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Radius and Altitudes (Posted on 2020-05-08) Difficulty: 3 of 5
Prove the following theorem:
The radius r of the circle inscribed in any triangle, say ABC ,
equals 1/3 of the harmonic average of the 3 altitudes,
i.e.
1/r = 1/h(A)+1/h(B)+1/h(C)

No Solution Yet Submitted by Ady TZIDON    
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solution and correction | Comment 1 of 3
Call the center of the inscribed circle O and the sides a,b,c.  Then r is the altitude for triangle OAB with base c which has area = r*c/2 and the area of ABC = r*(a+b+c)/2.

But the area of ABC can also be expressed as h(C)*c/2.

Equating and simplifying gives 1/h(C) = c*(1/r)*(1/(a+b+c). 

Doing the same for the other two triangles and adding gives
1/h(A) + 1/h(B) + 1/h(C) = (a+b+c)*(1/r)*(1/(a+b+c) = 1/r. 

That matches the formula following 'i.e.' in the problem but not the preceding text which I believe should read 'the inradius equals the reciprocal of 1/3 the harmonic average of the 3 altitudes'.



  Posted by xdog on 2020-05-08 08:52:40
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