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Let triangle ABC have side lengths AB=13, AC=14, and BC=16. There are two circles located inside angle BAC which are tangent to rays AB, AC, and segment BC. Compute the distance between the centers of these two circles.

 Feb 15, 2022
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Compute the distance between the centers of these two circles.

 

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s=a+b+c2=16+14+132=21.5ρ=(sa)(sb)(sc)s=(21.516)(21.514)(21.513)21.5=4.0383cosα=b2+c2a22bc=142+13216221413α=arccos142+13216221413α=72.5754o

mf(x)=tanα2=tan72.57542=0.7342425

f(x)=0.7342425x0.7342425x1=ρ0.7342425x1=4.0383x1=5.5

P1(5.5, 4.0383)

r=0.7342425x2x2=rm(ρ+r)2=(rρ)2+(x2x1)2(ρ+r)2=(rρ)2+(rmx1)2(4.0383+r)2=(r4.0383)2+(r0.73424255,5)2r=15.75 wolfram alpha

 

The distance between the centers of these two circles is

4.0383 + 15.75 = 19.788

laugh  !

 Feb 16, 2022
edited by asinus  Feb 16, 2022

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