In the triangle $ABC$ on the side $AC$ the points $F$ and $L$ are selected so that $AF = LC <\frac{1}{2} AC$. Find the angle $ \angle FBL $ if $A {{B} ^ {2}} + B {{C} ^ {2}} = A {{L} ^ {2}} + L {{C } ^ {2}}$ (Zhidkov Sergey)
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Tags: geometry, angles
30.06.2020 21:39
I suggest you post these questions in "High school Olympiads"
30.06.2020 22:23
in HSO I usually post harder Olympiads meaning National, Regional, International, TST, final round of multi-round olympiads this is the third City Olympiad from which I post geometries in aops (most people know Moscow City MO and St. Petersburg City MO), all that I post are collected below: $\bullet \bullet$ Almaty City Olympiad - geometry (Kazakhstan) $\bullet \bullet$ Kharkiv City Olympiad - geometry (Ukraine) $\bullet \bullet$ Kyiv City Olympiad - geometry (Ukraine) PS.1. I couldn't find online the problems from Minsk City MO (Belarus) to post them also PS.2. After I finish with the All-Siberian, I shall start posting the Ukraine NMO and Kyiv TSTs in HSO forum, so there will always be interesting geometries in both HSM and HSO forum (in HSM I shall post geometries from Teacher's Olympiads later)
13.02.2022 19:54
Really interesting, we will use Stewart's Theorem. Let $BA = c$, $BC = a$, $AF = LC = x$, $AL = CF = y$. $BF^2 = \frac{xa^2 + yc^2}{x+y}-xy$. $BL^2 = \frac{xc^2 + ya^2}{x+y}-xy$. $BL^2 + BF^2 = c^2 + a^2 - 2xy = x^2 + y^2 - 2xy = (y-x)^2 = FL^2$ so $ \angle FBL = 90 $.