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Consider $ABC$ is triangle and $a$ , $b$ and $c$ are sides of it. Can we say $a = x+y$ , $b = x+z$ and $c = y+z$ where $x,y,z > 0$ ? Is it true ?

Note : I know in triangle we have $a+b > c$ , $a+c > b$ and $b+c > a$ but I don't know that first statement is true or not.

Please Help!

3 Answers 3

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Yes.

  1. (Algebraic answer) just solve the system of equations you just written up with $a,b,c$ as known and $x,y,z$ as unknowns. E.g. adding the first two, we have $a+b=2x+y+z$. Combining with the 3rd equation, $x$ will be just expressed by $a,b,c$.

  2. (Geometric answer) Draw the inscribed circle in the triangle, and take the segments on the sides.

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    Your geometric answer is perfect but can you explain algebraic answer completely ? – 2017-01-07
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    No. Math is very much like *you-must-get-it-through-yourself*. Anyway, the next step will be $a+b=2x+c$. Now express $x$ from here. From here, $y$, and $z$ can be easily expressed. – 2017-01-07
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    Okay , Thanks a lot. – 2017-01-07
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    Note that meanwhile you have proved that the segments on the picture you have drawn are all of the form $\frac{a+b-c}2$. – 2017-01-07
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Hint. Consider the incentre of the triangle and the points at which the incentre touches the sides.

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    Can you provide a simple picture ? – 2017-01-07
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    @S.H.W Why not draw your own? It is not difficult. – 2017-01-07
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Intuitively, this gives three isosceles triangles, one constructed on each point using the sides of the triangle, which meet at their bases.

To see that this is possible, find the midpoint of the shortest side and use this to form the base of isosceles triangles using the triangle sides to the adjacent two points. Then use the intermediate point generated on the second-shortest side to make an isosceles triangle on the remaining point. This leaves a gap on the longest side, which can be closed by reducing the legs of the isosceles triangle on the opposite point by half the size of the gap. This increases the size of the other two isosceles triangles by the same amount, closing the gap on the longest side.

To put this into formulaic form, say $a\le b\le c$ are the side lengths. Then the first step above produces $2$ triangle of side length $a/2$ then one of side length $b-a/2$. Then the gap on the longest side is $c-(b-a/2)-a/2 = c-b$ and the adjustment is $(c-b)/2$. This gives:

$x=a/2-(c-b)/2 = (a+b-c)/2 \\ y=a/2+(c-b)/2 = (a-b+c)/2 \\ z=(b-a/2)+(c-b)/2 = (-a+b+c)/2 $

and $a=x+y, b=x+z, c=y+z$ as required. Note that the triangle inequality guarantees that these numbers are positive.