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math challenge II A geometry 4.29

 
 
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math challenge II A geometry 4.29
by Jialin Wang - Wednesday, 29 September 2021, 9:14 PM
 

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(sa)(sb)(sc).

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?

 
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Re: math challenge II A geometry 4.29
by John Lensmire - Thursday, 30 September 2021, 12:49 PM
 

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?