论文部分内容阅读
经过三角形ABC(?)顶点A引一直线AM交边BC于M,设∠AMC=2θ,O和I是△ABC的外接圆(O)和内切圆(I)的中心,以ω_1、ω_2为中心,以ρ_1ρ_2为半径的圆(ω_1)、(ω_2)都与圆(O)相切,且(ω_1)与∠AMC的两边相切,(ω_2)与∠AMB的两边相切。求证①连接ω_1、ω_2的直线经过I; ②点I分线段ω_1ω_2成比tg~2θ:1,且ρ_1+ρ_2=r~2sec~2θ,其中r是圆(I)的半径(见图)。上述问题在40多年前由多产和锐敏的几何学家V、Th′ebault提出,长期没有解决,直到最近才由英格兰的
After the triangle ABC(?) vertex A leads the line AM to the intersection BC to M, set ∠AMC=2?, O and I are the centers of the circumcircle (O) and inscribed circle (I) of ?ABC, to ω_1, ω_2 At the center, circles (ω_1) and (ω_2) with ρ_1ρ_2 as radius are all tangent to the circle (O), and (ω_1) is tangent to both sides of ∠AMC, and (ω_2) is tangent to both sides of ∠AMB. Proof 1 The straight line connecting ω_1 and ω_2 goes through I; the two-point I subsection ω_1ω_2 is compared to tg~2θ:1, and ρ_1+ρ_2=r~2sec~2θ, where r is the radius of circle (I) (see figure). The above problem was raised by prolific and agile geometer V, Th’ebault more than 40 years ago. It has not been solved for a long time until recently.