let $ ABC $ an equilateral triangle with circum circle $ w $ let $ P $ a point on arc $ BC $ ( point $ A $ is on the other side ) pass a tangent line $ d $ through point $ P $ such that $ P \cap AB = F $ and $ AC \cap d = L $ let $ O $ the center of the circle $ w $ prove that $ \angle LOF > 90^{0} $
Problem
Source:
Tags: IGO, geometry
andria
03.09.2015 20:44
Trivial problem: Let $M,N$ be the midpoints of $AB,AC$ respectively. then by angle casing we get $\angle LOF-\angle A=\angle MPN\Longrightarrow \angle LOF=\angle 60+\angle MPN$ so it's enough to prove that $\angle MPN>30$. note that in cyclic quadrilateral $MBCN$ arc $MN$ equals $60$ and it's clear that arc $BC$ of $\omega$ lies inside $\odot(MBCN)$ hence $MPN>MCN=30$. Q.E.D
QuadraticReciprocity
06.09.2015 14:47
Or alternatively we can use sine rule to represent the segments $PL$ and $PF$, which was how I solved it during the exam. btw the version I read didnt have $F$ but had $K$ in the position of $F$ instead
MRF2017
20.09.2015 17:37
Proposed by Iman Maghsoudi,IRAN
anantmudgal09
02.09.2016 18:52
Let $\omega$ be the circumcircle of triangle $ABC$. Let the tangents at $B$ and $C$ meet the tangent at $A$ to the circle $\omega$ at points $C'$ and $B'$, respectively. Notice that $K$ is the intersection of the polars of points $C'$ and $P$ wrt $\omega$. Thus, by La Hire's theorem, $PC'$ is the polar of point $K$ wrt $\omega$. Similarly $PB'$ is the polar of point $L$ wrt $\omega$. Therefore, $\angle KOL=180^{\circ}-\angle B'PC'>90^{\circ}$ since $\angle B'PC'<90^{\circ}$ because the point $P$ lies outside the circle with diameter $B'C'$ (this circle passes through points $B$ and $C$ and $P$ lies on the arc $BC$) and the result follows.
Note: As per the official wording, I took $K$ instead of $F$.
Mahdi_Mashayekhi
27.12.2021 09:29
Let D and E be midpoints of AC and AB. we want to prove ∠OLT + ∠OTL ≤ 90. ∠DPE = 120 - ∠ODP - ∠OEP = 120 - ∠OLT - ∠OTL so we can instead prove that ∠DPE ≥ 30. for this we have to prove P is inside of circle CDEB. ∠CPB = 120 but arc CDEB = 90 so P is inside circle CDEB. we're Done.
