Let $OFT$ and $NOT$ be two similar triangles (with the same orientation) and let $FANO$ be a parallelogram. Show that \[\vert OF\vert \cdot \vert ON\vert=\vert OA\vert \cdot \vert OT\vert.\]
Problem
Source: Germany 2021, Problem 4
Tags: geometry, similar triangles, parallelogram, geometry proposed, lengths
16.06.2021 11:14
Tintarn wrote: Let $OFT$ and $NOT$ be two similar triangles (with the same orientation) and let $FANO$ be a parallelogram. Show that \[\vert OF\vert \cdot \vert ON\vert=\vert OA\vert \cdot \vert OT\vert.\] Let $X:=NA\cap TF$. Angle chasing gives that $OXTN$ and $OFXA$ are cyclic and $OAN\sim OFT$. Thus $\frac{OF}{OT}=\frac{OA}{ON}$.
20.06.2021 14:18
let $z_O=0$, suppose $z_N-z_T=x(z_O-z_T)$, for the similarity we have $z_F-z_T=\frac 1 x (z_O-z_T)$,then $z_N=(1-x)z_T, z_F=(1-\frac 1 x)z_T$ from parallelogram FANO we have $z_A=z_F+z_N=(1-x)z_T(1-\frac 1 x)$ so $\frac {z_A}{z_F}=1-\frac 1 x =\frac {z_N}{z_T}$ $\therefore z_A \cdot z_T=z_F \cdot z_N$ $\therefore | z_A| \cdot |z_T|=|z_F| \cdot |z_N|$ which equals \[\vert OF\vert \cdot \vert ON\vert=\vert OA\vert \cdot \vert OT\vert.\]and we are done.
21.06.2021 00:47
Consider a composition of spiral similarity and translation $\mathcal{T}(\overrightarrow{FA})\circ \mathcal{H}(T,NO\mapsto OF):N\mapsto N,O\mapsto A$ so $\triangle ONA\sim \triangle OTF$ and $\frac{|ON|}{|OA|}=\frac{|OT|}{|OF|}$.
24.06.2021 02:56
Perform a $\sqrt{bc}$ inversion on triangle $OFN$. Since $(ONT)$ is tangent to $OF$ and $(OFT)$ is tangent to $ON$, $T$ goes to a point lying on the lines parallel to $OF$ and $ON$ passing through $N$ and $F$ respectively. But this point is $A$, so in fact $T$ and $A$ swap. The desired equality follows.
11.03.2022 23:56
We will use the notation of measured angles $\measuredangle$ mod $180^\circ$. We have $\measuredangle TON=\measuredangle TFO$ and by the inverse of the secant-tangent theorem, $NO$ is a tangent to the circumcircle of $\triangle TFO$ and thus $\measuredangle OTF=\measuredangle(NO,FO)$. But since $FANO$ is a parallelogram, we find $\measuredangle(NO,FO)=\measuredangle AFO=\measuredangle ONA$. Now, let $X\coloneq TF\cap NA$. We have $\measuredangle OTX=\measuredangle OTF=\measuredangle ONA=\measuredangle ONX$. Therefore $X$ lies on the circumcircle of $\triangle NOT$. Since $\measuredangle ONT=\measuredangle FOT$, $FO$ is tangent to $(AOTX)$. But $X$ is a point over arc $ON$, so $\measuredangle NXO=\measuredangle(NO,FO)=\measuredangle AFO$, so $AOFX$ is also cyclic. Hence, $\measuredangle XFO=\measuredangle NAO$, meaning that $\triangle ANO\sim \triangle XFO$. We deduce $$\frac{|OA|}{|ON|}=\frac{|OF|}{|OT|}$$and rearranging gives the result.
27.03.2022 18:29