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 Incenter and a Circumcircle (Posted on 2007-10-28)
Let I be the incenter of ABC. AI, BI, CI intersect the circumcircle of ABC again at A', B', C' respectively. Show that the area of A'B'C' >= the area of ABC

 See The Solution Submitted by Praneeth Rating: 3.0000 (2 votes)

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 Solution | Comment 5 of 6 |

Let the measure of angle CAB be denoted as a, ABC as b, and BCA as c.

Since I is the incenter of ABC, angle ABI = b/2.
Similarily angle BAI = a/2, angle ACI = c/2, angle CAI = a/2, angle BCI = c/2, and angle CBI = b/2.

Since angles ABB' and AA'B' both subtend arc AB, angle AA'B' = angle ABB'( = angle ABI) = b/2.
Similarily angle AA'C' = c/2, angle BB'A' = a/2, angle BB'C' = c/2, angle CC'A' = a/2, and angle CC'B' = b/2.

Let J be the mutual circumcenter of ABC and A'B'C'.
Then angle AJB = 2 * angle ACB = 2*c.  Similarily angle BJC = 2*a and angle CJA = 2*b.

Angles A'B'C', B'C'A', and C'A'B' are (a+c)/2, (a+b)/2, and (b+c)/2.  Then A'JC' = a+c, B'JC' = b+c, and C'JA' = a+c.

If the circumradius is denoted as r, then area of each triangle ABC and A'B'C' can be calculated as the sum of three triangles with vertex J:
area ABC = (r^2/2)*(sin(2*a) + sin(2*b) + sin(2*c))
area A'B'C' = (r^2/2)*(sin(b+c) + sin(a+c) + sin(a+b))

Combining the areas with the suggested inequality area(A'B'C') >= area(ABC) gives:
sin(b+c) + sin(a+c) + sin(a+b) >= sin(2*a) + sin(2*b) + sin(2*c)

Expanding the sines using addition and double angle formulas gives:
sin(b)*cos(c) + sin(c)*cos(b) + sin(a)*cos(c) + sin(c)*cos(a) + sin(b)*cos(a) + sin(a)*cos(b) >= 2*sin(a)*cos(a) + 2*sin(b)*cos(b) + 2*sin(c)*cos(c)

The inequality can be rearranged and factored to yield:
(sin(a) - sin(b))*(cos(b) - cos(a)) + (sin(a) - sin(c))*(cos(c) - cos(a)) + (sin(c) - sin(b))*(cos(b) - cos(c)) >= 0

With the constraints a,b,c >= 0 and a+b+c = 180, each product is always nonnegative as follows:
a>b implies sin(a) - sin(b) > 0 and cos(b) - cos(a) > 0, which makes the product greater than 0
a=b implies sin(a) - sin(b) = 0 and cos(b) - cos(a) = 0, which makes the product equal to 0
b>a implies 0 > sin(a) - sin(b) and 0 > cos(b) - cos(a), which makes the product greater than 0

Since all three products are nonnegative, their sum is also nonnegative so the inequality is true.  Equality holds only when a=b=c=60.

Edited on November 10, 2007, 1:39 am
 Posted by Brian Smith on 2007-11-10 01:38:32

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