Maths Olympiad Prep

Track / Stage 7 / 68 of 300 #1468 of 1964

Problem 1468

National olympiad second round; IMO P1/P4
Algebra Difficulty 7.1 Prove it

28. P,QP, Q are any two points on the plane of △A1A2A3\triangle A_{1} A_{2} A_{3}, then
PA1⋅QA1sin⁡A1+PA2⋅QA2sin⁡A2+PA3⋅QA3sin⁡A3⩾2Δ or PA1⋅QA1⋅a1+PA2⋅QA2⋅a2+PA3⋅QA3⋅a3⩾a1a2a3\begin{array}{ll} & P A_{1} \cdot Q A_{1} \sin A_{1}+P A_{2} \cdot Q A_{2} \sin A_{2}+P A_{3} \cdot Q A_{3} \sin A_{3} \geqslant 2 \Delta \\ \text { or } & P A_{1} \cdot Q A_{1} \cdot a_{1}+P A_{2} \cdot Q A_{2} \cdot a_{2}+P A_{3} \cdot Q A_{3} \cdot a_{3} \geqslant a_{1} a_{2} a_{3} \end{array}

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

28. P,QP, Q are any two points on the plane of △A1A2A3\triangle A_{1} A_{2} A_{3}, then

or
PA1⋅QA1sin⁡A1+PA2⋅QA2sin⁡A2+PA3⋅QA3sin⁡A3⩾2ΔPA1⋅QA1⋅a1+PA2⋅QA2⋅a2+PA3⋅QA3⋅a3⩾a1a2a3\begin{array}{l} P A_{1} \cdot Q A_{1} \sin A_{1}+P A_{2} \cdot Q A_{2} \sin A_{2}+P A_{3} \cdot Q A_{3} \sin A_{3} \geqslant 2 \Delta \\ P A_{1} \cdot Q A_{1} \cdot a_{1}+P A_{2} \cdot Q A_{2} \cdot a_{2}+P A_{3} \cdot Q A_{3} \cdot a_{3} \geqslant a_{1} a_{2} a_{3} \end{array}

Brief proof: Let x1,x2,x3,xx_{1}, x_{2}, x_{3}, x be any complex numbers, and x1,x2,x3x_{1}, x_{2}, x_{3} are distinct, then the equation holds:
x1(x−x1)(x1−x2)(x1−x3)+x2(x−x2)(x2−x1)(x2−x3)+x3(x−x3)(x3−x1)(x3−x2)=1\frac{x_{1}\left(x-x_{1}\right)}{\left(x_{1}-x_{2}\right)\left(x_{1}-x_{3}\right)}+\frac{x_{2}\left(x-x_{2}\right)}{\left(x_{2}-x_{1}\right)\left(x_{2}-x_{3}\right)}+\frac{x_{3}\left(x-x_{3}\right)}{\left(x_{3}-x_{1}\right)\left(x_{3}-x_{2}\right)}=1

Let PP correspond to the complex number 0,Q0, Q correspond to the complex number x,A1,A2,A3x, A_{1}, A_{2}, A_{3} correspond to the complex numbers x1,x2,x3x_{1}, x_{2}, x_{3}, respectively, then
PA1⋅QA1a2a3+PA2⋅QA2a1a3+PA3⋅QA3a1a2=∣x1∣∣x−x1∣∣x1−x2∣∣x1−x3∣+∣x2∣∣x−x2∣∣x2−x1∣∣x2−x3∣+∣x3∣∣x−x3∣∣x3−x1∣∣x3−x2∣⩾∣x1(x−x1)(x1−x2)(x1−x3)+x2(x−x2)(x2−x1)(x2−x3)+x3(x−x3)(x3−x1)(x3−x2)∣=1\begin{array}{l} \frac{P A_{1} \cdot Q A_{1}}{a_{2} a_{3}}+\frac{P A_{2} \cdot Q A_{2}}{a_{1} a_{3}}+\frac{P A_{3} \cdot Q A_{3}}{a_{1} a_{2}}= \\ \frac{\left|x_{1}\right|\left|x-x_{1}\right|}{\left|x_{1}-x_{2}\right|\left|x_{1}-x_{3}\right|}+\frac{\left|x_{2}\right|\left|x-x_{2}\right|}{\left|x_{2}-x_{1}\right|\left|x_{2}-x_{3}\right|}+\frac{\left|x_{3}\right|\left|x-x_{3}\right|}{\left|x_{3}-x_{1}\right|\left|x_{3}-x_{2}\right|} \geqslant \\ \left|\frac{x_{1}\left(x-x_{1}\right)}{\left(x_{1}-x_{2}\right)\left(x_{1}-x_{3}\right)}+\frac{x_{2}\left(x-x_{2}\right)}{\left(x_{2}-x_{1}\right)\left(x_{2}-x_{3}\right)}+\frac{x_{3}\left(x-x_{3}\right)}{\left(x_{3}-x_{1}\right)\left(x_{3}-x_{2}\right)}\right|=1 \end{array}

Note: (1) In the proof of inequalities, sometimes the application of complex number methods is particularly concise.
(2) This problem can be extended to nn-sided polygons.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.