Smallest angle in triangle

[man111]

The radius of escribed circles of [tex]\triangle ABC[/tex] are [tex]r_{a},r_{b},r_{c}[/tex] respectively.

If [tex]r_{a}+r_{b} = 3R[/tex] and [tex]r_{b}+r_{c} = 2R,[/tex] Then the smallest angle of Triangle is

[Davids]

Do we need to know which of the angles is smallest or do we search for its measure?

[man111]

Actually, we have to search its measure., Thanks

[nevrodermit]

Look at the sketch sketch1.
Let [tex]CM=CN=x\Rightarrow BN=BP=a-x\Rightarrow AM=b+x, AP=a+c-x[/tex] but [tex]AM=AP[/tex] as tangents and so [tex]b+x=a+c-x\Rightarrow 2x=a+c-b\Rightarrow 2AM=2p\Rightarrow AM=AP=p[/tex].
The area of [tex]API_aM[/tex] equals on one hand to [tex]2S_{\Delta AI_aM}=pr_a[/tex] and on the other hand to the sum [tex]S_{\Delta ABC}+S_{CBPI_AM}=s+2S_{\Delta CI_aB}=s+r_aa[/tex]. So [tex]pr_a=s+r_aa\Rightarrow r_a=\frac{s}{p-a}[/tex].

So
[tex]3R=r_a+r_b=\frac{s}{p-a}+\frac{s}{p-b}=s\frac{c}{(p-a)(p-b)}=\frac{sc(p-c)p}{s^2}=\frac{c(p-c)p}{s}[/tex]
[tex]2R=\cdots = \frac{a(p-a)p}{s}[/tex]

Subtracting them we get
[tex]R=\frac{p}{s}(cp-c^2-ap+a^2)=\frac{p}{s}(p(c-a)-(c+a)(c-a))=\frac{p}{s}(c-a)(p-c-a)[/tex]

Since [tex]4Rs=abc[/tex] putting this in [tex]2R=\cdots[/tex] we get [tex]bc=2p(p-a)[/tex] and after expanding
[tex]b^2+c^2=a^2[/tex]

In the same way but for [tex]3R=\cdots[/tex] we get[tex]3ab=(a+b+c)(a+b-c)[/tex]
[tex]ab=a^2+b^2-c^2[/tex] and combining it with the previous result we get [tex]a=2b[/tex] and so [tex]\angle B=30^o[/tex].

[nevrodermit]

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