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### Given a Triangle with Angles a,b,c

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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
```

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

Hi, Tom,

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/
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Associated Topics:
High School Coordinate Plane Geometry
High School Geometry
High School Triangles and Other Polygons
High School Trigonometry

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