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
Problem
Source: ELMO Shortlist 2023 G6
Tags: Elmo, geometry
DottedCaculator
29.06.2023 04:50
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 $J$.
MarkBcc168
29.06.2023 14:30
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 $$\angle BIP = \angle BB_1I = \angle BDI = \angle PDI.$$
Mahdi_Mashayekhi
30.06.2023 17:15
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 $BF$.
GeoKing
30.06.2023 18:35
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 $(A'ID)$. 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 $(A'ID)$. 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. 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 $Q'=Q$.
VZH
04.07.2023 09:18
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.
We invert the cute diagram wrt. the incircle $\odot (I)$ of $\triangle ACE$ to get an ugly nine-point circle configuration. It is phrased as follows, with $X'$ denoting the image of $X$ under inversion. (Note that the statement is symmetric with respect to $P'$ and $Q'$.)
Mr. Inversion wrote:
Let $XYZ$ be a triangle with Euler line $l$ and nine-point circle $\omega$. Denote by $A',C',E'$ the midpoints of $XY,YZ,ZX$ respectively. Suppose $l$ meets $\omega$ at $B', F'$ and the reflection of $A'$ wrt. $l$ is $D'$. If $P'$ is the intersection of $\odot (IB'D')$ and $\odot (IC'E')$($P'\ne I$), prove that $P'D'$ is parallel to $l$.
Proof: We simplify the problem by taking $P' = I+B'-A'$, and from here we only need to prove $\angle ZPI'= 90^{\circ}$, which is equivalent to proving $\frac{IP'}{IZ}=cos \angle ZIP'$, but the left hand side is exactly $cos \angle A'B'F'$ since $IZ=B'F'$ and $IP'=A'B'$. Now all that's left is to show $\angle ZIP' = \angle A'B'F'$ which is trivial by angle chasing.$\blacksquare$
Let $BF$ meet the incircle $\odot (I)$ of $\triangle ACE$ at $X,Y$ s.t. $X$ lies between $B$ and $Y$. Denote by $O$ the circumcenter of $\triangle ACE$.
Claim: $PX$ and $QY$ are tangent to $\odot (I)$.
Proof: Define $P'$ as $XX\cap CE$ and $B'$ as the Miquel point of the quadrilateral formed by $CE,AE,P'X,AC$. Now let us recall a well-known property of this configuration, namely, $(C,P'X\cap AE), (E, P'X \cap AC), (A, P')$ are swapped under inversion fixing $I$ centered at $B'$ followed by reflection wrt. $B'I$. But since $O$ is sent to the reflection of $B'$ wrt. $P'X$ (denote by $B_1$), we have $B'I$ bisects $\angle B_1B'O$, which means $B',I,F$ are collinear. This forces $B'=B$, and since $A,D$ are symmetrical wrt. $BF$, we conclude that $P=P'. \blacksquare$
To finish, observe that $\angle PIQ = 90^{\circ}$, $\angle PIX = \angle PQI$ and $\angle QIY = \angle QPI$. Hence we are done. $\blacksquare$
Marinchoo
24.05.2024 09:23
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.