MHB Prove Equilateral Triangle from Cosine Equality

AI Thread Summary
The discussion centers on proving that the equality \(2\cos A \cos B \cos C + \cos A \cos B + \cos B \cos C + \cos C \cos A = 1\) indicates that triangle ABC is equilateral. The proof involves manipulating the cosine values of the angles and leveraging properties of triangles. Participants share various approaches and insights into the geometric implications of the equality. The consensus is that the equality uniquely characterizes equilateral triangles due to the symmetry in angle measures and cosine values. The conclusion reinforces that this relationship is a definitive condition for equilateral triangles.
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Prove that if in a triangle $ABC$ we have the following equality that holds

$2\cos A \cos B \cos C + \cos A \cos B + \cos B \cos C + \cos C \cos A = 1$

then the triangle will be an equilateral triangle.
 
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anemone said:
Prove that if in a triangle $ABC$ we have the following equality that holds

$S=2\cos A \cos B \cos C + \cos A \cos B + \cos B \cos C + \cos C \cos A = 1---(1)$

then the triangle will be an equilateral triangle.
my solution:
if $(1)$ is true,it is easy to see that triangle $ABC$ must be an acute triangle
(if one of those 3 angles A,or B, or C $\geq 90^o$ then (1) will fail)
that is $0<cos A,cos B,cos C<1$
using $AP\geq GP$
$S\geq 4\sqrt[4]{2cos^3Acos^3Bcos^3C}=1$
(equality occurs at $A=B=C=60^o$)
so it is an equilateral triangle
 
Last edited:
Thanks for participating, Albert!

My solution:

In any triangle $ABC$, we have the following equality that holds:

$1-2\cos A \cos B \cos C=\cos^2 A+\cos^2 B+\cos^2 C$

This turns the given equality to become

$ \cos^2 A+\cos^2 B+\cos^2 C=1-2\cos A \cos B \cos C=\cos A \cos B + \cos B \cos C + \cos C \cos A$

For any angle where $0\lt A,\,B,\,C\lt 180^\circ$, the relation $\cos A \gt \cos B \gt \cos C$ must hold, Now, by applying the rearrangement inequality on $\cos^2 A+\cos^2 B+\cos^2 C$ leads to:

$\cos^2 A+\cos^2 B+\cos^2 C\ge \cos A \cos B + \cos B \cos C + \cos C \cos A$, equality occurs iff $\cos A=\cos B=\cos C$, i.e. $A=B=C$, when that triangle is equilateral, and we're hence done with the proof.
 
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