Pentagon \(ABCDE\) is inscribed in a circle. Its diagonals \(AC\) and \(BD\) intersect at \(F\). The bisectors of \(\angle BAC\) and \(\angle CDB\) intersect at \(G\). Let \(AG\) intersect \(BD\) at \(H\), let \(DG\) intersect \(AC\) at \(I\), and let \(EG\) intersect \(AD\) at \(J\). If \(FHGI\) is cyclic and \[JA \cdot FC \cdot GH = JD \cdot FB \cdot GI,\]prove that \(G\), \(F\) and \(E\) are collinear.
Problem
Source: Philippines MO 2016/5
Tags: geometry
ThE-dArK-lOrD
21.01.2017 22:34
When $E$ move along arc $\overarc{AD}$ not containing $B,C,D$, $\frac{JA}{JD}$ is an increasing function. So it's enough to prove that if $J'=GF\cap AD$, then $\frac{J'A}{J'D}\cdot \frac{FC}{FB}\cdot \frac{GH}{GI}=1$ Note that $G$ is midpoint of arc $BC$ not containing $A,D$, so $\angle{GHF}=\angle{GIF}\rightarrow =90^{\circ}$ We get that $F$ is orthocenter of $\triangle{ADG}$, the rest follows easily.
Forevermore
21.01.2017 22:40
Sniped... but I'll post this anyways. We introduce the point $K$, defined as the intersection of $GF$ and $\odot(ABCDE)$. Consider the following equivalent problem: Equivalent Problem wrote: In $\triangle GDA$, let $J',H,I$ be the feet of the altitudes from $G,D,A$, respectively, and let $F$ be the orthocenter. If $J$ lies on segment $\overline{AD}$ such that \[\frac{JA}{JD} = \frac{FB}{FC}\cdot\frac{GI}{GH},\]then $J=J'$.
All angles are directed in the following proof.
Since $G$ is the midpoint of $\overarc{BC}$, \[ \measuredangle HAI = \measuredangle GAC = \measuredangle BDG = \measuredangle HDI, \]so quadrilateral $ADIH$ is cyclic. Since $GHFI$ is cyclic as well, we have \[ \measuredangle DAK = \measuredangle DGK = \measuredangle IGF = \measuredangle IHF = \measuredangle IHD = \measuredangle IAD = \measuredangle CAD, \]implying that $D$ is the midpoint of $\overarc{CK}$. Similarly, $A$ is the midpoint of $\overarc{BK}$, implying that $F$ is the incenter of $\triangle BCK$.
It is well-known that the incenter $I$ of a triangle $ABC$ is the orthocenter of the triangle $DEF$ formed by the midpoints of minor arcs $\overarc{BC},\overarc{AC},\overarc{AB}$, so $F$ is the orthocenter of $\triangle GDA$ and the two problems are indeed equivalent.
Now the equivalent problem is not hard to check: note that $\frac{FB}{FC}\cdot\frac{GI}{GH}$ is fixed, so there is at most one value of $J$ that satisfies the equation. The rest is a computation using the Law of Sines to verify that $J'$ works.
rafaello
04.12.2020 00:42
Let $J'=FG\cap AD$. Since $\angle FAH=\angle HAB$ and $\angle AFH=\angle AGD=\angle ABH$, we get that $\triangle FAH\sim\triangle BAH$ and since they share same side $AH$, we have that $\triangle FAH\cong\triangle BAH$. Since, $F,H,B$ are collinear, we conclude that $\angle FHG=90^\circ$. Similarly, we have that $FI=IC$ and $\angle FIG=90^\circ$. Thus, $F$ is the orthocentre of $\triangle ADG$. Now, we have given that $JA \cdot FC \cdot GH = JD \cdot FB \cdot GI$. We also can show that $J'A \cdot FC \cdot GH = J'D \cdot FB \cdot GI$. This comes from the following: $\frac{J'A}{J'G}=\tan{\angle AGF}$ and $\frac{J'D}{J'G}=\tan{\angle DGF}$, thus $$\frac{J'A}{J'D}=\frac{\tan{\angle AGF}}{\tan{\angle DGF}}.$$We also have that $\tan{\angle DGF}=\frac{FI}{IG}$ and $\tan{\angle AGF}=\frac{FH}{HG}$ and since $2FH=FB$ and $FC=2FI$, we conclude that $$\frac{J'A}{J'D}=\frac{\frac{FH}{HG}}{\frac{FI}{IG}}=\frac{FH\cdot IG}{HG\cdot FI}=\frac{FB\cdot IG}{HG\cdot FC}\implies J'A \cdot FC \cdot GH = J'D \cdot FB \cdot GI.$$Thus,$$\frac{J'A}{J'D}=\frac{JA}{JD},$$which means that $J\equiv J'$, but since $J'$ lies on $EG$ and $J$ lies on $FG$, we directly have that $G$, $F$, $J$ and $E$ are collinear.