In triangle ABC, the inscribed circle touches sides BC, AC, and AB at points K₁, K₂, and K₃, respectively.
The excircle (Iₐ, rₐ=IₐT₂) touches the lines BC, AC, and AB at points T₁, T₂, and T₃, respectively.
Given that AI=3, r=1, and T₁K₁=2:
Let D be the intersection of K₁K₃ and T₁T₂, and let E be the intersection of K₁K₂ and T₁T₃.
Find DE.
Is it possible that K1K2 ∩ T1T2 = D
should be replaced by K1K3 ∩ T1T2 = D
At least this is what the picture shows ...
... thanks
Basically, we can solve for the lengths a, b, c, and the semiperimeter s, of triangle ABC ... online - from the three given lengths.
And, then every other length and angle can be found.
Technically,
2nd equation in the system comes from
2 = | T_1 K_1 | = CK_1 - CT_1 = s-c - (s-b),
where the lengths of the line segments CK_1 and CT_1 are well-known tangent lines length to incircle and excircle;
3rd equation is r*s = area of ABC = Heron's formula;
4th equation is 3 = AI = b*sin(C/2) / cos(B/2)
which comes from the sine rule for triangle ACI,
and the half angle formulas are
https://d138zd1ktt9iqe.cloudfront.net/media/seo_landing_files/half-angle-formulas-using-semiperimeter-1628074842.png
Consecutively, we have
sin(A/2) = 1/3
sin(B/2) = sqrt(17 - 2 sqrt(2) - sqrt(17) + 2 sqrt(34)) / 6
sin(C/2) = sqrt(17 + 2 sqrt(2) - sqrt(17) - 2 sqrt(34)) / 6
"Finally",
DE comes from cosine rule for triangle DEK_1,
DK_1 comes from sine rule for triangle DK_1T_1, and
EK_1 comes from sine rule for triangle EK_1T_1.
That is,
DK_1 / sin(C/2) = TK_1 / sin(A/2) = 2 / sin(A/2), where angle T_1DK_1 = A/2 = because it is 'pi' minus the other two angles in triangle DK_1T_1 at vertices K_1 and T_1 = pi - C/2 - (pi/2+B/2),
so,
DK_1 = sqrt( 17 + 2 sqrt(2) - sqrt(17) - 2 sqrt(34) )
Similarly,
EK_1 = sqrt( 17 - 2 sqrt(2) - sqrt(17) + 2 sqrt(34) )
Hence,
DE^2 = DK1^2 + EK1^2 - 2*DK1*EK1*cos(EK1D),
where angle EK1D = pi/2+C/2 + pi/2-B/2 = pi/2 + A/2,
i.e., cos(EK1D) = - sin(A/2).
Thus DE^2 = 32.
Please let me know about any questions.
Thanks
P.S. Of course, this is the most tasteless and messy solution to a geometric problem.
Steftcho P. Dokov Thank you for the excellent solution to the problem!
I want to provide a geometric hint for solving the problem
The difference between these two segments provides the answers to the problem's questions.
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