AoPS - Problem

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Problem

Source: Tournament of Towns Spring 2003 - Junior O-Level - Problem 3

Tags: geometry, trapezoid, geometry proposed

Amir Hossein

14.06.2011 14:50

Points $K$ and $L$ are chosen on the sides $AB$ and $BC$ of the isosceles $\triangle ABC$ ( $AB = BC$ ) so that $AK +LC = KL$ . A line parallel to $BC$ is drawn through midpoint $M$ of the segment $KL$ , intersecting side $AC$ at point $N$ . Find the value of $\angle KNL$ .

oneplusone

14.06.2011 15:33

Note that there is a circle with center $E$ that is tangent to $KL$ and tangent to $BA,BC$ at $A,C$ . Let $EL$ intersect $AC$ at $N'$ . By angle chasing, we get $\angle KAN'=\angle KEN'$ , so $KAEN'$ is cyclic, so $\angle KN'E=\angle KAE=90$ . So $MK=ML=MN'$ , and this implies that $MN'\parallel BC$ . So $N\equiv N'$ and we have $\angle KNL=90$ .

leader

02.04.2013 19:50

let the parallel through $K$ to $BC$ intersect $AC$ at $S$ than $AK/KS=AB/BC=1$ so in trapezoid $KSCL$ $MN=(KS+LC)/2=(AK+LC)/2=KL/2=MK=ML$ so $\angle KNL=90$ .

HamstPan38825

31.07.2023 14:27

Very cute problem. Let $B'$ be the reflection of $B$ over the perpendicular line to $\overline{BC}$ through $K$ . Hence $MN = \frac 12(KB'+LC) = \frac 12(KB+LC) = KM = ML,$ and thus $\angle KNL = 90^\circ$ .