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Source: ELMO Shortlist 2023 G6

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1740 posts

#1 h Jun 29, 2023, 1:40 AM • 1 Y

Y by ImSh95

Let $ABCDEF$ be a convex cyclic hexagon such that quadrilateral $ABDF$ is a square, and the incenter of $\triangle ACE$ lines on $\overline{BF}$ . Diagonal $CE$ intersects diagonals $BD$ and $DF$ at points $P$ and $Q$ , respectively. Prove that the circumcircle of $\triangle DPQ$ is tangent to $\overline{BF}$ .

Proposed by Elliott Liu

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DottedCaculator

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DottedCaculator

#2 h Jun 29, 2023, 1:50 AM • 1 Y

Y by ImSh95

Let $a=1$ , $b=i$ , $d=-1$ , and $f=1$ . Let $c=x^2$ and $e=y^2$ . Then, $-x-y-xy$ is imaginary, so $x^2y^2+x^2y+xy^2+x+y+1=0$ . We have
$\begin{align*} p+ce\overline p&=c+e\\ p+bd\overline p&=b+d\\ p+x^2y^2\overline p&=x^2+y^2\\ p-i\overline p&=i-1\\ \overline p(x^2y^2+i)&=x^2+y^2+1-i\\ \overline p&=\frac{x^2+y^2+1-i}{x^2y^2+i}\\ p&=\frac{x^2+y^2+(1-i)x^2y^2}{1+ix^2y^2}.\\ \end{align*}$
Similarly, $q=\frac{x^2+y^2+(1+i)x^2y^2}{1-ix^2y^2}$ . Therefore, the midpoint of $PQ$ is $\frac{x^2+y^2+x^2y^2-x^4y^4}{x^4y^4+1}.$ We claim that the circumcircle of $DPQ$ is tangent to $BF$ at the incenter $J$ of $ACE$ .

We have $j=-x-y-xy$ , so
$\frac{p+q}2-j=\frac{x^2+y^2+x^2y^2-x^4y^4+x^5y^4+x^4y^5+x^5y^5+x+y+xy}{x^4y^4+1}.$ Since $x^2y^2+x^2y+xy^2+x+y+1=0$ , we get $x^5y^5+x^5y^4+x^4y^5+x^4y^3+x^3y^4+x^3y^3=0$ , so $\frac{p+q}2-j=\frac{x^2y^2+x^2+y^2+xy+x+y-x^4y^4-x^4y^3-x^3y^4-x^3y^3}{x^4y^4+1}$ . Since $x^4y^4+x^4y^3+x^3y^4+x^3y^2+x^2y^3+x^2y^2=0$ , we get $\frac{p+q}2-j=\frac{x^2y^2+x^2+y^2+xy+x+y+x^3y^2+x^2y^3+x^2y^2-x^3y^3}{x^4y^4+1}$ .

Since $\frac{p-q}2=\frac{-ix^2y^2(x^2+1)(y^2+1)}{x^4y^4+1}$ , we get that the magnitude of $\frac{p-q}2$ is equal to $\frac{xy(x^2+1)(y^2+1)}{x^4y^4+1}$ since this expression is real.
We need to show
$\begin{align*} j-\frac{p+q}2&=\left|\frac{p-q}2\right|\\ -\frac{x^2y^2+x^2+y^2+xy+x+y+x^3y^2+x^2y^3+x^2y^2-x^3y^3}{x^4y^4+1}&=\frac{xy(x^2+1)(y^2+1)}{x^4y^4+1}\\ x^2y^2+x^2+y^2+xy+x+y+x^3y^2+x^2y^3+x^2y^2-x^3y^3&=-(x^3y^3+xy^3+x^3y+xy)\\ x^2y^2+x^2+y^2+2xy+x+y+x^3y^2+x^2y^3+x^2y^2+xy^3+x^3y&=0\\ (x+y)(x^2y^2+x^2y+xy^2+x+y+1)&=0. \end{align*}$
Therefore, $\frac{p+q}2-j$ is real and equal in magnitude to the radius of the circumcircle of $DPQ$ , so the circumcircle of $DPQ$ is tangent to $BF$ at $J$ .

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1594 posts

#3 h Jun 29, 2023, 11:30 AM • 2 Y

Y by GeoKing, ImSh95

Fun radical axis solution.

Let $\Omega$ be the circumcircle of $ABCDEF$ . Extend $AI$ to meet $\Omega$ again at $M$ , and reflect $I$ over $MB$ and $MF$ to obtain points $B_1$ and $F_1$ , respectively.

Claim: $M, D, B_1, F_1$ are colinear.

Proof. We have $\angle IMF_1 = 2\angle IMF = 2\angle AMF = 90^\circ$ . Similarly, $\angle IMB_1 = 90^\circ$ . Moreover, $\angle IMD = \angle AMD =90^\circ$ . Thus, $M, D, B_1, F_1$ all lie on a line perpendicular to $IM$ . $\blacksquare$

Claim: $C, I, E, B_1, F_1$ are concyclic on $\omega$ .

Proof: By Fact 5, $MC=MI=ME$ . Moreover, since $\triangle IB_1F_1$ is right isosceles with $IM$ being the altitude, we must have $MI=MB_1=MF_1$ , done. $\blacksquare$

Claim: $B,B_1,I,D$ are concyclic.

Proof: Follows from $\angle IB_1D = 45^\circ = \angle DBI$ . $\blacksquare$

Claim: $I, P, B_1$ are colinear. (Similarly, $I,Q,F_1$ are colinear.)

Proof: By radical center theorem on $\odot(IBB_1D)$ , $\Omega\equiv\odot(BMCE)$ , and $\omega\equiv\odot(IB_1CE)$ , lines $IB_1$ , $BM$ , $CE$ are concurrent. This point must be $P$ . $\blacksquare$

Thus, $\angle PIQ = \angle B_1IF_1 = 90^\circ$ , implying that $I\in\odot(PDQ)$ . To finish the problem, just notice that
$\angle BIP = \angle BB_1I = \angle BDI = \angle PDI.$

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#5 h Jun 30, 2023, 2:15 PM • 2 Y

Y by GeoKing, ImSh95

Let $I$ be incenter of $ACE$ which lies on $BF$ . Let $AI$ meet $ABCDEF$ at $T$ . Note that $TC=TE=TI$ . Let circle with center $T$ and radius $TI$ meet $DT$ at $X,Y$ such that $T$ lies between $X,D$ .
Claim $: IBXD$ and $IFYD$ are cyclic.
Proof $:$ Note that $\angle IXD = \angle IXT = \angle 45 = \angle IBD$ . we prove the other one with same approach.
Now using Radical Axis on circles $ABCDEF$ and $IBXD$ and $ICXE$ we have $BD,CE,IX$ are concurrent so $IX$ passes through $P$ . using same approach we have $IY$ passes through $Q$ . now since $\angle PIT = \angle XIT = \angle 45 = \angle ITB$ we have that $IP \perp BT$ and with same approach $IQ \perp FT$ so $\angle PIQ = \angle 180 - \angle FTB = \angle 90$ so $DPIQ$ is cyclic and $\angle BIP = \angle BIX = \angle BXI = \angle BDI = \angle PDI$ so $DPIQ$ is tangent to $BF$ .

This post has been edited 1 time. Last edited by Mahdi_Mashayekhi, Jun 30, 2023, 2:38 PM

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515 posts

#6 h Jun 30, 2023, 3:35 PM • 2 Y

Y by ImSh95, LoloChen

Sol:-Let $O$ be the midpoint of $AD$ , $I$ be the incenter of $ACE$ , let $AI$ meet $(ACE)$ again at $M$ . $A'$ be the reflection of $A$ across $I$ .Since $IA'=ID$ and $IO$ is external angle bisector of $\angle A'ID$ , $BF$ is tangent to $(A'ID)$ .

https://cdn.discordapp.com/attachments/961202142809563140/1124345498556436531/image.png

Let $DM$ meet $(A'ID)$ again at $J$ .
Step 1Line $EC$ is a diameter of $(A'ID)$
Proof $IO \parallel A'D$ so $\measuredangle AIO=\measuredangle IA'D=\measuredangle IJM$ . $90^\circ=\angle AOI=\angle AMD=\angle IMJ$ .Hence $\Delta AOI \sim \Delta IMJ$ . Since $\angle MIJ=\angle MAD$ , $IJ$ is parallel to isogonal of $AO$ i.e. $A-$ altitude $\implies IJ \perp CE$ . Let $r,R$ be the inradius and circumradius of $ACE$ . $\frac{IJ}{IM}=\frac{AI}{AO}=\frac{AI}{R} \implies IJ \cdot R=AI \cdot IM=2rR \implies IJ=2r$ . So $J$ is reflection of $I$ across $CE \implies$ line $EC$ is a diameter of $(A'ID)$ .

https://cdn.discordapp.com/attachments/961202142809563140/1124355608922038443/image.png

Let $(A'DI)$ meet $CE$ at $P',Q'$ and $AD$ meet $(A'ID)$ again at $T$ .
Step 2 $A'P'Q'T'$ is a square.
Proof Since $\angle A'DT=90^\circ$ , $A'T$ is diameter of $(A'DI)$ . $\angle IMD=90^\circ=\angle IOD$ so $IODM$ is cyclic. By reims theorem $MO \parallel A'T$ so $A'T \perp P'Q'$ . Since $A'T$ and $P'Q'$ are $2$ perpendicular diameters so $A'P'Q'T'$ is a square.

https://cdn.discordapp.com/attachments/961202142809563140/1124360478353784842/image.png

Step 3 $P'=P$ and $Q'=Q$ .
Proof $\angle TDP'=45^\circ=\angle TDB$ . So $T,B,P'$ are collinear and hence $P'=P$ and similarly $Q'=Q$ .

https://cdn.discordapp.com/attachments/961202142809563140/1124362336606617720/image.png

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#7 h Jul 4, 2023, 6:18 AM • 1 Y

Y by GeoKing

We will prove a slightly more generalized problem, where $ABDF$ is a kite (with $A,D$ symmetrical wrt. $BF$ ) instead of a square.
Here are two solutions I found.
Sol 1, Inversion
Sol 2, Tangents to the incircle

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407 posts

#8 h May 24, 2024, 6:23 AM

Y by

Denote by $I$ and $M$ the incenter of triangle $ACE$ and the second intersection of $AI$ at the circumcircle of triangle $ACE$ . Let $A=i$ , $B=-1$ , $D=-i$ , $F=1$ , $M=m$ . By chord intersection formula $I = \frac{m+i}{mi+1}$ . As $MI = MC = ME$ , then $c, e$ are the roots of
$(m-z)(1/m-1/z)=(m-I)(1/m-I).$ Hence, from Vieta's, $ce=m^2$ and $c+e=-m((m-I)(1/m-I)-2)=m+I+m^2I-mI^2$ . Hence, by the chord intersection formula:
$p-I=\frac{bd(c+e)-ce(b+d)}{bd-ce}-I=\frac{i(m+I+m^2I-mI^2)+m^2(1+i)}{i-m^2}-I=\frac{m(1+I)(m+mi+i-iI)}{i-m^2}.$ $q-I=\frac{ce(d+f)-df(c+e)}{ce-df}-I=\frac{m^2(1-i)+i(m+I+m^2I-mI^2)}{m^2+i}-I=\frac{m(1-I)(m-mi+i+iI)}{m^2+i}.$ The idea is to show that $\frac{p-I}{q-I}$ is pure imaginary, implying that $DPIQ$ is cyclic. However, this is obvious as $\frac{1+I}{1-I}$ is real and
$\frac{m+mi+i-iI}{m-mi+i+iI}=\frac{m+mi+i-\frac{mi-1}{mi+1}}{m-mi+i+\frac{mi-1}{mi+1}}=\frac{i(m^2-i)}{m^2+i}.$ What remains to be shown is that the circumcircle of triangle $DPQ$ touches $BF$ at $I$ , which is equivalent to $p+q-2I$ being pure imaginary:
$\begin{align*} p+q-2I&=\frac{m(1+I)(m+mi+i-iI)}{i-m^2}+\frac{m(1-I)(m-mi+i+iI)}{m^2+i}\\ &=\frac{m(1+I)(m+mi+i-iI)}{i-m^2}+i\frac{m(1-I)(m+mi+i-iI)}{i-m^2}\\ &=\frac{m(m+mi+i-iI)}{i-m^2}\cdot(1+I+i-iI)\\ &=\frac{m}{(i-m^2)(mi+1)^2}\cdot((mi+1)(m+mi+i)-(mi-1))((mi+1)+(m+i)+i(mi+1)-i(m+i))\\ &=\frac{m}{(i-m^2)(mi+1)^2}\cdot(i-1)(m^2-i)(2+2i)\\ &=\frac{4m}{(mi+1)^2} \end{align*}$ The last expression shows that $\overline{(p+q-2I)}=-(p+q-2I)$ , so we're done.

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1471 posts

#9 h Mar 27, 2025, 2:28 AM

Y by

Let the circumcircle of $ABCDEF$ be $\omega$ and consider $AI \cap \omega=N$ also consider reflections of $I$ over $NF,BB$ be $I_1,I_2$ respectively then due to reflection angles (square is OP) and I-E Lemma we easly get $I_1,M,D,I_2$ colinear and $I_1EICI_2$ cyclic wtih center $M$ .
Now because arcs $FA, AB$ are equal we have $NI$ bisects $\angle FNB$ so distances from $I$ to $FN,NB$ are equal and his means that $II_1=II_2$ so added with $\angle I_1II_2=90$ this gives $\angle DI_1I=\angle DFI=45$ which gives $I,Q,I_1$ colinear by radax and similarily $I,P,I_2$ colinear and thus clearly $QIPD$ cyclic.
To finish just notice that $IF=I_1F$ and thus by shooting Lemma we have $IF^2=FQ \cdot FD$ which means that $(QIPD)$ is tangent to $BF$ as desired thus we are done :cool:

:cool:

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