Let $ABC$ be a triangle and let $a,b,c$ be the lengths of the sides $BC, CA, AB$ respectively. Consider a triangle $DEF$ with the side lengths $EF=\sqrt{au}$, $FD=\sqrt{bu}$, $DE=\sqrt{cu}$. Prove that $\angle A >\angle B >\angle C$ implies $\angle A >\angle D >\angle E >\angle F >\angle C$.
Problem
Source:
Tags: inequalities, geometry proposed, geometry, geometry unsolved
huashiliao2020
09.05.2023 07:08
The angle A > angle D and angle F > angle C doesn't seem correct. For example, if it is a 3-4-5 triangle and u=sufficiently large (like 1,000,000), then $EF-FD=1000(\sqrt{5}-\sqrt{4})>BC-CA=1$, meaning BC and CA have closer angles together A and B than the abs value of the difference between EF and FD, or angles D and E. However, part of it is true. Note that a>b>c since the angles subtending them in the triangle are in this order as well. Then $\sqrt{au}>\sqrt{bu}>\sqrt{cu}$, so $\angle D>\angle E>\angle F$
huashiliao2020
21.05.2023 03:05
Sorry, I'll try to get it fixed. u doesn't matter because they are all similar triangles with lengths of a triangle $\sqrt{a}, \sqrt{b}, \sqrt{c}$. We know that $\sin D/\sin F=\sqrt{a}/\sqrt{c}<a/c=\sin A/\sin C$, meaning that $\angle D/\angle F<\angle A/\angle C$, and because $\angle A/\angle C>1$, the ratio between angles in DEF is closer to 1 (which represents an equilateral triangle) than in ABC, meaning that angle D is closer to 60 degrees than angle A. Continuing analogously on other cases for pairwise angles A,B,C vs. D,E,F, the result follows. Note that if A is obtuse, the sine angle formulas to angle inequalities still hold because both of angles A and D will either be obtuse or we will be done, with angle A obtuse and angle D nonobtuse.