MHB How Does the Incenter Position Relate to Triangle Side Lengths?

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The incenter \( I \) of triangle \( ABC \) is defined in relation to the triangle's sides \( a \), \( b \), and \( c \) opposite vertices \( A \), \( B \), and \( C \) respectively. The distances from the incenter to the vertices are denoted as \( IA = x \), \( IB = y \), and \( IC = z \). The key proof to establish is that \( ax^2 + by^2 + cz^2 = abc \). This relationship highlights how the lengths from the incenter to the vertices are influenced by the triangle's side lengths. Understanding this connection is crucial for deeper insights into triangle geometry.
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Point $I$ is the incenter of $\triangle ABC $
giving :$BC=a , \, AC=b,\, AB=c$
$IA=x, \, IB=y, \, IC=z$
prove :$ax^2+by^2+cz^2=abc$
 
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Albert said:
Point $I$ is the incenter of $\triangle ABC $
giving :$BC=a , \, AC=b,\, AB=c$
$IA=x, \, IB=y, \, IC=z$
prove :$ax^2+by^2+cz^2=abc$
soluton :
 

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Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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