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math challenge II A geometry 4.29

 
 
WangJialin的头像
math challenge II A geometry 4.29
WangJialin - 2021年09月29日 Wednesday 21:14
 

I'm confused about 4.29. it says if a triangle has side lengths a,b,ca,b,c semiperimeter ss, inradius rr, and circumradius RR we have Rr=abc4(s−a)(s−b)(s−c).

It contains 3 formulas and it's been compressed into 2 sides of the equations. I have no idea so I decided to solve it in reverse i.e. trying to get the formulas out of whatever this expression above is. I got (s-a)(s-b)(s-c)/r=abc/4R. What should I do next? Or is there a better way to solve it?

 
LensmireJohn的头像
Re: math challenge II A geometry 4.29
LensmireJohn - 2021年09月30日 Thursday 12:49
 

You already have that the RHS (right-hand side) of your expression, abc/(4R) is one way to calculate the area of the triangle.

Thus you can focus on the LHS. Here's a hint for there. The (s-a)(s-b)(s-c) reminds us of something we recognize from Heron's formula, except Heron's formula has an extra s in front. What happens if we multiply the numerator and denominator of that side by s?


 
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