MathLabs

Problem 6

Let ABCABC be an acute triangle with circumcircle Γ\Gamma. Let ℓ\ell be a tangent line to Γ\Gamma, and let ℓa\ell_a, ℓb\ell_b and ℓc\ell_c be the lines obtained by reflecting ℓ\ell in the lines BCBC, CACA and ABAB, respectively. Show that the circumcircle of the triangle determined by the lines ℓa,ℓb,ℓc\ell_a,\ell_b,\ell_c is tangent to the circle Γ\Gamma.
Step 4 of 6: Find the circumcircle of the new triangle
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In plain words

Fitting the general circle equation to the three vertices turns into another linear solve, and after grouping terms with the symmetric abbreviations, both defining constants collapse to short formulas.

U=−(tS−P)2t2Q,1+V=2(tS−P)(E−t)tQU=-\dfrac{(tS-P)^2}{t^2Q},\quad 1+V=\dfrac{2(tS-P)(E-t)}{tQ}
Detailed analysis

Write E=a+b+cE=a+b+c and Q=(a+b)(b+c)(c+a)Q=(a+b)(b+c)(c+a) (note Q=ES−PQ=ES-P). Any circle can be written zzˉ+Uˉz+Uzˉ+V=0z\bar z+\bar Uz+U\bar z+V=0 for a complex number UU (its center is −U-U) and real number VV. Substituting the three points A′,B′,C′A',B',C' of the previous step and solving the resulting linear system in U,Uˉ,VU,\bar U,V gives, after simplification, U=−(tS−P)2t2Q,1+V=2(tS−P)(E−t)tQU=-\dfrac{(tS-P)^2}{t^2Q},\quad 1+V=\dfrac{2(tS-P)(E-t)}{tQ}.