## viernes, 24 de abril de 2020

### 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.

## jueves, 23 de abril de 2020

### 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.

## sábado, 18 de abril de 2020

### 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.

## domingo, 12 de abril de 2020

### 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.

$\square$

## sábado, 11 de abril de 2020

### 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.

## martes, 7 de abril de 2020

### 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:

## sábado, 4 de abril de 2020

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!

## miércoles, 1 de abril de 2020

### 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).