Question Number 220486 by hardmath last updated on 13/May/25
$$\mathrm{z}\:\in\:\mathbb{C}\:\:\:\mathrm{and}\:\:\:\lambda\:>\:\mathrm{0} \\ $$$$\mathrm{Then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\mid\mathrm{z}\:+\:\mathrm{2}\lambda\mid\:+\:\mid\mathrm{z}\:+\:\lambda\mid\:\geqslant\:\mid\mathrm{z}\:+\:\frac{\mathrm{3}\lambda\:−\:\lambda\sqrt{\mathrm{3}}\mathrm{i}}{\mathrm{2}}\mid \\ $$
Commented by MrGaster last updated on 14/May/25
The original problem is equivalent to: In a plane, the sum of the distances from any point to two vertices of an equilateral triangle is greater than the distance from that point to the third vertex. This can be easily proven. Then Ptolemy's theorem can be applied.
Commented by hardmath last updated on 14/May/25
$$ \\ $$I wanted to see your beautiful solution, my dear professor
Answered by mr W last updated on 14/May/25
Commented by mr W last updated on 14/May/25
$${z}={re}^{{i}\theta} \\ $$$$\mid{z}+\mathrm{2}\lambda\mid={OB}' \\ $$$$\mid{z}+\lambda\mid={OA}' \\ $$$$\mid{z}+\frac{\mathrm{3}\lambda−\sqrt{\mathrm{3}}\lambda{i}}{\mathrm{2}}\mid={OC}' \\ $$$$\Delta{A}'{B}'{C}'\equiv\Delta{ABC}={equilateral}\:{with} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{side}\:{length}\:\lambda \\ $$$${it}\:{is}\:{to}\:{prove}\:{that}\:{OA}'+{OB}'\geqslant{OC}' \\ $$$${in}\:{quadrilateral}\:{OA}'{B}'{C}'\:{we}\:{have} \\ $$$${OA}'×{B}'{C}'+{OC}'×{A}'{B}'\geqslant{OB}'×{A}'{C}' \\ $$$${since}\:{A}'{B}'={B}'{C}'={A}'{C}'=\lambda \\ $$$${OA}'+{OC}'\geqslant{OB}' \\ $$$${similarly} \\ $$$${OA}'+{OB}'\geqslant{OC}' \\ $$$${OB}'+{OC}'\geqslant{OA}' \\ $$
Commented by mr W last updated on 14/May/25
Commented by hardmath last updated on 14/May/25
$$\mathrm{perfect}\:\mathrm{my}\:\mathrm{darling}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you} \\ $$