Relationships of Radii

Geometry Level 3

A triangle has three exradii 4, 6, 12. Find the area of the triangle.


The answer is 24.

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4 solutions

Michael Ng
Dec 25, 2014

We use the Radii Relationships in the Incircles and Excircles Wiki page. So: 1 r = 1 4 + 1 6 + 1 12 r = 2 \frac{1}{r} = \frac{1}{4} + \frac{1}{6} + \frac{1}{12} \\ \implies r = 2 Now we use the area formula: A r e a = r r 1 r 2 r 3 = 2 × 4 × 6 × 12 = 24 \mathrm{Area} = \sqrt{rr_1r_2r_3} = \sqrt{2 \times 4 \times 6 \times 12} = \boxed{24} For completeness, the 6 , 8 , 10 6, 8, 10 triangle is such a triangle that exists.

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.

Calvin Lin Staff - 6 years, 5 months ago

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Thank you. I have added that in.

Michael Ng - 6 years, 5 months ago

did the same! :)

Nihar Mahajan - 6 years, 4 months ago

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Doesn't the area not being zero imply the existence of such a triangle?

Bug Menot - 5 years, 3 months ago

Let us write the equations for the area of a triangle given one of its exradius:

S = r A ( s a ) S = r_{A}(s - a)

S = r B ( s b ) S = r_{B}(s - b)

S = r C ( s c ) 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:

S r A = ( s a ) \frac{S}{r_{A}} = (s - a)

S r B = ( s b ) \frac{S}{r_{B}} = (s - b)

S r C = ( s c ) \frac{S}{r_{C}} = (s - c)

Adding the three equations above and noting that a + b + c = 2 s a + b + c = 2s :

S 4 + S 6 + S 12 = 3 s 2 s \frac{S}{4} + \frac{S}{6} + \frac{S}{12} = 3s - 2s

Thus: S 2 = s \frac{S}{2} = s , or S = 2 s S = 2s

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 ) 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 ) 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 ) S^{4} = r_{A}*r_{B}*r_{C}*2s*(s - a)(s - b)(s - c)

Since S = s ( s a ) ( s b ) ( s c ) S = \sqrt{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 12 2 = 576 S = 24 S^{4} = r_{A}*r_{B}*r_{C}*2*S^{2} \rightarrow S^{2} = 4*6*12*2 = 576 \rightarrow S = 24

Now that the conditions are set, we must prove that such triangle exists. From S = 2 s S = 2s , we find that s = 12 s = 12 ; now, using the first three equations, we have:

24 = 4 ( 12 a ) a = 6 24 = 4*(12 - a) \rightarrow a = 6

24 = 6 ( 12 b ) b = 8 24 = 6*(12 - b) \rightarrow b = 8

24 = 12 ( 12 c ) c = 10 24 = 12*(12 - c) \rightarrow c = 10

This is a valid triangle, and a right one (note that a 2 + b 2 = c 2 a^{2} + b^{2} = c^{2} ). Thus, the solution sought is that the area of the triangle is 24.

Pratyush Pandey
Apr 29, 2017

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.

Ramiel To-ong
Sep 7, 2015

same analysis

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