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 6 of 6: Confirm the tangency identity
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In plain words

Substituting the equation of the unit circle into the new circle's equation collapses it to a single real line, and that line touches the unit circle at exactly one point exactly when a short algebraic identity holds — which the clean formulas for U and its conjugate confirm on the spot.

(1+V)2=4UUˉ(1+V)^2=4U\bar U
Detailed analysis

Substituting zzˉ=1z\bar z=1 (the equation of Γ\Gamma) into zzˉ+Uˉz+Uzˉ+V=0z\bar z+\bar Uz+U\bar z+V=0 gives the radical line Uˉz+Uzˉ=−(1+V)\bar Uz+U\bar z=-(1+V); a real line of this form is tangent to the unit circle exactly when (1+V)2=4UUˉ(1+V)^2=4U\bar U, which is precisely the condition for the two circles to meet in a single point, i.e. to be tangent. Multiplying the two boxed formulas of steps 4 and 5, 4UUˉ=4⋅(tS−P)2t2Q2⋅(E−t)2=(2(tS−P)(E−t)tQ)2=(1+V)24U\bar U=4\cdot\dfrac{(tS-P)^2}{t^2Q^2}\cdot(E-t)^2=\left(\dfrac{2(tS-P)(E-t)}{tQ}\right)^2=(1+V)^2, exactly matching the boxed value of 1+V1+V from step 4. So (1+V)2=4UUˉ(1+V)^2=4U\bar U holds identically, and the circumcircle of A′B′C′A'B'C' is tangent to Γ\Gamma, as required.