Let $ABCD$ be a convex quadrilateral with angles $\sphericalangle A, \sphericalangle C\geq90^{\circ}$. On sides $AB,BC,CD$ and $DA$, consider the points $K,L,M$ and $N$ respectively. Prove that the perimeter of $KLMN$ is greater than or equal to $2\cdot AC$.
Problem
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Tags: geometry, quadrilateral, romania, Romanian TST
trigadd123
07.06.2021 09:50
A possible source would be Belarus MO.
Miquel-point
19.05.2022 13:17
For each $X$ let $\vec{X}=\overrightarrow{AX}$. From $\angle A\ge 90^\circ$ it follows that $\cos\angle KAN\le 0$, which implies $\overrightarrow{AK}\cdot\overrightarrow{AN}\le 0$. Therefore \[ KN^2=\overrightarrow{KN}^2=\left(\overrightarrow{AN}-\overrightarrow{AK}\right)^2={\overrightarrow{AN}}^2+{\overrightarrow{AK}}^2-2\cdot \overrightarrow{AN}\cdot \overrightarrow{AK}\ge{\overrightarrow{AN}}^2+{\overrightarrow{AK}}^2+2\cdot \overrightarrow{AN}\cdot \overrightarrow{AK} =\left(\overrightarrow{AN}+\overrightarrow{AK}\right)^2=\left(\vec{K}+\vec{N}\right)^2.\]SImilarly we get $LM\ge |\overrightarrow{CL}+\overrightarrow{CM}|=|\vec{M}+\vec{L}-2\vec{C}|$, which implies \begin{align*} P_{KLMN}&=NK+LM+KL+MN\ge |\vec{K}+\vec{N}|+|\vec{M}+\vec{L}-2\vec{C}|+|\overrightarrow{KL}|+|\overrightarrow{NM}|\\ &= |\vec{K}+\vec{N}|+|2\vec{C}-\vec{M}-\vec{L}|+|\vec{L}-\vec{K}|+|\vec{M}-\vec{N}|\ge |2\vec{C}|=2\cdot AC. \end{align*}by using Triangle's inequality. Thus, we have solved the problem (one year after the contest end )