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Theorem 1 If \( \diamondsuit {ABCD} \) is a convex quadrilateral, then the four betweenness relations \( \left( {\overrightarrow{AB}\overrightarrow{AC}\overrightarrow{AD}}\right) ,\left( {\overrightarrow{BA}\overrightarrow{BD}\overrightarrow{BC}}\right) ,\left( {\overrightarrow{CB}\overrightarrow{CA}\overrightarrow{CD}... | Proof It suffices to prove the first of these, namely \( \left( {\overrightarrow{AB}\overrightarrow{AC}\overrightarrow{AD}}\right) \) , since by relabeling points, the others follow from this one case. By definition, \( C \) and \( D \) lie on the same side of line \( {AB} \), the line containing ray \( {AB} \equiv h \... | Yes |
Theorem 2 If \( \diamondsuit {ABCD} \) has the property that \( C \) and \( D \) lie on the same side of line \( {AB}, D \) and \( A \) lie on the same side of line \( {BC} \), and \( A \) and \( B \) lie on the same side of line \( {CD} \), then points \( B \) and \( C \) lie on the same side of line \( {DA} \), and \... | Proof Suppose \( B \) and \( C \) lie on opposite sides of line \( {AD} \) . Then there exists \( E \) on \( \overrightarrow{AD} \), such that \( \left( {BEC}\right) \) . By Theorem 2, Section 1.8, either (AED), \( \left( {ADE}\right) ,\left( {DAE}\right) \), or \( \left( {A{E}^{ * }D}\right) \) (when \( \alpha < \inft... | Yes |
In any Saccheri quadrilateral, the summit angles are congruent, as are the diagonals. The summit is not congruent to the base unless the quadrilateral is a rectangle. | Proof Using the notation of Figure 4.10, we have \( \overline{AD} \cong \overline{BC},\angle A \cong \angle B \) , \( \overline{AB} \cong \overline{BA},\angle B \cong \angle A \), and \( \overline{BC} \cong \overline{AD} \) . Therefore, regarding \( \diamondsuit {ABCD} \) and \( \diamond {BADC} \) as separate quadrilat... | Yes |
Theorem 3 If \( M \) and \( N \) are the respective midpoints of the base and summit of a Saccheri quadrilateral \( \diamondsuit {ABCD} \), then convex quadrilaterals \( \diamond {BMND} \) and \( \diamond {AMNC} \) are formed. Furthermore, line \( {MN} \) is the perpendicular bisector of both the base and summit (see F... | Proof Observing the hypothesis, with \( {AM} = {MB},\angle A \cong \angle B,{AD} = {BC} \) , \( \angle D \cong \angle C \), and \( {DN} = {NC} \), we note that \( \diamond {AMND} \cong \diamond {BMNC} \) by SASAS and therefore \( \angle {AMN} \cong \angle {BMN},\angle {DNM} \cong \angle {CNM} \) . Therefore, \( \overli... | Yes |
Theorem 1 Suppose that the foot \( C \) of point \( O \) on line \( {AB} \equiv l \) lies between \( A \) and \( B \) (Figure 4.28). Further, let \( {OA} \) and \( {OB} \) be each less than \( \alpha /2 \) . Then for any point \( X \) on line \( l \), either (1) \( {OX} > {OA} > {OC} \) if \( \left( {XAC}\right) ,\left... | Figure 4.28 Proof of Theorem 1. | No |
Theorem 2 Given any circle and three distinct points \( A, B \), and \( C \) on it, no point of \( \mathbf{C} \) can lie in the four shaded regions of Figure 4.30a and therefore \( \mathbf{C} \) (not including \( A, B \), or \( C \) ) lies in the complementary set, the shaded regions of Figure 4.30b. | Proof Recall the definition of an arc: the intersection of a circle and a closed half plane. It suffices to consider only the interior of the arc since we have only to add the endpoints to obtain the desired result. Let \( \angle {ABC} \) be an inscribed angle (Figure 4.31) and Int \( \overset{⏜}{AC} \equiv \) \( \math... | Yes |
Theorem 6 Suppose that two circles \( \mathbf{C} \) and \( \mathbf{D} \) centered at \( O \) and \( Q \) have radii \( r \) and \( s \) such that \( s \leq r \) and \( r - s < {OQ} < r + s \) . Then the circles intersect at two distinct points \( P \) and \( R \), having a common secant \( {PR} \) . | [Hint: Let \( {AB} \) be the diameter of \( C \) lying on line \( {OQ} \) ; with \( Q \) on ray \( {OB} \), define \( c \) as the least upper bound of \( \{ m\angle {XOB} \) : \( X \in \) Int \( \mathbf{D}, X \in \mathbf{C}\} \), construct \( \angle {COB} \) having measure \( c \) , \( C \in \mathbf{C} \), and show tha... | No |
Theorem 1 If \( \alpha < \infty \), each pair of lines intersects. | Proof Let \( l \) and \( m \) be any two lines. Choose \( A, B \), and \( C \) any three points on \( m \) such that \( {AB} = \alpha \), and let \( D \) be any point on \( l \) . (See Figure 5.2, left.) Then it follows that \( \left( {ACB}\right) \) and \( \left( {ADB}\right) \) . Thus, \( {AC} < {AB} = \alpha \) and ... | Yes |
Theorem 2 If \( \alpha = \infty \), parallel lines exist. In particular, if two lines are cut by a transversal such that a pair of alternate interior angles is congruent, the lines are parallel. | Proof Let \( l \) be a given line and \( A \) any point not on \( l \), and choose any line \( t \) passing through \( A \) and a point \( B \) on \( l \) . With \( F \) any point on \( l \) , \( F \neq B \), construct ray \( {AC} \) on the opposite side of \( t \) as \( F \) such that \( \angle {CAB} \cong \) \( \angl... | Yes |
Theorem 3 If \( \alpha < \infty \), the angle-sum of any right triangle is greater than 180. | Proof This has already been proven for right triangles having legs of length less than \( \alpha /2 \) (Saccheri’s theorem, Corollary A, Section 4.3), so let \( \bigtriangleup {ABC} \) be a right triangle with right angle at \( C \) and at least one leg, say \( {BC} \), of length \( \geq \alpha /2 \), as shown in Figur... | Yes |
Theorem 3 If in \( \bigtriangleup {ABC} \) a point \( D \) is taken on the interior of side \( {BC} \) , with \( {\varepsilon }_{1} = \varepsilon \left( {\Delta ABD}\right) \) and \( {\varepsilon }_{2} = \varepsilon \left( {\Delta ADC}\right) \), then \( {\varepsilon }_{1} + {\varepsilon }_{2} = \varepsilon \left( {\De... | Proof Using the notation shown in Figure 5.9, we have\n\n\[ \n{e}_{1} = \angle 1 + \angle 2 + \angle 3 - {180}\;{e}_{2} = \angle 4 + \angle 5 + \angle 6 - {180} \n\]\n\nHence\n\n\[ \n{e}_{1} + {e}_{2} = \left( {\angle 1 + \angle 2 + \angle 3 - {180}}\right) + \left( {\angle 4 + \angle 5 + \angle 6 - {180}}\right) \n\]\... | Yes |
Theorem 6 If the sides of a triangle become arbitrarily small, the angle-sum of the triangle converges to 180. (More precisely: Given \( p > 0 \) there exists \( q > 0 \) such that if the sides of \( \bigtriangleup {ABC} \) are of length less than \( q \), the angle-sum \( S \) of \( \bigtriangleup {ABC} \) differs fro... | Proof Consider any right triangle \( {PQR} \) having sides less than \( \pi /2 \) , with right angle at \( Q \) . For any integer \( n > 0 \), construct \( n \) points \( {Q}_{1} \) , \( {Q}_{2},{Q}_{3},\ldots ,{Q}_{n} = R \) on segment \( {QR} \) such that the angles determined at \( P \) are congruent: \( \angle {QP}... | Yes |
Theorem 1 The family of lines perpendicular to any given line \( l \) pass through the same two extremal points \( P \) and \( {P}^{ * } \), and every line passing through \( P \) and \( {P}^{ * } \) is perpendicular to \( l \) . | The points \( P \) and \( {P}^{ * } \) mentioned in the previous theorem are called the poles of line \( l \), and line \( l \) is called the polar of the two extremal points \( P \) and \( {P}^{ * } \). We have thus shown that every line has a unique pair of poles. Conversely, suppose we are given two extremal points ... | Yes |
Theorem 3 If \( P \) is a pole of line \( l \), then the polar of any point on \( l \) passes through \( P \), and dually, the poles of any line passing through \( P \) lie on \( l \) . | 1. Let \( Q \) be any point on \( l \) (Figure 5.14) and determine \( {Q}^{ * } \in l \) such that \( Q{Q}^{ * } = \pi \) . The perpendicular \( m \) to \( l \) at the point \( R \) midway between \( Q \) and \( {Q}^{ * } \) is thus the polar of \( Q \) and \( {Q}^{ * } \) (by construction given earlier). Since \( m \b... | Yes |
Theorem 1 If \( \left( {BDC}\right) \) in \( \bigtriangleup {ABC} \), with \( {\delta }_{1} = \delta \left( {\bigtriangleup {ABD}}\right) \) and \( {\delta }_{2} = \delta \left( {\bigtriangleup {ADC}}\right) \), then\n\n\[ \n{\mathrm{d}}_{1} + {\mathrm{d}}_{2} = \mathrm{d}\left( {\Delta ABC}\right) \n\] | The proof is virtually the same as that for the additivity of excess, so it will be omitted (you should try showing this). | No |
Theorem 1 Let \( {AB} \bot h \) at \( B \) (Figure 5.29). The ray \( {AW} \equiv w \) is asymptotically parallel to ray \( h \) iff the following is true:\n\n(a) \( w \) lies on the same side of line \( {AB} \) as \( h \) ,\n\n(b) \( w \) is parallel to \( h \), and\n\n(c) every ray \( u \) between \( w \) and \( v \eq... | Proof Certainly the asymptotic parallel \( {k}^{ * } \) has the three properties mentioned. Conversely, suppose ray \( w \) has these properties. If \( w \neq {k}^{ * } \) , then either \( \left( {w{k}^{ * }v}\right) \) as in Figure 5.29a, or \( \left( {{k}^{ * }{wv}}\right) \) as in Figure 5.29b. But in either case, t... | Yes |
Theorem 2 The function \( p \) defined above is nonincreasing. That is, given \( a < b, p\left( a\right) \geq p\left( b\right) \) . | Proof In Figure 5.34 let \( X \) and \( Y \) be points on the perpendicular to \( h \) at \( B \) such that \( \left( {BXY}\right) \), with \( {BX} = a < b = {BY} \) . Also, let rays \( k = {XK} \) and \( {k}^{\prime } = Y{K}^{\prime } \) be the asymptotic parallels to \( h \), with \( \gamma = p\left( a\right) \) and ... | Yes |
Theorem 1 If \( \bigtriangleup {AB\Omega } \) is an isosceles asymptotic triangle, the base angles are acute. | Proof If \( A = B > {90} \), then construct perpendiculars \( {AE} \) and \( {BF} \) to line \( {AB} \), as in Figure 5.38. Ray \( {AE} \) then lies between rays \( {AC} \) and \( {AB} \), and \( {AF} \) lies between \( {AB} \) and \( {AD} \) . Since \( \overrightarrow{AC}{\parallel }_{a}\overrightarrow{BD} \), ray \( ... | Yes |
Theorem 2 If \( \bigtriangleup {AB\Omega } \) is an asymptotic triangle, and \( P \) and \( Q \) are any two points lying on lines \( {AC} \) and \( {BD} \), respectively, then \( \bigtriangleup {PQ\Omega } \) is also an asymptotic triangle. | Proof Suppose line \( {PQ} \) is a perpendicular in common with lines \( {AC} \) and \( {BD} \), with \( P \) and \( Q \) on \( {AC} \) and \( {BD} \). Then by Theorem \( 2,{\Delta PQ\Omega } \) is an asymptotic triangle having supplementary base angles, contradicting (1). D | No |
Theorem 3 If \( \overrightarrow{AC}{\parallel }_{a}\overrightarrow{BD} \) and \( \angle {CAB} \cong \angle {ABD} \), then \( \overrightarrow{BD}{\parallel }_{a}\overrightarrow{AC} \) . | Proof Let \( u \equiv \overrightarrow{BW} \) be any ray between rays \( {BD} \) and \( {BA} \), to prove that \( u \) meets ray \( {AC} \) (Figure 5.44). On the \( C \) -side of line \( {AB} \) construct \( \angle {BAF} \cong \angle {ABW} \) . Since \( m\angle {BAF} = m\angle {ABW} < m\angle {ABD} \), then \( m\angle {... | Yes |
Theorem 1 If parallel lines are cut by a transversal, each pair of alternate interior angles thus formed are congruent. | Proof In Figure 5.49 we have \( l \backsim m \) and transversal \( t \) intersecting \( l \) and \( m \) at \( A \) and \( B \), respectively, with alternate interior angles \( \angle {CAB} \) and \( \angle {ABF} \) . Construct ray \( A{C}^{\prime } \) on the \( C \) -side of line \( {AB} \) such that \( \angle {C}^{\p... | Yes |
Theorem 1 The midpoint of the hypotenuse of a right triangle is equidistant from the three vertices. Conversely, if the midpoint of a side of a triangle is equidistant from the three vertices, the triangle is a right triangle. | Proof Suppose \( \bigtriangleup {ABC} \) is a right triangle with right angle at \( C \) , and with \( M \) the midpoint of hypotenuse \( {AB} \) . Since we already have \( {AM} = {BM} \) it remains to prove \( {CM} = {AM} \) . First, suppose that \( {CM} > {AM} \), as shown in Figure 5.57. Then by the scalene inequali... | Yes |
Theorem 4 If two circles \( \left\lbrack {O, r}\right\rbrack \) and \( \left\lbrack {Q, s}\right\rbrack \) intersect at \( A \) and \( B \), the locus of points \( P \) whose powers with respect to the two circles are equal is line containing the common chord \( {AB} \) . | Proof\n\n1. By (1) Section 7.3 we have, as illustrated in Figure 7.24,\n\n\[ \text{Power}P\left( {\text{circle}O}\right) = {PA} \cdot {PB} \]\n\n\[ \text{Power}\mathrm{P}\left( {\text{circle}Q}\right) = {PA} \cdot {PB} \]\n\nThat is, the power of \( P \) with respect to the two circles is the same.\n\n2. Conversely, su... | Yes |
Theorem 1 Let \( A, B, C \), and \( D \) be any four distinct collinear points. Then the 24 permutations of \( A, B, C \), and \( D \) result in at most six distinct values for the cross ratio of these four points, as given by the following list, where \( \lambda = \left\lbrack {{AB},{CD}}\right\rbrack \neq 0,1 \) : | \[ \{ 1,1/1,1 - 1,1 - 1/1,1/\left( {1 - 1}\right) ,1/\left( {1 - 1}\right) \} \] | No |
Theorem 2 Consider collinear points \( A, B, C \), and \( D \), with \( O \) the midpoint of segment \( {AB} \), as shown in Figure 7.29b. Then \( \left\lbrack {{AB},{CD}}\right\rbrack = \) \( - 1 \) iff\n\n\[{OC} \cdot {OD} = O{B}^{2}\] | Proof The relation \( \left\lbrack {{AB},{CD}}\right\rbrack = - 1 \) implies, by definition, \( {AC} \cdot {BD} = \) \( - {AD} \cdot {BC} \) or\n\n\[{AC} \cdot {BD} = {AD} \cdot {CB}\]\n\nIn this expression, insert \( O \) between the pairs \( \left( {A, C}\right) ,\left( {B, D}\right) ,\left( {A, D}\right) \), and \( ... | Yes |
Theorem 3 Points \( C \) and \( D \) are harmonic conjugates of \( A \) and \( B \) iff \( {AB} \) is the harmonic mean of \( {AC} \) and \( {AD} \), that is, iff\n\n\[ \frac{1}{AB} = \frac{1}{2}\left( {\frac{1}{AC} + \frac{1}{AD}}\right) \] | Proof Begin with the relation \( \left\lbrack {{AB},{CD}}\right\rbrack = - 1 \) . Then \( {AC} \cdot {BD} = - {AD} \) . \( {BC} = {AD} \cdot {CB} \) . Also, by (3), \( \left\lbrack {{AC},{BD}}\right\rbrack = 1 - \left\lbrack {{AB},{CD}}\right\rbrack = 2 \) . Thus,\n\n\[ \frac{{AB} \cdot {CD}}{{AD} \cdot {CB}} = 2\;\tex... | Yes |
Theorem 1 Suppose that \( \left\lbrack {O, r}\right\rbrack \) is a given circle having diameter \( {AB} \) , and that for two points \( C \) and \( D \) on line \( {AB}\left\lbrack {{AB},{CD}}\right\rbrack = - 1 \) . Then any circle \( \left\lbrack {Q, s}\right\rbrack \) passing through \( C \) and \( D \) is orthogona... | Proof (See Figure 7.44.) Since \( O \) is the midpoint of \( {AB} \), by Theorem 2 in Section 7.5, we have \( O{B}^{2} = {OC} \cdot {OD} \) . That is,\n\n\[ \n{r}^{2} = {OC} \cdot {OD} = \operatorname{Power}O\left( {\operatorname{circle}\left\lbrack {Q, s}\right\rbrack }\right) = O{Q}^{2} - {s}^{2} \n\]\n\nor\n\n\[ \n{... | Yes |
Theorem 2 The locus of points whose powers with respect to two non-concentric circles \( \left\lbrack {O, r}\right\rbrack \) and \( \left\lbrack {Q, s}\right\rbrack \) are equal is a line, and this line is perpendicular to the line of centers \( m \) at the unique point \( W \) on \( m \) in that locus. If the circles ... | Proof The existence of \( W \) follows by showing, by algebra, that the condition \( W{O}^{2} - {r}^{2} = W{Q}^{2} - {s}^{2} \) (powers of \( W \) are equal) leads to \( {WO} = \) \( \left( {Q{O}^{2} + {r}^{2} - {s}^{2}}\right) /{2QO} \) (left to the reader). Since \( {WO} \) is directed distance, \( W \) is unique on ... | No |
Theorem 3 Two circles \( {\mathbf{C}}_{1} \) and \( {\mathbf{C}}_{2} \) represented by the respective general equations \( {x}^{2} + {y}^{2} + {a}_{i}x + {b}_{i}y + {c}_{i} = 0\left( {i = 1,2}\right) \) are orthogonal iff\n\n\[ {a}_{1}{a}_{2} + {b}_{1}{b}_{2} = 2{c}_{1} + 2{c}_{2} \] | Proof Using condition (C) on orthogonal circles, if \( {O}_{1} \) and \( {O}_{2} \) are the centers of \( {\mathbf{C}}_{1} \) and \( {\mathbf{C}}_{2} \) respectively, and \( {r}_{1} \) and \( {r}_{2} \) their radii, then \( {\mathbf{C}}_{1} \bot {\mathbf{C}}_{2} \) iff \( {\left( {O}_{1}{O}_{2}\right) }^{2} = {r}_{1}^{... | Yes |
Theorem 2 An affine transformation \( f \) preserves ratios of collinear or parallel segments. That is, if \( {AB} \) and \( {CD} \) are either collinear or parallel segments, then \( {AB}/{CD} = {A}^{\prime }{B}^{\prime }/{C}^{\prime }{D}^{\prime } \) . | Proof It suffices to show that if \( C \) lies between \( A \) and \( B \), then \( {AC}/{AB} = \) \( {A}^{\prime }{C}^{\prime }/{A}^{\prime }{B}^{\prime } \) . (One can see that this implies the general case by the use of parallelograms.) Consider (ACB), as shown in Figure 8.9. Suppose that \( {AC}/{AB} \neq {A}^{\pri... | Yes |
Theorem 3 Given any two triangles \( {ABC} \) and \( {DEF} \), there exists an affine transformation mapping \( \bigtriangleup {ABC} \) onto \( \bigtriangleup {DEF} \) consisting of the product of three, or fewer, affine reflections. | Proof We have already proven that a product of two or fewer affine reflections \( {f}_{1} \) and \( {f}_{2} \) will map \( {AB} \) to \( {DE} \), with \( {B}^{\prime } \notin \) line \( {DE} \) . If \( {C}^{\prime \prime } = F \) , we are finished. Otherwise, consider line \( {C}^{\prime \prime }F \) . If \( {C}^{\prim... | Yes |
Theorem 1 If a circle \( \mathbf{D} \) is orthogonal to the circle of inversion, it maps to itself. That is, \( {\mathbf{D}}^{\prime } = \mathbf{D} \) . | Proof Let \( \mathbf{D} \equiv \left\lbrack {D, s}\right\rbrack \) be orthogonal to \( \mathbf{C} \equiv \left\lbrack {O, r}\right\rbrack \), the circle of inversion, as shown in Figure 8.43. We must show that the image \( {\mathbf{D}}^{\prime } \) of \( \mathbf{D} \) under inversion coincides with \( \mathbf{D} \) . I... | Yes |
Theorem 2 The inversion mapping is conformal, that is, the angle measure between any two curves remains unchanged under inversion. | Proof We must show that if \( {\mathbf{C}}_{\mathbf{1}} \) and \( {\mathbf{C}}_{\mathbf{2}} \) are any two curves intersecting at \( P \), and \( {t}_{1} \) and \( {t}_{2} \) are their tangent lines at \( P \), respectively, the angle \( \theta \) between \( {t}_{1} \) and \( {t}_{2} \) equals that of the corresponding... | Yes |
Theorem 3 The generalized cross ratio of four points is an invariant under a circular inversion. | Proof Let the images of \( A, B, C \), and \( D \) be, respectively, \( {A}^{\prime },{B}^{\prime },{C}^{\prime } \), and \( {D}^{\prime } \) . By (3), \( \bigtriangleup {OAC} \sim \bigtriangleup O{C}^{\prime }{A}^{\prime } \) (Figure 8.40) and\n\n\[ \n\frac{AC}{{A}^{\prime }{C}^{\prime }} = \frac{OC}{O{A}^{\prime }}\n... | Yes |
Theorem 2 Let \( \bigtriangleup {ABC} \) be any triangle in non-Euclidean geometry, with \( \Delta {A}^{\prime }{B}^{\prime }{C}^{\prime } \) the reference triangle whose sides have the same lengths as those of \( \bigtriangleup {ABC} \) . Then, in standard notation, \( A \prec {A}^{\prime }, B \prec {B}^{\prime } \), ... | Proof Let \( \Delta {A}^{\prime }{B}^{\prime }{C}^{\prime } \) be the reference triangle. Then, the Euclidean law of cosines is valid for \( \Delta {A}^{\prime }{B}^{\prime }{C}^{\prime } \), whose sides are \( a, b \), and \( c \) . By Theorem 2, and since cosine is decreasing on \( \left\lbrack {0,\pi }\right\rbrack ... | No |
Lemma 1 Let \( A, B, D \), and \( X \) be distinct, collinear points. In elliptic geometry, if\n\n\[ \frac{\sin {AX}}{\sin {BX}} = \frac{\sin {AD}}{\sin {BD}} \]\n\nand no two of \( A, B \), and \( D \) are extremal opposites, then either \( X = D \) or \( X = {D}^{ * } \) . In hyperbolic geometry, if\n\n\[ \frac{\sinh... | Proof (for elliptic geometry) For convenience, let \( x = {AX}, y = {BX} \) , \( a = {AD} \), and \( b = {BD} \) . The hypothesis then reads \( \sin x/\sin y = \sin a/\sin b \) . Since \( {AB} = {AD} + {DB} = {AX} + {XB} \), then \( x - y = {AB} = a - b \) . Thus, if \( x - y = \theta \), then \( y = x - \theta \) and ... | No |
Lemma 2 If \( D, E \), and \( F \) are points of triangle \( {ABC} \) lying on the extended sides \( \overrightarrow{BC},\overrightarrow{CA} \) and \( \overrightarrow{AB} \) respectively, and the linearity number of \( D, E \), and \( F \) with respect to \( \bigtriangleup {ABC} \) equals -1, then two of these points d... | Proof There is nothing to prove in elliptic geometry, so we turn to hyperbolic geometry. Since the linearity number is negative, and, for example, \( \left( {BDC}\right) \) iff \( {BD} \) and \( {DC} \) have the same sign (and the same for sinh \( {BD} \) and \( \sinh {DC} \) ), either (1) two of the points lie interio... | Yes |
Lemma 3 Suppose that lines \( \overrightarrow{AD},\overrightarrow{BE} \) and \( \overrightarrow{CF} \) are cevians of \( \bigtriangleup {ABC} \) , such that \( \overrightarrow{AD} \cap \overrightarrow{BE} = P,\overrightarrow{BE} \cap \overrightarrow{CF} = Q \), and \( \overrightarrow{CF} \cap \overrightarrow{AD} = R \)... | Proof Suppose neither of the desired intersections exist. That is,\n\n\[ \overrightarrow{CP} \cap \overrightarrow{AB} = \varnothing ,\;\overrightarrow{AQ} \cap \overrightarrow{BC} = \varnothing ,\;\text{ and }\;\overrightarrow{BR} \cap \overrightarrow{AC} = \varnothing \]\n\nHence neither \( P, Q \), nor \( R \) can li... | Yes |
Theorem 1 Given any family of lines concurrent at \( \Omega \) and point \( A \) lying on one of those lines, there exists a unique point \( {A}^{\prime } \) lying on any other such line in the family, such that \( \angle A{A}^{\prime }\Omega \cong \angle {A}^{\prime }{A\Omega } \) . | Proof Since this is obvious if \( \Omega \) is an ordinary point \( P \), let \( \Omega \) be either an ideal or ultra-ideal point. If \( \Omega \) is an ideal point, use the construction for an isosceles asymptotic triangle (Figure 5.45, Section 5.8), and if \( \Omega \) is an ultra-ideal point, construct a Saccheri q... | No |
Theorem 2 As with the case of an ordinary circle orthogonal to its radii, the curve \( \mathcal{C}\left\lbrack {A,\Omega }\right\rbrack \) is an orthogonal trajectory of the family of lines passing through \( \Omega \) . That is, the curve \( \mathcal{C}\left\lbrack {A,\Omega }\right\rbrack \) makes right angles with e... | Intuitive proof Let \( {A}^{\prime } \) approach \( A \) as limit, \( A \in l \) and \( {A}^{\prime } \in {l}^{\prime } \) . \n\n1. If \( \Omega \) is an ideal point (Figure 9.35a), then the angle of parallelism of \( l \) with respect to \( A{A}^{\prime } \) tends to \( \pi /2 \) as the distance \( A{A}^{\prime } \rig... | No |
Theorem 3 In the extended hyperbolic plane, the perpendicular bisectors of the sides of any triangle are concurrent. | Proof Since the theorem is clear if any two of the perpendicular bisectors meet at an ordinary point, assume that two of them, say \( l \) and \( m \) (the perpendicular bisectors of sides \( {BC} \) and \( {AC} \) ), meet at an ultra-ideal point \( {G}^{* * } \) (Case 1).\n\n1. Let \( l \) and \( m \) have a common pe... | Yes |
Theorem 4 The perpendicular bisector of any chord \( {AB} \) of a cycle \( \mathcal{C}\left\lbrack {D,\Omega }\right\rbrack \) passes through \( \Omega \) . | Proof If \( \Omega \) is an ordinary point, then \( \mathcal{C}\left\lbrack {D,\Omega }\right\rbrack \) is a circle centered at \( \Omega \) and the conclusion follows from the familiar theorem concerning the chord of a circle. If \( \Omega \) is an ideal point, the result was already observed earlier in Problem 20, Se... | Yes |
Theorem 5 Given any three ordinary non-collinear points \( A, B \) , and \( C \) in the extended hyperbolic plane, there exists a unique locus \( \mathcal{C}\left\lbrack {D,\Omega }\right\rbrack \) (a circle or a cycle) passing through them. | Proof Consider the perpendicular bisectors of the sides of \( \bigtriangleup {ABC} \), which are concurrent at some point \( \Omega \) . A candidate for a cycle containing \( A, B \) , and \( C \) is therefore \( \mathcal{C}\left\lbrack {A,\Omega }\right\rbrack \) ; we proceed to show that \( B, C \in \mathcal{C}\left\... | No |
Lemma 3.1 \( \left( i\right) ,\ell \left( {\widetilde{\alpha }}_{t}\right) \geq \ell \left( \widetilde{\beta }\right) \) | On the other hand, \[ \ell \left( {\bar{\alpha }}_{t}\right) = \ell \left( {\alpha }_{t}\right) \leq \ell \left( \beta \right) = \ell \left( \widetilde{\beta }\right) = t \] hence \( \ell \left( {\alpha }_{t}\right) = t \), as was claimed. Since \( t \) is arbitrary, \( M \) is not bounded, which concludes the proof of... | No |
Theorem 4. Consider the group structure on a pivotal isocubic with the pivot \( F \) as neutral element. The constant point is \( N = {F}_{t} \). | (1) \( F \) is the pivot, so \( P,{P}^{ * } \) and \( F \) are collinear.\n\n(2) Put \( P = F \) in (1).\n\n(3) \( P + {P}^{ * } = \left( {P \cdot {P}^{ * }}\right) \cdot F = F \cdot F = {F}_{t} \).\n\n(4) \( P + Q = \left( {P \cdot Q}\right) \cdot F = {\left( P \cdot Q\right) }^{ * } \) . (use (1))\n\n(5) This is Coro... | Yes |
Theorem 8. A pivotal isocubic of pivot \( \left( {u : v : w}\right) \) has Salmon cross ratio | \[ \frac{{q}^{2}\left( {{r}^{2}{u}^{2} - {p}^{2}{w}^{2}}\right) }{{r}^{2}\left( {{q}^{2}{u}^{2} - {p}^{2}{v}^{2}}\right) } \] | Yes |
Let \( \Gamma \) be the circumcircle of acute triangle \( {ABC} \) . Points \( D \) and \( E \) are on segments \( {AB} \) and \( {AC} \) respectively such that \( {AD} = {AE} \) . The perpendicular bisectors of \( \overline{BD} \) and \( \overline{CE} \) intersect minor arcs \( {AB} \) and \( {AC} \) of \( \Gamma \) a... | First solution, by constructing parallelograms Construct points \( P \) and \( Q \) on \( \Gamma \) such that \( {ABFP} \) and \( {ACGQ} \) are isosceles trapezoids, and let \( M \) and \( N \) be the midpoints of minor arcs \( {AB} \) and \( {AC} \) respectively. It is obvious that \( M \) and \( N \) are the midpoint... | Yes |
In acute triangle \( {ABC} \), let \( E \) and \( F \) be the feet of the altitudes from \( B \) and \( C \), respectively. Let \( M, K, L \) denote the midpoints of \( \overline{BC},\overline{ME},\overline{MF} \), respectively. Line \( {KL} \) intersects the line through \( A \) parallel to \( \overline{BC} \) at \( T... | Let \( \omega \) be the circle centered at \( M \) with radius 0 . It is well-known that \( \overline{TA},\overline{ME} \), and \( \overline{MF} \) are tangent to \( \left( {AEF}\right) \), so line \( {KL} \) is the radical axis of \( \left( {AEF}\right) \) and \( \omega \) . It follows that \( T{A}^{2} = \) \( \operat... | No |
Let \( {AXYZB} \) be a convex pentagon inscribed in a semicircle of diameter \( {AB} \) . Denote by \( P \) , \( Q, R, S \) the feet of the perpendiculars from \( Y \) onto lines \( {AX},{BX},{AZ},{BZ} \), respectively. Prove that the acute angle formed by lines \( {PQ} \) and \( {RS} \) is half the size of \( \angle {... | Let \( T \) be the projection of \( Y \) onto \( \overline{AB} \) . Notice that \( T \) lies on the Simson Line \( \overline{PQ} \) from \( Y \) to \( \bigtriangleup {AXB} \), and the Simson Line \( \overline{RS} \) from \( Y \) to \( \bigtriangleup {AZB} \) . Hence, \( T = \overline{PQ} \cap \overline{RS} \), so it su... | Yes |
Let \( {ABC} \) be a triangle with incenter \( I \), and let \( D \) be a point on line \( {BC} \) satisfying \( \angle {AID} = \) \( {90}^{ \circ } \) . Denote by \( E \) and \( F \) the feet of the altitudes from \( B \) and \( C \), respectively.\n\nLet the excircle of triangle \( {ABC} \) opposite the vertex \( A \... | Solution to part (a) It is easy to check that \( \bigtriangleup {ABC} \) is acute, say by repeating the argument of JMO 2019/4.\n\n\n\nDenote the intouch points by \( {A}_{0},{B}_{0},{C}_{0} \), the orthocenter by \( H... | Yes |
Let \( {ABC} \) be a triangle with circumcircle \( \Gamma \) and let \( M \) be an arbitrary point on \( \Gamma \) . Suppose that the tangents from \( M \) to the incircle of \( {ABC} \) intersect \( \overline{BC} \) at two distinct points \( {X}_{1} \) and \( {X}_{2} \) . Prove that the circumcircle of triangle \( M{X... | First solution, by inversion Let the incircle touch \( \overline{BC},\overline{CA},\overline{AB} \) at \( D, E, F \), respectively, and let \( \overline{M{X}_{1}} \) and \( \overline{M{X}_{2}} \) touch the incircle at \( {Y}_{1} \) and \( {Y}_{2} \), respectively. Denote by \( I, K, T, H, N \) the incenter of \( \bigtr... | Yes |
Let \( {ABCD} \) be a cyclic quadrilateral. Prove that if the symmedian points of \( \bigtriangleup {ABC} \) , \( \bigtriangleup {BCD},\bigtriangleup {CDA},\bigtriangleup {DAB} \) are concyclic, then quadrilateral \( {ABCD} \) has two parallel sides. | Let \( {K}_{A},{K}_{B},{K}_{C},{K}_{D} \) be the symmedian points of \( \bigtriangleup {BCD},\bigtriangleup {CDA},\bigtriangleup {DAB},\bigtriangleup {ABC} \), and let \( E = \overset{―}{AA} \cap \overset{―}{BB},\;F = \overset{―}{BB} \cap \overset{―}{CC},\;G = \overset{―}{CC} \cap \overset{―}{DD},\;H = \overset{―}{DD} ... | Yes |
A convex quadrilateral \( {ABCD} \) satisfies \( {AB} \cdot {CD} = {BC} \cdot {DA} \) . Point \( X \) lies inside \( {ABCD} \) so that\n\n\[ \angle {XAB} = \angle {XCD}\;\text{ and }\;\angle {XBC} = \angle {XDA}. \]\n\nProve that \( \angle {BXA} + \angle {DXC} = {180}^{ \circ } \) . | First solution, by inversion We first require the following two lemmas.\n\nLemma 1\n\nIf two quadril | No |
If two quadrilaterals have the same angles and both obey \( {AB} \cdot {CD} = {BC} \cdot {DA} \), then they are similar. | Proof. Omitted. | No |
Let \( \omega \) be a circle with center \( O \), and let \( T \) be a point outside of \( \omega \) . Points \( B \) and \( C \) lie on \( \omega \) such that \( \overline{TB} \) and \( \overline{TC} \) are tangent to \( \omega \) . Select two points \( K \) and \( H \) on \( \overline{TB} \) and \( \overline{TC} \) ,... | First solution to part (a), by angle chasing Ignore the collinearities for now. Let \( \overline{{B}^{\prime }K} \) and \( \overline{{C}^{\prime }H} \) intersect \( \omega \) again at \( P \) and \( Q \), and let \( \overline{BC} \) intersect \( \overline{K{K}^{\prime }} \) at \( X \) and \( \overline{H{H}^{\prime }} \... | Yes |
Theorem 1.1 (The optical property of the ellipse). Suppose a line \( l \) is tangent to an ellipse at a point \( P \) . Then \( l \) is the bisector of the exterior angle \( {\mathrm{F}}_{1}{\mathrm{{PF}}}_{2} \) (Figure 1.10). | Proof. Let \( X \) be an arbitrary point of \( l \) different from \( P \) . Since \( X \) is outside the ellipse, we have \( X{F}_{1} + X{F}_{2} > P{F}_{1} + P{F}_{2} \), i.e., of all the points of \( l \) the point \( P \) has the smallest sum of the distances to \( {F}_{1} \) and \( {F}_{2} \) . This means that the ... | Yes |
Theorem 1.2. Suppose the chord PQ contains a focus \( {F}_{1} \) of the ellipse and \( R \) is the intersection of the tangents to the ellipse at \( P \) and \( Q \) . Then \( R \) is the center of an excircle of the triangle \( {F}_{2}{PQ} \), and \( {F}_{1} \) is the tangency point of that circle and the side \( {PQ}... | Proof. By the optical property, \( {PR} \) and \( {QR} \) are the bisectors of the exterior angles of the triangle \( {F}_{2}{PQ} \) . Therefore \( R \) is the center of an excircle. The tangency point (call it \( {F}_{1}^{\prime } \) ) of the excircle and the corresponding side and the point \( {F}_{2} \) cut the peri... | Yes |
Theorem 1.3. From any point \( P \) outside an ellipse draw two tangents to the ellipse, with tangency points \( X \) and \( Y \) . Then the angles \( {F}_{1}{PX} \) and \( {F}_{2}{PY} \) are equal \( \left( {F}_{1}\right. \) and \( \left. {F}_{2}\right) \) are the foci of the ellipse \( ) \) . | Proof. Let \( {F}_{1}^{\prime },{F}_{2}^{\prime } \) be the reflections of \( {F}_{1} \) and \( {F}_{2} \) in \( {PX} \) and \( {PY} \), respectively (Figure 1.15).\n\nThen \( P{F}_{1}^{\prime } = P{F}_{1} \) and \( P{F}_{2}^{\prime } = P{F}_{2} \) . Moreover, the points \( {F}_{1}, Y \) and \( {F}_{2}^{\prime } \) lie... | Yes |
Theorem 1.5. The locus of points from which a given ellipse is seen at a right angle (i.e., the tangents to the ellipse drawn from such a point are perpendicular) is a circle centered at the center of the ellipse (Figure 1.18). | Proof. Let \( {F}_{1} \) and \( {F}_{2} \) be the foci of the ellipse and suppose that the tangents to the ellipse at \( X \) and \( Y \) intersect in \( P \) . Reflecting \( {F}_{1} \) in \( {PX} \) we have a point \( {F}_{1}^{\prime } \) . It follows from Theorem 1.3 that \( \angle {XPY} = \angle {F}_{1}^{\prime }P{F... | Yes |
Theorem 1.6. Suppose a string is put on an ellipse \( \alpha \) and then pulled tight using a pencil. If the pencil is rotated about the ellipse, it will traverse another ellipse confocal with \( \alpha \) (Figure 1.19). | Proof. Clearly, the new figure (call it \( {\alpha }_{1} \) ) has a smooth boundary. We shall show that at each point \( X \) on \( {\alpha }_{1} \) the tangent to the new curve coincides with the bisector of the exterior angle \( {F}_{1}X{F}_{2} \) .\n\nLet \( {XM} \) and \( {XN} \) be the tangents to \( \alpha \) . T... | No |
Lemma 1.1. If the focus of a parabola is reflected in a tangent, then its image will be on the directrix. That image is the projection of the point where the tangent touches the parabola (Figure 1.24). | Proof. Suppose a line \( l \) touches the parabola at \( P \) and let \( {P}^{\prime } \) be the projection of \( P \) to the directrix. Since the triangle \( {FP}{P}^{\prime } \) is isosceles and \( l \) is the bisector of the angle \( P, l \) is an axis of symmetry of the triangle. Hence the reflection \( {P}^{\prime... | Yes |
Lemma 1.2. Suppose the tangents to the parabola at points \( X \) and \( Y \) intersect at a point \( P \) . Then \( P \) is the center of the circumcircle of the triangle \( F{X}^{\prime }{Y}^{\prime } \), where \( {X}^{\prime } \) and \( {Y}^{\prime } \) are the projections of \( X \) and \( Y \) to the directrix of ... | Proof. By Lemma 1.1, these two tangents are midpoint perpendiculars to the segments \( F{X}^{\prime } \) and \( F{Y}^{\prime } \) . Therefore their point of intersection is the center of the circumcircle of the triangle \( F{X}^{\prime }{Y}^{\prime } \) (Figure 1.26). | Yes |
Theorem 1.7. The set of points \( P \) where the parabola is seen at a right angle is the directrix of the parabola. Moreover, if \( {PX} \) and \( {PY} \) are tangent to the parabola, then \( {XY} \) contains \( F \) and \( {PF} \) is a height of the triangle PXY (Figure 1.28). | Proof. Suppose \( P \) lies on the directrix, and let \( {X}^{\prime } \) and \( {Y}^{\prime } \) be the projections of \( X \) and \( Y \) to the directrix. Then the triangles \( {PXF} \) and \( {PX}{X}^{\prime } \) are equal (since they are symmetric with respect to \( {PX} \) ). Hence \( \angle {PFX} = \) \( \angle ... | Yes |
Theorem 1.8. The set of points from which a parabola is seen at an angle \( \phi \) or \( {180}^{ \circ } - \phi \) is a hyperbola with focus \( F \) and directrix \( l \) (Figure 1.29). | Proof. Indeed, suppose the tangents \( {PX} \) and \( {PY} \) to the parabola drawn from \( P \) form an angle \( \phi \) . We first consider the case when \( \phi > {90}^{ \circ } \) .\n\nLet \( {X}^{\prime } \) and \( {Y}^{\prime } \) be the projections of \( X \) and \( Y \) to the directrix. Clearly, \( \angle {X}^... | Yes |
Theorem 1.9. Let \( {PX} \) and \( {PY} \) be the tangents to the parabola passing through \( P \), and let \( l \) be the line passing through \( P \) parallel to the axis of the parabola. Then the angle between the lines \( {PY} \) and \( l \) is equal to \( \angle {XPF} \) and the triangles \( {XFP} \) and \( {PFY} ... | Proof. Let \( {X}^{\prime } \) and \( {Y}^{\prime } \) be the projections of \( X \) and \( Y \) to the directrix. Then, by Theorem 1.2, the points \( F,{X}^{\prime } \), and \( {Y}^{\prime } \) lie on a circle centered at \( P \) . Hence \( \angle {X}^{\prime }{Y}^{\prime }F = \frac{1}{2}\angle {X}^{\prime }{PF} = \an... | Yes |
Theorem 1.10. Suppose a triangle ABC is circumscribed about a parabola (i.e., the lines \( {AB},{BC},{CA} \) are tangent to the parabola). Then the focus of the parabola lies on the circumcircle of the triangle \( {ABC} \) . | Proof. By the Corollary of Lemma 1.1, the projections of the focus to the sides all lie on a straight line (which is parallel to the directrix and lies at half the distance from the focus). Now we can use Simson's lemma. | No |
Lemma 1.3 (Simson). The projections of \( P \) to the sides of a triangle \( {ABC} \) lie on a line if and only if \( P \) lies on the circumcircle of the triangle. | Proof. Let \( {P}_{a},{P}_{b} \) and \( {P}_{c} \) be the projections of \( P \) to \( {BC},{CA} \) and \( {AB} \) , respectively. We consider the case shown in Figure 1.31; the remaining cases are argued similarly.\n\nThe quadrilateral \( {PC}{P}_{b}{P}_{a} \) is inscribed, hence \( \angle P{P}_{b}{P}_{a} = \angle {PC... | Yes |
Lemma 1.4. Suppose a point \( P \) lies on the circumcircle of a triangle \( {ABC} \) . Choose a point \( {B}^{\prime } \) on the circumcircle such that the line \( P{B}^{\prime } \) is perpendicular to \( {AC} \) . Then \( B{B}^{\prime } \) is parallel to Simson’s line of \( P \) (Figure 1.33). | Proof. Consider the case shown in Figure 1.33; the remaining cases are argued similarly. Let \( {P}_{c} \) and \( {P}_{b} \) be the projections of \( P \) to the sides \( {AB} \) and \( {AC} \), respectively. Then \( \angle {AB}{B}^{\prime } = \angle {AP}{B}^{\prime } \) as the angles subtending the arc \( A{B}^{\prime... | Yes |
Corollary 2. Simson’s line of \( P \) relative to a triangle \( {ABC} \) cuts the segment \( {PH} \) (where \( H \) is the orthocenter of the triangle \( {ABC} \) ) in half (Figure 1.34). | Proof. It is easy to see that \( \angle {AHC} = {180}^{ \circ } - \angle {ABC} \), and therefore the reflection \( {H}^{\prime } \) of \( H \) in \( {AC} \) lies on the circumcircle of the triangle \( {ABC} \) . Since the lines \( P{B}^{\prime } \) and \( B{H}^{\prime } \) are perpendicular to \( {AC} \), the quadrilat... | Yes |
Theorem 2.1 (Feuerbach). The nine-point circle is tangent to the incircle and the excircles of the triangle (if the triangle is equilateral, it coincides with the incircle) (Figure 2.3). | Proof. Let \( {G}_{a},{G}_{b} \) and \( {G}_{c} \) be the tangency points of the incircle and the sides of the triangle. Let \( {A}_{1} \) be the foot of the bisector of the angle \( A \) and \( {C}^{\prime } \) the reflection of \( C \) in \( A{A}_{1} \) . Let \( P \) be the intersection of \( A{A}_{1} \) and \( C{C}^... | Yes |
Theorem 2.2. If points \( {A}_{1},{B}_{1},{C}_{1} \) lie on a line \( {l}_{1} \) and \( {A}_{2},{B}_{2},{C}_{2} \) lie on a line \( {l}_{2} \), then the intersections of the lines \( {A}_{1}{B}_{2} \) and \( {A}_{2}{B}_{1},{B}_{1}{C}_{2} \) and \( {B}_{2}{C}_{1} \) , \( {C}_{1}{A}_{2} \) and \( {C}_{2}{A}_{1} \) lie on... | To see this, just move the intersections of the pairs \( {A}_{1}{B}_{2} \) and \( {A}_{2}{B}_{1} \) , \( {B}_{1}{C}_{2} \) and \( {B}_{2}{C}_{1} \) to infinity and use Thales’ theorem. | No |
Theorem 2.3. The lines \( {A}_{1}{A}_{2},{B}_{1}{B}_{2},{C}_{1}{C}_{2} \) connecting the corresponding vertices of triangles \( {A}_{1}{B}_{1}{C}_{1} \) and \( {A}_{2}{B}_{2}{C}_{2} \) intersect in one point if and only if the intersections of lines \( {A}_{1}{B}_{1} \) and \( {A}_{2}{B}_{2},{B}_{1}{C}_{1} \) and \( {B... | Here use a projective transformation to move the intersections of pairs of lines \( {A}_{1}{B}_{1} \) and \( {A}_{2}{B}_{2},{B}_{1}{C}_{1} \) and \( {B}_{2}{C}_{2} \) to infinity and again use Thales’ theorem. | No |
Theorem 2.4. The intersection points of the opposite sides of an inscribed hexagon are all on one straight line (Figure 2.7). | Proof. Let \( {ABCDEF} \) be the inscribed hexagon. Using a projective transformation, move the intersections of the pairs of lines \( {AB} \) and \( {DE},{BC} \) and \( {EF} \) to infinity. Then \( {AB}\parallel {DE} \) and \( {BC}\parallel {EF} \) ; we need to show that \( {CD}\parallel {FA} \) . But this is not diff... | Yes |
Theorem 2.5. The principal diagonals of a circumscribed hexagon meet at one point (Figure 2.8). | Proof. Move the intersection of two diagonals to the center of the circle. We need to show that the third diagonal passes through the center.\n\nThus let the hexagon \( {ABCDEF} \) be circumscribed about a circle centered at \( O \) and suppose that the diagonals \( {AD} \) and \( {BE} \) pass through \( O \) . Let \( ... | Yes |
Theorem 2.6. The pedal triangle degenerates (i.e., the projections lie on a line) \( \Leftrightarrow P \) lies on the circumcircle of the triangle \( {ABC} \) . | (This is just a reformulation of Simson's lemma.) | No |
Lemma 2.5. The pedal and the circumcevian triangles of \( P \) with respect to a triangle \( {ABC} \) are similar and have the same orientation. | Proof. We consider the case shown in Figure 2.17. The remaining cases are proved similarly.\n\nThe points \( {P}_{A},{P}_{B},{P}_{C} \) are the vertices of the pedal triangle, and \( {A}^{\prime } \) , \( {B}^{\prime },{C}^{\prime } \) are the vertices of the circumcevian triangle. We have \( \angle A{A}^{\prime }{C}^{... | No |
Lemma 2.6. Suppose points \( P \) and \( Q \) are inverse to each other with respect to a circle \( \omega \) centered at \( O \) and the segment \( {PQ} \) intersects \( \omega \) at a point \( R \) . Then for any point \( A \) on \( \omega ,{RA} \) is the bisector of the angle \( {PAQ} \) . | Proof. Since \( P \) and \( Q \) are inverse to each other, the triangles \( {OAP} \) and \( {OQA} \) are similar (Figure 2.18), and therefore \( \angle {OQA} = \angle {OAP} \) . Since \( O \) is the center of \( \omega \), the triangle \( {AOR} \) is isosceles and therefore \( \angle {OAR} = \angle {ORA} \) . Thus\n\n... | Yes |
Theorem 2.8. The pedal circles of two points coincide if and only if the points are isogonally conjugate. | Proof. Indeed, if \( P \) and \( {P}^{\prime } \) are isogonally conjugate, then the pedal circle for each of them is the circle centered at the midpoint of \( P{P}^{\prime } \) of radius \( \frac{{P}^{\prime }{P}_{a}}{2} = \frac{P{P}_{a}^{\prime }}{2} \), where \( {P}_{a} \) and \( {P}_{a}^{\prime } \) are the reflect... | Yes |
Theorem 2.9 (Pascal). Suppose points \( A, B, C, D, E \) and \( F \) lie on a conic. Then the intersections of the lines \( {AB} \) and \( {DE},{BC} \) and \( {EF},{CD} \) and \( {FA} \) lie on a line. | Proof. We consider only one case of the relative positions of the points on the circle (or a conic). The other cases are treated similarly.\n\nUsing a projective transformation we can transform the conic into a circle. We then have the following configuration (Figure 2.24).\n\nThe points \( A, B, C, D, E \) and \( F \)... | Yes |
Theorem 2.10. Let \( p \) and \( q \) be the roots of the derivative of the polynomial \( P\left( z\right) = \left( {z - {z}_{a}}\right) \left( {z - {z}_{b}}\right) \left( {z - {z}_{c}}\right) \) . Then the points of the complex plane corresponding to \( p \) and \( q \) are isogonally conjugate with respect to the tri... | Proof. We prove that \( \angle {BAP} = \angle {CAQ} \), where the points \( P \) and \( Q \) correspond to \( p \) and \( q \) . Without loss of generality we may assume that \( {z}_{a} = 0 \) , since subtracting \( {z}_{a} \) from \( {z}_{a},{z}_{b} \) and \( {z}_{c} \) we will change the roots of the derivative of \(... | Yes |
Lemma 2.7. Suppose a line passing through \( P \) intersects the circle in \( X \) and \( Y \) . Then \( {PX} \cdot {PY} \) does not depend on the line and equals the absolute value of the power of \( P \) with respect to the circle. | Proof. Suppose we have two lines passing through \( P \) such that the first intersects the circle at points \( A \) and \( B \) and the second at points \( C \) and \( D \) . We prove that \( {PA} \cdot {PB} = {PC} \cdot {PD} \) . Clearly, the triangles \( {PAC} \) and \( {PDB} \) are similar and therefore\n\n\[ \frac... | Yes |
Theorem 2.11. The set of points whose powers with respect to two given nonconcentric circles are equal is a line. This line is called the radical axis of the circles. | Proof. Suppose that the two circles intersect. Draw the line through the intersection points. We claim that this line is the radical axis. Suppose the circles intersect at points \( A \) and \( B \) . Take an arbitrary point \( P \) on the line \( {AB} \) . Then the absolute value of the power of \( P \) with respect t... | Yes |
Theorem 2.12. Given two circles \( {\omega }_{1} \) and \( {\omega }_{2} \), the locus where the ratio of powers with respect to those two circles is constant is a circle belonging to the pencil formed by \( {\omega }_{1} \) and \( {\omega }_{2} \). | Proof. Suppose that \( {\omega }_{1} \) and \( {\omega }_{2} \) intersect at \( A \) and \( B \) (Figure 2.40). Let \( {O}_{1} \) and \( {O}_{2} \) be their centers and, respectively, \( {r}_{1} \) and \( {r}_{2} \) their radii. Let \( {A}_{1} \) and \( {A}_{2} \) be the reflections of \( A \) in \( {O}_{1} \) and \( {... | Yes |
Theorem 2.13 (Poncelet). Suppose circles \( {\omega }_{i} \) belong to the same pencil and \( {A}_{0} \) is a point on \( {\omega }_{0} \) . The tangent to \( {\omega }_{1} \) from \( {A}_{0} \) intersects \( {\omega }_{0} \) again at \( {A}_{1} \), the tangent to \( {\omega }_{2} \) from \( {A}_{1} \) intersects \( {\... | Proof. It suffices to show that \( {A}_{i}{B}_{i} \) is tangent to some fixed circle from the pencil. Indeed, if \( {A}_{0} \) coincides with \( {A}_{n} \), then the tangent to the circle passing through \( {A}_{0} \) coincides with the tangent passing through \( {A}_{n} \) (assuming the tangents run in the appropriate... | Yes |
Theorem 2.14 (Brianchon). Suppose lines \( {l}_{i}, i = 1,\ldots ,6 \), are tangent to the same conic, and let \( {A}_{ij} \) be the intersection of \( {l}_{i} \) and \( {l}_{j} \) . Then the lines \( {A}_{12}{A}_{45},{A}_{23}{A}_{56} \) and \( {A}_{34}{A}_{61} \) intersect at a single point. | Proof. Using a projective transformation we make the conic into a circle. We assume that we have a hexagon circumscribed about the circle, and the lines in question are its main diagonals. Our arguments can easily be adapted to the cases where the lines form other configurations.\n\nThus, given a circumscribed hexagon ... | Yes |
Theorem 3.1 (The converse of Pascal’s theorem). For any six points \( {X}_{i}, i = 1,\ldots ,6 \), such that the intersections of the lines \( {X}_{1}{X}_{2} \) and \( {X}_{4}{X}_{5} \) , \( {X}_{2}{X}_{3} \) and \( {X}_{5}{X}_{6},{X}_{3}{X}_{4} \) and \( {X}_{6}{X}_{1} \) are on a straight line, there is a conic passi... | Proof. We use the fact that for any five points in general position, there is a unique conic containing them. Let \( \alpha \) be such a conic for \( {X}_{i}, i = 1,\ldots ,5 \) . Let \( A, B, C \) be the intersections of the lines \( {X}_{1}{X}_{2} \) and \( {X}_{4}{X}_{5},{X}_{2}{X}_{3} \) and \( {X}_{5}{X}_{6},{X}_{... | Yes |
Theorem 3.2 (The converse of Brianchon’s theorem). Given any six lines \( {l}_{i}, i = 1,\ldots ,6 \), let \( {A}_{ij} \) be the intersection of the lines \( {l}_{i} \) and \( {l}_{j} \) . If the lines \( {A}_{12}{A}_{45},{A}_{23}{A}_{56} \) and \( {A}_{34}{A}_{61} \) intersect at a single point, then there is a conic ... | The proof of this theorem is similar to that of the previous one. However, here we first need to show that there is a unique conic tangent to five given lines. Using Brianchon's theorem, we can construct the tangency points of those lines with the conic. But only one conic can pass through five points. The construction... | No |
Theorem 3.3. Let \( R\left( \gamma \right) \) be the dual curve of a curve \( \gamma \) . Then \( R\left( {R\left( \gamma \right) }\right) = \gamma \) . | Proof. Suppose a point \( X \) moves along \( \gamma \) toward \( A \) . Then, clearly, the intersections of the tangents at \( X \) and \( A \) (call them \( x \) and \( a \), respectively)\n\n\n\nFIGURE 3.8\n\ntend t... | Yes |
Theorem 3.4. The polar curve of a circle with respect to another circle is a conic. | Proof. Let \( \omega \) be a circle with center \( O \) and \( {\omega }_{1} \) a circle with center \( {O}_{1} \) (for convenience, assume that \( O \) lies inside \( {\omega }_{1} \) ; see Figure 3.10). Suppose that the inversion with respect to \( \omega \) transforms \( {\omega }_{1} \) into a circle \( {\omega }_{... | Yes |
Theorem 3.6. A focus and the corresponding directrix of a conic are polar to each other (Figure 3.11). | Proof. Consider a focus \( {F}_{1} \) and its polar \( l \) . We prove that for any two points \( X \) and \( Y \) on the conic, the ratios of the distances to \( {F}_{1} \) and \( l \) are equal. Let \( S \) be the intersection of the lines \( {XY} \) and \( l \) . Let \( Z \) denote the intersection of the tangents t... | Yes |
Theorem 3.7. A triangle ABC is self-polar (i.e., its sides are the polars of the corresponding vertices) with respect to a conic if and only if it is the Ceva triangle of a point on the conic with respect to a triangle inscribed in the conic. | Proof. We begin by moving the vertices \( B \) and \( C \) of the triangle to points at infinity with perpendicular directions. Then our conic will, obviously, become a conic centered at \( A \) (since the center of the conic is the pole of the line at infinity). Consider the rectangle inscribed in the conic with sides... | Yes |
Theorem 3.8. Suppose we are given a triangle ABC and a point \( Z \) . For an arbitrary line passing through \( Z \), let \( {A}^{\prime } \) and \( {B}^{\prime } \) be its intersection points with \( {BC} \) and \( {AC} \) . Then the locus of the intersections of the lines \( A{A}^{\prime } \) and \( B{B}^{\prime } \)... | Proof. Apply a projective transformation making the triangle \( {ABC} \) into a right isosceles triangle \( \left( {{AC} = {BC}}\right) \) and sending the point \( Z \) to infinity in the direction perpendicular to \( {AB} \) . Then the triangles \( A{A}^{\prime }P \) and \( {B}^{\prime }{BP} \) , where \( P \) is the ... | Yes |
Theorem 3.9. Suppose we are given a triangle ABC and points \( P \) and \( Q \) , and suppose that the lines \( {AP},{BP} \) and \( {CP} \) intersect the respective sides of the triangle at points \( {A}_{1},{B}_{1},{C}_{1} \) and that the lines \( {AQ},{BQ},{CQ} \) intersect the respective sides at points \( {A}_{2},{... | Proof. The points \( A, B, C, P, Q \) determine a conic. Without loss of generality we may assume that the conic is a circle and \( {PQ} \) is its diameter. Since \( {A}_{1},{B}_{1},{C}_{1} \) are the intersections of the opposite sides and the diagonals of the inscribed quadrilateral \( {ABCP} \), we have that \( {A}_... | Yes |
Theorem 3.11 (The four conics theorem). Suppose we are given three conics \( {\alpha }_{1},{\alpha }_{2},{\alpha }_{3} \), and let \( {P}_{1},{Q}_{1},{P}_{1}^{\prime },{Q}_{1}^{\prime } \) be the intersections of \( {\alpha }_{2} \) and \( {\alpha }_{3};{P}_{2},{Q}_{2},{P}_{2}^{\prime },{Q}_{2}^{\prime } \) the interse... | Proof. We first need the following auxiliary result. | No |
Theorem 3.12 (The three conics theorem). Suppose three conics have two common points. Then their common chords passing through the remaining intersections of each pair meet at a single point (Figure 3.29). | To prove this claim it suffices to transform the common points of the conics into the intersection points of the line at infinity and circles. Then all three conics will transform into circles and the desired assertion will follow from the existence of the radical center.\n\nNow let \( {\alpha }_{0} \) be the conic pas... | Yes |
Theorem 3.15. Let \( A, B, C, D \) be four distinct points, and \( X, Y, Z \) the intersections of the lines \( {AB} \) and \( {CD},{AC} \) and \( {BD},{AD} \) and \( {BC} \) . Let \( P \) be a point different from \( X, Y \) and \( Z \) . Then the polars of \( P \) with respect to all the conics of the pencil determin... | Here is an interesting special case. If the points \( A, B, C \) and \( D \) form an orthocentric quadruple (i.e., each point is the orthocenter of the triangle formed by the remaining points), then the obtained point is isogonally conjugate to \( P \) with respect to the triangle \( {XYZ} \) . Indeed, the polar of \( ... | Yes |
Theorem 3.16. Suppose we are given four lines \( {l}_{i}, i = 1,\ldots ,4 \), and let \( {X}_{ij} \) be the intersection of the lines \( {l}_{i} \) and \( {l}_{j} \) . Then the locus of the poles of any line different from \( {X}_{12}{X}_{34},{X}_{13}{X}_{24},{X}_{14}{X}_{23} \) with respect to the conics of the pencil... | Proof. Using a projective transformation, we move the line in question to infinity. It follows from the assumptions that in this case the lines \( {l}_{i} \) form a quadrilateral \( {ABCD} \) which is not a parallelogram. We prove that the centers of the conics inscribed in it lie on the so-called Gauss line passing th... | Yes |
Theorem 3.17. The poles of a fixed line with respect to all the conics of the pencil defined by points \( A, B, C \) and \( D \) form a conic. | Proof. Transform the given line into the line at infinity. Then its poles will be the centers of the conics of the pencil. It follows from the converse to Pascal’s theorem that the midpoints \( K, L, M \) and \( N \) of the sides of the quadrilateral \( {ABCD} \) belong to the set of the centers. Hence it suffices to s... | Yes |
Lemma 3.8. 1. Suppose points \( A, B \) and \( C \) lie on a conic \( {\alpha }_{0} \), the line \( {AB} \) is tangent to the conic \( {\alpha }_{1} \) at \( K \), and the line \( {AC} \) is tangent to the conic \( {\alpha }_{2} \) at \( L \) . Then there is a point \( D \) on \( {\alpha }_{0} \) such that \( {\alpha }... | Proof. 1. Let \( M \) be the second intersection point of \( {\alpha }_{1} \) and \( {KL} \), and let \( \mathcal{G} \) be the pencil containing \( {\alpha }_{1} \) and the degenerate curve consisting of the lines \( {CM} \) and \( {AB} \) . On the line \( {AC} \), the pencils \( \mathcal{F} \) and \( \mathcal{G} \) gi... | Yes |
Theorem 3.21. Suppose conics \( {\alpha }_{1} \) and \( {\alpha }_{2} \) are dual with respect to a conic \( \alpha \) (assuming that all of them are from \( \mathcal{Q} \) ). Then \( {\epsilon }_{1}{\epsilon }_{2} = {\epsilon }^{2} \), where \( \epsilon ,{\epsilon }_{1} \) and \( {\epsilon }_{2} \) are the eccentricit... | Proof. Let \( {F}_{l} \) be the projection of \( F \) to the line \( l \) . Let \( X, Y \) and \( Z \) be the points where the segment \( F{F}_{l} \) intersects the conics \( \alpha ,{\alpha }_{1} \) and \( {\alpha }_{2} \), respectively. We need to show that\n\n\[ \frac{FY}{{F}_{l}Y} \cdot \frac{FZ}{{F}_{l}Z} = \frac{... | Yes |
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