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c2c3bcd | 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 | # 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
\[
\boxed{OM=ON}.
\]
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