Correction again. That should be [math]-v\bar{P}Y=-v\bar{P}\bar{u}M+v\bar{P}\bar{u}A-v\bar{P}u\bar{M}+v\bar{P}u\bar{A}[/math]
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geometry problem
- Thread starter stephanson12
- Start date
Using the complex–vector dictionary:Incomplete for now, but sharing what I have so far.
We can simplify the LHS again by compressing the brackets
[math]-\Im(\bar{A}M)\left((\Im(H\bar{P})-\Im(H\bar{Q})+\Im(P\bar{Q})\right)=-\Im(\bar{A}M)\left(\Im\left(H(\bar{P}-\bar{Q})+P\bar{Q}\right)\right)[/math]Since [imath]v=P-Q, \quad \bar{v}=\bar{P}-\bar{Q}[/imath] and [imath]Q=P-v[/imath] we have
[math]P\bar{Q}=P(\bar{P}-\bar{v})=\left|P\right|^2-P\bar{v}[/math]But [imath]\left|P\right|^2[/imath] is real so its imaginary part vanishes. Hence
[math]LHS=-\Im(\bar{A}M)\left(\Im\left(H(\bar{P}-\bar{Q})+P\bar{Q}\right)\right)=-\Im(\bar{A}M)\Im\left(H\bar{v}-P\bar{v}\right)=-\Im(\bar{A}M)\Im\left((H-P)\bar{v}\right)=-\Im(\bar{A}M)\Im\left(u\bar{v}\right)[/math][math]LHS=\Im(\bar{A}M)\Im\left(\bar{u}v\right)[/math][math][/math]Now we can use the identity [imath]2\Im\alpha\Im\beta=\Re(\alpha\bar{\beta})-\Re(\alpha\beta)[/imath] again. Choose [imath]\alpha=\bar{A}M[/imath] and [imath]\beta=\bar{u}v[/imath]. Then
[math]2LHS=2\Im(\bar{A}M)\Im(\bar{u}v)=\Re\left((\bar{A}M)u\bar{v}\right)-\Re\left((\bar{A}M)\bar{u}v\right)[/math]
I noticed an error in my reasoning. The identity missed the fraction [imath]\frac{1}{2}[/imath]. It comes from working with the scaled triangle [imath]H' = 2H, P' = 2P, Q' = 2Q[/imath]: the perpendicular-bisector constants pick up a factor [imath]2[/imath], and when you translate them back to [imath](H,P,Q)[/imath], a global factor [imath]2[/imath] appears on the right side. Accounting for that should yields the correct equality:
[math]-\Im(\overline{A}M) \left[ \Im(H\overline{P}) -\Im(H\overline{Q}) +\Im(P\overline{Q}) \right] = \frac12 \left[ (|H|^2 - |P|^2)\,\Re\big(\overline{M-A}\,v\big) - (|P|^2 - |Q|^2)\,\Re\big(\overline{u}\,(M-A)\big) \right].[/math]While I don't yet see a path to prove this, it may require techniques beyond my current mathematical toolkit.
[math] x\cdot y = \Re(\overline{x}y), \qquad x\times y = \Im(\overline{x}y) [/math]Left hand side
[math] -\Im(\overline{A}M)\big[\Im(H\overline P)-\Im(H\overline Q)+\Im(P\overline Q)\big]=\Im(\bar{A}M)\Im\left(\bar{u}v\right)=(A\times M)(u\times v) [/math]
By the 2-D Lagrange identity
[math] (X\times Y)(U\times V) = (X\cdot U)(Y\cdot V) - (X\cdot V)(Y\cdot U) [/math]with [imath]X=A, Y=M, U=u, V=v[/imath], we obtain
[math] \text{LHS} = (A\cdot u)(M\cdot v) - (A\cdot v)(M\cdot u) [/math]
Reformulating the right-hand side
[math] \tfrac12\Big[(|H|^2-|P|^2)\Re(\overline{M-A}v) - (|P|^2-|Q|^2)\Re(\overline u(M-A))\Big] [/math]
Note that
[math] |X|^2-|Y|^2 = (X+Y)\cdot(X-Y) [/math]
Also
[math] \Re(\overline{M-A}v) = (M-A)\cdot v \qquad \Re(\overline u(M-A)) = u\cdot(M-A) [/math]
Thus
[math] 2\text{RHS} = (H+P)\cdot u(M-A)\cdot v - (P+Q)\cdot vu\cdot(M-A) [/math]
Expanding:
[math] = (H+P)\cdot uM\cdot v - (H+P)\cdot uA\cdot v - (P+Q)\cdot vu\cdot M + (P+Q)\cdot vu\cdot A [/math]
We now calculate the difference between the right-hand side (RHS) and the left-hand side (LHS), aiming to demonstrate that the result is zero.
Let
[math] D = 2\text{RHS} - 2\text{LHS} [/math]We define scalars:
[math]s_{1} = (H+P)\cdot u, \quad s_{2} = (P+Q)\cdot v[/math][math]a_{1} = A\cdot u, \quad a_{2} = A\cdot v[/math][math]m_{1} = M\cdot v,\quad m_{2} = M\cdot u[/math]
Start with
[math] D = s_{1}m_{1} - s_{1}a_{2} - s_{2}m_{2} + s_{2}a_{1} - 2(a_{1}m_{1} - a_{2}m_{2}) [/math]
Group by [imath]m_{1},m_{2},a_{1},a_{2}[/imath]:
[math] D = m_{1}(s_{1} - 2a_{1}) - m_{2}(s_{2} - 2a_{2}) - s_{1}a_{2} + s_{2}a_{1} [/math]
Now rewrite the last part by adding and subtracting [imath]2a_{1}a_{2}[/imath]:
[math] D = m_{1}(s_{1} - 2a_{1}) - a_{2}(s_{1} - 2a_{1}) - m_{2}(s_{2} - 2a_{2}) + a_{1}(s_{2} - 2a_{2}) [/math]
Factor pairs:
[math] D = (s_{1} - 2a_{1})(m_{1} - a_{2}) - (s_{2} - 2a_{2})(m_{2} - a_{1}) [/math]
Check by expansion:
[math] (s_{1} - 2a_{1})(m_{1} - a_{2}) = s_{1}m_{1} - s_{1}a_{2} - 2a_{1}m_{1} + 2a_{1}a_{2} [/math][math] -(s_{2} - 2a_{2})(m_{2} - a_{1}) = -s_{2}m_{2} + s_{2}a_{1} + 2a_{2}m_{2} - 2a_{1}a_{2} [/math]Adding cancels [imath]\pm 2a_{1}a_{2}[/imath], recovering the original [imath]D[/imath].
So the exact factorization is
[math] D = (s_{1} - 2a_{1})(m_{1} - a_{2}) - (s_{2} - 2a_{2})(m_{2} - a_{1}) [/math]
Returning to vector notation:
[math] D = \big((H+P)\cdot u - 2A\cdot u\big)(M-A)\cdot v - \big((P+Q)\cdot v - 2A\cdot v\big)(M-A)\cdot u [/math]
Thus the identity [imath]LHS = RHS[/imath] holds provided
[math] (H+P-2A)\cdot u = 0, \qquad (P+Q-2A)\cdot v = 0 [/math]
Equivalently,
[math] (H+P-2A)\perp (H-P), \qquad (P+Q-2A)\perp (P-Q) [/math]
This conclude the algebraic portion of the proof. In the next section, we return to the specific problem at hand, incorporating the definitions of [imath]A, B, C, H, P, Q[/imath]. This step will involve a considerable amount of elementary algebra—entirely manageable, though admittedly quite tedious. For now, I’ll be pausing work on this proof, as the process has been mentally exhausting.
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A test with [imath]A=–i, B=1, C=i[/imath], showed that [imath]LHS = RHS = –1/2[/imath], so the identity holds but the vectors are not perpendicular. What failed was only the claim that [imath]D=0[/imath] follows from those “orthogonality” equations. So we can’t guarantee [imath]D=0[/imath] from that reasoning. This means the orthogonality condition approach is flawed: it adds a condition that isn’t always satisfied.Using the complex–vector dictionary:
[math] x\cdot y = \Re(\overline{x}y), \qquad x\times y = \Im(\overline{x}y) [/math]Left hand side
[math] -\Im(\overline{A}M)\big[\Im(H\overline P)-\Im(H\overline Q)+\Im(P\overline Q)\big]=\Im(\bar{A}M)\Im\left(\bar{u}v\right)=(A\times M)(u\times v) [/math]
By the 2-D Lagrange identity
[math] (X\times Y)(U\times V) = (X\cdot U)(Y\cdot V) - (X\cdot V)(Y\cdot U) [/math]with [imath]X=A, Y=M, U=u, V=v[/imath], we obtain
[math] \text{LHS} = (A\cdot u)(M\cdot v) - (A\cdot v)(M\cdot u) [/math]
Reformulating the right-hand side
[math] \tfrac12\Big[(|H|^2-|P|^2)\Re(\overline{M-A}v) - (|P|^2-|Q|^2)\Re(\overline u(M-A))\Big] [/math]
Note that
[math] |X|^2-|Y|^2 = (X+Y)\cdot(X-Y) [/math]
Also
[math] \Re(\overline{M-A}v) = (M-A)\cdot v \qquad \Re(\overline u(M-A)) = u\cdot(M-A) [/math]
Thus
[math] 2\text{RHS} = (H+P)\cdot u(M-A)\cdot v - (P+Q)\cdot vu\cdot(M-A) [/math]
Expanding:
[math] = (H+P)\cdot uM\cdot v - (H+P)\cdot uA\cdot v - (P+Q)\cdot vu\cdot M + (P+Q)\cdot vu\cdot A [/math]
We now calculate the difference between the right-hand side (RHS) and the left-hand side (LHS), aiming to demonstrate that the result is zero.
Let
[math] D = 2\text{RHS} - 2\text{LHS} [/math]We define scalars:
[math]s_{1} = (H+P)\cdot u, \quad s_{2} = (P+Q)\cdot v[/math][math]a_{1} = A\cdot u, \quad a_{2} = A\cdot v[/math][math]m_{1} = M\cdot v,\quad m_{2} = M\cdot u[/math]
Start with
[math] D = s_{1}m_{1} - s_{1}a_{2} - s_{2}m_{2} + s_{2}a_{1} - 2(a_{1}m_{1} - a_{2}m_{2}) [/math]
Group by [imath]m_{1},m_{2},a_{1},a_{2}[/imath]:
[math] D = m_{1}(s_{1} - 2a_{1}) - m_{2}(s_{2} - 2a_{2}) - s_{1}a_{2} + s_{2}a_{1} [/math]
Now rewrite the last part by adding and subtracting [imath]2a_{1}a_{2}[/imath]:
[math] D = m_{1}(s_{1} - 2a_{1}) - a_{2}(s_{1} - 2a_{1}) - m_{2}(s_{2} - 2a_{2}) + a_{1}(s_{2} - 2a_{2}) [/math]
Factor pairs:
[math] D = (s_{1} - 2a_{1})(m_{1} - a_{2}) - (s_{2} - 2a_{2})(m_{2} - a_{1}) [/math]
Check by expansion:
[math] (s_{1} - 2a_{1})(m_{1} - a_{2}) = s_{1}m_{1} - s_{1}a_{2} - 2a_{1}m_{1} + 2a_{1}a_{2} [/math][math] -(s_{2} - 2a_{2})(m_{2} - a_{1}) = -s_{2}m_{2} + s_{2}a_{1} + 2a_{2}m_{2} - 2a_{1}a_{2} [/math]Adding cancels [imath]\pm 2a_{1}a_{2}[/imath], recovering the original [imath]D[/imath].
So the exact factorization is
[math] D = (s_{1} - 2a_{1})(m_{1} - a_{2}) - (s_{2} - 2a_{2})(m_{2} - a_{1}) [/math]
Returning to vector notation:
[math] D = \big((H+P)\cdot u - 2A\cdot u\big)(M-A)\cdot v - \big((P+Q)\cdot v - 2A\cdot v\big)(M-A)\cdot u [/math]
Thus the identity [imath]LHS = RHS[/imath] holds provided
[math] (H+P-2A)\cdot u = 0, \qquad (P+Q-2A)\cdot v = 0 [/math]
Equivalently,
[math] (H+P-2A)\perp (H-P), \qquad (P+Q-2A)\perp (P-Q) [/math]
This conclude the algebraic portion of the proof. In the next section, we return to the specific problem at hand, incorporating the definitions of [imath]A, B, C, H, P, Q[/imath]. This step will involve a considerable amount of elementary algebra—entirely manageable, though admittedly quite tedious. For now, I’ll be pausing work on this proof, as the process has been mentally exhausting.
A test with [imath]A=–i, B=1, C=i[/imath], showed that [imath]LHS = RHS = –1/2[/imath], so the identity holds but the vectors are not perpendicular. What failed was only the claim that [imath]D=0[/imath] follows from those “orthogonality” equations. So we can’t guarantee [imath]D=0[/imath] from that reasoning. This means the orthogonality condition approach is flawed: it adds a condition that isn’t always satisfied.
Okay, I think I’ve got the right idea now. Let [math]D=\left(M-A\right)\cdot\left(\alpha v-\beta u\right)[/math]where [imath]\alpha, \beta[/imath] are defined byUsing the complex–vector dictionary:
[math] x\cdot y = \Re(\overline{x}y), \qquad x\times y = \Im(\overline{x}y) [/math]Left hand side
[math] -\Im(\overline{A}M)\big[\Im(H\overline P)-\Im(H\overline Q)+\Im(P\overline Q)\big]=\Im(\bar{A}M)\Im\left(\bar{u}v\right)=(A\times M)(u\times v) [/math]
By the 2-D Lagrange identity
[math] (X\times Y)(U\times V) = (X\cdot U)(Y\cdot V) - (X\cdot V)(Y\cdot U) [/math]with [imath]X=A, Y=M, U=u, V=v[/imath], we obtain
[math] \text{LHS} = (A\cdot u)(M\cdot v) - (A\cdot v)(M\cdot u) [/math]
Reformulating the right-hand side
[math] \tfrac12\Big[(|H|^2-|P|^2)\Re(\overline{M-A}v) - (|P|^2-|Q|^2)\Re(\overline u(M-A))\Big] [/math]
Note that
[math] |X|^2-|Y|^2 = (X+Y)\cdot(X-Y) [/math]
Also
[math] \Re(\overline{M-A}v) = (M-A)\cdot v \qquad \Re(\overline u(M-A)) = u\cdot(M-A) [/math]
Thus
[math] 2\text{RHS} = (H+P)\cdot u(M-A)\cdot v - (P+Q)\cdot vu\cdot(M-A) [/math]
Expanding:
[math] = (H+P)\cdot uM\cdot v - (H+P)\cdot uA\cdot v - (P+Q)\cdot vu\cdot M + (P+Q)\cdot vu\cdot A [/math]
We now calculate the difference between the right-hand side (RHS) and the left-hand side (LHS), aiming to demonstrate that the result is zero.
Let
[math] D = 2\text{RHS} - 2\text{LHS} [/math]We define scalars:
[math]s_{1} = (H+P)\cdot u, \quad s_{2} = (P+Q)\cdot v[/math][math]a_{1} = A\cdot u, \quad a_{2} = A\cdot v[/math][math]m_{1} = M\cdot v,\quad m_{2} = M\cdot u[/math]
Start with
[math] D = s_{1}m_{1} - s_{1}a_{2} - s_{2}m_{2} + s_{2}a_{1} - 2(a_{1}m_{1} - a_{2}m_{2}) [/math]
Group by [imath]m_{1},m_{2},a_{1},a_{2}[/imath]:
[math] D = m_{1}(s_{1} - 2a_{1}) - m_{2}(s_{2} - 2a_{2}) - s_{1}a_{2} + s_{2}a_{1} [/math]
Now rewrite the last part by adding and subtracting [imath]2a_{1}a_{2}[/imath]:
[math] D = m_{1}(s_{1} - 2a_{1}) - a_{2}(s_{1} - 2a_{1}) - m_{2}(s_{2} - 2a_{2}) + a_{1}(s_{2} - 2a_{2}) [/math]
Factor pairs:
[math] D = (s_{1} - 2a_{1})(m_{1} - a_{2}) - (s_{2} - 2a_{2})(m_{2} - a_{1}) [/math]
Check by expansion:
[math] (s_{1} - 2a_{1})(m_{1} - a_{2}) = s_{1}m_{1} - s_{1}a_{2} - 2a_{1}m_{1} + 2a_{1}a_{2} [/math][math] -(s_{2} - 2a_{2})(m_{2} - a_{1}) = -s_{2}m_{2} + s_{2}a_{1} + 2a_{2}m_{2} - 2a_{1}a_{2} [/math]Adding cancels [imath]\pm 2a_{1}a_{2}[/imath], recovering the original [imath]D[/imath].
So the exact factorization is
[math] D = (s_{1} - 2a_{1})(m_{1} - a_{2}) - (s_{2} - 2a_{2})(m_{2} - a_{1}) [/math]
Returning to vector notation:
[math] D = \big((H+P)\cdot u - 2A\cdot u\big)(M-A)\cdot v - \big((P+Q)\cdot v - 2A\cdot v\big)(M-A)\cdot u [/math]
Thus the identity [imath]LHS = RHS[/imath] holds provided
[math] (H+P-2A)\cdot u = 0, \qquad (P+Q-2A)\cdot v = 0 [/math]
Equivalently,
[math] (H+P-2A)\perp (H-P), \qquad (P+Q-2A)\perp (P-Q) [/math]
This conclude the algebraic portion of the proof. In the next section, we return to the specific problem at hand, incorporating the definitions of [imath]A, B, C, H, P, Q[/imath]. This step will involve a considerable amount of elementary algebra—entirely manageable, though admittedly quite tedious. For now, I’ll be pausing work on this proof, as the process has been mentally exhausting.
[math]\alpha:=(H+P-2A)\cdot u=(H+P-2A)\cdot(H-P)[/math][math]\beta=(P+Q-2A)\cdot v =(P+Q-2A)\cdot (P-Q)[/math]So proving the original identity is equivalent to proving [math]\left(M-A\right)\cdot\left(\alpha v-\beta u\right)=0[/math]Thus there are two routes to make [imath]D=0[/imath].
(i) Prove the vector [imath]\alpha v-\beta u[/imath] is the zero vector.
(ii) Prove [imath]\alpha v-\beta u[/imath] is orthogonal to [imath]M-A[/imath], i.e. [imath]\alpha v-\beta u[/imath] is proportional to [imath]i(M-A)[/imath], equivalently a 90° rotation of [imath]M-A[/imath], which immediately implies the dot product with [imath]M-A[/imath] vanishes.
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We use the unit‑circle model and fix the circumcenter at the origin. Take
[math]A=-i,\qquad B=e^{i\beta},\qquad C=e^{i\gamma}[/math][math]H=A+B+C,\qquad M=\frac{B+C}{2}[/math][math]P=-i\,p,\qquad Q=-i\,q\qquad(p,q\in\mathbb R)[/math]Define the direction vectors used in the perpendicular‑bisector forms:
[math]u:=H-P,\qquad v:=P-Q[/math]
Finally define the two scalar quantities (these came from the factorization of the long difference ):
[math]\alpha:=(H+P-2A)\cdot u,\qquad \beta:=(P+Q-2A)\cdot v[/math]
Our goal is to show the compact identity
[math](M-A)\cdot(\alpha v-\beta u)=0[/math]
Decompose along [imath]OA[/imath] and its perpendicular
Write with
[math]R=\Re(B+C)=\cos\beta+\cos\gamma,\qquad S=\Im(B+C)=\sin\beta+\sin\gamma[/math][math]H = A + B + C = -i + (R + iS) = R + i(S-1)[/math]
Using we get
[math]u = H - P = R + i(S-1+p) = R + i\,u_0,\qquad u_0 := S-1+p\in\mathbb R[/math]
Also
[math]H+P-2A = (R) + i(S+1-p) = R + i\alpha_0,\qquad \alpha_0 := S+1-p[/math][math]P+Q-2A = i(2-p-q) = i\beta_0,\qquad \beta_0:=2-p-q[/math]
Finally write
[math]M-A = \frac{B+C}{2} - (-i) = \frac R2 + i\Big(\frac S2 + 1\Big) =: X + iY[/math]
Compute and in these coordinates using the dot product :
[math]\alpha=(H+P-2A)\cdot u= \Re\big( (R-i\alpha_0)(R+i u_0)\big)= R^2 + \alpha_0 u_0[/math]
[math]\beta= (P+Q-2A)\cdot v= \Re\big( (i\beta_0)\cdot(-i v_0)\big)= -\beta_0 v_0[/math]
So compactly,
[math]\alpha = R^2 + \alpha_0 u_0,\qquad \beta = -\beta_0 v_0[/math]
Form [imath]\alpha v - \beta u[/imath] and take dot with [imath]M-A[/imath] using [imath]v=-iv_0,u=R+iu_0[/imath]
[math]\alpha v - \beta u = \alpha(-i v_0) - (-\beta_0 v_0)(R+i u_0) = -\beta R + i\big(-\alpha v_0 - \beta u_0\big)[/math]
Now compute the dot product again.
[math](M-A)\cdot(\alpha v-\beta u)= X(-\beta R) + Y(-\alpha v_0 - \beta u_0)= -\beta R X - Y(\alpha v_0 + \beta u_0)[/math]
Substitute and to factor out (we may assume because the degenerate case [imath]v_0=0[/imath] means and the statement is trivial (in this case [imath]P=Q[/imath]):
[math](M-A)\cdot(\alpha v-\beta u) = v_0\Big( Y(\alpha - \beta_0 u_0) - \frac{\beta_0}{2}R^2 \Big)[/math]
Thus the orthogonality condition is equivalent to the single scalar identity
[math]Y(\alpha - \beta_0 u_0) \;=\; \frac{\beta_0}{2}R^2\,\quad (1)[/math]
So we have reduced the problem to verifying [imath](1)[/imath]
Substitute explicit altitude intersection formulas from the geometry (altitude from [imath]B[/imath] perpendicular to [imath]AC[/imath], intersecting the imaginary axis), one computes
[math]p = -\sin\beta - \frac{\cos\beta\cos\gamma}{1+\sin\gamma}\qquad q = -\sin\gamma - \frac{\cos\gamma\cos\beta}{1+\sin\beta}[/math]
Using these formulas one can get explicit expressions for [imath]u_0,\alpha_0,\beta_0[/imath] in terms of [imath]\beta,\gamma[/imath] and for [imath]R,S,Y[/imath] in terms of [imath]\beta,\gamma[/imath]. Expanding numerator and denominator and clearing the common positive denominator reduces the check to verifying that a certain polynomial is identically zero using basic trig identities.
[math]A=-i,\qquad B=e^{i\beta},\qquad C=e^{i\gamma}[/math][math]H=A+B+C,\qquad M=\frac{B+C}{2}[/math][math]P=-i\,p,\qquad Q=-i\,q\qquad(p,q\in\mathbb R)[/math]Define the direction vectors used in the perpendicular‑bisector forms:
[math]u:=H-P,\qquad v:=P-Q[/math]
Finally define the two scalar quantities (these came from the factorization of the long difference ):
[math]\alpha:=(H+P-2A)\cdot u,\qquad \beta:=(P+Q-2A)\cdot v[/math]
Our goal is to show the compact identity
[math](M-A)\cdot(\alpha v-\beta u)=0[/math]
Decompose along [imath]OA[/imath] and its perpendicular
Write with
[math]R=\Re(B+C)=\cos\beta+\cos\gamma,\qquad S=\Im(B+C)=\sin\beta+\sin\gamma[/math][math]H = A + B + C = -i + (R + iS) = R + i(S-1)[/math]
Using we get
[math]u = H - P = R + i(S-1+p) = R + i\,u_0,\qquad u_0 := S-1+p\in\mathbb R[/math]
Also
[math]H+P-2A = (R) + i(S+1-p) = R + i\alpha_0,\qquad \alpha_0 := S+1-p[/math][math]P+Q-2A = i(2-p-q) = i\beta_0,\qquad \beta_0:=2-p-q[/math]
Finally write
[math]M-A = \frac{B+C}{2} - (-i) = \frac R2 + i\Big(\frac S2 + 1\Big) =: X + iY[/math]
Compute and in these coordinates using the dot product :
[math]\alpha=(H+P-2A)\cdot u= \Re\big( (R-i\alpha_0)(R+i u_0)\big)= R^2 + \alpha_0 u_0[/math]
[math]\beta= (P+Q-2A)\cdot v= \Re\big( (i\beta_0)\cdot(-i v_0)\big)= -\beta_0 v_0[/math]
So compactly,
[math]\alpha = R^2 + \alpha_0 u_0,\qquad \beta = -\beta_0 v_0[/math]
Form [imath]\alpha v - \beta u[/imath] and take dot with [imath]M-A[/imath] using [imath]v=-iv_0,u=R+iu_0[/imath]
[math]\alpha v - \beta u = \alpha(-i v_0) - (-\beta_0 v_0)(R+i u_0) = -\beta R + i\big(-\alpha v_0 - \beta u_0\big)[/math]
Now compute the dot product again.
[math](M-A)\cdot(\alpha v-\beta u)= X(-\beta R) + Y(-\alpha v_0 - \beta u_0)= -\beta R X - Y(\alpha v_0 + \beta u_0)[/math]
Substitute and to factor out (we may assume because the degenerate case [imath]v_0=0[/imath] means and the statement is trivial (in this case [imath]P=Q[/imath]):
[math](M-A)\cdot(\alpha v-\beta u) = v_0\Big( Y(\alpha - \beta_0 u_0) - \frac{\beta_0}{2}R^2 \Big)[/math]
Thus the orthogonality condition is equivalent to the single scalar identity
[math]Y(\alpha - \beta_0 u_0) \;=\; \frac{\beta_0}{2}R^2\,\quad (1)[/math]
So we have reduced the problem to verifying [imath](1)[/imath]
Substitute explicit altitude intersection formulas from the geometry (altitude from [imath]B[/imath] perpendicular to [imath]AC[/imath], intersecting the imaginary axis), one computes
[math]p = -\sin\beta - \frac{\cos\beta\cos\gamma}{1+\sin\gamma}\qquad q = -\sin\gamma - \frac{\cos\gamma\cos\beta}{1+\sin\beta}[/math]
Using these formulas one can get explicit expressions for [imath]u_0,\alpha_0,\beta_0[/imath] in terms of [imath]\beta,\gamma[/imath] and for [imath]R,S,Y[/imath] in terms of [imath]\beta,\gamma[/imath]. Expanding numerator and denominator and clearing the common positive denominator reduces the check to verifying that a certain polynomial is identically zero using basic trig identities.
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We use the unit‑circle model and fix the circumcenter at the origin. Take
[math]A=-i,\qquad B=e^{i\beta},\qquad C=e^{i\gamma}[/math][math]H=A+B+C,\qquad M=\frac{B+C}{2}[/math][math]P=-i\,p,\qquad Q=-i\,q\qquad(p,q\in\mathbb R)[/math]Define the direction vectors used in the perpendicular‑bisector forms:
[math]u:=H-P,\qquad v:=P-Q[/math]
Finally define the two scalar quantities (these came from the factorization of the long difference ):
[math]\alpha:=(H+P-2A)\cdot u,\qquad \beta:=(P+Q-2A)\cdot v[/math]
Our goal is to show the compact identity
[math](M-A)\cdot(\alpha v-\beta u)=0[/math]
Decompose along [imath]OA[/imath] and its perpendicular
Write with
[math]R=\Re(B+C)=\cos\beta+\cos\gamma,\qquad S=\Im(B+C)=\sin\beta+\sin\gamma[/math][math]H = A + B + C = -i + (R + iS) = R + i(S-1)[/math]
Using we get
[math]u = H - P = R + i(S-1+p) = R + i\,u_0,\qquad u_0 := S-1+p\in\mathbb R[/math]
Also
[math]H+P-2A = (R) + i(S+1-p) = R + i\alpha_0,\qquad \alpha_0 := S+1-p[/math][math]P+Q-2A = i(2-p-q) = i\beta_0,\qquad \beta_0:=2-p-q[/math]
Finally write
[math]M-A = \frac{B+C}{2} - (-i) = \frac R2 + i\Big(\frac S2 + 1\Big) =: X + iY[/math]
Compute and in these coordinates using the dot product :
[math]\alpha=(H+P-2A)\cdot u= \Re\big( (R-i\alpha_0)(R+i u_0)\big)= R^2 + \alpha_0 u_0[/math]
[math]\beta= (P+Q-2A)\cdot v= \Re\big( (i\beta_0)\cdot(-i v_0)\big)= -\beta_0 v_0[/math]
So compactly,
[math]\alpha = R^2 + \alpha_0 u_0,\qquad \beta = -\beta_0 v_0[/math]
Form [imath]\alpha v - \beta u[/imath] and take dot with [imath]M-A[/imath] using [imath]v=-iv_0,u=R+iu_0[/imath]
[math]\alpha v - \beta u = \alpha(-i v_0) - (-\beta_0 v_0)(R+i u_0) = -\beta R + i\big(-\alpha v_0 - \beta u_0\big)[/math]
Now compute the dot product again.
[math](M-A)\cdot(\alpha v-\beta u)= X(-\beta R) + Y(-\alpha v_0 - \beta u_0)= -\beta R X - Y(\alpha v_0 + \beta u_0)[/math]
Substitute and to factor out (we may assume because the degenerate case [imath]v_0=0[/imath] means and the statement is trivial (in this case [imath]P=Q[/imath]):
[math](M-A)\cdot(\alpha v-\beta u) = v_0\Big( Y(\alpha - \beta_0 u_0) - \frac{\beta_0}{2}R^2 \Big)[/math]
Thus the orthogonality condition is equivalent to the single scalar identity
[math]Y(\alpha - \beta_0 u_0) \;=\; \frac{\beta_0}{2}R^2\,\quad (1)[/math]
So we have reduced the problem to verifying [imath](1)[/imath]
Substitute explicit altitude intersection formulas from the geometry (altitude from [imath]B[/imath] perpendicular to [imath]AC[/imath], intersecting the imaginary axis), one computes
[math]p = -\sin\beta - \frac{\cos\beta\cos\gamma}{1+\sin\gamma}\qquad q = -\sin\gamma - \frac{\cos\gamma\cos\beta}{1+\sin\beta}[/math]
Using these formulas one can get explicit expressions for [imath]u_0,\alpha_0,\beta_0[/imath] in terms of [imath]\beta,\gamma[/imath] and for [imath]R,S,Y[/imath] in terms of [imath]\beta,\gamma[/imath]. Expanding numerator and denominator and clearing the common positive denominator reduces the check to verifying that a certain polynomial is identically zero using basic trig identities.
We already had
[math]\alpha=R^2+\alpha_0u_0,\qquad \beta=-\beta_0 v_0,\qquad \alpha_0:=S+1-p,\ \ \beta_0:=2-p-q[/math]
Our goal reduces to the scalar identity
[math]Y(\alpha-\beta_0 u_0)=\frac{\beta_0}{2}R^2\qquad (\star)[/math]
Key simplifications
Use the classical “tangent half‑angle” trick in its gentlest form:
[math]\frac{\cos\gamma}{1+\sin\gamma}=\frac{1-\sin\gamma}{\cos\gamma},\qquad \frac{\cos\beta}{1+\sin\beta}=\frac{1-\sin\beta}{\cos\beta}[/math]
[math]t_\gamma:=\frac{1-\sin\gamma}{\cos\gamma},\qquad t_\beta:=\frac{1-\sin\beta}{\cos\beta}[/math]
[math]p= -\sin\beta - \cos\beta\,t_\gamma,\qquad q= -\sin\gamma - \cos\gamma\,t_\beta[/math]
A clean formula for [imath]u_0[/imath]
[math]u_0=S-1+p = (\sin\beta+\sin\gamma)-1 -\sin\beta - \cos\beta\,t_\gamma = -(1-\sin\gamma)\Big(1+\frac{\cos\beta}{\cos\gamma}\Big)[/math][math][/math][math]\,u_0 = -\,t_\gamma\,(\cos\beta+\cos\gamma)\,= -\,t_\gamma\,R[/math]
Handy expressions for [imath]\alpha_0[/imath] and [imath]\beta_0[/imath].
[math]\alpha_0=S+1-p = (\sin\beta+\sin\gamma)+1 +\sin\beta +\cos\beta\,t_\gamma = 1+2\sin\beta+\sin\gamma+\cos\beta\,t_\gamma[/math]
[math]\beta_0=2-p-q =2+\sin\beta+\cos\beta t_{\gamma}+\sin \gamma+\cos \gamma t_{\beta}=2+S+\cos\beta t_{\gamma}+\cos \gamma t_{\beta}[/math]
Rewrite [imath]\alpha-\beta_0u_0[/imath] using [imath]\alpha=R^2+\alpha_0u_0[/imath] and [imath]u_0=-Rt_\gamma[/imath]
[math]\alpha-\beta_0 u_0 = R^2 + u_0(\alpha_0-\beta_0) = R^2 - R t_\gamma\big(\alpha_0-\beta_0\big)[/math]
[math]\alpha_0-\beta_0 = (1+2\sin\beta+\sin\gamma+\cos\beta t_\gamma) - (2+\sin\beta+\sin\gamma+\cos\beta t_\gamma+\cos\gamma t_\beta) = \sin\beta -1 - \cos\gamma\,t_\beta[/math]
[math]\alpha-\beta_0 u_0 = R^2 - R t_\gamma\big(\sin\beta-1-\cos\gamma t_\beta\big) = R^2 + R t_\gamma\big(1-\sin\beta+\cos\gamma t_\beta\big)[/math]
Multiply by [imath]Y[/imath] and compare with [imath](\star)[/imath]
[math]Y(\alpha-\beta_0 u_0) = YR^2 + YR\,t_\gamma\big(1-\sin\beta+\cos\gamma t_\beta\big)[/math]
[math](Y-\tfrac{\beta_0}{2})R^2 + YR\,t_\gamma\big(1-\sin\beta+\cos\gamma t_\beta\big)=0[/math]Expand the expression
[math]\frac{\beta_0}{2}-Y=\frac12\big(\cos\beta\,t_\gamma+\cos\gamma\,t_\beta\big)\quad\Longrightarrow\quad Y-\frac{\beta_0}{2}=-\frac12\big(\cos\beta\,t_\gamma+\cos\gamma\,t_\beta\big)[/math]
[math]-\frac12 R\big(\cos\beta\,t_\gamma+\cos\gamma\,t_\beta\big) \;+\; Y\,t_\gamma\big(1-\sin\beta+\cos\gamma t_\beta\big)=0[/math]
[math]-\,R\big(\cos\beta\,t_\gamma+\cos\gamma\,t_\beta\big) \;+\;(2+S)\,t_\gamma\big(1-\sin\beta+\cos\gamma t_\beta\big)=0[/math]
Insert the values of [imath]t_\beta[/imath] and [imath]t\gamma[/imath]
[math]-(\cos\beta+\cos\gamma)\Big(\cos\beta\,\frac{1-\sin\gamma}{\cos\gamma} +\cos\gamma\,\frac{1-\sin\beta}{\cos\beta}\Big)+(2+\sin\beta+\sin\gamma)\,\frac{1-\sin\gamma}{\cos\gamma}\Big((1-\sin\beta)+\cos\gamma\,\frac{1-\sin\beta}{\cos\beta}\Big)=0[/math].
Now note that [math](1-\sin\beta)(1+\sin\beta)=\cos^2\beta[/math][math](1-\sin\gamma)(1+\sin\gamma)=cos^2\gamma[/math]
Compute each part separately, we start with the righthand side of the expression
Multiply by [imath]1+\sin\beta[/imath]
[math](1+\sin\beta)\frac{(1-\sin\gamma)}{\cos\gamma}(1-\sin\beta)(1+\frac{\cos\gamma}{\cos\beta})=\cos^2\beta t_\gamma+\cos\beta(1-\sin\gamma)[/math]
And similarly multiply by [imath]1+\sin\gamma[/imath]
[math](1+\sin\gamma)\frac{(1-\sin\gamma)}{\cos\gamma}(1-\sin\beta)(1+\frac{\cos\gamma}{\cos\beta})=\cos^2\gamma t_\beta+\cos\gamma(1-\sin\beta)[/math]
Hence
[math](2+S)\,t_\gamma\big(1-\sin\beta+\cos\gamma t_\beta\big) =\cos^2\beta t_\gamma+\cos^2\gamma t_\beta+\cos\beta(1-sin\gamma)+cos\gamma(1-\sin\beta)[/math]
Remains to compute the lefthand side
[math]-R(\cos\beta t_\gamma+\cos\gamma t_\beta)=-(\cos\beta+\cos\gamma)\left(\cos\beta\frac{1-\sin\gamma}{\cos\gamma}+\cos\gamma\frac{1-\sin\beta}{\cos\beta}\right)[/math][math]-R(\cos\beta t_\gamma+\cos\gamma t_\beta)=-\cos^2\beta t_\gamma-\cos^2\gamma t_\beta-\cos\beta(1-\sin\gamma)-\cos\gamma(1-\sin\beta)[/math]
Now observe the sweet cancellation! This matches exactly the negative of the other expression and we see the cancellation is transparent when we add back everything vanishes. Therefore [imath](\star)[/imath] holds, and thus [imath](M-A)\cdot(\alpha v-\beta u)=0[/imath]. This implies the circumcenter of [imath]\Delta HPQ[/imath] lies on the line [imath]AM[/imath] as required [imath]\square[/imath].
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Since the proof turned out to be quite extensive, I believe a summary is in order.
Part 1 [imath]\text{—}[/imath] Reduction to a constant term
Reformulated the collinearity condition [imath]A,O,M[/imath] as vanishing of a real-linear form [imath]W(Z)[/imath] at the circumcenter [imath]O'[/imath]. Reduced the problem to showing a constant term involving coefficients [imath]s[/imath] and [imath]t[/imath] vanishes.
Part 2 [imath]\text{—}[/imath] Solving for coefficients
Solved for [imath]s[/imath] and [imath]t[/imath] using Cramer's rule and expressed the constant term in terms of [imath]u,v,H,P,Q,A,M[/imath]. Reduced the verification to a single scalar identity involving real and imaginary parts.
Part 3 [imath]\text{—}[/imath] Vector reformulation
Translated the scalar identity into vector form using dot and cross products. Factored the difference to the compact expression [math]D=(M-A)\cdot(\alpha v-\beta u)[/math]
Part 4 [imath]\text{—}[/imath] Unit circle coordinates
Specialized to the unit circle model [imath](A=-i, B=e^{i\beta},C=e^{i\gamma})[/imath] and computed explicit coordinates for all relevant points and vectors. Reduced the problem to a single scalar trigonometric identity [imath](\star)[/imath] in terms of [imath]R,S,t_\beta,t_\gamma,u_0,\alpha_0,\beta_0[/imath].
Part 5 [imath]\text{—}[/imath] Trigonometric verification
Simplified the identity [imath](\star)[/imath] using tangent half-angle formulas. This allowed us to express all quantities explicitly in terms of [imath]\beta[/imath] and [imath]\gamma[/imath]. The identity simplifies to a trigonometric expression where the left- and right-hand sides cancel perfectly. Therefore, the scalar identity holds, and it follows that the circumcenter of [imath]\Delta HPQ[/imath] lies on the line [imath]AM[/imath].
I am also curious whether others might find a more synthetic or elegant solution that avoids so many technical calculations and long algebraic manipulations. The current approach relies heavily on complex coordinates, trigonometric substitutions, and careful term-by-term cancellation, which, while rigorous, can obscure the underlying geometric intuition. It would be interesting to see if there is a method that leverages more geometric properties, symmetry, or classical theorems to reach the same conclusion in a more conceptual or concise manner, perhaps making the proof shorter and highlighting the key ideas more transparently.
Part 1 [imath]\text{—}[/imath] Reduction to a constant term
Reformulated the collinearity condition [imath]A,O,M[/imath] as vanishing of a real-linear form [imath]W(Z)[/imath] at the circumcenter [imath]O'[/imath]. Reduced the problem to showing a constant term involving coefficients [imath]s[/imath] and [imath]t[/imath] vanishes.
Part 2 [imath]\text{—}[/imath] Solving for coefficients
Solved for [imath]s[/imath] and [imath]t[/imath] using Cramer's rule and expressed the constant term in terms of [imath]u,v,H,P,Q,A,M[/imath]. Reduced the verification to a single scalar identity involving real and imaginary parts.
Part 3 [imath]\text{—}[/imath] Vector reformulation
Translated the scalar identity into vector form using dot and cross products. Factored the difference to the compact expression [math]D=(M-A)\cdot(\alpha v-\beta u)[/math]
Part 4 [imath]\text{—}[/imath] Unit circle coordinates
Specialized to the unit circle model [imath](A=-i, B=e^{i\beta},C=e^{i\gamma})[/imath] and computed explicit coordinates for all relevant points and vectors. Reduced the problem to a single scalar trigonometric identity [imath](\star)[/imath] in terms of [imath]R,S,t_\beta,t_\gamma,u_0,\alpha_0,\beta_0[/imath].
Part 5 [imath]\text{—}[/imath] Trigonometric verification
Simplified the identity [imath](\star)[/imath] using tangent half-angle formulas. This allowed us to express all quantities explicitly in terms of [imath]\beta[/imath] and [imath]\gamma[/imath]. The identity simplifies to a trigonometric expression where the left- and right-hand sides cancel perfectly. Therefore, the scalar identity holds, and it follows that the circumcenter of [imath]\Delta HPQ[/imath] lies on the line [imath]AM[/imath].
I am also curious whether others might find a more synthetic or elegant solution that avoids so many technical calculations and long algebraic manipulations. The current approach relies heavily on complex coordinates, trigonometric substitutions, and careful term-by-term cancellation, which, while rigorous, can obscure the underlying geometric intuition. It would be interesting to see if there is a method that leverages more geometric properties, symmetry, or classical theorems to reach the same conclusion in a more conceptual or concise manner, perhaps making the proof shorter and highlighting the key ideas more transparently.
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