Let $ABC$ be an acute triangle. Let $D$ be a point inside side $BC$. Let $E$ be the foot from $D$ to $AC$, and let $F$ be a point on $AB$ so that $FE\perp AB$. It is given that the lines $AD, BE, CF$ concur. $M_A, M_B, M_C$ are the midpoints of sides $BC, AC, AB$ respectively, and $O$ is the circumcenter of $ABC$. Moreover, we define $P=EF\cap M_AM_B, S=DE\cap M_AM_C$. Prove that $O, P, S$ are collinear.
Problem
Source: 2024 Israel Olympic Revenge P4
Tags: geometry, olympic revenge
a22886
21.06.2024 05:28
$~~~~~~~~~$
Phorphyrion
21.06.2024 06:13
Yes, your D isn't on BC.
zhihanpeng
21.06.2024 07:25
Sketch for a calculation method: 1) BS⊥AB 2) Use this lemma: If $E$, $F$, $G$ are points outside of two intersecting lines, their projections onto the first line are $H$, $J$, $L$ and onto the second line are $I$, $K$, $M$, then $E$, $F$, $G$ are collinear $\Leftrightarrow$ $\frac{HJ}{JL} = \frac{IK}{KM}$. In this case we prove $\frac{BF}{BM_C} = \frac{EL}{EM_B}$, simplifying to $\frac{BF}{AB} = \frac{sin^2A}{2}$, which is easy.
Attachments:
Sourorange
21.06.2024 14:19
a22886 wrote: Is there anything wrong? Maybe inside $BC$ means on the segment $BC$, not above $BC$
a22886
21.06.2024 15:02
Right. Thought is was "inside $ABC$". Here goes another proof:
This is because as $D$ moves along $BC$, the length of $BF$ is a linear function of $BD$. The concurrence relation imposes another relation $\frac{BD}{DC}\frac{CE}{EA}\frac{AF}{FB}=1$, and $\frac{DC}{CE},\frac{EA}{AF}$ are constants, so there is another linear relation. Then either all points on $BC$ satisfy(which is never true; take $D\to B$.), either there's only one such $D$.P.S.The point $D=C$ also satisfies, but this is only because $DC=CE=0$ so it is illegal to bring them out of the fraction.
We then construct $D$ as follows: Consider the foot $H$ from $B$ to $M_AM_C$. We let $F$ be the foot from $H$ to $AB$, $E,D$ constructed respectively by $EF\perp AB,DE\perp AC$. It is easy, then, by calculating, to show that $D$ is the required point. Let the foot from $M_C$ to $AC$ be $G$, notice we have $GH\parallel AB$. Pascal's theorem (inverse) on the hexagon $EM_BM_AM_C\infty_{\perp AC}\infty_{\perp AB}$ shows that the hexagon is inscribed in a conic; Pascal again on $M_CM_AM_B\infty_{\perp AC}E\infty_{\perp AB}$ gives the desired result.
There should be a nicer way to get to $GH'\parallel AB$ where $H'=M_AM_C\cap EF$, rather than this stinky way of proving uniqueness blah. Nevertheless, to me, this solution suffices.
axolotlx7
26.08.2024 14:32
Let $I$ be on $AB$ such that $IE \parallel BC$, let $J = ES \cap AB$, and $K = EM_A \cap AB$. Then $(A,B;F,J) = (EC,EB;EF,ED)=-1$, and $(A,B;I,K)=(EC,EB;EI,EM_a)=-1$. Similar to #6, we want to show that $M_CM_AM_B\infty_{\perp AC}E\infty_{\perp AB}$ lies on a common conic. We will do so using DIT: Consider the involution that sends any $P_\infty \in \ell_\infty$ to the second intersection of $\ell_\infty$ and the conic passing through $P_\infty, M_A, M_B, M_C, E$. This involution fixes $\infty_{AC}$, swaps $\infty_{BC}$ with $\infty_{M_AE}$, and we wish to show that it also swaps $\infty_{\perp AC}$ with $\infty_{\perp AB}$. Projecting from $E$ on $AB$, it suffices to show that there is an involution fixing $A$, swapping $I$ with $K$ and swapping $J$ with $F$. From the harmonic bundles proved above, this involution is simply the inversion w.r.t. the circle with diameter $AB$, which proves the claim. Finally, note that $S = M_CM_A \cap \infty_{\perp AC}E$, $P = M_AM_B \cap E\infty_{\perp AB}$ and $O = M_B\infty_{\perp AC} \cap \infty_{\perp AB}M_C$, so Pascal finishes.