP,AD$ are isogonals WRT $\angle BAC.$
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Lemma: $\triangle ABC$ is acute and $\triangle A'B'C'$ is its tangential triangle. $\omega_A$ is the circle pasing through $B',C'$ tangent to $\odot(ABC)$ at $V.$ Then $AV$ is the isogonal of the isotomic of the A-altitude of $\triangle ABC.$

Let $M,N,L$ be the midpoints of $BC,CA,AB$ and let $D$ be the foot of the A-altitude. Inversion WRT $\odot(ABC)$ takes $B',C'$ into $N,L,$ hence by conformity $\omega_A$ is transformed into the circle through $N,L$ and tangent to $\odot(ABC)$ at $V.$ From IMO Shortlist 2011 (G4), $VD$ passes through the centroid $G$ of $\triangle ABC.$ Thus if $U \equiv VDG \cap NL,$ then $(UM \parallel AD) \perp BC,$ so by symmetry $AU$ is the isotomic of $AD;$ reflection of $UD$ WRT $UM.$ Thus by obvious symmetry $AU$ cuts $\odot(ABC)$ again at the reflection of $V$ across $UM$ $\Longrightarrow$ $AU$ is the isogonal of $AV.$
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Back to the main problem. Let $\triangle D'E'F'$ be the antimedial triangle of $\triangle DEF.$ The circle through $B,C$ tangent to $(I)$ touches this at $T_A$ $\Longrightarrow$ $T \in DT_A$ (well-known), so applying the previous lemma in the acute $\triangle DEF,$ we deduce that $T$ is the isogonal conjugate of the isotomic conjugate of the orthocenter of $\triangle DEF,$ i.e. isogonal conjugate of the symmedian point $K'$ of $\triangle D'E'F'$ WRT $\triangle DEF.$ Thus since $D'$ is the isogonal conjugate of $A$ WRT $\triangle DEF,$ then it follows that $D'K'$ is the isogonal of $\mathcal{H}_A$ WRT $\triangle DEF.$ As $D'K'$ is parallel to the D-symmedian $DA$ of $\triangle DEF,$ then $\mathcal{H}_A$ cuts $(I)$ again at $X,$ such that $DA,DX$ are isogonals WRT $\triangle DEF,$ i.e. $DX$ is the D-median of $\triangle DEF$ $\Longrightarrow$ $AX$ is the isogonal of $AD$ WRT $\angle BAC.$

On the other hand let $Q \equiv FN \cap EM.$ Then
$F(E,D,X,N)=E(A,D,X,N)$ and similarly $E(F,D,X,M)=F(A,D,X,M).$ But $E(A,D,X,N)=E(A,D,X,J)=F(A,D,X,J)=F(A,D,X,M)$ $\Longrightarrow$ $F(E,D,X,N)=E(F,D,X,M)$ $\Longrightarrow$ $D,X,Q$ are collinear, i.e. $\triangle XMN$ is the circumcevian triangle of $Q$ WRT $\triangle DEF,$ so by Steinbart theorem $P \in AX$ $\Longrightarrow$ $AP,AD$ are isogonals WRT $\angle BAC.$

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