# Solution to Problem 2 We give an analytic proof whose final step is a polynomial identity. This form is especially suitable for formal verification. ## 1. Coordinates and parameters Scale the figure so that \(AC=1\), and choose coordinates \[ A=(0,0),\qquad B=(\rho,0),\qquad C=(c,s), \] where \[ c^2+s^2=1, \qquad \rho=\frac{AB}{AC}>0. \] Let \[ \alpha=\angle KBA=\angle ACL, \qquad x=\angle BAK, \qquad z=\angle LAC, \] and write \[ a=\tan\alpha, \qquad p=\tan x, \qquad q=\tan z. \] For the generic case in which these tangents are finite, intersecting the corresponding rays gives \[ K=\frac{a\rho}{a+p}(1,p), \tag{1} \] and \[ L=\frac{a}{a+q}(c+qs,\ s-qc). \tag{2} \] Indeed, \((1,p)\) is the direction of \(AK\), while \((-1,a)\) is the direction of \(BK\). Similarly, \((c+qs,s-qc)\) is the direction of \(AL\). The midpoint coordinates are \[ M=\left(\frac\rho2,0\right), \qquad N=\left(\frac c2,\frac s2\right). \tag{3} \] All equations below can be cleared of denominators. Thus the exceptional right-angle cases are covered by the same homogeneous polynomial identities, using direction vectors \((\cos t,\sin t)\) instead of tangents. ## 2. Translating the two remaining angle conditions From (2), the vectors \(NC\) and \(NL\), expressed in the orthonormal basis \[ (c,s),\qquad (s,-c), \] show that \[ \tan\angle LNC=\frac{2aq}{a-q}. \tag{4} \] Likewise, from (1), \[ \tan\angle BMK=\frac{2ap}{a-p}. \tag{5} \] Define \[ F(t)= c\bigl((1+2a^2)t^2+a^2\bigr) +s\bigl(a(t^2+1)-(a^2+1)t\bigr). \tag{6} \] For two vectors \(u,v\), use \[ \tan\angle(u,v)=\frac{\det(u,v)}{u\cdot v} \] and clear denominators. The equality \(\angle LBK=\angle LNC\), together with (4), becomes \[ \boxed{\rho(a+q)^2=F(q)}. \tag{7} \] Similarly, \(\angle LCK=\angle BMK\), together with (5) and \(c^2+s^2=1\), becomes \[ \boxed{\rho F(p)=(a+p)^2}. \tag{8} \] For example, a direction vector of \(BK\) is \((-1,a)\). Substituting the coordinates of \(L\) into \[ (a-q)\det(BL,BK)=2aq\,(BL\cdot BK) \] and simplifying gives exactly (7). Equation (8) follows analogously from the vectors \(CL\) and \(CK\). ## 3. Equations for the circumcentre Write \[ O=(u,v). \] Since \(O\) is the circumcentre of \(AKL\) and \(A=(0,0)\), \[ 2O\cdot K=|K|^2, \qquad 2O\cdot L=|L|^2. \] Using (1) and (2), these equations become \[ 2(a+p)(u+pv)=a\rho(1+p^2), \tag{9} \] and \[ 2(a+q)\bigl(u(c+qs)+v(s-qc)\bigr)=a(1+q^2). \tag{10} \] Set \[ \Delta=s(1-pq)-c(p+q). \tag{11} \] This is the determinant of the direction vectors of \(AK\) and \(AL\), so \(\Delta\ne0\) because \(A,K,L\) are not collinear. Solving (9)--(10) for the linear expression needed later gives \[ \begin{aligned} 4\bigl((\rho-c)u-sv\bigr) =\frac{2a}{(a+p)(a+q)\Delta} \Bigl[&\rho(1+p^2)(a+q)\bigl(\rho(s-qc)+q\bigr)\\ &-(1+q^2)(a+p)\bigl(s+p(\rho-c)\bigr) \Bigr]. \end{aligned} \tag{12} \] ## 4. Polynomial certificate Introduce \[ H_q=\rho(a+q)^2-F(q), \qquad H_p=\rho F(p)-(a+p)^2, \qquad H_0=c^2+s^2-1. \] By (7), (8), and the definition of \(c,s\), \[ H_q=H_p=H_0=0. \tag{13} \] Let \[ \begin{aligned} T={}&2a\Bigl[ \rho(1+p^2)(a+q)\bigl(\rho(s-qc)+q\bigr)\\ &\hspace{28mm}-(1+q^2)(a+p)\bigl(s+p(\rho-c)\bigr) \Bigr]\\ &-(\rho^2-1)(a+p)(a+q)\Delta. \end{aligned} \tag{14} \] Define \[ U_p=a+p+\rho\bigl(-2acp^2-ac+aps+cp+p^2s\bigr), \] \[ V_q=\rho(a+q)-2acq^2-ac+aqs+cq+q^2s. \] A direct expansion gives the identity \[ \boxed{ T=U_pH_q+V_qH_p -\rho(p-q)\bigl(2a^2pq-a^2-ap-aq+pq\bigr)H_0. } \tag{15} \] This is a pure polynomial identity; it can be verified by expansion, or by Lean's `ring` tactic. From (13) and (15), we obtain \(T=0\). Comparing this with (12) yields \[ 4\bigl((\rho-c)u-sv\bigr)=\rho^2-1. \tag{16} \] ## 5. Comparing the two distances Using (3), \[ \begin{aligned} OM^2-ON^2 &=\left(u-\frac\rho2\right)^2+v^2 -\left(u-\frac c2\right)^2 -\left(v-\frac s2\right)^2\\ &=-(\rho-c)u+sv+\frac{\rho^2-(c^2+s^2)}4\\ &=-(\rho-c)u+sv+\frac{\rho^2-1}{4}. \end{aligned} \] Equation (16) shows that the last expression is zero. Hence \[ OM^2=ON^2. \] Both distances are nonnegative, so \[ \boxed{OM=ON}. \]