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Given a Triangle with Angles a,b,c


Date: 05/03/99 at 22:42:33
From: Tom Kellogg
Subject: Trigonometry

Given a triangle with angles a,b,c, 
show that cos(a)+cos(b)+cos(c) <= 3/2

This problem appears simple enough, yet it has caused me many hours
of frustration. I've tried using the law of cosines, numerous
trigonometric identities and several other approaches, but all seem 
to lead nowhere.


Date: 05/12/99 at 03:50:29
From: Doctor Floor
Subject: Re: Trigonometry

Hi, Tom,

Thanks for your question.

For the solution we will need something we know about the graph of 
f(x) = cos(x). We know that this graph is convex on the interval 
[0, pi/2]. This means that if we have two unequal acute angles a and 
b, then the midpoint of (a,cos(a)) and (b,cos(b)) is below the graph.

When one of the angles of triangle ABC is obtuse, say a > pi/2, then 
we know that b = pi-a-c, so b < pi-a. We can use this fact to show 
that for obtuse a and acute b, too, the midpoint of (a,cos(a)) and 
(b,cos(b)) is below the graph.

To see this, note that the midpoint of (a,cos(a)) and (pi-a,cos(pi-a)) 
is (pi/2,0). Now we shift (pi-a, cos(pi-a)) to (b, cos(b)) = 
(pi-a-c,cos(b)). We find that the midpoint of (b, cos(b)) and 
(a,cos(a)) gets x-coordinate=(pi-c)/2. The y-coordinate of this 
midpoint has become higher by the shift; it becomes exactly [cos(b) - 
cos(pi-a)]/2. 

We now see, again using the convexity of cos(x) on [0,pi/2], that 

  cos((pi-c)/2)= cos((pi-c)/2) - cos(pi/2)
               > cos(pi-a-c/2) - cos(pi-a)
               > [cos(pi-a-c) - cos(pi-a)]/2
               = [cos(b) - cos(pi-a)]/2
               = y-coordinate of midpoint (b, cos(b)) and (a,cos(a))

This shows that if a is obtuse, then the midpoint of (a,cos(a)) and 
(b,cos(b)) is below the graph of f(x) = cos(x), too.

We can conclude, that if a,b,c are the unequal angles of a triangle, 
then the midpoints of (a,cos(a)), (b,cos(b)) and (c,cos(c)) are below 
the graph. This means that the centroid (the point of intersection of 
the medians) of the triangle formed by these three points is below 
the graph. This centroid can be calculated by taking the mean values 
of the x- and y-coordinates of the three points, hence

      (cos(a)+cos(b)+cos(c))/3 < cos((a+b+c)/3) = cos(pi/3) = 1/2
      cos(a)+cos(b)+cos(c) < 3/2

This proves the inequality for a triangle with three unequal angles.

If two of the three angles are equal, then one midpoint lies on the 
graph, but the centroid lies below the graph. And we have the same 
conclusion.

When the triangle is equilateral, then it is easy to verify that 

       cos(a)+cos(b)+cos(c) = 3/2.

So we have proven the inequality.

I hope this helped!

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Coordinate Plane Geometry
High School Geometry
High School Triangles and Other Polygons
High School Trigonometry

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