A triangle has three exradii 4, 6, 12. Find the area of the triangle.
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For completeness, you should also show that such a triangle can exists.
Just because some variables satisfy a certain algebraic condition doesn't imply that there is a geometric interpretation of them.
did the same! :)
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Doesn't the area not being zero imply the existence of such a triangle?
Let us write the equations for the area of a triangle given one of its exradius:
S = r A ( s − a )
S = r B ( s − b )
S = r C ( s − c )
Firstly, let us derive a direct relationship between the area S and the semiperimeter s. Dividing the area by the radius, we obtain the following:
r A S = ( s − a )
r B S = ( s − b )
r C S = ( s − c )
Adding the three equations above and noting that a + b + c = 2 s :
4 S + 6 S + 1 2 S = 3 s − 2 s
Thus: 2 S = s , or S = 2 s
Now, taking the three original equations and multiplying them together yield:
S 3 = r A ∗ r B ∗ r C ∗ ( s − a ) ( s − b ) ( s − c )
Multiplying both sides by S:
S 4 = r A ∗ r B ∗ r C ∗ S ∗ ( s − a ) ( s − b ) ( s − c )
Thus: S 4 = r A ∗ r B ∗ r C ∗ 2 s ∗ ( s − a ) ( s − b ) ( s − c )
Since S = s ( s − a ) ( s − b ) ( s − c ) , we can write:
S 4 = r A ∗ r B ∗ r C ∗ 2 ∗ S 2 → S 2 = 4 ∗ 6 ∗ 1 2 ∗ 2 = 5 7 6 → S = 2 4
Now that the conditions are set, we must prove that such triangle exists. From S = 2 s , we find that s = 1 2 ; now, using the first three equations, we have:
2 4 = 4 ∗ ( 1 2 − a ) → a = 6
2 4 = 6 ∗ ( 1 2 − b ) → b = 8
2 4 = 1 2 ∗ ( 1 2 − c ) → c = 1 0
This is a valid triangle, and a right one (note that a 2 + b 2 = c 2 ). Thus, the solution sought is that the area of the triangle is 24.
Realise that these are twice the ex-radii of the famous 3-4-5 right angled triangle, having ex-radii 2,3,6 respectively. We use scaling to understand that since linear dimension increased by a factor of 2 in the new triangle, area will increase by a factor of 4. So area is 4 times (1/2) (3) (4) which is 24.
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We use the Radii Relationships in the Incircles and Excircles Wiki page. So: r 1 = 4 1 + 6 1 + 1 2 1 ⟹ r = 2 Now we use the area formula: A r e a = r r 1 r 2 r 3 = 2 × 4 × 6 × 1 2 = 2 4 For completeness, the 6 , 8 , 1 0 triangle is such a triangle that exists.