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Proposition 2. \( {O}_{a}{A}^{\prime } \) is parallel to \( {OH} \) . | Proof. Let \( {X}_{a},{B}_{a}^{\prime },{C}_{a}^{\prime } \) be the pedals of \( {O}_{a} \) on \( {BC},{CA},{AB} \) respectively. From\n\n\[ \frac{A{M}_{b}}{A{B}_{a}^{\prime }} = \frac{AO}{A{O}_{a}} = \frac{A{M}_{c}}{A{C}_{a}^{\prime }} \]\n\nwe have \( {B}_{a}^{\prime }{C}_{a}^{\prime }//{M}_{b}{M}_{c}//{BC} \) . Ther... | Yes |
Theorem 3. The pedal circle of \( {A}^{\prime } \) (and \( {O}_{a} \) ) is the image of the nine-point circle of \( {ABC} \) under the homothety \( \mathrm{h}\left( {A,{t}_{a}}\right) \), where \( {t}_{a} = \frac{2\sin \beta \sin \gamma }{\cos \alpha } \) . | Proof. The circle \( {B}_{a}{B}_{a}^{\prime }{C}_{a}^{\prime } \) is homothetic to the nine-point circle \( {H}_{b}{M}_{b}{M}_{c} \) at \( A \) since\n\n\[ \frac{A{B}_{a}}{A{H}_{b}} = \frac{A{A}^{\prime }}{AH} = \frac{A{O}_{a}}{AO} = \frac{A{B}_{a}^{\prime }}{A{M}_{b}} = \frac{A{C}_{a}^{\prime }}{A{M}_{c}}. \]\n\nThe r... | Yes |
Theorem 4. \( {N}^{a}{N}^{b}{N}^{c} \) is the anticevian triangle of the nine-point \( N \) . | Proof. Since \( {N}^{a} \) is the midpoint of \( {O}_{a}{A}^{\prime } \) and the nine-point center \( N \) is the midpoint of \( {OH} \), by Proposition \( 2, A, N,{N}_{a} \) are collinear. Similarly, \( {N}^{b} \) and \( {N}^{c} \) are on the cevians \( {BN} \) and \( {CN} \) respectively. We show that the line \( {N}... | Yes |
Theorem 5. The incenter of triangle \( {A}^{\prime \prime }{B}^{\prime \prime }{C}^{\prime \prime } \) is the orthocenter of the orthic triangle \( {\mathrm{H}}_{a}{\mathrm{H}}_{b}{\mathrm{H}}_{c} \), and the incircle touches the sides at the midpoints \( {P}_{a},{P}_{b},{P}_{c} \) of the segments \( {B}_{a}{C}_{a},{C}... | Proof. We first claim that the segments \( {B}_{a}{C}_{a},{C}_{b}{A}_{b},{A}_{c}{B}_{c} \) have equal lengths. Note that the homothety \( \mathrm{h}\left( {A,{t}_{a}}\right) \) maps \( {H}_{b},{H}_{c} \) to \( {B}_{a},{C}_{a} \) respectively. Hence,\n\n\[ \n{B}_{a}{C}_{a} = {t}_{a} \cdot {H}_{b}{H}_{c} = \frac{\sin \be... | Yes |
Theorem 8. The triangles \( {ABC},{A}^{\prime \prime }{B}^{\prime \prime }{C}^{\prime \prime } \), and \( {P}_{a}{P}_{b}{P}_{c} \) are perspective at the symmedian point of triangle \( {ABC} \) | Proof. (1) Since \( A{A}_{c}{A}_{b} \) and \( A{B}_{c}{C}_{b} \) are isosceles triangles, \( {B}_{c}{C}_{b} \) and \( {A}_{c}{A}_{b} \) are parallel, and the triangles \( A{C}_{b}{B}_{c} \) and \( {ABC} \) are homothetic (see Figure 5). Now,\n\n\[ \n\angle {A}^{\prime \prime }{B}_{c}{C}_{b} = \angle {A}^{\prime \prime ... | Yes |
Lemma 9. Let \( {J}_{a} \) be the midpoint of \( {OA} \). The line \( {J}_{a}{M}_{a} \) is perpendicular to \( {Q}_{b}{Q}_{c} \) and contains the midpoint of \( {ON} \). | Proof. Since \( N \) is the midpoint of \( {OH} \), the segment \( {J}_{a}N \) is parallel to \( {AH} \) and therefore to \( O{M}_{a} \) . Furthermore, \( {J}_{a}N = \frac{1}{2}{AH} = O{M}_{a} \) . It follows that \( {J}_{a}{M}_{a} \) intersects \( {ON} \) at its midpoint. | No |
Proposition 10. Given triangle \( {ABC} \) with incentral triangle \( {DEF} \), extend \( {AB} \) and \( {AC} \) to \( P \) and \( Q \) such that \( {BP} = {BC} = {CQ} \) . Let \( T \) be the midpoint \( {PQ} \), and \( M \) the midpoint of the arc \( {BAC} \) of the circumcircle.\n\n(a) The line TM is perpendicular EF... | Proof. (a) By the angle bisector theorem, \( {AE} = \frac{bc}{a + b},{AF} = \frac{bc}{a + c} \) . Therefore, \( \frac{AE}{AF} = \frac{a + c}{a + b} = \frac{AQ}{AP} \), showing that \( {PQ} \) is parallel to \( {EF} \) (see Figure 8). On the other hand, the circumcircles of \( {ABC} \) and \( {APQ} \) intersect at \( A ... | Yes |
Theorem 11. The orthocenter of triangle \( {Q}_{a}{Q}_{b}{Q}_{c} \) is the midpoint of \( {ON} \) . | Proof. It is enough to prove that \( {Q}_{a}{M}_{a} \) is parallel to \( {AN} \) .\n\nLet \( {D}_{a} = {AH} \cap {H}_{b}{H}_{c},{D}_{b} = {BH} \cap {H}_{c}{H}_{a},{D}_{c} = {CH} \cap {H}_{a}{H}_{b} \) . We claim that \( {D}_{b}{D}_{c} \) is perpendicular to \( {AN} \) . The points \( {D}_{b} \) and \( {D}_{c} \) have e... | Yes |
Lemma 2. If \( {a}_{i},{b}_{i} \) are positive real numbers, \( 2 \leq n \) and \( 1 \leq i \leq n - 1 \), and \( \frac{{a}_{i}}{{b}_{i}} = \frac{{a}_{i + 1}}{{b}_{i + 1}} \) then \( \frac{{a}_{i}}{{b}_{i}} = \frac{{a}_{1} + {a}_{2} + \cdots + {a}_{n}}{{b}_{1} + {b}_{2} + \cdots + {b}_{n}} \) . | Proof. \( \frac{{a}_{i}}{{b}_{i}} = \frac{{a}_{i + 1}}{{b}_{i + 1}} \) implies that \( \frac{{a}_{i}}{{b}_{i}} = \frac{{a}_{j}}{{b}_{j}} \) for \( 1 \leq i, j \leq n \) and \( {a}_{i}{b}_{j} = {b}_{i}{a}_{j} \) . Thus,\n\n\[ {a}_{i}{b}_{1} + {a}_{i}{b}_{2} + \cdots + {a}_{i}{b}_{n} = {b}_{i}{a}_{1} + {b}_{i}{a}_{2} + \... | Yes |
Lemma 3. If \( 0 \leq x \leq \frac{\pi }{3} \) then \( f\left( x\right) = \sin \left( {x + \frac{\pi }{3}}\right) + \sin x \) has a minimum value of \( \frac{\sqrt{3}}{2} \) at \( x = 0 \) and a maximum value of \( \sqrt{3} \) at \( x = \frac{\pi }{3} \) . | Proof. For \( 0 \leq x \leq \frac{\pi }{3} \) , \[ {f}^{\prime }\left( x\right) = \cos \left( {x + \frac{\pi }{3}}\right) + \cos x = \sqrt{3}\cos \left( {x + \frac{\pi }{6}}\right) \geq 0, \] which implies that \( f \) is an increasing function on \( \left\lbrack {0,\frac{\pi }{3}}\right\rbrack \) . Thus, \( f\left( 0\... | Yes |
Proposition 4. Assume triangle \( {ABC} \) has sides \( a, b, c \) and opposite vertices \( A \) , \( B, C \), respectively. If \( \angle C = {60}^{ \circ } \), then \( \frac{1}{2} \leq \frac{c}{a + b} < 1 \) . | Proof. Using the law of cosines, \( {c}^{2} = {a}^{2} + {b}^{2} - {2ab}\cos {60}^{ \circ } = {a}^{2} + {b}^{2} - {ab} \) . Now,\n\n\( {\left( 2c\right) }^{2} - {\left( a + b\right) }^{2} = 4\left( {{a}^{2} - {ab} + {b}^{2}}\right) - \left( {{a}^{2} + {2ab} + {b}^{2}}\right) = 3{a}^{2} - {6ab} + 3{b}^{2} = 3{\left( a - ... | Yes |
Theorem 1 ([2,§2.12],[5,§8.3]). \( P/Q = {Q}^{\prime } \) if and only if \( P/{Q}^{\prime } = Q \) . | This is equivalent to \( P/\left( {P/Q}\right) = Q \) . It can be proved by direct verification with coordinates. We offer an indirect proof, with the advantage of an explicit construction, for given \( Q \) and \( {Q}^{\prime } \), of a point \( P \) with \( P/Q = {Q}^{\prime } \) and \( P/{Q}^{\prime } = Q \) .\n\nFo... | Yes |
Proposition 2. Let \( P \) be a fixed point. If \( Q \) moves along a line \( \mathcal{L} \), then the quotient \( P/Q \) traverses the bicevian conic of \( P \) and the trilinear pole of \( \mathcal{L} \) . | Proof. Let \( P = \left( {u : v : w}\right) \) and \( Q \) move along the line \( \mathcal{L} \) with line coordinates \( \left\lbrack {p : q : r}\right\rbrack \) . If \( {Q}^{\prime } = P/Q = \left( {x : y : z}\right) \), then \( Q = P/{Q}^{\prime } \) is on the line \( \mathcal{L} \), and\n\n\[ \n{px}\left( {-\frac{x... | Yes |
Proposition 4. Let \( P \) be a fixed point. If \( Q \) moves along a line \( \mathcal{L} \), then the cevian quotient \( Q/P \) traverses the circumconic of the anticevian triangle of \( P \) with perspector \( {P}_{\mathcal{L}}/P \), where \( {P}_{\mathcal{L}} \) is the trilinear pole of \( \mathcal{L} \) . | Proof. Let \( P = \left( {u : v : w}\right) \) and \( Q \) move along the line \( \mathcal{L} \) with line coordinates \( \left\lbrack {p : q : r}\right\rbrack \) . If \( {Q}^{\prime \prime } = Q/P = \left( {x : y : z}\right) \), then \( Q = \left( {\frac{1}{{wy} + {vz}} : \frac{1}{{uz} + {wx}} : \frac{1}{{vx} + {uy}}}... | Yes |
Proposition 5. Let \( P \) be a fixed point. The locus of \( Q \) for which the line joining \( \left( {P/Q}\right) \) to \( \left( {Q/P}\right) \) is parallel to \( {PQ} \) is the union of the cevian lines \( {AP},{BP},{CP} \) and a conic \( \Gamma \left( P\right) \n\n(1) homothetic to the circumconic with perspector ... | Proof. If \( P = \left( {u : v : w}\right) \) and \( Q = \left( {x : y : z}\right) \), the line joining \( P/Q \) and \( Q/P \) contains the infinite point of \( {PQ} \) if and only if\n\n\[ \left| \begin{matrix} x\left( {-\frac{x}{u} + \frac{y}{v} + \frac{z}{w}}\right) & y\left( {\frac{x}{u} - \frac{y}{v} + \frac{z}{w... | Yes |
Theorem 1 (Power of a point). Let \( \Gamma \) be a circle, and \( P \) a point. Let a line through \( P \) meet \( \Gamma \) at points \( A \) and \( B \), and let another line through \( P \) meet \( \Gamma \) at points \( C \) and \( D \). Then\n\n\[ {PA} \cdot {PB} = {PC} \cdot {PD}\text{.} \] | Proof. There are two configurations to consider, depending on whether \( P \) lies inside the circle or outside the circle. In the case when \( P \) lies inside the circle, as the left diagram in Figure 1. we have \( \angle {PAD} = \angle {PCB} \) and \( \angle {APD} = \angle {CPB} \), so that triangles \( {PAD} \) and... | Yes |
Theorem 2 (Converse to power of a point). Let \( A, B, C, D \) be four distinct points. Let lines \( {AB} \) and \( {CD} \) intersect at \( P \) . Assume that either (1) \( P \) lies on both line segments \( {AB} \) and \( {CD} \), or (2) \( P \) lies on neither line segments. Then \( A, B, C, D \) are concyclic if and... | Proof. The expression \( {PA} \cdot {PB} = {PC} \cdot {PD} \) can be rearranged as \( \frac{PA}{PD} = \frac{PC}{PB} \) . In both configurations described in the statement of the theorem, we have \( \angle {APD} = \angle {CPB} \) . It follows by angles and ratios that triangles \( {APD} \) and \( {CPB} \) are similar (w... | Yes |
Theorem 3 (Radical axis theorem). Given three circles, no two concentric, the three pairwise radical axes are either concurrent or all parallel. | Proof. Denote the three circles by \( {\Gamma }_{1},{\Gamma }_{2} \), and \( {\Gamma }_{2} \), and denote the radical axes of \( {\Gamma }_{i} \) and \( {\Gamma }_{j} \) by \( {\ell }_{ij} \) . Suppose that the radical axes are not all parallel. Let \( {\ell }_{12} \) and \( {\ell }_{13} \) meet at \( X \) . Since \( X... | Yes |
Proposition 0.1.4. 令 \( \mathcal{C} \) 为一圆锥曲線，則\n\n1. \( \mathcal{C} \) 为椭圆若且唯若存在两点 \( {F}_{1},{F}_{2} \) 及正实数 \( a > \overline{{F}_{1}{F}_{2}}/2 \) 使得\n\n\[ \mathcal{C} = \left\{ {P \mid \overline{{F}_{1}P} + \overline{{F}_{2}P} = {2a}}\right\} \]\n\n2. \( \mathcal{C} \) 为抛物線若且唯若存在一點 \( F \) 及一線 \( L \) 使得\n\n\[ \math... | Proof. 只證明橢圓的情形，其餘類似。由圓錐曲線定義知，存在頂點為 \( V \) 的直圆锥面 \( s \) 及平面 \( E \) 满足 \( \mathcal{C} = \mathcal{S} \cap E \) ,令 \( {\mathcal{B}}_{1},{\mathcal{B}}_{2} \) 为与 \( \mathcal{S} \) 及 \( E \) 皆相切的两球,並設 \( \odot \left( {O}_{1}\right) = \mathcal{S} \cap {\mathcal{B}}_{1}, \odot \left( {O}_{2}\right) = \mathcal{S} \cap {\math... | Yes |
Proposition 0.1.5. 令 \( \mathcal{C} \) 为一圆锥曲線,一直線 \( \ell \) 与 \( \mathcal{C} \) 只有一個交點若且唯若 \( \ell \in T\mathcal{C} \) 。 | Proof. 設 \( \mathcal{C} \) 位於平面 \( E \) 上且以 \( \odot \left( O\right) \) 为準線的直圓錐面 \( \mathcal{S} \) 满足 \( \mathcal{C} = \mathcal{S} \cap E \) 。令 \( {E}_{1} \) 为过 \( \mathcal{S} \) 的顶点及 \( \ell \) 的平面, \( {E}_{2} \) 为 \( \odot \left( O\right) \) 所在的平面,則\n\n\[ 1 = \left| {\ell \cap \mathcal{C}}\right| = \left| {E \cap {E}... | No |
Proposition 0.1.6. 令非抛物線的圆锥曲線 \( \mathcal{C} \) 的焦點為 \( {F}_{1},{F}_{2} \) ,則對於 \( \mathcal{C} \) 上任意一點 \( P,{T}_{p}\mathcal{C} \) 为 \( \angle {F}_{1}P{F}_{2} \) 的分角線。 | Proof. 只證明椭圆的情形,雙曲線則類似。令 \( {T}_{P}^{\prime } \) 為 \( \angle {F}_{1}P{F}_{2} \) 的外角平分線, \( {F}_{2}^{\prime } \) 为 \( {F}_{2} \) 關於 \( {T}_{P}^{\prime } \) 的對稱點,則 \( {F}_{1}, P,{F}_{2}^{\prime } \) 共線,所以對於任意一點 \( Q \in {T}_{P}^{\prime } \smallsetminus \{ P\} \)\n\n\[ \overline{{F}_{1}Q} + \overline{{F}_{2}Q} = \overline... | Yes |
Proposition 0.1.7. 令抛物線 \( \mathcal{P} \) 的焦點為 \( F \) ,準線為 \( L \) ,對於 \( \mathcal{P} \) 上任意一點 \( P, P \) 關於 \( L \) 的垂足為 \( K \) ,則 \( {T}_{p}\mathcal{P} \) 為 \( \angle {FPK} \) 的内角平分線且為 \( \overline{FK} \) 的中垂線。 | Proof. 令 \( {T}_{P}^{\prime } \) 为 \( \angle {FPK} \) 的内角平分線,由 \( \overline{FP} = \overline{KP} \) 知 \( {T}_{P}^{\prime } \) 为 \( \overline{FK} \) 的中垂線, 所以對於任意一點 \( Q \in {T}_{P}^{\prime } \smallsetminus \{ P\} \) ,令 \( {K}_{Q} \) 为 \( Q \) 關於 \( L \) 的垂足,我們有\n\n\[ \overline{{K}_{Q}Q} < \overline{KQ} = \overline{FQ} \R... | Yes |
Proposition 0.1.9. 對於非拋物線的圓錐曲線 \( \mathcal{C} \) ,令 \( O \) 為其中心，則 \( \mathcal{C} \) 關於 \( O \) 點對稱。 | Proof. 由焦點的定義。 | No |
Theorem 0.2.2. 给定共點四線 \( {\ell }_{1},{\ell }_{2},{\ell }_{3},{\ell }_{4} \) ,若非無窮遠線的直線 \( L ∌ \bigcap {\ell }_{i} \) 分别交 \( {\ell }_{i} \) 於 \( {P}_{i} \) ,則 \( \left( {P}_{i}\right) 为定值 (与 \( L \) 的選取無關)。當 \( \bigcap {\ell }_{i} \notin {\mathcal{L}}_{\infty } \) 时, | Proof. 令 \( {L}^{\prime } ∌ \bigcap {\ell }_{i} \) 为另外一線且分別交 \( {\ell }_{i} \) 於 \( {P}_{i}^{\prime } \) ,若 \( \bigcap {\ell }_{i} \in {\mathcal{L}}_{\infty } \) ,則顯然有\n\n\[ \frac{{P}_{1}{P}_{3}}{{P}_{3}{P}_{2}} = \frac{{P}_{1}^{\prime }{P}_{3}^{\prime }}{{P}_{3}^{\prime }{P}_{2}^{\prime }},\frac{{P}_{1}{P}_{4}}{{P}_{4... | Yes |
對於共線三點 \( {P}_{1},{P}_{2},{P}_{3} \in \ell \) 及 \( \ell \) 外雨點 \( A, B \) ,若點 \( {P}_{4} \) 滿足 \( {P}_{4} \notin {AB} \) ,則 \( A\left( {P}_{i}\right) = B\left( {P}_{i}\right) \) 若且唯若 \( {P}_{4} \in \ell \) 。 | 證回來不難，留給讀者。令 \( {P}_{i}^{\prime } = {P}_{i}, i = 1,2,3,{P}_{4}^{\prime } = A{P}_{4} \cap \ell \) ,則\n\n\[ B\left( {P}_{i}^{\prime }\right) = \left( {P}_{i}^{\prime }\right) = A\left( {P}_{i}^{\prime }\right) = A\left( {P}_{i}\right) = B\left( {P}_{i}\right) ,\]\n\n因此 \( {P}_{4}^{\prime } \in B{P}_{4} \) ,所以 \( {P}_{4}^... | No |
Proposition 0.2.10. 對於三點 \( {P}_{1},{P}_{2},{P}_{3} \) 及兩點 \( A, B \notin \bigcup {P}_{i}{P}_{j} \) , \[ A\left( {{P}_{1},{P}_{2};{P}_{3}, B}\right) = B\left( {{P}_{1},{P}_{2};{P}_{3}, A}\right) \] 若且唯若 \( {P}_{1},{P}_{2},{P}_{3} \) 共線。 | Proof. 證回來一樣不難，留給讀者。令 \( {P}_{4} = {AB} \cap {P}_{2}{P}_{3} \) ,則 \( {P}_{2},{P}_{3},{P}_{4} \) 共線且 \[ A\left( {{P}_{1},{P}_{2};{P}_{3},{P}_{4}}\right) = A\left( {{P}_{1},{P}_{2};{P}_{3}, B}\right) = B\left( {{P}_{1},{P}_{2};{P}_{3}, A}\right) = B\left( {{P}_{1},{P}_{2};{P}_{3},{P}_{4}}\right) , \] 因此 \( {P}_{1} \in {P... | No |
Theorem 0.2.13 (迪沙格定理/Desargues'). 對於兩個三角形 \( \bigtriangleup {A}_{1}{B}_{1}{C}_{1} \) 與 \( \bigtriangleup {A}_{2}{B}_{2}{C}_{2},{B}_{1}{C}_{1} \cap {B}_{2}{C}_{2},{C}_{1}{A}_{1} \cap {C}_{2}{A}_{2},{A}_{1}{B}_{1} \cap {A}_{2}{B}_{2} \) 共線若且唯若 \( {A}_{1}{A}_{2},{B}_{1}{B}_{2} \) , \( {C}_{1}{C}_{2} \) 共點。 | Proof. 令 \( X = {B}_{1}{B}_{2} \cap {C}_{1}{C}_{2},{P}_{BC} = {B}_{1}{C}_{1} \cap {B}_{2}{C}_{2},{P}_{CA} = {C}_{1}{A}_{1} \cap {C}_{2}{A}_{2},{P}_{AB} = \) \( {A}_{1}{B}_{1} \cap {A}_{2}{B}_{2},{A}_{1}{A}_{2} \) 分别交 \( {B}_{i}{C}_{i} \) 於 \( {Z}_{i} \) 。注意到 \( X \in {A}_{1}{A}_{2} \) 等價於 \( X \in {Z}_{1}{Z}_{2} \) ,即\... | Yes |
Proposition 0.2.14. 给定共圆四點 \( {P}_{1},{P}_{2},{P}_{3},{P}_{4} \in \Omega \) ,若 \( A \) 为 \( \Omega \) 上異於 \( {P}_{i} \) 的點，則 \( A\left( {P}_{i}\right) \) 為定值 (與 \( A \) 的選取無關)。 | Proof. 令 \( {A}^{\prime } \) 为 \( \Omega \) 上異於 \( {P}_{i} \) 的點,由圓周角性質知 \( \measuredangle {P}_{i}A{P}_{j} = \measuredangle {P}_{i}{A}^{\prime }{P}_{j} \) 。因此由 (0.2.2) 知 \( A\left( {P}_{i}\right) = {A}^{\prime }\left( {P}_{i}\right) \) ,故 \( A\left( {P}_{i}\right) \) 为定值。 | Yes |
Theorem 0.2.16. 给定任意一圆 \( \odot \left( O\right) \) ,令共點四線 \( {\ell }_{1},{\ell }_{2},{\ell }_{3},{\ell }_{4} \) 分别为共線四點 \( {P}_{1},{P}_{2},{P}_{3},{P}_{4} \) 關於 \( \odot \left( O\right) \) 的極線，則 \( \left( {P}_{i}\right) = \left( {\ell }_{i}\right) \) 。 | Proof. 若 \( O \notin \overline{{P}_{1}{P}_{2}{P}_{3}{P}_{4}} \) ,由配極性質知 \( O{P}_{i} \bot {\ell }_{i} \) ,因此\n\n\[ \measuredangle {P}_{i}O{P}_{j} = \measuredangle \left( {O{P}_{i}, O{P}_{j}}\right) = \measuredangle \left( {{\ell }_{i},{\ell }_{j}}\right) ,\]\n\n所以由角度關係知\n\n\[ \left( {P}_{i}\right) = O\left( {P}_{i}\righ... | Yes |
Corollary 0.2.17. 给定四線 \( {\ell }_{1},{\ell }_{2},{\ell }_{3},{\ell }_{4} \in T \odot \left( O\right) \) ,若 \( L \in T \odot \left( O\right) \) 为異於 \( {\ell }_{i} \) 的直線，則 \( L\left( {\ell }_{i}\right) \) 為定值 (與 \( L \) 的選取無關)。 | Proof. 若 \( {L}^{\prime } \in T \odot \left( O\right) \) 为異於 \( {\ell }_{i} \) 的直線,則由 (0.2.16) 知\n\n\[ \nL\left( {\ell }_{i}\right) = {T}_{L} \odot \left( O\right) \left( {{T}_{{\ell }_{i}} \odot \left( O\right) }\right) = {T}_{{L}^{\prime }} \odot \left( O\right) \left( {{T}_{{\ell }_{i}} \odot \left( O\right) }\right... | Yes |
Theorem 0.3.3. \( {E}_{1},{E}_{2} \) 为两平面, \( \varphi : {E}_{1} \rightarrow {E}_{2} \) 为一透视變换,則對於 \( {E}_{1} \) 上共線四點 \( {P}_{1},{P}_{2},{P}_{3},{P}_{4},\left( {P}_{i}\right) = \left( {\varphi \left( {P}_{i}\right) }\right) \) 。 | Proof. 令 \( M \) 为 \( \varphi \) 的中心,則 \( M,\ell ,\varphi \left( \ell \right) \) 共平面,所以有\n\n\[ \left( {P}_{i}\right) = M\left( {P}_{i}\right) = M\left( {\varphi \left( {P}_{i}\right) }\right) = \left( {\varphi \left( {P}_{i}\right) }\right) . \] | No |
Theorem 0.3.4. \( {E}_{1},{E}_{2} \) 为两平面, \( \varphi : {E}_{1} \rightarrow {E}_{2} \) 为一透视变换,則對於 \( {E}_{1} \) 上共點四線 \( {\ell }_{1},{\ell }_{2},{\ell }_{3},{\ell }_{4},\left( {\ell }_{i}\right) = \left( {\varphi \left( {\ell }_{i}\right) }\right) \) 。 | Proof. 令 \( M \) 为 \( \varphi \) 的中心,過 \( M \) 作任意一不過 \( \bigcap {\ell }_{i} \) 的平面 \( {E}_{3} \) ,設 \( {R}_{i} = {\ell }_{i} \cap {E}_{3} \) , 則\n\n\[ \left( {\ell }_{i}\right) = \left( {R}_{i}\right) = \left( {\varphi \left( {R}_{i}\right) }\right) = \left( {\varphi \left( {\ell }_{i}\right) }\right) . \] | Yes |
Theorem 0.3.6 (圆锥曲線基本定理/Fundamental Theorem of Conic Sections). 令 \( {P}_{1},{P}_{2},{P}_{3},{P}_{4}, A,{A}^{\prime } \) 为平面上六點且任三點不共線，則 \( {P}_{1},{P}_{2},{P}_{3},{P}_{4}, A,{A}^{\prime } \) 共一圆锥曲線若且唯若 \( A\left( {P}_{i}\right) = {A}^{\prime }\left( {P}_{i}\right) \) 。 | Proof. 令 \( \mathcal{C} = \left( {{P}_{1}{P}_{2}{P}_{3}A{A}^{\prime }}\right) \) 且 \( E \) 为 \( \mathcal{C} \) 所在平面,則存在一平面 \( {E}_{1} \) 、一圆 \( \Omega \) 在 \( {E}_{1} \) 上及一透視變換 \( \varphi : E \rightarrow {E}_{1} \) 使 \( \varphi \left( \mathcal{C}\right) = \Omega \) 。\n\n\( \left( \Rightarrow \right) \) 由 \( \varphi \l... | Yes |
Theorem 0.3.7. 令 \( {\ell }_{1},{\ell }_{2},{\ell }_{3},{\ell }_{4}, L,{L}^{\prime } \) 为平面上六線且任三線不共點,則 \( {\ell }_{1},{\ell }_{2} \) , \( {\ell }_{3},{\ell }_{4}, L,{L}^{\prime } \) 共切圆锥曲線若且唯若 \( L\left( {\ell }_{i}\right) = {L}^{\prime }\left( {\ell }_{i}\right) \) 。 | 證明與點的版本類似。於是我們就可以好好的在圓錐曲線上 (或切線) 定義交比了。 | No |
Proposition 0.3.11. 给定任意圆锥曲線 \( \mathcal{C} \) ,四點 \( {P}_{1},{P}_{2},{P}_{3},{P}_{4} \in \mathcal{C} \) ,令 \( {P}_{1},{P}_{2} \) 分别關於 \( \mathcal{C} \) 的切線交於 \( A \) ,則 \( {P}_{1}{P}_{3}{P}_{2}{P}_{4} \) 为 \( \mathcal{C} \) 上的調和四邊形若且唯若 \( A \) 在 \( {P}_{3}{P}_{4} \) 上。 | Proof. 由\n\n\[ \n{\left( {P}_{i}\right) }_{\mathcal{C}} = - 1 \Leftrightarrow {P}_{1}\left( {{P}_{1},{P}_{2};{P}_{3},{P}_{4}}\right) = {P}_{2}\left( {{P}_{2},{P}_{1};{P}_{3},{P}_{4}}\right) \n\] \n\n及 (0.2.10) 可得 \( {P}_{1}{P}_{2}{P}_{3}{P}_{4} \) 为 \( \mathcal{C} \) 上的调和四邊形若且唯若 \( A \in {P}_{3}{P}_{4} \) 。 | Yes |
Theorem 0.4.1 (帕斯卡定理/Pascal's). 令 \( {P}_{1},{P}_{2},{P}_{3},{P}_{4},{P}_{5},{P}_{6} \) 为任三點不共線的六點，則 \( {P}_{1},{P}_{2},{P}_{3},{P}_{4},{P}_{5},{P}_{6} \) 共一圓錐曲線若且唯若\n\n\[ \n{P}_{1}{P}_{2} \cap {P}_{4}{P}_{5},{P}_{2}{P}_{3} \cap {P}_{5}{P}_{6},{P}_{3}{P}_{4} \cap {P}_{6}{P}_{1} \n\]\n\n共線。 | Proof. 令 \( {Q}_{1} = {P}_{1}{P}_{2} \cap {P}_{4}{P}_{5},{Q}_{2} = {P}_{2}{P}_{3} \cap {P}_{5}{P}_{6},{Q}_{3} = {P}_{3}{P}_{4} \cap {P}_{6}{P}_{1} \) ,則\n\n\[ \n{P}_{1}\left( {{P}_{2},{P}_{3};{P}_{4},{P}_{6}}\right) = {Q}_{1}\left( {{P}_{1},{P}_{3};{P}_{4},{Q}_{3}}\right) \n\]\n\n\[ \n{P}_{5}\left( {{P}_{2},{P}_{3};{P}... | Yes |
Theorem 0.4.2 (布里昂雄定理/Brianchon's). 令 \( {\ell }_{1},{\ell }_{2},{\ell }_{3},{\ell }_{4},{\ell }_{5},{\ell }_{6} \) 为任三線不共點的六線,則 \( {\ell }_{1},{\ell }_{2},{\ell }_{3},{\ell }_{4},{\ell }_{5},{\ell }_{6} \) 切一圓錐曲線若且唯若\n\n\[ \left( {{\ell }_{1} \cap {\ell }_{2}}\right) \left( {{\ell }_{4} \cap {\ell }_{5}}\right) ,\left... | 其證明與帕斯卡定裡類似。 | No |
Theorem 0.4.3 (卡諾圓錐曲線定理/Carnot's). 給定任意 \( \bigtriangleup {ABC} \) ,點對們 \( \left( {{D}_{1},{D}_{2}}\right) ,\left( {{E}_{1},{E}_{2}}\right) ,\left( {{F}_{1},{F}_{2}}\right) \) 分别位於 \( {BC},{CA},{AB} \) 上,則 \( {D}_{1},{D}_{2},{E}_{1},{E}_{2},{F}_{1},{F}_{2} \) 共一圆锥曲線若且唯若 \( A{D}_{1}, A{D}_{2}, B{E}_{1}, B{E}_{2}, C{F}_{... | Proof. 只證 \( {D}_{1},{D}_{2},{E}_{1},{E}_{2},{F}_{1},{F}_{2} \) 共一圓錐曲線若且唯若\n\n\[ \frac{B{D}_{1}}{{D}_{1}C} \cdot \frac{B{D}_{2}}{{D}_{2}C} \cdot \frac{C{E}_{1}}{{E}_{1}A} \cdot \frac{C{E}_{2}}{{E}_{2}A} \cdot \frac{A{F}_{1}}{{F}_{1}B} \cdot \frac{A{F}_{2}}{{F}_{2}B} = 1, \]\n\n另一個等價證明類似。令 \( {E}_{2}{F}_{1},{F}_{2}{D}_{... | Yes |
Theorem 0.4.6 (龐色列閉合/Poncelet’s Closure Theorem). 設 \( \bigtriangleup {A}_{1}{B}_{1}{C}_{1} \) 及 \( \bigtriangleup {A}_{2}{B}_{2}{C}_{2} \) 为平面上雨個三角形,則 \( \bigtriangleup {A}_{1}{B}_{1}{C}_{1},\bigtriangleup {A}_{2}{B}_{2}{C}_{2} \) 共一外接圆锥曲線若且唯若 \( \bigtriangleup {A}_{1}{B}_{1}{C}_{1},\bigtriangleup {A}_{2}{B}_{2}{C}_{2... | Proof. 注意到\n\n\[ \n{A}_{1}\left( {{B}_{1},{C}_{1};{B}_{2},{C}_{2}}\right) = \left( {{B}_{2}{C}_{2}}\right) \left( {{A}_{1}{B}_{1},{C}_{1}{A}_{1};{A}_{2}{B}_{2},{C}_{2}{A}_{2}}\right) , \n\]\n\n\[ \n{A}_{2}\left( {{B}_{2},{C}_{2};{B}_{1},{C}_{1}}\right) = \left( {{B}_{1}{C}_{1}}\right) \left( {{A}_{2}{B}_{2},{C}_{2}{A}_... | Yes |
Theorem 0.4.8. 對於任意完全四點形 \( \mathcal{Q} = \left( {{P}_{1},{P}_{2},{P}_{3},{P}_{4}}\right) \) ,取一線 \( \ell \) ,定義 \( {Q}_{ij} = {P}_{i}{P}_{j} \cap L \) 。在 \( {P}_{i}{P}_{j} \) 上取 \( {R}_{ij} \) 使得\n\n\[ \left( {{P}_{i},{P}_{j};{Q}_{ij},{R}_{ij}}\right) = - 1 \]\n\n那麼 \( {R}_{23},{R}_{14},{R}_{31},{R}_{24},{R}_{12},{R}_... | Proof. 令 \( {P}_{2}{P}_{3},{P}_{3}{P}_{1},{P}_{1}{P}_{2} \) 分别交 \( {P}_{1}{P}_{4},{P}_{2}{P}_{4},{P}_{3}{P}_{4} \) 於 \( X, Y, Z \) ,由西瓦定理與孟氏定理可得\n\n\[ \frac{{P}_{2}X}{X{P}_{3}} \cdot \frac{{P}_{3}Y}{Y{P}_{1}} \cdot \frac{{P}_{1}Z}{Z{P}_{2}} \cdot \frac{{P}_{2}{R}_{23}}{{R}_{23}{P}_{3}} \cdot \frac{{P}_{3}{R}_{31}}{{R}_... | Yes |
Theorem 0.4.12 (牛頓一號/Newton’s Theorem I). 對於任意完全四線形 \( \mathcal{Q} \) ,其三個對角線段的中點共線。 | Proof. 令 \( \mathcal{Q} = \left( {{\ell }_{1},{\ell }_{2},{\ell }_{3},{\ell }_{4}}\right) ,{P}_{ij} = {\ell }_{i} \cap {\ell }_{j},\bigtriangleup {M}_{1}{M}_{2}{M}_{3} \) 为 \( \bigtriangleup {P}_{23}{P}_{31}{P}_{12} \) 的中點三角形, \( {R}_{1},{R}_{2},{R}_{3} \) 分别为 \( \overline{{P}_{23}{P}_{14}},\overline{{P}_{31}{P}_{24}},... | Yes |
Theorem 0.4.14 (牛頓二號/Newton's Theorem II). 對於完全四線形 \( \mathcal{Q} \) ,若圓錐曲線 \( \mathcal{C} \) 与 \( \mathcal{Q} \) 中的四線皆相切,则 \( \mathcal{C} \) 的中心位於 \( \mathcal{Q} \) 的牛頓線上。反之,若 \( O \) 為 \( \mathcal{Q} \) 的牛顿線上一點，則存在一以 \( O \) 為中心的圓錐曲線 \( \mathcal{C} \) 與 \( \mathcal{Q} \) 相切。 | Proof. 不妨假設 \( {\ell }_{1} \) 不平行於 \( {\ell }_{4} \) 。令 \( \mathcal{Q} = \left( {{\ell }_{1},{\ell }_{2},{\ell }_{3},{\ell }_{4}}\right) ,{P}_{ij} = {\ell }_{i} \cap {\ell }_{j},{R}_{2},{R}_{3} \) 分别为 \( \overline{{P}_{31}{P}_{24}},\overline{{P}_{12}{P}_{34}} \) 的中點, \( O \) 为 \( \mathcal{C} \) 的中心。作 \( {\ell }_{1},{\e... | Yes |
Theorem 0.4.15 (牛頓三號/Newton's Theorem III). 對於完全四線形 \( \mathcal{Q} = \left( {\ell }_{1}\right. \) , \( \left. {{\ell }_{2},{\ell }_{3},{\ell }_{4}}\right) \) ,定義 \( {P}_{ij} = {\ell }_{i} \cap {\ell }_{j} \) ,若圆锥曲線 \( \mathcal{C} \) 分别与 \( {\ell }_{1},{\ell }_{2},{\ell }_{3},{\ell }_{4} \) 切於 \( {Q}_{1},{Q}_{2},{Q}_{3}... | Proof. 考虑六切線組 \( \left( {{\ell }_{1}{\ell }_{1}{\ell }_{2}{\ell }_{3}{\ell }_{3}{\ell }_{4}}\right) \) 與 \( \left( {{\ell }_{1}{\ell }_{2}{\ell }_{2}{\ell }_{3}{\ell }_{4}{\ell }_{4}}\right) \) ,由 (0.4.2) 知 \( {Q}_{1}{Q}_{3},{P}_{12}{P}_{34} \) , \( {P}_{23}{P}_{41} \) 与 \( {Q}_{2}{Q}_{4},{P}_{12}{P}_{34},{P}_{23}{P}_{... | Yes |
Corollary 1.1.10. 给定圆锥曲線 \( \mathcal{C} \) ,任意六點 \( {P}_{1},{P}_{2},{P}_{3},{P}_{4}, A, B \) 共一圆锥曲線若且唯若六線 \( {\mathfrak{p}}_{\mathcal{C}}\left( \left\{ {{P}_{1},{P}_{2},{P}_{3},{P}_{4}, A, B}\right\} \right) \) 切一圓錐曲線。 | Proof. 由\n\n\[ \nA\left( {P}_{i}\right) = {\mathfrak{p}}_{\mathcal{C}}\left( A\right) \left( {\ell }_{i}\right), B\left( {P}_{i}\right) = {\mathfrak{p}}_{\mathcal{C}}\left( B\right) \left( {\ell }_{i}\right) , \n\] \n\n知 \( {P}_{1},{P}_{2},{P}_{3},{P}_{4}, A, B \) 共一圓錐曲線若且唯若 \n\n\[ \nA\left( {P}_{i}\right) = B\left( {P... | Yes |
Proposition 1.1.12. 令 \( \mathcal{C} \) 为一圆锥曲線, \( O \) 为 \( \mathcal{C} \) 的中心,則 \( {\mathfrak{p}}_{\mathcal{C}}\left( O\right) = {\mathcal{L}}_{\infty } \) 。 | Proof. 過 \( O \) 作兩條與 \( \mathcal{C} \) 有交點的直線 \( {\ell }_{1}{\ell }_{2} \) ,設其分別交 \( \mathcal{C} \) 於 \( \left( {{P}_{1},{Q}_{1}}\right) ,\left( {{P}_{2},{Q}_{2}}\right) \) , 則 \( \overline{{P}_{1}{Q}_{1}},\overline{{P}_{2}{Q}_{2}} \) 的中點們重合於 \( O \) ,所以 \( {\infty }_{{P}_{1}{Q}_{1}},{\infty }_{{P}_{2}{Q}_{2}} \in {\m... | Yes |
Corollary 1.1.13 (平行弦定理). 令 \( \mathcal{C} \) 为一圆锥曲線、 \( O \) 为 \( \mathcal{C} \) 的中心, \( {P}_{1},{P}_{2} \) 为平面上雨點,則 \( O,{P}_{1},{P}_{2} \) 共線若且唯若 \( {\mathfrak{p}}_{\mathcal{C}}\left( {P}_{1}\right) \) 平行於 \( {\mathfrak{p}}_{\mathcal{C}}\left( {P}_{2}\right) \) 。 | Proof. 注意到 \( O,{P}_{1},{P}_{2} \) 共線若且唯若 \( {\mathfrak{p}}_{\mathcal{C}}\left( O\right) = {\mathcal{L}}_{\infty },{\mathfrak{p}}_{\mathcal{C}}\left( {P}_{1}\right) ,{\mathfrak{p}}_{\mathcal{C}}\left( {P}_{2}\right) \) 共點,即 \( {\mathfrak{p}}_{\mathcal{C}}\left( {P}_{1}\right) \) 平行於 \( {\mathfrak{p}}_{\mathcal{C}}\left... | Yes |
Corollary 1.1.14. 令 \( \mathcal{C} \) 为一圆锥曲線, \( O \) 为 \( \mathcal{C} \) 的中心, \( \overline{AB} \) 为 \( \mathcal{C} \) 上的弦,則 \( O{\mathfrak{p}}_{\mathcal{C}}\left( {AB}\right) \) 平分 \( \overline{AB} \) 。 | Proof. 令 \( T = {AB} \cap O{\mathfrak{p}}_{\mathcal{C}}\left( {AB}\right), U = {\infty }_{AB} \) ,則 \( O{\mathfrak{p}}_{\mathcal{C}}\left( {AB}\right) = {\mathfrak{p}}_{\mathcal{C}}\left( U\right) \) ,因此由定義有 \( \left( {U, T;A, B}\right) = - 1 \) ,故 \( T \) 为 \( \overline{AB} \) 中點。 | Yes |
Theorem 1.1.15. 给定 \( \bigtriangleup {ABC} \) ,令 \( P, Q \) 为两點, \( \bigtriangleup {P}_{a}{P}_{b}{P}_{c},\bigtriangleup {Q}_{a}{Q}_{b}{Q}_{c} \) 分别为 \( P, Q \) 關於 \( \bigtriangleup {ABC} \) 的反西瓦三角形,那麼 \( P,{P}_{a},{P}_{b},{P}_{c}, Q,{Q}_{a},{Q}_{b},{Q}_{c} \) 八點共一圓錐曲線。 | Proof. 令 \( \mathcal{C} \) 为经过 \( P,{P}_{a},{P}_{b},{P}_{c}, Q \) 的圆锥曲線,則 \( {\mathfrak{p}}_{\mathcal{C}}\left( A\right) ,{\mathfrak{p}}_{\mathcal{C}}\left( B\right) ,{\mathfrak{p}}_{\mathcal{C}}\left( C\right) \) 为 \( \bigtriangleup {ABC} \) 的三邊，令 \( {AQ} \) 與 \( \mathcal{C} \) 的另一個交點為 \( {Q}_{a}^{\prime } \) 則\n\n\[ ... | Yes |
Proposition 1.2.4. 令 \( X \) 为一可交比化的集合, \( \varphi \in \operatorname{Aut}\left( X\right) \smallsetminus \left\{ {\mathrm{{id}}}_{X}\right\} \) 且存在一點 \( A \in X \) 使得 \( \varphi \left( A\right) \neq A,\varphi \left( {\varphi \left( A\right) }\right) = A \) ,則 \( \varphi \) 为一对合变换。 | Proof. 令 \( P \in \mathcal{C} \) ,則\n\n\[ \left( {A,\varphi \left( A\right) ;P,\varphi \left( P\right) }\right) = \left( {\varphi \left( A\right), A;\varphi \left( P\right) ,\varphi \left( {\varphi \left( P\right) }\right) }\right) = \left( {A,\varphi \left( A\right) ;{\varphi }^{2}\left( P\right) ,\varphi \left( P\rig... | Yes |
Theorem 1.2.5 (迪沙格對合定理/Desargues' Involution Theorem). 给定任意完全四點形 \( \mathcal{Q}\left( {{P}_{1},{P}_{2},{P}_{3},{P}_{4}}\right) \) 及任意一線\n\n\[ \ell \notin \left\{ {{P}_{2}{P}_{3},{P}_{1}{P}_{4},{P}_{3}{P}_{1},{P}_{2}{P}_{4},{P}_{1}{P}_{2},{P}_{3}{P}_{4}}\right\} \]\n\n定義 \( {Q}_{ij} = {P}_{i}{P}_{j} \cap \ell \) ,則\n\n(... | Proof. 定義 \( \varphi : \ell \rightarrow \ell \) 为一射影變换使得 \( {Q}_{23} \mapsto {Q}_{14},{Q}_{14} \mapsto {Q}_{23},{Q}_{12} \mapsto {Q}_{34} \) ,則\n\n(i) 由交比性質知\n\n\[ \left( {{Q}_{23},{Q}_{14};{Q}_{12},{Q}_{31}}\right) = \left( {{Q}_{23},{P}_{1}{P}_{4} \cap {P}_{2}{P}_{3};{P}_{2},{P}_{3}}\right) \]\n\n\[ = \left( {{Q}_{23... | Yes |
Proposition 1.2.8. 延續迪沙格定理 (1.2.5) 的標號,令 \( \mathcal{C} \) 为 \( \ell \) 關於 \( \mathcal{Q} \) 的九點圓錐曲線，那麼 \( \left( {A, B}\right) \) 為關於 \( \varphi \) 的相互對若且唯若 \( A, B \) 關於 \( \mathcal{C} \) 共轭。 | Proof. 令 \( {\varphi }^{\prime } = \left\lbrack {A \mapsto {\mathfrak{p}}_{\mathcal{C}}\left( A\right) \cap \ell }\right\rbrack \in \operatorname{Aut}\left( \ell \right) \) ,取 \( {R}_{ij} \) 使得 \( \left( {{P}_{i},{P}_{j};{Q}_{ij},{R}_{ij}}\right) = - 1 \) 。那麼我們知道\n\n\[ \n{Q}_{23} = {R}_{12}{R}_{31} \cap {R}_{24}{R}_{34... | Yes |
Proposition 1.2.12. 令 \( \ell \) 为一直線, \( \varphi : \ell \rightarrow \ell \) 为一變換,則 \( \varphi \) 為一對合變換若且唯若 \( \varphi \) 为反演变换 (這邊視對稱變換為反演變換)。 | Proof. 只證過去，證回來由同一法即可。令 \( A = \varphi \left( {\infty }_{\ell }\right) \) ,這邊分兩個情形:\n\n(i) \( A \neq {\infty }_{\ell } \) ,那麼對於 \( \varphi \) 的任兩對相互對 \( \left( {P,{P}^{\prime }}\right) ,\left( {Q,{Q}^{\prime }}\right) \) ,我們有\n\n\[ \left( {A,{\infty }_{\ell };P, Q}\right) = \left( {{\infty }_{\ell }, A;{P}^{\prime },{Q... | No |
Proposition 1.2.13. 令 \( \mathcal{C} \) 为一圆锥曲線、 \( \varphi : \mathcal{C} \rightarrow \mathcal{C} \) 为一變换,則 \( \varphi \) 为一对合變換若且唯若存在恰一點 \( A \notin \mathcal{C} \) 使得 \( \forall P \in \mathcal{C}, A, P,\varphi \left( P\right) \) 共線。 | Proof. 只證過去，證回來由同一法即可。對於 \( \varphi \) 的兩對相互對 \( \left( {P,{P}^{\prime }}\right) ,\left( {Q,{Q}^{\prime }}\right) \) , 定義 \( A = P{P}^{\prime } \cap Q{Q}^{\prime } \) ,則對於任意一對相互對 \( \left( {R,{R}^{\prime }}\right) \) ,我們有\n\n\[ P\left( {{P}^{\prime }, R;Q,{Q}^{\prime }}\right) = {\left( {P}^{\prime }, R;Q,{Q}^{\prime }... | No |
Proposition 1.2.16. 延續 (1.2.5) 的標號。設 \( {F}_{1},{F}_{2} \) 为 \( \varphi \) 的不動點，那麼對於任意過 \( \mathcal{Q} \) 的圆锥曲線 \( \mathcal{C},{F}_{1},{F}_{2} \) 關於 \( \mathcal{C} \) 共轭。 | Proof. 令 \( {F}_{1}{P}_{1},{F}_{1}{P}_{2} \) 分别与 \( \mathcal{C} \) 交另一點於 \( X, Y \) ,定義 \( {X}_{ij},{Y}_{ij} \) 分别为 \( {F}_{1}{P}_{1},{F}_{1}{P}_{2} \) 与 \( {P}_{i}{P}_{j} \) 的交点, \( A = {P}_{1}{P}_{4} \cap {P}_{2}{P}_{3}, B = {P}_{2}{P}_{4} \cap {P}_{3}{P}_{1} \) 。分别在 \( {F}_{1}{P}_{1},{F}_{1}{P}_{2} \) 上取點 \( {X}^{ *... | Yes |
Proposition 1.3.3 (等共轭變换基本定理). 给定 \( \bigtriangleup {ABC} \) ,一圆锥曲線族 \( \mathcal{F} \) 由其中的雨個相異元素 \( {\mathcal{C}}_{0} \) 與 \( {\mathcal{C}}_{\infty } \) 决定。事實上,所有 \( \mathcal{F} \) 中的元素皆可寫成\n\n\[ \n{\mathcal{C}}_{t} = {\mathcal{C}}_{0} + t \cdot {\mathcal{C}}_{\infty } \n\]\n\n也就是說 \( \mathcal{F} \) 是一個圓錐曲線束 (Pencil)。 | Proof. 我們只證明第一部分,也就是對於所有 \( P \notin {BC} \cup {CA} \cup {AB},{\mathfrak{p}}_{{\mathcal{C}}_{0}}\left( P\right) \neq \) \( {\mathfrak{p}}_{{\mathcal{C}}_{\infty }}\left( P\right) \) 。考虑變换 \( \varphi \left( P\right) = {\mathfrak{p}}_{{\mathcal{C}}_{\infty }}\left( {{\mathfrak{p}}_{{\mathcal{C}}_{0}}\left( P\right) }\rig... | Yes |
Proposition 1.3.4. 给定任意 \( \bigtriangleup {ABC} \) ,令 \( \varphi \) 为一點等共轭變換, \( \ell \) 为一直線\n\n(i) 若 \( \ell \cap \{ A, B, C\} = X \) ,那麼 \( \varphi \left( \ell \right) \) 为过 \( X \) 的直線。\n\n(ii) 若 \( \ell \cap \{ A, B, C\} = \varnothing \) ,那麼 \( \varphi \left( \ell \right) \) 为 \( \bigtriangleup {ABC} \) 的外接圆锥曲線。 | Proof. 令 \( \mathcal{F} \) 为 \( \varphi \) 的对角圆锥曲線族,易知存在兩圓錐曲線 \( {\mathcal{C}}_{1},{\mathcal{C}}_{2} \in \mathcal{F} \) 使得 \( \ell \) 關於 \( {\mathcal{C}}_{1},{\mathcal{C}}_{2} \) 的極點 \( {P}_{1},{P}_{2} \) 不重合,令 \( M \in \ell \) 为一點。\n\n(i) 不妨假設 \( \ell \cap \{ A, B, C\} = A \) ,令 \( \ell \) 交 \( {BC} \) 於 \( D \) ,由配極變... | Yes |
Proposition 1.3.5. 给定任意 \( \bigtriangleup {ABC} \) ,令 \( \varphi \) 为 \( \bigtriangleup {ABC} \) 上的點等共轭變換,則对於所有顶点 \( X \in \{ A, B, C\} \) , \( \left\lbrack {{XP} \mapsto {X\varphi }\left( P\right) }\right\rbrack \) 是一对合变换。 | Proof. 易知其階為 2,因此只要證明它保交比即可。令 \( \mathcal{C} \) 为 \( \varphi \) 的其中一個對角圓錐曲線, \( O \) 為其中心, \( M \) 為一無窮遠點，那麼由配極變換為保交比變換知\n\n\[ \n{\left( A, B;C,\varphi \left( M\right) \right) }_{\varphi \left( {\mathcal{L}}_{\infty }\right) } = O\left( {A, B;C,\varphi \left( M\right) }\right) = \left( {{\infty }_{BC},{\infty }_{CA};{\... | Yes |
Corollary 1.3.6. 给定任意 \( \bigtriangleup {ABC} \) ,令 \( \varphi \) 为 \( \bigtriangleup {ABC} \) 上的點等共轭變換,對於\n\n任兩點 \( P, Q \) ，令 \( R = {PQ} \cap \varphi \left( P\right) \varphi \left( Q\right), S = {P\varphi }\left( Q\right) \cap \varphi \left( P\right) Q \) ，則 \( S = \varphi \left( R\right) \) 。 | Proof. 由上述性質知\n\n\[ A\left( {P, Q;\varphi \left( P\right), S}\right) = P\left( {A, R;\varphi \left( P\right) ,\varphi \left( Q\right) }\right) \]\n\n\[ = A\left( {P, R;\varphi \left( P\right) ,\varphi \left( Q\right) }\right) \]\n\n\[ = A\left( {P, Q;\varphi \left( P\right) ,\varphi \left( R\right) }\right) ,\]\n\n故 \(... | Yes |
Proposition 1.3.7. 给定任意 \( \bigtriangleup {ABC},\varphi \) 为 \( \bigtriangleup {ABC} \) 上的點等共轭變換。則對於任意一點 \( P,\varphi \left( P\right) \) 为 \( \bigtriangleup {ABC} \) 与 \( {\mathfrak{p}}_{\varphi \left( {\mathfrak{t}}_{p}\right) }\left( {\bigtriangleup {ABC}}\right) \) 的透视中心,其中 \( {\mathfrak{t}}_{P} \) 为 \( P \) 關於 \( \... | Proof. 由 (1.3.5),\n\n\[ \left( {{A\varphi }\left( P\right) ,{T}_{A}\varphi \left( {\mathfrak{t}}_{P}\right) ;{AB},{AC}}\right) = A\left( {P,{\mathfrak{t}}_{P} \cap {BC};C, B}\right) = - 1, \]\n\n所以 \( {\mathfrak{p}}_{\varphi \left( {\mathbf{t}}_{P}\right) }\left( A\right) = {T}_{A}\varphi \left( {\mathbf{t}}_{P}\right)... | Yes |
Proposition 1.3.9. 若一變換\n\n\[ \n\varphi : {\mathbb{P}}_{\mathbb{R}}^{2} \smallsetminus {BC} \cup {CA} \cup {AB} \rightarrow {\mathbb{P}}_{\mathbb{R}}^{2} \smallsetminus {BC} \cup {CA} \cup {AB} \n\]\n\n滿足：對於所有頂點 \( X \in \{ A, B, C\} \) ,\n\n\[ \nX,{P}_{1},{P}_{2}\text{共線} \Rightarrow X,\varphi \left( {P}_{1}\right) ,\... | Proof. 顯然地, \( {\varphi }^{2} = \mathrm{{id}} \) 。取任意一點 \( {P}_{0} \) 使得對於所有頂點 \( X \in \{ A, B, C\} \) , \( X{P}_{0} \neq {X\varphi }\left( {P}_{0}\right) \) ,那麼 \( \varphi \) 由 \( {P}_{0},\varphi \left( {P}_{0}\right) \) 决定 :\n\n\[ \nX\left( {Y, Z;{P}_{0}, P}\right) = X\left( {Z, Y;\varphi \left( {P}_{0}\right) ,\var... | Yes |
Theorem 1.3.10. 给定 \( \bigtriangleup {ABC} \) ,對於任意一不過頂點的直線 \( \mathcal{L} \) ,我們有如下的一一對應：\n\n\( \{ \bigtriangleup {ABC} \) 上的點等共軛變換 \( \} \leftrightarrow \{ \bigtriangleup {ABC} \) 的外接圓錐曲線 \( \} \)\n\n\[ \varphi \mapsto \varphi \left( \mathcal{L}\right) \] | Proof. 若有兩個等共軛變換 \( \varphi ,\psi \) 使得 \( \varphi \left( \mathcal{L}\right) = \psi \left( \mathcal{L}\right) \) 。令 \( P \) 为 \( \mathcal{L} \) 關於 \( \bigtriangleup {ABC} \) 的三線性極點，由 (1.3.8) 我們知道 \( \varphi \left( P\right) = \psi \left( P\right) \) 。對於任意一點 \( Q \),\n\n\[ A\left( {B, C;\varphi \left( G\right) ,\varphi \... | Yes |
Proposition 1.3.12. 给定任意 \( \bigtriangleup {ABC},\varphi \) 为 \( \bigtriangleup {ABC} \) 上的點等共軛變換, \( \mathcal{F} \) 为 \( \varphi \) 的对角圆锥曲線族,若 \( P \) 为 \( \varphi \) 的不动點,則\n\n\[ P \in \mathop{\bigcap }\limits_{{\mathcal{C} \in \mathcal{F}}}\mathcal{C} \] | Proof. 對於任意 \( \mathcal{C} \in \mathcal{F} \) ,由 \( P, P \) 關於 \( \mathcal{C} \) 共轭知 \( P \in \mathcal{C} \) 。 | Yes |
Proposition 1.3.13. 给定任意 \( \bigtriangleup {ABC},\varphi \) 为 \( \bigtriangleup {ABC} \) 上的點等共轭變换,則 \( \varphi \) 的不動點數量至多為 4。 | Proof. 令 \( \mathcal{F} \) 为 \( \varphi \) 的对角圆锥曲線族。若存在五個不動點 \( {P}_{1},\ldots ,{P}_{5} \) ,則對於所有 \( \mathcal{C} \in \mathcal{F},{P}_{1},\ldots ,{P}_{5} \in \mathcal{C} \) ,而 \( \left| \mathcal{F}\right| > 1 \) ,矛盾。 | Yes |
Proposition 1.3.14. 给定任意 \( \bigtriangleup {ABC} \) ,令 \( P \notin {BC} \cup {CA} \cup {AB} \) 为任意一點, \( \varphi \) 为一點等共軛變換， \( \mathcal{F} \) 為 \( \varphi \) 的對角圓錐曲線族，令 \( \bigtriangleup {P}_{a}{P}_{b}{P}_{c} \) 為 \( P \) 關於 \( \bigtriangleup {ABC} \) 的反西瓦三角形,若 \( P \) 为 \( \varphi \) 的不動點,則 \( {P}_{a},{P}_{b},{P}_{c... | Proof. 令 \( {Q}_{a} = {AP} \cap {BC},\mathcal{C} \in \mathcal{F} \) ,則 \( A,{Q}_{a} \) 關於 \( \mathcal{C} \) 共轭,因此由\n\n\[ \left( {A,{Q}_{a};P,{P}_{a}}\right) = - 1 \]\n\n及 \( P \in \mathcal{C} \) 知 \( {P}_{a} \in \mathcal{C} \) ,故 \( {P}_{a},{P}_{a} \) 關於 \( \mathcal{C} \) 共轭,即 \( {P}_{a} \) 也為 \( \varphi \) 的不動點,同理可知, ... | Yes |
Proposition 1.3.15. 给定任意 \( \bigtriangleup {ABC} \) 及一點 \( P \) ,令 \( \bigtriangleup {P}_{a}{P}_{b}{P}_{c} \) 为 \( P \) 關於 \( \bigtriangleup {ABC} \) 的反西瓦三角形, \( \mathcal{F} \) 为所有經過 \( P,{P}_{a},{P}_{b},{P}_{c} \) 的圓錐曲線的集合, 對於任意一點 \( M \notin \{ A, B, C\} \) ,存在一點 \( {M}^{ * } \) 滿足\n\n\[ \n- 1 = \left( {{AM}, A{M}^{ ... | Proof. 取點 \( {M}^{ * } \) 满足 \( \left( {{BM}, B{M}^{ * };P{P}_{b},{P}_{c}{P}_{a}}\right) = \left( {{CM}, C{M}^{ * };P{P}_{c},{P}_{a}{P}_{b}}\right) = - 1 \) ,考慮完全四點形 \( \left( {P,{P}_{a},{P}_{b},{P}_{c}}\right) \) 与直線 \( M{M}^{ * } \) ,存在一射影对合变换 \( \varphi \in \operatorname{Aut}\left( {M{M}^{ * }}\right) \) 使得\n\n\[ \l... | Yes |
Proposition 1.3.17. 给定任意 \( \bigtriangleup {ABC} \) ,令 \( \varphi \) 为一點等共轭變换且 \( P,{P}_{a},{P}_{b} \) , \( {P}_{c} \) 为 \( \varphi \) 的不动點,那麼對於任意直線 \( \ell ,\varphi \left( \ell \right) \) 为 \( \ell \) 關於完全四點形 \( \left( {P,{P}_{a},{P}_{b},{P}_{c}}\right) \) 的九點圓錐曲線 (0.4.8)。 | Proof. 令 \( \mathcal{C} \) 为 \( \ell \) 關於完全四點形 \( \left( {P,{P}_{a},{P}_{b},{P}_{c}}\right) \) 的九點圓錐曲線, \( P{P}_{a}, P{P}_{b}, P{P}_{c} \) , \( {P}_{b}{P}_{c},{P}_{c}{P}_{a},{P}_{a}{P}_{b} \) 分别与 \( \ell \) 交於 \( {Q}_{a},{Q}_{b},{Q}_{c},{Q}_{a}^{\prime },{Q}_{b}^{\prime },{Q}_{c}^{\prime } \) ,取 \( {R}_{a} \) 使得\n\n\[... | Yes |
Theorem 1.3.19. 對於任意完全四點形 \( \left( {{P}_{1},{P}_{2},{P}_{3},{P}_{4}}\right) \) 及一線 \( \ell ,\ell \) 關於所有經過 \( {P}_{1},{P}_{2},{P}_{3},{P}_{4} \) 的圆锥曲線的極點的軌跡為 \( \ell \) 關於 \( \left( {{P}_{1},{P}_{2},{P}_{3},{P}_{4}}\right) \) 的九點圓錐曲線。 | Proof. 令 \( \bigtriangleup {ABC} \) 为 \( \left( {{P}_{1},{P}_{2},{P}_{3},{P}_{4}}\right) \) 的西瓦三角形, \( \varphi \) 为 \( \bigtriangleup {ABC} \) 上並且以 \( {P}_{1},{P}_{2} \) , \( {P}_{3},{P}_{4} \) 为不动點的點等共軛變換。\n\n\( \left( \Rightarrow \right) \) 令 \( \mathcal{C} \) 为过 \( {P}_{1},{P}_{2},{P}_{3},{P}_{4} \) 的圆锥曲線,那麼\n\n\[ \... | Yes |
Proposition 1.4.5. 给定任意 \( \bigtriangleup {ABC} \) ,則對於任意一點 \( P \) ,存在一點 \( {P}^{ * } \) 使得 \( {P}^{ * } \) 为 \( P \) 關於 \( \bigtriangleup {ABC} \) 的等角共轭點。 | Proof. 令 \( \bigtriangleup {P}_{a}{P}_{b}{P}_{c} \) 为 \( P \) 關於 \( \bigtriangleup {ABC} \) 的圆西瓦三角形, \( \bigtriangleup {DEF} \) 为 \( P \) 關於 \( \bigtriangleup {P}_{a}{P}_{b}{P}_{c} \) 的佩多三角形，則\n\n\[ \measuredangle {EDF} = \measuredangle {EDP} + \measuredangle {PDF} = \measuredangle {P}_{a}{P}_{c}C + \measuredangle B{P}... | Yes |
Proposition 1.4.6. 给定任意 \( \bigtriangleup {ABC} \) ,則對於任意一點 \( P \) ,點 \( Q \) 為 \( P \) 關於 \( \bigtriangleup {ABC} \) 的等角共轭點若且唯若\n\n\[ \measuredangle {CPA} + \measuredangle {CQA} = \measuredangle {CBA}\text{ 及 }\measuredangle {APB} + \measuredangle {AQB} = \measuredangle {ACB}. \] | Proof.\n\n\( \left( \Rightarrow \right) \) 簡單的算角度練習。\n\n\( \left( \Leftarrow \right) \) 令 \( {P}^{ * } \) 为 \( P \) 關於 \( \bigtriangleup {ABC} \) 的等角共轭點。熟知滿足 \( \measuredangle {CPA} + \measuredangle {CRA} = \measuredangle {CBA} \) 的 \( R \) 的軌跡為 \( \odot \left( {C{P}^{ * }A}\right) \smallsetminus \{ C, A\} \) ,同樣地，滿足 \... | No |
對於任意完全 \( n \) 線形 \( \mathcal{N}\left( {{\ell }_{1},{\ell }_{2},\ldots ,{\ell }_{n}}\right) \) 及一點 \( P \) ,定義 \( {P}_{i} \) 為 \( P \) 關於 \( {\ell }_{i} \) 的垂足,則 \( P \) 存在關於 \( \mathcal{N} \) 的等角共軛點若且唯若 \( {P}_{1},{P}_{2},\ldots ,{P}_{n} \) 共圆 (佩多圆)。此时,若 \( {P}^{ * } \) 为 \( P \) 關於 \( \mathcal{N} \) 的等角共轭點,定義 \( {P}_... | Proof. 令 \( {A}_{ij} = {\ell }_{i} \cap {\ell }_{j} \) 。\n\n\( \left( \Rightarrow \right) \) 令 \( {P}^{ * } \) 为 \( P \) 關於 \( \mathcal{N} \) 的等角共轭點, \( {P}_{i}^{ * } \) 为 \( {P}^{ * } \) 關於 \( {\ell }_{i} \) 的垂足,則由 \( {A}_{ij}P{P}_{i}{P}_{j} \) 与 \( {A}_{ij}{P}^{ * }{P}_{j}^{ * }{P}_{i}^{ * } \) 负向相似知 \( {P}_{i},{P}_{... | Yes |
Theorem 1.4.9. 對於任意完全 \( n \) 線形 \( \mathcal{N}\left( {{\ell }_{1},{\ell }_{2},\ldots ,{\ell }_{n}}\right) \) 及不在線上的兩點 \( P \) , \( {P}^{ * } \) ,则 \( \left( {P,{P}^{ * }}\right) \) 为 \( \mathcal{N} \) 的一对等角共轭點對若且唯若存在以 \( P,{P}^{ * } \) 为焦点的圆锥曲線 \( \mathcal{C} \) 使得 \( \mathcal{C} \) 與 \( \mathcal{N} \) 相切。 | Proof. 令 \( {P}_{1}^{ * },{P}_{2}^{ * },\ldots ,{P}_{n}^{ * } \) 分别为 \( {P}^{ * } \) 關於 \( {\ell }_{1},{\ell }_{2},\ldots ,{\ell }_{n} \) 的對稱點。\n\n\( \left( \Rightarrow \right) \) 則 \( {P}_{1}^{ * },{P}_{2}^{ * },\ldots ,{P}_{n}^{ * } \) 共圆且圆心为 \( P \) ,令 \( {T}_{i} \) 为 \( P{P}_{i}^{ * } \) 与 \( {\ell }_{i} \) 的交點,則 \... | Yes |
Corollary 1.4.11. 對於任意完全四線形 \( \mathcal{Q} \) ,其所有等角共轭點對的中點軌跡為 \( \mathcal{Q} \) 的牛頓線。 | Proof. 结合牛頓第二定理 (0.4.14). | No |
Proposition 1.4.12. 给定任意完全四線形 \( \mathcal{Q} = \left( {{AB},{BC},{CD},{DA}}\right) \) ,則對於任意一點 \( P, P \) 有關於 \( \mathcal{Q} \) 的等角共轭點若且唯若\n\n\[ \measuredangle {APB} + \measuredangle {CPD} = {0}^{ \circ }.\] | Proof. 令 \( P \) 關於 \( {AB},{BC},{CD},{DA} \) 的垂足分别为 \( W, X, Y, Z \) ,則 \( P \) 有關於 \( \mathcal{Q} \)\n\n的等角共轭點若且唯若 \( W, X, Y, Z \) 共圆,等價於\n\n\[ \measuredangle {WXP} + \measuredangle {PXY} + \measuredangle {YZP} + \measuredangle {PZW} = {0}^{ \circ } \Leftrightarrow \measuredangle {APB} + \measuredangle {CPD} = {0}^{... | Yes |
Theorem 1.4.13. 给定任意完全四線形 \( \mathcal{Q} \) ,令 \( M \) 为 \( \mathcal{Q} \) 的密克點, \( \tau \) 为 \( \mathcal{Q} \) 的牛頓線，則 \( \left( {M,{\infty }_{\tau }}\right) \) 為關於 \( \mathcal{Q} \) 的等角共軛點對。 | Proof. 令 \( \mathcal{Q} = \left( {{\ell }_{1},{\ell }_{2},{\ell }_{3},{\ell }_{4}}\right) ,{M}_{i} \) 为 \( M \) 關於 \( {\ell }_{i} \) 的對稱點,則知 \( {M}_{1},{M}_{2},{M}_{3},{M}_{4} \) 共於垂心線 \( L \) 且垂直於 \( \tau \) ,因此以 \( M \) 為焦點, \( L \) 為準線的拋物線與 \( {\ell }_{i} \) 相切且另一個焦點為 \( {\infty }_{\tau } \) ,故 \( \left( {M,{\infty ... | Yes |
Proposition 1.4.14. 给定任意 \( \bigtriangleup {ABC} \) ,将一點 \( P \) 映射至其關於 \( \bigtriangleup {ABC} \) 的等角共轭點 \( {P}^{ * } \) 的變換為一點等共軛變換 (稱為等角共軛變換)。 | Proof. 令 \( I,{I}_{a},{I}_{b},{I}_{c} \) 为 \( \bigtriangleup {ABC} \) 的内心与 \( A, B, C \) 分别關於 \( \bigtriangleup {ABC} \) 的旁心,考慮所有經過 \( I,{I}_{a},{I}_{b},{I}_{c} \) 的圆锥曲線族定義的點等共軛變換 \( \varphi \) ,則由\n\n\[ \left( {{AP}, A{P}^{ * };I{I}_{a},{I}_{b}{I}_{c}}\right) = \left( {{BP}, B{P}^{ * };I{I}_{b},{I}_{c}{I}_{a}}\right) ... | Yes |
Corollary 1.4.15. 给定任意 \( \bigtriangleup {ABC} \) ,令 \( \varphi \) 为 \( \bigtriangleup {ABC} \) 上的等角共轭变换,則 \( \varphi \left( {\odot \left( {ABC}\right) }\right) = {\mathcal{L}}_{\infty } \) 。 | 這個算角度也是顯然的。 | No |
Proposition 1.4.17. 给定 \( \\bigtriangleup {ABC} \) ,令 \( O \) 为 \( \\bigtriangleup {ABC} \) 的外心, \( \\left( {P,{P}^{ * }}\\right) \) 为關於 \( \\bigtriangleup {ABC} \) 的等角共轭點對, \( {P}^{{ * }^{\prime }} \) 為 \( {P}^{ * } \) 關於 \( O \) 的對稱點。令 \( \\bigtriangleup {P}_{a}{P}_{b}{P}_{c} \) 为 \( P \) 關於 \( \\bigtriangleup {ABC} ... | Proof. 我們依序證明命題 (i), (ii), (iii)。\n\n(i) 注意到 \( {P}^{ * } \) 關於 \( \\bigtriangleup {P}_{a}^{ * }{P}_{b}^{ * }{P}_{c}^{ * } \) 的佩多圓為 \( \\odot \\left( {ABC}\\right) \) ,所以 \( {P}^{ * } \) 關於 \( \\bigtriangleup {P}_{a}^{ * }{P}_{b}^{ * }{P}_{c}^{ * } \) 的等角共轭點為 \( {P}^{ * } \) 關於 \( \\odot \\left( {ABC}\\right) \) 的圓心的對稱... | Yes |
Proposition 2.1.1. 令 \( \mathcal{P} \) 为一個以 \( F \) 为焦点、 \( L \) 为準線的拋物線,設 \( \bigtriangleup {ABC} \) 为与 \( \mathcal{P} \) 相切的三角形,則 \( F \in \odot \left( {ABC}\right) \) 且 \( L \) 經過 \( \bigtriangleup {ABC} \) 的垂心。 | Proof. 令 \( F \) 關於 \( {BC},{CA},{AB} \) 的對稱點分別為 \( {F}_{a},{F}_{b},{F}_{c} \) , \( \mathcal{P} \) 分别與 \( {BC},{CA} \) , \( {AB} \) 切於 \( {T}_{a},{T}_{b},{T}_{c} \) ,那麼由拋物線的光學性質可得 \( {F}_{a},{F}_{b},{F}_{c} \) 分別為 \( {T}_{a},{T}_{b},{T}_{c} \) 關於 \( L \) 的垂足,因此由 \( {F}_{a},{F}_{b},{F}_{c} \) 共線於 \( L \) 及施坦納定理知 \( F \i... | Yes |
Proposition 2.1.2. 令 \( \mathcal{P} \) 为一個以 \( F \) 为焦点、 \( L \) 为準線的拋物線,設 \( \bigtriangleup {ABC} \) 為 \( \mathcal{P} \) 的自共轭三角形,則 \( F \) 位於 \( \bigtriangleup {ABC} \) 的九點圓上且 \( L \) 經過 \( \bigtriangleup {ABC} \) 的外心。 | Proof. 令 \( U \) 为垂直於 \( L \) 方向上的無窮遠點，則 \( U \in \mathcal{P} \) 且 \( U \) 為 \( O \) 的中心，令 \( {M}_{a},{M}_{b},{M}_{c} \) 分别为 \( \overline{BC},\overline{CA},\overline{AB} \) 的中點, \( {AU} \) 分别交 \( {BC},{M}_{b}{M}_{c} \) 於 \( P, T \) ,則由 \( \left( {A, U;P, T}\right) = - 1 \) 知 \( T \in \mathcal{P} \) ,由 \( {M}_{b}{M}_{c}... | Yes |
Proposition 2.1.4. 给定任意非直角三角形 \( \bigtriangleup {ABC} \) ,設其垂心為 \( H,\mathcal{H} \) 为 \( \bigtriangleup {ABC} \) 的外接圆锥曲線，則 \( H \in \mathcal{H} \) 若且唯若 \( \mathcal{H} \) 为等軸雙曲線。 | Proof. 考虑 \( \bigtriangleup {ABC} \) 上的等角共轭變換 \( \varphi \) ,令 \( O \) 为 \( \bigtriangleup {ABC} \) 的外心,則 \( H \in \mathcal{H} \) 若且唯若 \( O \in \varphi \left( \mathcal{H}\right) \) ,設 \( \varphi \left( \mathcal{H}\right) \) 与 \( \odot \left( {ABC}\right) \) 交於 \( X, Y \) 兩點 (若無交點,則 \( \mathcal{H} \) 與 \( {\mathcal{L}}_... | Yes |
给定任意直角三角形 \( \bigtriangleup {ABC} \) ,設 \( {BC} \) 为其斜邊, \( {AD} \) 為高, \( \mathcal{H} \) 为 \( \bigtriangleup {ABC} \) 的外接圆锥曲線,則 \( {AD} \) 與 \( \mathcal{H} \) 相切若且唯若 \( \mathcal{H} \) 为等軸雙曲線。 | 考虑 \( \bigtriangleup {ABC} \) 上的等角共轭變換 \( \varphi \) ,令 \( O \) 为 \( \bigtriangleup {ABC} \) 的外心,則 \( {AD} \) 與 \( \mathcal{H} \) 相切若且唯若 \( \varphi \left( \mathcal{H}\right) ,{AO},{BC} \) 共點若且唯若 \( O \in \varphi \left( \mathcal{H}\right) \) 。与上述的性質證明相同， \( O \in \varphi \left( \mathcal{H}\right) \) 若且唯若 \( \mathcal{H} ... | Yes |
Proposition 2.1.6. 令 \( \mathcal{H} \) 为以 \( O \) 为中心的等軸雙曲線,設 \( A, B, C \in \mathcal{H} \) ,則 \( O \) 位於 \( \bigtriangleup {ABC} \) 的九點圓上。 | Proof. 令 \( H \) 为 \( \bigtriangleup {ABC} \) 的垂心,則 \( H \in \mathcal{H} \) ,因此由 (1.3.18) 知 \( O \) 位於完全四點形 \( \left( {A, B, C, H}\right) \) 的九點圓錐曲線上，即 \( \bigtriangleup {ABC} \) 的九點圓上。 | Yes |
Proposition 2.1.7. 令 \( \mathcal{H} \) 为以 \( O \) 为中心的等轴双曲線,则对於任意兩點 \( {P}_{1} \) , \( {P}_{2} \) ，\n\n\[ \measuredangle \left( {{\mathfrak{p}}_{\mathcal{H}}\left( {P}_{1}\right) ,{\mathfrak{p}}_{\mathcal{H}}\left( {P}_{2}\right) }\right) + \measuredangle {P}_{1}O{P}_{2} = 0. \] | Proof. 令 \( {U}_{i} = O{P}_{i} \cap {\mathcal{L}}_{\infty },{V}_{i} = {\ell }_{\mathcal{H}}\left( {P}_{i}\right) \cap {\mathcal{L}}_{\infty } \) ,由平行弦定理知 \( {\mathfrak{p}}_{\mathcal{H}}\left( {U}_{i}\right) \parallel {\mathfrak{p}}_{\mathcal{H}}\left( {P}_{i}\right) \) 。令 \( {W}_{1},{W}_{2} \) 为 \( \mathcal{H} \) 与 \( ... | Yes |
Corollary 2.1.8. 令 \( \mathcal{H} \) 为以 \( O \) 为中心的等軸雙曲線,設 \( \bigtriangleup {ABC} \) 为 \( \mathcal{H} \) 的自共軛三角形，則 \( O \in \odot \left( {ABC}\right) \) 。 | Proof. 由 (2.1.7) 可得 \( \measuredangle {BOC} = - \measuredangle \left( {{CA},{AB}}\right) = \measuredangle {BAC} \) ，即 \( O \in \odot \left( {ABC}\right) \) 。 | Yes |
Proposition 2.1.9. 令 \( \mathcal{H} \) 为以 \( O \) 为中心的等軸雙曲線,設 \( \bigtriangleup {ABC} \) 为 \( \mathcal{H} \) 的自共轭三角形,則 \( \bigtriangleup {ABC} \) 的内心与三個旁心 \( I,{I}_{a},{I}_{b},{I}_{c} \in \mathcal{H} \) 。 | Proof. 令 \( P \in \mathcal{H} \smallsetminus \left\{ {I,{I}_{a},{I}_{b},{I}_{c}}\right\} \) ,設 \( \bigtriangleup {P}_{a}{P}_{b}{P}_{c} \) 为 \( P \) 關於 \( \bigtriangleup {ABC} \) 的反西瓦三角形, 因为 \( \bigtriangleup {ABC} \) 为 \( \mathcal{H} \) 的自共轭三角形,所以 \( {P}_{a},{P}_{b},{P}_{c} \in \mathcal{H} \) 。由 \( \bigtriangleup {I}_{... | Yes |
Proposition 2.1.10. 若 \( \overline{BC} \) 为一個等軸雙曲線 \( \mathcal{H} \) 上的直徑，那麼對於任意 \( A \in \mathcal{H}, A \) 關於 \( \odot \left( {ABC}\right) \) 的對徑點 \( {A}^{ * } \in \mathcal{H} \) ,且 \( {T}_{A}\mathcal{H} \) 為 \( \bigtriangleup {ABC} \) 的 \( A \) -共轭中線。 | Proof. 考慮關於 \( \bigtriangleup {ABC} \) 的等角共軛變換 \( \varphi \) ,設 \( L = \varphi \left( \mathcal{H}\right) ,{L}^{\prime } = \varphi \left( {{T}_{A}\mathcal{H}}\right) \) ,那麼 \( \varphi \left( {L \cap {L}^{\prime }}\right) \in \mathcal{H} \cap {T}_{A}\mathcal{H} = A \) ,所以 \( L \cap {L}^{\prime } \in {BC} \) 。注意到 \( \bigt... | Yes |
Theorem 2.1.12. 令 \( T \) 为完全四點形 \( \mathcal{Q} = \left( {{P}_{1},{P}_{2},{P}_{3},{P}_{4}}\right) \) 的龐色列點,則 \( \bigtriangleup {P}_{i - 1}{P}_{i}{P}_{i + 1} \) 的九點圓, \( {P}_{i} \) 關於 \( \bigtriangleup {P}_{i + 1}{P}_{i + 2}{P}_{i + 3} \) 的佩多圓及 \( \mathcal{Q} \) 的西瓦圓，這九個圓共點，且該點為 \( T \) 。 | Proof. 令 \( \mathcal{H} \) 为通过 \( {P}_{1},{P}_{2},{P}_{3},{P}_{4} \) 的等軸雙曲線, \( \bigtriangleup {XYZ} \) 为 \( \mathcal{Q} \) 的西瓦三角形, 則 \( \bigtriangleup {XYZ} \) 为 \( \mathcal{H} \) 的自共轭三角形,因此由上述性質知 \( T \) 位於 \( \bigtriangleup {P}_{i - 1}{P}_{i}{P}_{i + 1} \) 的九點圓上且位於 \( \mathcal{Q} \) 的西瓦圓上。令 \( {P}_{4} \) 關於 \( \bigt... | Yes |
Proposition 2.1.14. 给定任意 \( \bigtriangleup {ABC} \) ,若 \( {P}^{ * } \) 为 \( P \) 關於 \( \bigtriangleup {ABC} \) 的 antigonal conjugate，则 \( \overline{P{P}^{ * }} \) 中點 \( T \) 为完全四點形 \( \left( {A, B, C, P}\right) \) 的龐色列點。 | Proof. 令 \( P \) 關於 \( {BC},{CA},{AB} \) 的對稱點分別為 \( {P}_{A},{P}_{B},{P}_{C} \) ,則 \( B, C,{P}_{A},{P}^{ * } \) 及其輪換分別共圓。注意到 \( \odot \left( {{BC}{P}_{A}}\right) \) 關於 \( P \) 位似 \( 1/2 \) 倍的像為 \( \bigtriangleup {BPC} \) 的九點圓 \( \odot \left( {N}_{a}\right) \) ,所以 \( T \in \odot \left( {N}_{a}\right) \) ,同理有 \( T \in \od... | Yes |
Proposition 2.1.15. 给定任意 \( \bigtriangleup {ABC} \) ,令 \( O \) 为 \( \bigtriangleup {ABC} \) 的外心,對於 \( \bigtriangleup {ABC} \) 的任意一對等角共轭點對 \( \left( {P, Q}\right) \) ,令 \( T \) 为 \( \left( {A, B, C, P}\right) \) 的龐色列點,則 \( T \) 關於 \( \bigtriangleup {ABC} \) 的中點三角形的施坦納線為 \( {OQ} \) 。 | Proof. 令 \( \bigtriangleup {M}_{a}{M}_{b}{M}_{c} \) 为 \( \bigtriangleup {ABC} \) 的中點三角形, \( \mathcal{S} \) 为 \( T \) 關於 \( \bigtriangleup {M}_{a}{M}_{b}{M}_{c} \) 的施坦納線, \( \mathcal{H} \) 為經過 \( A, B, C, P \) 的等軸雙曲線, \( {H}^{ * } \) 為 \( H \) 關於 \( \mathcal{H} \) 的對徑點,則 \( {H}^{ * } \in \odot \left( {ABC}\right) \) ,考虑... | Yes |
Proposition 2.1.16. 给定任意 \( \bigtriangleup {ABC} \) ,令 \( O \) 为 \( \bigtriangleup {ABC} \) 的外心,對於 \( \bigtriangleup {ABC} \) 的任意一對等角共轭點對 \( \left( {P, Q}\right) \) ,令 \( T \) 为 \( \left( {A, B, C, P}\right) \) 的龐色列點,則 \( T \) 關於 \( P \) 關於 \( \bigtriangleup {ABC} \) 的佩多三角形的施坦納線平行於 \( {OQ} \) 。 | Proof. 令 \( \bigtriangleup {P}_{a}{P}_{b}{P}_{c} \) 为 \( P \) 關於 \( \bigtriangleup {ABC} \) 的佩多三角形, \( \bigtriangleup {M}_{a}{M}_{b}{M}_{c} \) 为 \( \bigtriangleup {ABC} \) 的中點三角形，由上一個性質知原命題等價於證明 \( {P}_{a}T \) 關於 \( \angle {P}_{b}{P}_{a}{P}_{c} \) 的等角線平行於 \( {M}_{a}T \) 關於 \( \angle {M}_{b}{M}_{a}{M}_{c} \) 的等角線，即\n\n\... | Yes |
Theorem 2.1.17. 给定任意 \( \bigtriangleup {ABC} \) 及一點 \( P \) ,那麼 \( P \) 關於 \( \bigtriangleup {ABC} \) 的佩多圓与 \( \bigtriangleup {ABC} \) 的九点圆在 \( \left( {A, B, C, P}\right) \) 的龐色列點的夾角為\n\n\[ \n{90}^{ \circ } + \measuredangle \left( {{BC},{AP}}\right) + \measuredangle \left( {{CA},{BP}}\right) + \measuredangle \left( {{A... | Proof. 令 \( \bigtriangleup {P}_{a}{P}_{b}{P}_{c} \) 为 \( P \) 關於 \( \bigtriangleup {ABC} \) 的佩多三角形, \( \bigtriangleup {M}_{a}{M}_{b}{M}_{c} \) 为 \( \bigtriangleup {ABC} \) 的中點三角形, \( T \) 为 \( \left( {A, B, C, P}\right) \) 的龐色列點,則 \( T \) 位於 \( \bigtriangleup {PCA} \) 與 \( \bigtriangleup {PAB} \) 的九點\n\n圆上。設 \( {L}_{1}... | Yes |
Theorem 2.1.18. 给定任意 \( \bigtriangleup {ABC} \) 与一外接等轴双曲線 \( \mathcal{H} \) ,令 \( P, Q \in \mathcal{H} \) ,那麼 \( P, Q \) 關於 \( \bigtriangleup {ABC} \) 的佩多圓在 \( \mathcal{H} \) 的中心的夾角為\n\n\[ \angle {QAP} + \angle {QBP} + \angle {QCP}\text{.} \] | Proof. 令 \( \bigtriangleup {P}_{a}{P}_{b}{P}_{c},\bigtriangleup {Q}_{a}{Q}_{b}{Q}_{c} \) 分别为 \( P, Q \) 關於 \( \bigtriangleup {ABC} \) 的佩多三角形, \( O \) 为 \( \mathcal{H} \) 的中心, \( {P}^{ * },{Q}^{ * } \) 分别为 \( P, Q \) 關於 \( \bigtriangleup {ABC} \) 的等角共轭點。設 \( {L}_{P} = {T}_{O} \odot \left( {{P}_{a}{P}_{b}{P}_{c}}\right) ... | Yes |
Theorem 2.1.19 (封腾一號/Fontené's Theorem I). 令 \( \bigtriangleup {M}_{a}{M}_{b}{M}_{c} \) 为 \( \bigtriangleup {ABC} \) 的中點三角形。對於任意一對等角共軛點對 \( \left( {P, Q}\right) \) ,令 \( T \) 為 \( \left( {A, B, C, P}\right) \) 的龐色列點, \( \bigtriangleup {Q}_{a}{Q}_{b}{Q}_{c} \) 为 \( Q \) 關於 \( \bigtriangleup {ABC} \) 的佩多三角形, \( {R}_{a} =... | Proof. 令 \( {T}^{\prime },{Q}_{a}^{\prime } \) 分别为 \( T,{Q}_{a} \) 關於 \( {M}_{b}{M}_{c} \) 的對稱點，則 \( {T}^{\prime } \in \odot \left( \overline{AO}\right) \) 且 \( {T}^{\prime } \in {OQ} \) , 因此有 \( \measuredangle A{T}^{\prime }Q = {90}^{ \circ } \) ,即 \( {T}^{\prime } \) 为完全四線形 \( \left( {{CA},{AB},{M}_{b}{M}_{c},{Q}_{b}... | Yes |
Corollary 2.1.20. 同樣定義 \( {R}_{b},{R}_{c} \) ,則 \( \bigtriangleup {R}_{a}{R}_{b}{R}_{c} \) 的垂心为 \( \bigtriangleup {Q}_{a}{Q}_{b}{Q}_{c} \) 的外心, 即 \( \overline{PQ} \) 中點。 | Proof. 注意到 \( \bigtriangleup {R}_{a}{R}_{b}{R}_{c} \) 为 \( T \) 關於 \( \bigtriangleup {Q}_{a}{Q}_{b}{Q}_{c} \) 的西瓦三角形,因此 \( \bigtriangleup {R}_{a}{R}_{b}{R}_{c} \) 为關於 \( \odot \left( {{Q}_{a}{Q}_{b}{Q}_{c}}\right) \) 的自共轭三角形,故 \( \bigtriangleup {R}_{a}{R}_{b}{R}_{c} \) 的垂心为 \( \odot \left( {{Q}_{a}{Q}_{b}{Q}_{c}}\right... | Yes |
Theorem 2.1.21 (封騰二號/Fontené's Theorem II). 给定任意 \( \bigtriangleup {ABC},\ell \) 为一個通過 \( \bigtriangleup {ABC} \) 的外心的直線, \( Q \) 為 \( \ell \) 上一動點,則 \( Q \) 關於 \( \bigtriangleup {ABC} \) 的佩多圓過一定點且該定點位於 \( \bigtriangleup {ABC} \) 的九點圓上 (該定點即為 \( \left( {A, B, C, P}\right) \) 的龐色列點， 其中 \( P \) 为 \( Q \) 關於 \( \bigtriang... | Proof. 令 \( P \) 为 \( Q \) 關於 \( \bigtriangleup {ABC} \) 的等角共轭點,那麼 \( Q \) 關於 \( \bigtriangleup {ABC} \) 的佩多圓, 即 \( P \) 關於 \( \bigtriangleup {ABC} \) 的佩多圓,經過 \( \left( {A, B, C, P}\right) \) 的龐色列點，而其即為 \( {OQ} \) 關於 \( \bigtriangleup {ABC} \) 的中點三角形的反施坦納點，故為定點。 | Yes |
Theorem 2.1.22 (封腾三號/Fontené's Theorem III). 给定任意 \( \bigtriangleup {ABC} \) , \( \left( {P, Q}\right) \) 為 \( \bigtriangleup {ABC} \) 的一對等角共轭點對，則 \( P \) 關於 \( \bigtriangleup {ABC} \) 的佩多圓與 \( \bigtriangleup {ABC} \) 的九點圓相切若且唯若 \( {PQ} \) 通過 \( \bigtriangleup {ABC} \) 的外心。 | Proof. 令 \( O, H, N \) 分别为 \( \bigtriangleup {ABC} \) 的外心、垂心、九點圓圓心, \( {T}_{1} \) 為 \( \left( {A, B, C, P}\right) \) 的龐色列點， \( {T}_{2} \) 為 \( \left( {A, B, C, Q}\right) \) 的龐色列點。\n\n\( \left( \Rightarrow \right) \) 若 \( P \) 關於 \( \bigtriangleup {ABC} \) 的佩多圓與 \( \bigtriangleup {ABC} \) 的九點圓相切,那麼切點為 \( {T}_{1} \) 也為 \... | Yes |
Theorem 2.2.1 (费爾巴哈定理/Feuerbach's). 给定任意 \( \bigtriangleup {ABC} \) ,其九點圓與内切圓及三個旁切圓相切。 | Proof. 由2.1.17及 \( \sum \measuredangle \left( {A{I}_{X},{BC}}\right) = {90}^{ \circ } \) 就有九點圓與内切圓及三個旁切圓的夾角是 \( {0}^{ \circ } \) ,其中 \( {I}_{X} \) 为内心或旁心。 | No |
Proposition 2.2.4. 令 \( X, Y \) 分别为 \( \odot \left( I\right) \) 及 \( \odot \left( O\right) \) 的内位似中心、外位似中心, 那麼 \( X, G,{Fe} \) 及 \( Y, H,{Fe} \) 分别共線。 | Proof. 注意到 \( G, H \) 分别为 \( \odot \left( N\right) \) 及 \( \odot \left( O\right) \) 的内位似中心、外位似中心,因此由 Monge 定理知 \( X, G,{Fe} \) 及 \( Y, H,{Fe} \) 分别共線。 | Yes |
Proposition 2.2.5. 令 \( {I}^{ * } \) 为 \( I \) 關於 \( \odot \left( O\right) \) 的反演點, \( \mathcal{E} \) 为 \( \bigtriangleup {ABC} \) 的欧拉線, 那麼 \( {I}^{ * }{Fe}\parallel \mathcal{E} \) 。 | Proof. 令 \( R, r \) 分别为 \( \odot \left( O\right) , \odot \left( I\right) \) 的半径長,注意到\n\n\[\n\frac{O{I}^{ * }}{OI} = \frac{{R}^{2}}{O{I}^{2}} = \frac{R/2}{R/2 - r} = \frac{NFe}{NI}\n\]\n\n所以 \( {I}^{ * }{Fe}\parallel {ON} = \mathcal{E} \) 。 | Yes |
Proposition 2.2.11. 令 \( P \) 为任意一點, \( Q \) 为 \( P \) 關於 \( \bigtriangleup {ABC} \) 的等角共轭點, \( T \) 为 \( \left( {A, B, C, Q}\right) \) 的龐色列點。設 \( {T}_{P},{T}_{O} \) 分别為 \( P, O \) 關於 \( {BC} \) 的垂足, \( e \) 为 \( {OP} \) 與 \( {BC} \) 的交點。若 \( {P}_{e},{O}_{e} \) 分别为 \( e \) 關於 \( {AP},{AO} \) 的垂足,那麼\n\n\[ \bigtriangleup... | Proof. 令 \( T,{T}_{P},{T}_{O} \) 關於平行於 \( {BC} \) 的中位線的對稱點分別為 \( {T}^{\prime },{T}_{P}^{\prime },{T}_{O}^{\prime } \) ,則由封腾一號的證明知 \( {T}^{\prime } \in {OP} \) 且 \( A, P,{T}^{\prime },{T}_{P}^{\prime } \) 及 \( A, O,{T}^{\prime },{T}_{O}^{\prime } \) 分别共圆,故\n\n\[ \measuredangle {OPA} = \measuredangle {T}^{\prime }{T}_{P}... | Yes |
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