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