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 3 of 6: Solve for the vertices of the new triangle
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In plain words

Each vertex is where two of the reflected lines cross; solving the two corresponding linear equations in z and z-bar together, and abbreviating the symmetric combinations of a, b, c, produces a strikingly compact formula.

A′=P(a−2t)+St2t2(b+c)A'=\dfrac{P(a-2t)+St^2}{t^2(b+c)}
Detailed analysis

Write S=ab+bc+caS=ab+bc+ca and P=abcP=abc for the elementary symmetric combinations. Solving the linear system ℓb∩ℓc\ell_b\cap\ell_c for zz (treating zz and zˉ\bar z as the two unknowns of two linear equations) and simplifying gives A′=ℓb∩ℓc=A'=\ell_b\cap\ell_c=P(a−2t)+St2t2(b+c)\dfrac{P(a-2t)+St^2}{t^2(b+c)}. Cycling a→b→c→aa\to b\to c\to a gives the analogous formulas for B′=ℓc∩ℓaB'=\ell_c\cap\ell_a and C′=ℓa∩ℓbC'=\ell_a\cap\ell_b.