Selectivo Argentina Cono Sur - Problema 3

Sea $\triangle{ABC}$  un triángulo e $I$ el punto de intersección de sus bisectrices. Sea $\tau$ la circunferencia con centro $I$ que es tangente a los tres lados del triángulo y sean $D$ en $BC$ y $E$ en $AC$ los puntos de tangencia de $\tau$ con $BC$ y $AC$. Sean $M$ y $N$ los puntos medios de $BC$ y $AB$, respectivamente. Demostrar que $AI$, $DE$ y $MN$ concurren en un punto común. 

Demostración. Supongamos que $ED$ y $MN$ se cortan en $P$. Si llamamos $X$ a la intersección de $EF$ con $MN$, donde $F$ es el punto de contacto de $\tau$ con $AB$, sabemos que $BXFD$ es cíclico y $NX=NB$ (ver demostración aquí). Claramente $\triangle{FMX}\sim{\triangle{AEF}}$ con $\angle{FXP}=\angle{AFE}=\angle{FDP}$, implicando que $P$ yace sobre $(BXFD)$. Al ser $\triangle{NXB}$ isósceles, $\angle{PFB}=\angle{PXB}=\frac{180^\circ-\angle{ACB}}{2}$. Dicho esto, no es difícil darse cuenta que $FP=EP$, significando que $PFAE$ es un deltoide y, como consecuencia, $AP$ es la bisectriz de $\angle{FAE}$. Esto demuestra que $MN$, $ED$ y $AI$ son concurrentes y hemos terminado.

Un lema sobre segmentos congruentes

Considera un triángulo $\triangle{ABC}$. Los puntos $E$ y $F$ son los puntos donde el incírculo toca los lados $AB$ y $AC$, respectivamente. $M$ y $N$ son los puntos medios de los lados $AB$ y $BC$, respectivamente. Llamemos $P$ es la intersección de $EF$ y $MN$.

Lema.
a) $BN=CN=NP$.
b) $CP$ es la bisectriz de $\angle{ACB}$.
c) $BPEID$ es cíclico ($I$ es el incentro de $\triangle{ABC}$).

Demostración a). Denotemos con $s$ el semiperímetro de $\triangle{ABC}$. Entonces tenemos

$$s=AE+BC$$

$$\frac{AB}{2}+\frac{BC}{2}+\frac{AC}{2}=AE+BC$$

$$\frac{AB}{2}+\frac{AC}{2}-AE=\frac{BC}{2}$$

Pero $\frac{AB}{2}-AE=EM$ y $\frac{AC}{2}=MN$. Además, es fácil notar que $\triangle{AEF}$ y $\triangle{EMP}$ son triángulos isósceles semejantes con $EM=MP$, de modo que  $\frac{AB}{2}+\frac{AC}{2}-AE=\frac{BC}{2}$ puede reescribirse como $MP+MN=\frac{BC}{2}$ y hemos terminado.

Demostración b). Note que $AC$ es paralela con $MN$ y $\triangle{PNC}$ es isósceles, de donde resulta que  $\angle{NCP}=\angle{NPC}=\frac{180^\circ-(180-\angle{ACB})}{2}=\frac{\angle{ACB}}{2}$. 

Demostración c). El cuadrilátero $BEID$ es cíclico puesto que es un deltoide recto. Note que como consecuencia de a) y b) $\angle{EPI}=\angle{EPM}-\frac{\angle{ACB}}{2}=\frac{180^\circ-\angle{BAC}}{2}-\frac{\angle{ACB}}{2}=\frac{\angle{ABC}}{2}$. Pero $\angle{EBI}=\frac{\angle{ABC}}{2}$, por lo tanto, $BPEID$ es cíclico.

Another Tran Viet Hung's problem

Lemma 1. Let $CD$ cuts circle $(D, DB)$ in $G$. Then, $BGIC$ is cyclic.

Proof. Notice that $\angle{BDC}=\angle{BAC}$, so 

$$\angle{GBI}=\frac{\angle{ABC}}{2}-(\angle{DBG}-\angle{DBA})=$$ 
$$\angle{GBI}=\frac{\angle{ABC}}{2}-(90^\circ-\frac{\angle{BAC}}{2})+\angle{DCA}.$$

Replacing $90^\circ$ by $\frac{\angle{ABC}}{2}+\frac{\angle{BAC}}{2}+\frac{\angle{BCA}}{2}$ and simplifying, we get $\angle{GBI}=\angle{DCA}-\frac{\angle{BCA}}{2}$. But $\angle{DCA}=\angle{BCA}-\angle{BCD}$, thus, $\angle{GBI}=\frac{\angle{BCA}}{2}-\angle{BCD}$. Moreover, $\angle{DCI}=\angle{GCI}=\frac{\angle{BCA}}{2}-\angle{BCD}$, hence,$\angle{GBI}=\angle{GCI}$ implying $BGIC$ is cyclic.

$\square$

Lemma 2. $IE=IG$.

Proof. Notice that $\angle{BEG}=\frac{\angle{BDG}}{2}=\frac{\angle{BAC}}{2}=\angle{BAI}$ which means that $EG\parallel{AI}$ implying that $ID\perp{EG}$ and as $DI$ passes through the center of circle $(D, DB)$ we deduce $DI$ is the perpendicular bisector of $EG$ meaning $IE=IG$.

$\square$

Back to our main problem.

$$\angle{IEG}=\angle{IGE}=360^\circ-\angle{EGF}-\angle{DGB}-\angle{BGC}-\angle{CGI}.$$

Combining lemmas 1 and 2 we get  $\angle{EGF}=\frac{\angle{BAC}}{2}+\angle{DCA}$; $\angle{DGB}=90^\circ-\frac{\angle{BAC}}{2}$; $\angle{BGC}=\angle{BIC}=180^\circ-\frac{\angle{ABC}}{2}-\frac{\angle{BCA}}{2}$; $\angle{CGI}=\frac{\angle{ABC}}{2}$.

Replacing and simplifying we get $\angle{IEG}=90^\circ+\angle{BCD}-\frac{\angle{BCA}}{2}$. Finally, $\angle{FEI}+\angle{FCI}=180^\circ+\angle{BCD}-\frac{\angle{BCA}}{2}+\frac{\angle{BCA}}{2}-\angle{BCD}=180^\circ$ and we are done.

Solution to problem 4975 in Romantics of Geometry

This problem was proposed by Tran Viet Hung (Vietnam). Here is my proof. 

Proof for a). 

Perform an inversion around circle, $w$, centred at $A$ with radius $AE=AD$. The incircle is fixed since it is orthogonal with $w$. Touch points $E$ and $D$ are also fixed since they are on the circumference of $w$. The inverted image of $I$ is the intersection of $AI$ with $ED$, denoted $I'$. $B$ and $C$ are transformed into $B'$ and $C'$ on lines $AB$ and $AC$, respectively. The circumcircle $(ABC)$ is sent to the line passing through $B'$, $C'$. Line $BC$ is sent to the circle $(AB'C')$ and is tangent to the incircle at $P'$, the inverted image of $P$. The point $M$ is sent to $M'$, the intersection of $AI$ and $B'C'$. The line $MP$ is sent to the circle passing through $A$, $M'$ and $P'$. Finally, the inverted image of $Q$ is the second intersection of line $B'M'$ with $(AM'P')$ (See the inverted diagram above).

As angles are preserved under inversion, we want to show that $Q'$ lies on $E'D'$, taking advantage of the fact that $\angle{AI'E'}=90^\circ$. Let's supose that $E'D'$ cuts $B'C'$ in $Q''$. It suffices to show $AM'P'Q''$ is cyclic. Indeed, if $N$ is the intersection of $AI'$ with arc $B'C'$,  then, $N$ is the midpoint of arc $B'C'$, and it is known that $Q''$, $P'$ and $N$ are collinear (notice that the incircle turns out to be the $A$-mixtilinear excircle of $\triangle{AB'C'}$). A simple angle chase shows that $\angle{M'AP'}=\angle{M'Q''P'}$, implying $AM'P'Q''$ is cyclic and that $Q''=Q'$, as desired.

$\square$

Proof for b).

Under the same inversion we perfom in part a), the $A$-mixtilinear excircle is sent to the incircle of $\triangle{AB'C'}$ which touches $B'C'$ in $X''$, the inverted image of $X'$. The intersection of $AI$ with $BC$ is sent to $N$, the same midpoint of arc $B'C'$ (not containing $A$) we described in part a). It suffices to show that $Q'NX''A$ is cyclic. Indeed, we proved already that $Q'$ lies on $B'C'$, so if we prove that $\angle{X''Q'N} = \angle{X''AN}$ we are done. But it is well known that $AX''$, $AP'$ are isogonal and $Q'$, $P'$ and $N$ are collinear, so $\angle{X''AN} = \angle{M'AP'} = \angle{M'Q'P'} = \angle{X''Q'N}$, implying that $Q'NX''A$ is cyclic, as desired.

$\square$

Collinearity in a Mixtilinear Configuration

Consider a triangle $\triangle{ABC}$, its Incircle, $\psi$, and its $B$-mixtilinear incircle, $\omega$. Let $T$ be the point of tangency of $\psi$ and $\omega$. Let $K$ and $L$ be the points of tangency of $\omega$ with the sides $AB$ and $BC$, respectively. Call $D$ the point of tangency of $\psi$ with $AC$. Call $E$ the  second intersection (further to $B$) of $\psi$ with $BT$. Call $F$ the second intersection of the circumcircle of $\triangle{EDI}$ with $BT$. Prove that $K$, $F$ and $I$ are collinear. 

Proof. $\angle{ATI}=\angle{CTI}$ and $\angle{ATB}=\angle{CTD}$ (well-known). We have $\angle{BTD}=\angle{ATC}-\angle{ATB}-\angle{CTD}$. Notice that $\angle{ATB}=\angle{CTD}=\angle{ACB}$ and $\angle{ATC}=180^\circ-\angle{ABC}$, then, $\angle{BTD}=\angle{BAC}-\angle{ACB}$. Moreover, $TI$ is an angle bisector of $\angle{BTD}$, therefore, $D$ is the reflection of $E$ around $TI$, which means $TE=TD$. Said this, and having in mind that $EDIF$ is cyclic, we find $\angle{DIF}=\frac{\angle{ABC}}{2}+\angle{ACB}$.

On the other hand, $K$, $I$ and $L$ are collinear (well-known), consequently, $\angle{AKL}=\angle{AKI}=90^\circ+\frac{\angle{ABC}}{2}$. Finally, some angle chase in quadrilateral $AKID$ give us $\angle{DIK}= \frac{\angle{ABC}}{2}+\angle{ACB}$. As $\angle{DIF}=\angle{DIK}$, we are done.

An appearance of harmonic division

Consider a semicircle $(AB)$ and a circle, $w$, tangent internally at $T$. Let $N$ be the point of tangency of $w$ and the diameter, $AB$. If $O$ is the center of $(AB)$ and $C$ is the midpoint of arc $AB$, let $F$ be the intersection of a parallel line to $CT$ passing through $D$ with $CO$. Denote $P$ the intersection of $CN$ with $DF$. Prove that $DP=FP$. (Carlos Hugo Olivera Diaz)

Proof. It is a well-known fact that $O$, $D$ and $T$ are collinear, so we have that $\triangle{OFD}\sim{\triangle{OCT}}$, as a consequence, $DT=CF=DN=r$, where $r$ is the radius of $w$. Since $CO\parallel{DN}$, $\angle{FCP}=\angle{DNP}$. Also, $\angle{FPC}=\angle{NPD}$, hence $\triangle{CFP}\cong{\triangle{DNP}}$ and the proof is complete. 

$\square$

Remark 1. If $CN$ meet $OT$ at $H$, then, $(O, D; H, T)=-1$.

Proof 1. If we project from $C$ onto the line $OT$, we have $-1=(F, D; P, P_\infty)=(O, D; H, T)$. 

$\square$

Proof 2. Let $N'$ be the second intersection of $ND$ with $w$. Then, from the first lemma in the Archimedes's Book of Lemmas, we know that $C$, $N'$ and $T$ are collinear. As $D$ is the midpoint of $NN'$ and $NN'$ is parallel to $CO$, if we project from $C$ onto the line $OT$, $-1=(N', N; D, P_{\infty})=(T, H; D, O)$.

$\square$

Notice that the original problem can also be proven using harmonic division. Indeed, projecting from $C$ onto the line $DF$, we have $-1=(N', N; D, P_{\infty})=(P_{\infty}, P, D, F)$, which means $P$ is the midpoint of $DF$.

External version.  Here $D'$ is the reflection of $D$ around $N$. Let $H$ be the intersection of $OD$ with $CD'$, then, $(O, T; H, D)=-1$.

Proof. As mentioned, it is known that $O$, $T$ and $D$ are collinear. Also, because of the external version of Archimedes's lemma, $C$, $T$ and $N$ are collinear. As $CO\parallel{DD'}$, if we project from $C$ onto the line $OD$, $-1=(D, D'; N, P_{\infty})=(D, H; T, O)$.

$\square$

Remark 2. In this case, let $H$ be the intersection of $CT$ with $AB$. Then, $(A, B; N, H)=-1$. (See problems 2 and 3 for applications)

Proof. From Archimedes's lemma we know that $C'$, $N$ and $T$ are collinear. The quadrilateral $ACBC'$ is harmonic, a square specifically, so if we project from $T$ onto the line $AB$ we have $-1=(A, B; C', C)=(A, B; N, H)$. 

$\square$

In general, if $N$ and $H$ are the intersections of $C'T$ with $AB$ and $CT$ with $AB$, respectively, $AB$ could be any chord perpendicular to $CC'$ and $T$ be any point on the circle $(ACBC')$ and the result still holds. This is because $ACBC'$ is always a harmonic quadrilateral (e.g. a square or a kite).

Applications.

Problem 1. Let $w$ and $v$, be two circles tangent internally at $T$. Let $N$ be the point of tangency of the smaller circle, $v$, and the diameter of $w$, $AB$. If $C$ is the midpoint of arc $AB$ (above $AB$), let's $C'$ be its antipode. Also, let $N'$ be the antipode of $N$. If $O$ is the center of circle $w$, prove that $CN$, $C'N'$ and $OT$ are concurrent at $P$.

Proof. Let's supose $CN$ meets $OT$ at $P$ and let $D$ be the center of $v$. From the remark discussed previously, we know that $(O, D; P, T)=-1$. We also know from Archimedes's lemma that $C$, $N'$ and $T$ are collinear. So if we project from $N'$ onto the line $CO$ we have that $T$ goes to $C$; $O$ remains the same and $D$ goes to the point at infinity, hence, $N'$, $P$ and  $C'$ must be collinear and the proof is complete.

The following is problem 310 in Gogeometry. I have provided two more proofs here (in Spanish).

Problem 2. Let $w$ and $v$, be two circles tangent internally at $T$. Let $N$ be the point of tangency of the smaller circle, $v$, and the diameter of $w$, $AB$. Prove that $\angle{ATN}=45^\circ$.

Proof. Let $C$ be the midpoint of arc $AB$ (containing $T$) and let $C'$ be its antipode. Because of Archimedes's lemma $C'$, $N$ and $T$ are collinear. If $H$ is the intersection of $CT$ with $AB$, from remark 2 we know that $(A, B; N, H)=-1$. Also, notice that $\angle{CTC'}=\angle{NTH}=90^\circ$, since $CC'$ is a diameter. Therefore, $NT$ is an angle bisector of $\angle{ATB}=90^\circ$ and the proof is complete.

Problem 3. Let $D$ be the center of a circle inscribed in a semicircle $(AB)$. Let $T$ and $N$ be the points where the circle touches $(AB)$ and $AB$, respectively. Call $C$ the midpoint of arc $AB$. Prove that $AC$, $ND$ and $BT$ are concurrent at a common point. 

Proof. Let $H$ be the intersection of $CT$ and $AB$ and $P$ the intersection of $AC$ and $BT$. From the remark 2 we know that $(A,B;N,H)=-1$. This implies that $AT$, $BC$ and $NP$ are concurrent at a common point. Since $\angle{ACB}=\angle{BTA}=90^\circ$, the segment $NP$ must be an altitude of $\triangle{ABP}$. As $\angle{BND}=\angle{BNP}=90^\circ$, we conclude that $AC$, $ND$ and $BT$ are concurrent at a common point and the proof is complete. 

Related material:

Harmonic Divisions (Lemmas)

Lozada's cyclologic triangles

Let $ABC$ be a triangle, $P$ a point, $A'B'C'$ the antipedal triangle of $P$ and $A''$, $B''$, $C''$ the reflections of $P$ in $A$, $B$, $C$, respectively. Then, for any $P$ distinct from the Incenter of $ABC$, $A'B'C'$ and $A''B''C''$ are cyclologic. (César E. Lozada)

Proof. Notice that as $B''$ and $A''$ are reflections around $A'C'$ and $B'C'$, respectively, $\angle{BC'P}=\angle{BC'B''}$ and $\angle{AC'P}=\angle{AC'A''}$. Since $\angle{BC'P}+\angle{AC'P}=\angle{A'C'B'}$ it follows that $\angle{B''C'A''}=2\angle{A'C'B'}$. Analogously, $\angle{B''A'C''}=2\angle{C'A'B'}$ and $\angle{C''B'A''}=2\angle{A'B'C'}$. 

Supose $Q$ is the second intersection of circles $(A'C''B'')$ and $(B'A''C'')$, then, $\angle{A''QB''}=\angle{C''QB''}-\angle{C''QA''}$. But $$\angle{C''QB''}=\angle{C''A'B''}=2\angle{B'A'C'}$$ and $$\angle{C''QA''}=180^\circ-2\angle{A'B'C'},$$ so $$\angle{A''QB''}=2\angle{B'A'C'}+2\angle{A'B'C'}-180^\circ.$$ Now,  $\angle{A''QB''}+\angle{B''C'A''}=2\angle{B'A'C'}+2\angle{A'B'C'}-180^\circ+2\angle{A'C'B'}=180^\circ$. Which means that $QB''C'A''$ is cyclic. 

$\square$

The concurrency of circles $(A'B'C'')$, $(A'C'B'')$ and $(B'C'A'')$ is guaranteed by the fact that the cyclologic theorem is symmetric. 

Remark: $Q$ lies on the circumcircle of $A'B'C'$.

Proof: The facts that $ABC$ is homothetic with $A''B''C''$ and $A'CPB$ and $B'CPA$ are cyclic quadrilaterals lead us to infer that

$$\angle{A'QB''}=\angle{A'C''B''}=(90^\circ-\angle{CA'P})-\angle{BA'P}=90^\circ-\angle{B'A'C'}.$$ 

Similarly,

$$\angle{A''QB'}=\angle{A''C''B'}=(90^\circ-\angle{CB'P})-\angle{AB'P}=90^\circ-\angle{A'B'C'}.$$

We already know that $\angle{A''QB''}=2\angle{B'A'C'}+2\angle{A'B'C'}-180^\circ$. As $\angle{A'QB'}=\angle{A'QB''}+\angle{B''QA''}+\angle{A''QB'}$, it follows that

$$\angle{A'QB'}+\angle{A'C'B'}=\angle{B'A'C'}+\angle{A'B'C'}+\angle{A'C'B'}=180^\circ.$$

$\square$ 

Related material: Cyclologic triangles!

Cyclologic triangles!

Two triangles $A_1B_1C_1$ and $A_2B_2C_2$ are Cyclologic if the circles $(A_1B_2C_2)$, $(B_1A_2C_2)$, and $(C_1A_2B_2)$ are concurrent in a common point. The point of concurrence is known as the Cyclologic center of $A_1B_1C_1$ with respect to $A_2B_2C_2$. If this is the case, then the circles $(A_2B_1C_1)$, $(B_2A_1C_1)$, $(C_2A_1B_1)$ also will be concurrent. The point of concurrence is known as the Cyclologic center of $A_2B_2C_2$ with respect to $A_1B_1C_1$.

Proposition. Let $ABC$ be a triangle. Consider two points, $P$ and $Q$, in the plane of $ABC$. Let $P_aP_bP_c$ and $Q_aQ_bQ_c$ be the pedal triangles of $P$ and $Q$, respectively.  Let $X$, $Y$ and $Z$ be the orthogonal projections of $P$ onto the sides $Q_aQ_b$, $Q_bQ_c$ and $Q_aQ_c$, respectively. Then, $P_aP_bP_c$ and $XYZ$ are cyclologic triangles. In other words, the circles $(P_bP_cY)$, $(P_aP_cZ)$ and $(P_aP_bX)$ are concurrent at a point, and so are circles $(P_cYZ)$, $(P_bXY)$ and $(P_aXZ)$.

Proof. Supose $G$ is the second intersection of circles $(P_bP_cY)$ and $(P_cZP_a)$, then, 

$$\angle{P_aGP_b}+\angle{P_bXP_a}=\angle{P_cGP_a}-\angle{P_cGP_b}+\angle{P_bXP}+\angle{PXP_a}.$$

Notice that $\angle{P_cGP_a}=\angle{P_cZP_a}$ and $\angle{P_cGP_b}=\angle{P_cYP}+\angle{PYP_b}$. By supplementary angles it is easy to realize that $PP_cQ_cZY$, $PZP_aQ_aX$ and $PYXQ_bP_b$ are cyclic pentagons. So, $\angle{P_cZP}=\angle{P_cYP}$,  $\angle{PYP_b}=\angle{P_bXP}$ and $\angle{P_aZQ_a}=\angle{Q_aXP_a}$. This allows us to re-write the above expression as follows

$$\angle{P_cZP_a}-\angle{P_cYP_b}+\angle{P_bXP_a}=$$

$$\angle{P_cZP}+90^\circ+\angle{P_aZQ_a}-\angle{P_cYP}-\angle{PYP_b}+\angle{P_bXP}+90^\circ-\angle{P_aZQ_a}=180^\circ.$$

Which means that $GP_aXP_b$ is cyclic, so done.

The cyclologic center of $XYZ$ with respect to $P_aP_bP_c$ is clearly $P$.

Remark: the triangle $Q_aQ_bQ_c$ can be arbitrary inscribed in $ABC$. The fact that $Q_aQ_bQ_c$ is a pedal triangle was never used in the proof.

Some more properties of this configuration can be found in the ETC: Cyclologic centers: X(37743) - X(37744).

See also Lozada's cyclologic triangles.