The point $P$ lies inside, or on the boundary of, the triangle $ABC$. Denote by $d_{a}$, $d_{b}$ and $d_{c}$ the distances between $P$ and $BC$, $CA$, and $AB$, respectively. Prove that $\max\{AP,BP,CP \} \ge \sqrt{d_{a}^{2}+d_{b}^{2}+d_{c}^{2}}$. When does the equality holds?
Problem
Source: Bulgarian IMO TST 2008, Day 1, Problem 2
Tags: inequalities, trigonometry, complex numbers, inequalities proposed, geometry
24.01.2015 23:45
A bit of a necro, but... Does anybody have any ideas for this problem? I have been working on it for a few days, but have had almost no successes. Neither trigonometry nor complex numbers seem to be working, and I don't know of any other method by which I could approach this problem. So yea, ideas, hints, or solutions would be very much appreciated.
30.01.2015 16:59
heron wrote: A bit of a necro, but... So yea, ideas, hints, or solutions would be very much appreciated. Here is the requested hint :
Attachments:
19.03.2015 12:32
This could be solve by simple calculating. if all of them is not right. then, by sine_law, d_a^2+d_c^2+2d_ad_ccosB<=d_a^2+d_b^2+d_c^2 it is equal to cosB<=(-d_ad_c+sqrt((d_b^2+d_a^2)(d_c^2+d_b^2))/(d_a^2+d_b^2+d_c^2)<=d_b^2/(d_a^2+d_b^2+d_c^2) sum all of them, cos A+ cos B+ cos C<=1, that is not true
18.01.2016 18:12
Does anyone have a good solution for this problem?
30.07.2019 23:27
Bump Bump
30.07.2019 23:43
Solution Idea: By Erdos-Mordell, we have $AP + BP + CP \ge 2(d_a + d_b + d_c)$ with equality holding when $P$ is the center of equilateral $\triangle ABC$, so it suffices to prove $$2(d_a + d_b + d_c) \ge \sqrt{d_a^2 + d_b^2 + d_c^2}.$$
31.07.2019 00:14
tworigami wrote: Solution Idea: By Erdos-Mordell, we have $AP + BP + CP \ge 2(d_a + d_b + d_c)$ with equality holding when $P$ is the center of equilateral $\triangle ABC$, so it suffices to prove $$2(d_a + d_b + d_c) \ge \sqrt{d_a^2 + d_b^2 + d_c^2}.$$ Do you mean $$2(d_a + d_b + d_c) \ge 3 \sqrt{d_a^2 + d_b^2 + d_c^2}.$$Because $AP + BP + CP \ge \sqrt{d_a^2 + d_b^2 + d_c^2}.$ doesn’t imply the required result.