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Theorem 4.1. The centers of all equilateral hyperbolas passing through the vertices of a triangle \( {ABC} \) lie on the Euler circle of the triangle. | Proof. Let \( D \) be the fourth (besides \( A, B \) and \( C \) ) intersection point of the hyperbola and the circumcircle of the triangle \( {ABC} \), and \( {A}^{\prime },{B}^{\prime },{C}^{\prime } \) and \( {D}^{\prime } \) the orthocenters of the triangles \( {BCD},{CDA},{DAB} \) and \( {ABC} \), respectively (Fi... | Yes |
Theorem 4.2. Suppose we are given a triangle ABC and a point \( P \) different from its orthocenter. Then the centers of the incircle and the excircles of the Ceva triangle of \( P \) with respect to the triangle \( {ABC} \) lie on an equilateral hyperbola passing through \( A, B, C \) and \( P \) . | Proof. This property is a special case of the following fact. | No |
Lemma 4.9. Suppose we are given two triangles \( {A}_{1}{B}_{1}{C}_{1} \) and \( {A}_{2}{B}_{2}{C}_{2} \), and let \( {A}^{\prime },{B}^{\prime } \) and \( {C}^{\prime } \) be the intersections of \( {B}_{1}{C}_{1} \) and \( {B}_{2}{C}_{2},{C}_{1}{A}_{1} \) and \( {C}_{2}{A}_{2} \) , \( {A}_{1}{B}_{1} \) and \( {A}_{2}... | Proof. Using an appropriate projective transformation, we can make the quadrilateral \( {A}_{1}{B}_{1}{C}_{1}{D}_{1} \) a square. Since the points \( {A}^{\prime } \) and \( {C}^{\prime } \) will go to infinity in perpendicular directions, the quadrilateral \( {A}_{2}{B}_{2}{C}_{2}{D}_{2} \) will become a rectangle who... | Yes |
Theorem 4.3. Suppose points \( A, B, C, D \) lie on an equilateral hyperbola. Then the Ceva circle of D with respect to the triangle ABC passes through the center of the hyperbola (Figure 4.4). | Proof. By Theorem 4.2, the centers \( {I}_{a}^{\prime },{I}_{b}^{\prime },{I}_{c}^{\prime } \) of the excircles of the Ceva triangle lie on the hyperbola. Since the Ceva circle is the nine-point circle of the triangle \( {I}_{a}^{\prime }{I}_{b}^{\prime }{I}_{c}^{\prime } \), it passes through the center of the hyperbo... | Yes |
Theorem 4.4. Suppose points \( A, B, C \) and \( D \) lie on an equilateral hyperbola. Then the pedal circle of \( D \) with respect to the triangle \( {ABC} \) passes through the center of the hyperbola. | Proof. Let \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) be the pedal triangle of \( D \), and \( {B}_{1},{C}_{1} \) the midpoints of the segments \( {BD} \) and \( {CD} \) (Figure 4.5).\n\n\n\nFigure 4.5\n\nTo prove... | Yes |
Theorem 4.5 (Emelyanov and Emelyanova). Let \( {A}_{1},{B}_{1} \) and \( {C}_{1} \) be the feet of the bisectors of a triangle \( {ABC} \) , \( {A}_{2},{B}_{2},{C}_{2} \) the feet of its heights, \( {C}^{ * },{B}^{ * },{A}^{ * } \) the intersections of the lines \( {A}_{1}{B}_{1} \) and \( {A}_{2}{B}_{2},{C}_{1}{A}_{1}... | Proof. Parts 1 and 2 follow from Theorem 3.9. We now prove part 3. Since the set of the centers of perspective of the triangles from the Feuerbach family lies on an equilateral hyperbola (it is called the Feuerbach hyperbola), their Ceva circles pass through the center of the hyperbola. Their pedal circles also pass th... | Yes |
Theorem 4.7. Let \( P \) be the perspector of the conic, \( Q \) its center, and \( M \) the centroid of the triangle. Then \( M \) lies on the segment \( {P}^{\prime }Q \), where \( {P}^{\prime } \) is the isotomic conjugate of \( P \) and \( {P}^{\prime }M = {2MQ} \) (Figure 4.7). | Proof. First, we note that if the conic is inscribed in a triangle \( {ABC} \), then the pole of the median \( C{M}_{c} \) lies on the line \( {c}_{m} \) passing through \( C \) and parallel to \( {AB} \) . Indeed, the cross-ratio of the lines \( {c}_{m}, C{M}_{c},{CA},{CB} \) equals 1 . Suppose now that a conic with c... | Yes |
Theorem 4.8. The center of an inscribed conic with perspector \( P \) is the pole of the line \( {PM} \) with respect to the conic passing through \( A, B, C, M \) and \( P \) (Figure 4.8). | ## Proof. This follows from the theorem just proved and Theorem 3.9. | No |
Theorem 4.11. Suppose we are given a complete quadrilateral. Then the orthocenters of the four triangles formed by its lines lie on a straight line. That line is perpendicular to the Gauss line of the quadrilateral. | Proof. As in the proof of the preceding theorem, consider the parabola tangent to the sides of the quadrilateral. By Theorem 1.11, the orthocenters of the corresponding triangles lie on the directrix of that parabola. We now prove the second part of the theorem. Using the corollary to Lemma 1.2 and Problem 10, one easi... | Yes |
Lemma 4.10 (Dual to Lemma 4.9). If two complete quadrilaterals have the same diagonals, then there is a conic tangent to all the sides of those quadrilaterals (obviously, such a conic is unique). | Proof of Theorem 4.12. Consider the midlines of the triangle formed by the diagonals of the quadrilateral and the line at infinity. Those four lines form a quadrilateral whose diagonals are nothing but the sides of the triangle. Therefore, by Lemma 4.10, there is a conic tangent to the sides of the quadrilateral, the l... | Yes |
Theorem 4.13 (Droz-Farny). Suppose a line \( {l}_{1} \) passing through the orthocenter \( H \) of a triangle \( {ABC} \) intersects its sides at points \( {A}_{1},{B}_{1} \) and \( {C}_{1} \). Another line \( {l}_{2} \), perpendicular to \( {l}_{1} \) and also passing through \( H \), intersects the sides of the trian... | Proof. Consider the parabola tangent to the sides of the triangle and the line \( {l}_{1} \). By Theorems 1.11 and 1.7, the tangent to that parabola passing through \( H \) and different from \( {l}_{1} \) is perpendicular to \( {l}_{1} \), and therefore it coincides with \( {l}_{2} \). By Theorem 1.10, the circumcircl... | Yes |
Theorem 4.14. Let \( {P}_{1},{P}_{2},{P}_{3},{P}_{4} \) be four points of a conic \( \alpha \) with center O. Suppose the normals at those points pass through a single point \( Q \) . Then \( {P}_{1},{P}_{2},{P}_{3},{P}_{4}, O, Q \) lie on an equilateral hyperbola whose asymptotes are parallel to the axes of the conic. | Proof. Given an arbitrary circle \( \omega \) with center \( Q \), consider the locus of the centers of the conics from the pencil generated by the circle and the conic \( \alpha \) . By Theorem 3.17, it is a conic and, since the pencil contains a circle, it is an equilateral hyperbola (see Problem 29). We denote it \(... | Yes |
Theorem 4.15. The points \( {P}_{1},{P}_{2},{P}_{3} \) and the point symmetric to \( {P}_{4} \) with respect to the center of the conic lie on a circle. | Proof. Consider the case where the given conic is an ellipse. Let \( {P}^{\prime } \) be the point symmetric to \( {P}_{4} \) with respect to \( O \) (Figure 4.16). The assertion of the theorem is equivalent to the statement that the conic of the centers of the pencil determined by \( {P}_{1},{P}_{2},{P}_{3},{P}^{\prim... | Yes |
Theorem 4.16. If a convex \( n \) -gon is inscribed in a given ellipse \( \alpha \) and has the longest perimeter among all such n-gons, then it is circumscribed about an ellipse \( {\alpha }_{n} \) confocal with \( \alpha \) . | Proof. Let \( {M}_{1}{M}_{2}\ldots {M}_{n} \) be a polygon with the longest perimeter. We shall prove that for each \( i = 1,2,\ldots, n \), the bisector of the exterior angle \( {M}_{i - 1}{M}_{i}{M}_{i + 1} \) is tangent to \( \alpha \) at \( {M}_{i} \) .\n\nSuppose this is not the case. Let \( {\alpha }^{\prime } \)... | Yes |
Theorem 4.17. A convex \( n \) -gon circumscribed about a given ellipse \( \alpha \) has the shortest perimeter among all such n-gons if and only if all of its vertices lie on an ellipse confocal with \( \alpha \) . | Proof. Fix the tangency points \( {M}_{i - 1},{M}_{i + 1} \) of the sides of the polygon and the ellipse. Let \( T \) be the intersection of the tangents at those points (Figure 4.20). For the sake of definiteness, suppose that the arc \( {M}_{i - 1}{M}_{i + 1} \) is less than half of the ellipse. We want to find a poi... | Yes |
Proposition 1.2 (Incidence)\n\nThe point \( \mathbf{x} \) lies on the line \( \ell \) if and only if \( \mathbf{x} \cdot \ell = 0 \) . | The above basically means: \( \mathbf{x} \) lies on \( \ell \) if and only if the vectors \( \mathbf{x},\ell \) are orthogonal. In particular, the cross product \( u \times v \) gives the vector orthogonal to both \( u, v \), so | No |
Theorem 1.4 (Desargues' theorem)\n\nLet \( {ABC},{XYZ} \) be triangles. Then \( \overline{AX},\overline{BY},\overline{CZ} \) concur if and only if \( \overline{BC} \cap \overline{YZ},\overline{CA} \cap \overline{ZX} \) , \( \overline{AB} \cap \overline{XY} \) are collinear. | I bet you can already smell the point-line duality in this one - the theorem itself is self-dual. To spell it out completely algebraically, consider the following two interpretations of the \ | No |
Theorem 1.9 (La Hire)\n\nLet \( P, Q \) be points and \( \mathcal{C} \) a conic. Then \( P \) lies on the polar of \( Q \) if and only if \( Q \) lies on the polar of \( P \) . | Proof. We want to show \( \mathbf{p} \cdot M\mathbf{q} = 0 \) if and only if \( \mathbf{q} \cdot M\mathbf{p} = 0 \) . This is obvious by \( M = {M}^{\top } \) . | Yes |
If \( f : \mathcal{P} \rightarrow \mathcal{P} \) preserves cross ratio and swaps some two points \( A, B \), then \( f \) is an involution. | Proof. For \( X \in \mathcal{P} \), let \( Y = f\left( X\right) \) and \( {X}^{\prime } = f\left( Y\right) \) . Then\n\n\[ \left( {{AB};{XY}}\right) \overset{f}{ = }\left( {{BA};Y{X}^{\prime }}\right) = \left( {{AB};{X}^{\prime }Y}\right) ,\n\nhence \( X = {X}^{\prime } \) . | Yes |
Any involution on a line \( \ell \) is an inversion of some nonzero power, possibly centered at infinity. | Proof. Let the involution swap \( O \) and infinity and have reciprocal pairs \( \left( {{X}_{1},{X}_{2}}\right) ,\left( {{Y}_{1},{Y}_{2}}\right) \) . Then\n\n\[ \left( {O\infty ;{X}_{1}{Y}_{1}}\right) = \left( {\infty O;{X}_{2}{Y}_{2}}\right) \Rightarrow \frac{O{X}_{1}}{O{Y}_{1}} = \frac{O{Y}_{2}}{O{X}_{2}}. \]\n\nThu... | Yes |
For any involution \( f \) on a conic \( \mathcal{C} \), there is a point \( P \) such that \( f \) sends each point \( A \) to the second intersection of line \( {PA} \) and \( \mathcal{C} \) . | Proof. The statement is purely projective, so send \( \mathcal{C} \) to a circle. Consider reciprocal pairs \( \left( {{A}_{1},{A}_{2}}\right) ,\left( {{B}_{1},{B}_{2}}\right) ,\left( {{C}_{1},{C}_{2}}\right) \), and take a point \( X \in \mathcal{C} \) . Invert about \( X \) sending \( \mathcal{C} \) to a line \( \ell... | Yes |
Theorem 5.1 (DIT)\n\nLet \( {ABCD} \) be a quadrilateral inscribed in a conic \( \mathcal{C} \) . A line \( \ell \) intersects lines \( {AB},{CD},{AD} \) , \( {BC},{AC},{BD} \) at points \( {X}_{1},{X}_{2},{Y}_{1},{Y}_{2},{Z}_{1},{Z}_{2};\ell \) intersects \( \mathcal{C} \) at \( {W}_{1},{W}_{2} \) . Then \( \left( {{W... | Proof. Take a homography \( f \) fixing \( \ell \) and sending \( {W}_{1},{W}_{2},{X}_{1} \) to \( {W}_{2},{W}_{1},{X}_{2} \) . Then \( f \) is an involution by Proposition 4.2, so \( \left( {{X}_{1},{X}_{2}}\right) \) is a reciprocal pair. It suffices to show (via symmetry) that \( \left( {{Y}_{1},{Y}_{2}}\right) \) i... | No |
Let \( P \) be a point in the plane of \( \bigtriangleup {ABC} \), and \( \gamma \) a line passing through \( P \) . Let \( {A}^{\prime },{B}^{\prime },{C}^{\prime } \) be the points where the reflections of lines \( {PA},{PB},{PC} \) with respect to \( \gamma \) intersect lines \( {BC},{AC},{AB} \) respectively. Prove... | Walkthrough. Define \( {C}_{1}^{\prime } = \overline{AB} \cap \overline{{A}_{1}{B}_{1}} \) . Show there is an involution swapping \( \left( {\overline{PA},\overline{P{A}_{1}}}\right) \) , \( \left( {\overline{PB},\overline{P{B}_{1}}}\right) ,\left( {\overline{PC},\overline{P{C}_{1}^{\prime }}}\right) \) ; then show thi... | No |
Problem 7.1 (ISL 2016 G5). Let \( D \) be the foot of perpendicular from \( A \) to the Euler line (the line passing through the circumcenter and the orthocenter) of an acute scalene triangle \( {ABC} \) . A circle \( \omega \) with center \( S \) passes through \( A \) and \( D \), and it intersects sides \( {AB} \) a... | Walkthrough. Verify the problem when \( S \) is the midpoint of \( \overline{AH} \) and \( \overline{AO} \) . Then move \( S \) linearly, and apply spiral similarity at \( D \) to show \( X, Y \) move linearly along \( \overline{AB},\overline{AC} \) . Since \( \bigtriangleup {XLY} \) has fixed shape ( \( L \) being cir... | No |
If \( f, g : {\mathcal{C}}_{1} \rightarrow {\mathcal{C}}_{2} \) are projective and \( f, g \) coincide at at least three values, then \( f \equiv g \) . | Proof. This is very much the fundamental theorem of algebra: if \( f\left( A\right) = g\left( A\right), f\left( B\right) = g\left( B\right) \) , \( f\left( C\right) = g\left( C\right) \) then for any point \( D \in {\mathcal{C}}_{1} \) , \n\n\[ \n\left( {f\left( A\right) f\left( B\right) ;f\left( C\right) f\left( D\rig... | No |
Given a triangle \( {ABC} \), with \( I \) as its incenter and \( \Gamma \) as its circumcircle, line \( {AI} \) intersects \( \Gamma \) again at \( D \) . Let \( E \) be a point on the arc \( {BDC} \), and \( F \) a point on the segment \( {BC} \), such that\n\n\[ \angle {BAF} = \angle {CAE} < \frac{1}{2}\angle {BAC}\... | Walkthrough. Animate \( E \), let \( {E}^{\prime } \) be reflection of \( E \) in \( \overline{AI} \) . Show the map \( E \mapsto {E}^{\prime } \mapsto F \mapsto G \) is projective. | No |
Lemma 10.1 (Zack's lemma)\n\nFor two moving points \( A, B \) coinciding at \( k \) points (counting multiplicity), the degree of line \( {AB} \) is at most \( \deg A + \deg B - k \) . | Proof of both. Let\n\n\[ A = \left( {{P}_{1}\left( t\right) : {Q}_{1}\left( t\right) : {R}_{1}\left( t\right) }\right) \]\n\n\[ B = \left( {{P}_{2}\left( t\right) : {Q}_{2}\left( t\right) : {R}_{2}\left( t\right) }\right) \]\n\nwhere \( A, B \) are either moving points or lines.\n\nThen line \( {AB} \) or the intersect... | Yes |
Proposition 10.4 (Conic doubling)\n\nIf \( {\mathcal{C}}_{1},{\mathcal{C}}_{2} \) are lines or conics and \( \phi : {\mathcal{C}}_{1} \rightarrow {\mathcal{C}}_{2} \) is a projective map, then for a moving point \( P \in {\mathcal{C}}_{1} \) with degree \( d \) ,\n\n\[ \deg \left( {\phi \left( P\right) }\right) = \left... | The proof requires work in \( \mathbb{C} \) that we will omit. (However the result still holds in \( \mathbb{R} \).) | No |
Let \( {ABC} \) be a non-equilateral triangle and let \( {M}_{a},{M}_{b},{M}_{c} \) be the midpoints of sides \( {BC} \) , \( {CA},{AB} \), respectively. Let \( S \) be a point lying on the Euler line. Denote by \( X, Y, Z \) the second intersections of lines \( {M}_{a}S,{M}_{b}S,{M}_{c}S \) with the nine-point circle.... | Walkthrough. Animate \( S \), and show the concurrence has degree 6 . Then find seven values of \( S \) . (This is very easy by triangle symmetry.) | No |
Let \( {ABCD} \) be a cyclic quadrilateral with circumcenter \( O \) . Let lines \( {AB} \) and \( {CD} \) meet at \( E \) and lines \( {AC} \) and \( {BD} \) meet at \( P \) . Furthermore, let lines \( {EP} \) and \( {AD} \) meet at \( K \), and let \( M \) be the midpoint of \( \overline{AD} \) . Prove that \( {BCMK}... | Work in \( {\mathbb{{CP}}}^{2} \) . Animate \( B \), and apply Pascal on hexagon IJBKMC. | No |
The sum of two rotations with centres \( {O}_{1} \) and \( {O}_{2} \) through angles \( \alpha \) and \( \beta \) respectively is a rotation through the angle \( \alpha + \beta \) around some centre O. | Consider Figure 115. The sum of the two rotations carries the centre \( {O}_{1} \) of the first into a point \( {O}_{1}^{\prime } \) such that \( {O}_{1}^{\prime }{O}_{2} = {O}_{1}{O}_{2} \) and \( \angle {O}_{1}{O}_{2}{O}_{1}^{\prime } = \beta \) . (The first rotation leaves \( {O}_{1} \) in place, and the second carr... | Yes |
If similar triangles ADB, CBE and FAC are erected outwardly on the sides of any triangle \( \mathrm{{ABC}} \), their incentres \( \mathrm{G},\mathrm{H} \) and \( \mathrm{I} \) form a triangle similar to the three triangles. | We shall mainly use the following special case of the Petersen-Schoute theorem (Coxeter & Greitzer, 1967:99):\n\n\ | No |
Consider Figure 4.63a with cyclic quadrilateral ABCD. Construct the angle bisector AF of \( \angle {BAC} \) with \( \mathrm{F} \) on the circle. Connect \( \mathrm{F} \) with \( \mathrm{D} \), and \( \mathrm{F} \) with \( \mathrm{B} \). Now we have \( \angle {BAC} = \angle {BDC} \) on chord BC, but \( \angle {BAF} = \a... | Construct a circle with \( \mathrm{F} \) as centre and \( \mathrm{{BF}} \) as radius, and label its intersection with \( \mathrm{{AF}} \) as P. Connect P with C. We now have \( \angle {AFB} = \angle {ACB} \) on chord \( \mathrm{{AB}} \), but \( \angle {PFB}\left( { = \angle {AFB}}\right) = 2\angle {PCB} \) on chord BP;... | Yes |
Theorem 3.12 [HL]. If the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and a leg of a second right triangle, then the triangles are congruent. | Proof. Consider right triangles \( \bigtriangleup {ABC} \) and \( \bigtriangleup {DEF} \) with right angles \( \angle B \) and \( \angle E \) , respectively. Assume that \( {AC} = {DF} \) and \( {AB} = {DE} \) . Extend the ray \( \overrightarrow{CB} \) to a point \( G \) such that \( {GB} = {EF} \) . Join \( {AG}\ldots | Yes |
Lemma 4.5. The total length of the sides of any spherical triangle on a sphere of radius \( R \) is less than the circumference of a great circle, that is, \( {2\pi R} \) . | Proof. Consider a spherical triangle \( \bigtriangleup {ABC} \) . Extend \( {AB} \) and \( {AC} \) so that they meet at \( {A}^{\prime } \), the antipodal point of \( A \) . Then \( \widehat{{AB}{A}^{\prime }} + \widehat{{AC}{A}^{\prime }} = {2\pi R} \) . Consider the spherical triangle \( \bigtriangleup {A}^{\prime }{... | Yes |
Lemma 4.7. On a sphere of radius \( R \), the area of a lune with angle \( \alpha \) radians is given by Area \( = {2\alpha }{R}^{2} \) . | Proof. A sphere of radius \( R \) has surface area \( {4\pi }{R}^{2} \) . A lune of angle \( \frac{\pi }{2} \) has one-quarter of the sphere’s total surface area, or \( \pi {R}^{2} \), since it is a proportion of \( \frac{\frac{\pi }{2}}{2\pi } = \frac{1}{4} \) of the total surface area. Thus, a lune of angle \( \alpha... | Yes |
Theorem 4.8. On a sphere of radius \( R \), the area of spherical triangle \( \bigtriangleup {ABC} \) with angles \( \alpha ,\beta \) and \( \gamma \) (measured in radians) is given by Area \( \left( {\bigtriangleup {ABC}}\right) = \left( {\alpha + \beta + \gamma - \pi }\right) {R}^{2} \) . | Proof. The sides of \( \bigtriangleup {ABC} \) create lunes of angles \( \alpha ,\beta \) and \( \gamma \), with areas of \( {2\alpha }{R}^{2},{2\beta }{R}^{2} \) and \( {2\gamma }{R}^{2} \), respectively. Together, these three lunes cover exactly half of the area of the sphere, or \( {2\pi }{R}^{2} \) . (Use your sphe... | Yes |
Theorem 4.10. On a sphere of radius \( R = 1 \), the area of the spherical quadrilateral region, \( Q \), formed by quadrilateral \( {ABCD} \) with vertex angles \( \alpha ,\beta ,\gamma \) and \( \delta \), where \( 0 < \) \( \alpha ,\beta ,\gamma ,\delta < \pi \), is given by\n\n\[ \operatorname{Area}\left( Q\right) ... | Proof. Let quadrilateral region, \( Q \), formed by \( {ABCD} \), be given as specified, and join \( {AC} \) . Since all vertex angles are less than \( \pi, Q \) is convex. Thus, \( {AC} \) lies entirely inside our quadrilateral region and, since \( R = 1 \), is less than \( \pi \) in length. Consider the two trilatera... | Yes |
Theorem 4.11. On a sphere of radius \( R = 1 \), the area of the spherical polygonal region, \( P \), formed by \( n \) -gon \( {A}_{1}{A}_{2}\ldots {A}_{n} \) with vertex angles \( {\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{n} \), where each \( {\alpha }_{i} < \pi \), is given by\n\n\[ \operatorname{Area}\left( P\... | When \( P \) is a convex polygonal region as specified in the previous theorem, then the sum of the angles must be less than \( {n\pi } \) . Thus, by Theorem 4.11, the area of \( P \) is less than \( {2\pi } \) . Since the area of a sphere of radius \( R = 1 \) is \( {4\pi } \), we can determine the area of the complem... | Yes |
Suppose we are given angle \( \alpha \), and two sides \( b \) and \( c \), of \( \bigtriangleup {ABC} \) as \( \alpha = \frac{\pi }{2}, b = \frac{2\pi }{3} \) and \( c = \frac{\pi }{2} \). To find the remaining angles and side of the triangle, using | \( \left\lbrack {C}_{1}\right\rbrack \) we have \( \cos a = - \frac{1}{2} \cdot 0 + 0 \), which gives \( a = \frac{\pi }{2} \). Using the alternative version of \( \left\lbrack {C}_{1}^{ * }\right\rbrack \) as above gives \( \cos \beta = \frac{-\frac{1}{2} - 0}{1 \cdot 1} = - \frac{1}{2} \). So, \( \beta = \frac{2\pi }... | Yes |
Suppose we are given three sides of spherical \( \bigtriangleup {ABC} \) as \( a = \frac{\pi }{2}, b = \frac{2\pi }{3} \) and \( c = \frac{\pi }{2} \). To find the angles, using \( \left\lbrack {C}_{1}^{ * }\right\rbrack \) we have \( \cos \alpha = 0 \) which gives \( \alpha = \frac{\pi }{2} \). We could continue to us... | As noted above, since arcsine only returns angles between \( - \frac{\pi }{2} \) and \( \frac{\pi }{2} \), we will never obtain an obtuse solution. In the given \( \bigtriangleup {ABC},\beta \) must be obtuse, and the obtuse angle satisfying \( \sin \beta = \frac{\sqrt{3}}{2} \) is \( \beta = \frac{2\pi }{3} \). The le... | Yes |
Theorem 4.15. To construct a length equal to a given angle \( \alpha \), where \( \alpha < \pi \) . | Proof. Let \( \angle {BAC} = \alpha \) . Extend \( {AB} \) and \( {AC} \) until they intersect at \( {A}^{\prime } \) which is antipodal to \( A \) . Bisect \( \widehat{{AB}{A}^{\prime }} \) and \( \widehat{{AC}{A}^{\prime }} \) . Label these points \( D \) and \( E \), respectively, and construct segment \( {DE} \) . ... | Yes |
Theorem 4.18 [Polar Triangle Involution Theorem]. The polar triangle of a polar triangle is the original triangle. | Proof. Let \( \bigtriangleup {ABC} \) be a given spherical triangle and \( \bigtriangleup {A}^{\prime }{B}^{\prime }{C}^{\prime } \) be its polar triangle. Since \( {C}^{\prime } \) is a pole of \( {AB} \) and \( {A}^{\prime } \) is a pole of \( {BC} \), both \( {A}^{\prime } \) and \( {C}^{\prime } \) are a length of ... | Yes |
Theorem 4.19 [Polar Duality Theorem]. The sides of a polar triangle are the supplements of the angles of the original triangle, and the angles of a polar triangle are the supplements of the sides of the original. [Note that the supplement of side \( a \) is \( \pi - a \) .] | Proof. Let \( \bigtriangleup {ABC} \) be a given spherical triangle with angles \( \alpha ,\beta \) and \( \gamma \) at vertices \( A \) , \( B \) and \( C \), respectively. Consider its polar triangle \( \bigtriangleup {A}^{\prime }{B}^{\prime }{C}^{\prime } \) . We will start by showing that \( {B}^{\prime }{C}^{\pri... | No |
Theorem 4.21. Given three angles between 0 and \( \pi \) whose sum is greater than \( \pi \), and such that the sum of any two angles is less than the remaining angle augmented by \( \pi \), to construct a triangle. | Proof. Let \( \alpha ,\beta \) and \( \gamma \) be the given angles. By assumption, \( \alpha ,\beta ,\gamma < \pi \) . By Theorem 4.15, construct a segment with length \( \alpha \) and label its endpoints \( A \) and \( B \) . Construct \( {A}^{\prime } \) , the antipodal point to \( A \) . Segment \( {A}^{\prime }B \... | Yes |
Theorem 6.1. All theorems resulting from an axiomatic system are true in any model of the system. | To help see this, we note that in an axiomatic system the proof of any theorem does not depend on any particular interpretation of the basic terms, but rather, its proof relies only on the relationships prescribed by the axioms (or previously proven theorems which, ultimately, rely on the axioms). For example, a proof ... | No |
Theorem 6.8. There are exactly six lines in Four Point geometry. | Proof. By Axiom 4P-1 we have exactly four points, say \( A, B, C \) and \( D \), and thus, \( \left( \begin{array}{l} 4 \\ 2 \end{array}\right) = \) \( \frac{4!}{2!2!} = 6 \) pairs of distinct points. Thus, by Axiom 4P-2 there must be at least 6 lines, namely \( {AB},{AC},{AD},{BC},{BD} \) and \( {CD} \) . How do we kn... | Yes |
Theorem 6.12 [Crossbar Theorem]. If \( D \) is in the interior of angle \( \angle {BAC} \), then ray \( \overrightarrow{AD} \) intersects segment \( {BC} \) . | Proof. Consider a point \( D \) that lies in the interior of angle \( \angle {BAC} \) . Construct \( E \) on the line \( \overrightarrow{AB} \) such that \( A \) lies between \( E \) and \( B \), as shown in Figure 6.5. Consider triangle \( \bigtriangleup {BEC} \) . Since \( \overrightarrow{AD} \) intersects \( {BE} \)... | "No" |
Theorem 6.13. Given a circle with center \( O \) and radius \( r \), any ray \( \overrightarrow{OA} \) emanating from O must intersect the circle. | Proof.\n\nCase 1. \( {OA} = r \) . In this case, \( A \) lies on the circle and is itself the point of intersection.\n\nCase 2. \( {OA} > r \) . Then \( A \) lies outside of the circle and the Circular Continuity Principle implies that \( {OA} \), and thus \( \overrightarrow{OA} \), must intersect the circle.\n\nCase 3... | Yes |
Theorem 6.14. The two circles in Proposition I. 1 intersect. | Proof. Note that \( A \) lies on the circle centered at \( B \) . By Axiom II.2, there exists a point \( E \) such that \( B \) is between \( A \) and \( E \) . By Theorem 6.13, ray \( \overrightarrow{BE} \) must intersect the circle centered at \( B \) at a point \( F \) . Now, \( B \) lies between \( A \) and \( F \)... | Yes |
Theorem 7.2 [Area of a Rectangle]. The area of a rectangular region is the product of its base and its altitude (height). | Proof. Consider a rectangle with base \( b \) and height \( h \) . We will start by constructing a square whose side has length \( b + h \) . We leave it to the reader to give a careful construction of the decomposition of this large square into four regions consisting of two squares and two copies of our original rect... | No |
Theorem 8.3 [I.45S]. To construct a spherical square given an angle between \( \frac{\pi }{2} \) and \( \pi \) . | Proof. Given angle \( \alpha \) between \( \frac{\pi }{2} \) and \( \pi \), by Theorem 4.21 we may construct isosceles triangle \( \bigtriangleup {ABC} \) with angles \( \alpha /2,\alpha /2 \) and \( \alpha = \angle {ABC} \) . Construct two circles of radius \( {AB} \), one centered at \( A \) and the other at \( C \) ... | Yes |
Theorem 8.4 [II.14S]. To construct a spherical square equal to a given spherical rectangle. | Proof. Suppose the given rectangle has angle \( \alpha \) . The area of this rectangle is \( {4\alpha } - {2\pi } \) . Construct a square with angle \( \alpha \) . It will have the same area. | No |
Theorem 8.5. To construct a spherical square equal to a given spherical triangle. | Proof. Let spherical triangle \( \bigtriangleup {ABC} \) be given with angles \( {\alpha }_{1},{\alpha }_{2} \) and \( {\alpha }_{3} \) . By Theorem 4.8, the area of \( \bigtriangleup {ABC} \) is equal to \( {\alpha }_{1} + {\alpha }_{2} + {\alpha }_{3} - \pi \) . Let\n\n\[ \beta = \left( {\mathop{\sum }\limits_{{i = 1... | Yes |
Theorem 8.6. To construct a spherical square equal to a convex spherical polygon. | Proof. In a convex spherical polygon, each vertex angle is less than \( \pi \) . Consider the spherical polygon with \( n \) sides and angles, \( {\alpha }_{i}\left( {1 \leq i \leq n}\right) \), where each \( {\alpha }_{i} \) is less than \( \pi \) . We will assume that \( n \geq 4 \) . By Theorem 4.11, the area of thi... | Yes |
Corollary 9.1. [AA]. If two triangles have two pairs of equal angles, then they are similar. | Proof. By I.32, the third pair of angles are also equal. Thus, we may apply \( \widetilde{AAA} \) . | No |
Theorem 9.2. If \( {P}_{1} \) and \( {P}_{2} \) are regular \( n \) -gons with sides, \( {s}_{1} \) and \( {s}_{2} \), respectively, then\n\n\[ \frac{\operatorname{Area}\left( {P}_{1}\right) }{\operatorname{Area}\left( {P}_{2}\right) } = \frac{{s}_{1}^{2}}{{s}_{2}^{2}} \] | Proof. We leave it to the reader to prove that the area of a regular \( n \) -gon, with a side of length \( s \), is given by\n\n\[ \operatorname{Area}\left( {\text{ regular }n\text{-gon }}\right) = \frac{n{s}^{2}}{4\tan \left( \frac{\pi }{n}\right) }.\]\n\nFor a fixed \( n \), the quantity \( \frac{n}{4\tan \left( \fr... | No |
Theorem 11.2. Given any triangle, the three angle bisectors of the triangle are concurrent. Their point of intersection is called the incenter. | Proof. Consider a triangle \( \bigtriangleup {ABC} \) . Let the angle bisectors of \( \angle {ABC} \) and \( \angle {ACB} \) intersect at a point \( D \) . Construct \( E, F, G \) on \( {AB},{BC},{AC} \), respectively such that \( {DE} \bot \) \( {AB},{DF} \bot {BC} \) and \( {DG} \bot {AC} \) . As \( D \) lies on the ... | Yes |
Theorem 11.3. Given any triangle, the three perpendicular bisectors of the sides of the triangle are concurrent. The point of concurrency is called the circumcenter. | Proof. Consider the triangle \( \bigtriangleup {ABC} \) . Let \( D, E \) be the midpoints of \( {AB},{AC} \), respectively. Let the perpendicular bisectors of \( {AB} \) and \( {AC} \) intersect at a point \( F \) . Since \( F \) lies on the perpendicular bisector of \( {AB},{AF} = {BF} \) . Similarly, \( F \) lies on ... | Yes |
Theorem 11.6. Given any triangle, the three altitudes of the triangle are concurrent at a point called the orthocenter. | Proof. Consider triangle \( \bigtriangleup {ABC} \) . The idea behind the proof is to create a larger triangle \( \bigtriangleup {DEF} \) in such a way that the altitudes of our original \( \bigtriangleup {ABC} \) are the perpendicular bisectors of our new \( \bigtriangleup {DEF} \) . We can then use Theorem 11.3 to ju... | No |
Theorem 11.7. Given any triangle, the three medians of the triangle are concurrent at a point called the centroid. | Proof. Consider triangle \( \bigtriangleup {ABC} \) . The idea behind the proof is to show that any two medians must intersect in a point that is two-thirds of the way from a vertex to its opposite side. Since this holds for any pair of medians, all three medians must coincide. We begin the proof, but leave it to the r... | No |
Theorem 11.9 [Euler Line Theorem]. Given any nonequilateral triangle, its circumcenter, orthocenter and centroid are collinear, lying on a line called the \( \\mathbf{{Euler}\\;{line}.} \) | Proof. Consider nonequilateral triangle \( \\bigtriangleup {ABC} \), with circumcenter \( F \) and centroid \( G \) . As noted above, and as the reader will show in Exercise 11.2.14, \( F \) and \( G \) must be distinct. There are two cases to consider.\n\nCase 1. \( \\bigtriangleup {ABC} \) is isosceles. The reader wi... | No |
Lemma 11.10 [Converse of Intersecting Secants Theorem]. Consider distinct segments \( {AC} \) and \( {AE} \) with common endpoint \( A \) . Let \( B \) be a point on \( {AC} \), and \( D \) a point on \( {AE} \) . If \( {AB} \cdot {AC} = {AD} \cdot {AE} \), then quadrilateral \( {BCED} \) is cyclic, that is, points \( ... | Proof. By Corollary 11.4, let \( c \) be the unique circle joining the three noncollinear points \( B, C \) and \( D \) . Since \( {AC} \) intersects \( c \) at two points, \( A \) must lie outside of \( c \) .\n\nClaim: The ray \( \overrightarrow{AD} \) will also intersect \( c \) at two points. For the sake of reachi... | Yes |
Theorem 11.11 [Nine-point Circle Theorem]. Given any triangle, the midpoints of the three sides, the feet of the three altitudes, and the three midpoints of the line segments connecting the orthocenter to the vertices lie on a circle, called the \( \\mathbf{{Nine}} \) -point \( \\mathbf{{circle}} \) . | Proof. Consider triangle \( \\bigtriangleup {ABC} \) . Let \( {A}_{T} \) be the foot of the altitude from \( A \) to \( {BC} \), and let \( {A}_{G} \) be the midpoint of the side opposite \( A \) . Define \( {B}_{T},{B}_{G},{C}_{T} \) and \( {C}_{G} \) similarly. Let \( H \) be the orthocenter, and let \( {A}_{H} \) be... | Yes |
Lemma 11.17. A regular polygon with perimeter \( P \) and apothem \( h \) has an area of \( \frac{1}{2}h \cdot P \) . | Proof of Lemma. Suppose the regular polygon has \( n \) sides, each of length \( s \) . Join each vertex to the center, \( O \) . The polygon is composed of \( n \) congruent triangles, each with area \( \frac{1}{2}{hs} \) . Hence, the area of the polygon is \( n \cdot \left( {\frac{1}{2}{hs}}\right) = \frac{1}{2}h \cd... | Yes |
Lemma 11.18. If \( {a}_{n} \) denotes the length of the side of a regular \( n \) -sided polygon inscribed in a circle of radius 1, then\n\n\[ \n{a}_{2n} = \sqrt{2 - \sqrt{4 - {a}_{n}^{2}}} \n\] | Proof of Lemma. Let \( A \) be the center of the given circle of radius \( 1,{BC} = {a}_{n} \) and \( {CD} = {a}_{2n} \), as shown in Figure 11.42. Note that \( {AD} \) bisects arc \( \overset{⏜}{BC} \), meeting \( {BC} \) at \( E \) . Since angle \( \angle {AEB} \) is a right angle (by proof of Theorem 11.14), then by... | Yes |
Theorem 12.5. Between any two points, there exists a unique hyperbolic line. | Proof. Suppose we are given points \( A = \left( {{x}_{1},{y}_{1}}\right) \) and \( B = \left( {{x}_{2},{y}_{2}}\right) \) . \n\nCase 1. If \( {x}_{1} = {x}_{2} = k \), then no semicircle centered at the \( x \) -axis can pass through these points. Therefore, \( x = k \) is the only hyperbolic line that passes through ... | Yes |
The linear path, shown as the dashed line in Figure 12.15, is given parametrically by \( x = 1 + t \) and \( y = 1 + {3t} \) for \( 0 \leq t \leq 1 \) . Using Equation (12.4) to determine the length of the path, we have | \[ s = {\int }_{0}^{1}\sqrt{{\left( 1\right) }^{2} + {\left( 3\right) }^{2}}{dt} = {\left. t\sqrt{10}\right| }_{0}^{1} = \sqrt{10} \approx {3.162}. \] | Yes |
The lower curved path, shown as the dotted curve in Figure 12.15, is given parametrically by \( x = 1 + t \) and \( y = 3{t}^{\frac{3}{2}} + 1 \) for \( 0 \leq t \leq 1 \) . Using Equation (12.4) to determine the length of the path, we have | \[ s = {\int }_{0}^{1}\sqrt{{\left( 1\right) }^{2} + {\left( 3 \cdot \frac{3}{2}{t}^{\frac{1}{2}}\right) }^{2}}{dt} = {\int }_{0}^{1}\sqrt{\left( 1 + \frac{81}{4}t\right) }{dt} \] \[ = {\left. \frac{4}{81} \cdot \frac{2}{3}{\left( 1 + \frac{81}{4}t\right) }^{\frac{3}{2}}\right| }_{0}^{1} \] \[ = \frac{-8 + {85}\sqrt{85... | Yes |
The middle curved path, shown as the solid black curve in Figure 12.15, is given parametrically by \( x = t \) and \( y = {t}^{2} \) for \( 1 \leq t \leq 2 \) . Using Equation (12.4), trigonometric substitution and integration by parts to determine the length of the path, we have | \[ s = {\int }_{1}^{2}\sqrt{{\left( 1\right) }^{2} + {\left( 2t\right) }^{2}}{dt} = \frac{1}{2}{\int }_{t = 1}^{t = 2}{\sec }^{3}{\theta d\theta } \]\n\[ = {\left. \frac{1}{2}\sec \theta \tan \theta \right| }_{t = 1}^{t = 2} - \frac{1}{2}{\int }_{t = 1}^{t = 2}\sec \theta {\tan }^{2}{\theta d\theta } \]\n\[ = {\left. \... | Yes |
For the path given parametrically by \( x = 1 + t \) and \( y = 1 + {3t} \) for \( 0 \leq t \leq 1 \), using Equation (12.5) to determine the length of the path in the hyperbolic model, we have | \[ {s}_{H} = {\int }_{0}^{1}\frac{1}{\left( 1 + 3t\right) }\sqrt{{\left( 1\right) }^{2} + {\left( 3\right) }^{2}}{dt} \] \[ = {\left. \frac{\sqrt{10}}{3}\ln \left( 1 + 3t\right) \right| }_{0}^{1} = \frac{\sqrt{10}}{3}\ln 4 \approx {1.4613}\text{.} \] | Yes |
For the path given parametrically by \( x = 1 + t \) and \( y = 3{t}^{\frac{3}{2}} + 1 \) for \( 0 \leq t \leq 1 \), using Equation (1.2.5) to determine the length of the path in the hyperbolic model, we have | \[ {s}_{H} = {\int }_{0}^{1}\frac{1}{3{t}^{\frac{3}{2}} + 1}\sqrt{{\left( 1\right) }^{2} + {\left( 3 \cdot \frac{3}{2}{t}^{\frac{1}{2}}\right) }^{2}}{dt} \] \[ = {\int }_{0}^{1}\frac{1}{3{t}^{\frac{3}{2}} + 1}\sqrt{1 + \frac{81}{4}t}{dt} \] \[ \approx \;{1.5114}\text{. } \] | Yes |
To calculate the distance between \( A = \left( {1,1}\right) \) and \( B = \left( {2,4}\right) \) | \[ {d}_{H}\left( {A, B}\right) = \left| {\ln \left( \frac{4\left( {9 + \sqrt{65} - 1}\right) }{1\left( {9 + \sqrt{65} - 2}\right) }\right) }\right| = \ln \left( \frac{4\left( {8 + \sqrt{65}}\right) }{7 + \sqrt{65}}\right) \approx {1.4506}. \] | Yes |
Theorem 13.1. Through a point not on a given straight line, there are infinitely many straight lines parallel to the given line. | Proof. Suppose that we given a line, \( \ell \), and a point, \( A \), not on it. By Euclid I.12, we can construct a line through \( A \) that is perpendicular to \( \ell \) . Let \( B \) be the intersection of this line with \( \ell \) . By Euclid I.11, we can construct a line, \( m \), through \( A \) that is perpend... | Yes |
Theorem 13.2. Given a point \( A \) not on a line \( \ell \), the angle of parallelism to \( \ell \) through \( A \) on the left equals the angle of parallelism to \( \ell \) through \( A \) on the right. | Proof. Given a point \( A \) not on a line \( \ell \), once again, construct the perpendicular from \( A \) to \( \ell \), and let the intersection with \( \ell \) be \( B \) . Suppose that \( k \) and \( m \) are the right- and left-sensed parallels to \( \ell \) through \( A \), respectively. Pick \( C \) on \( k \) ... | Yes |
Theorem 13.11 [Crossbar Theorem for Omega Triangles]. If \( C \) is in the interior of omega triangle \( \bigtriangleup {AB\Omega } \), then\n\n- ray \( \overrightarrow{AC} \) intersects ray \( \overrightarrow{B\Omega } \),\n\n- ray \( \overrightarrow{BC} \) intersects ray \( \overrightarrow{A\Omega } \), and\n\n- line... | Proof. Consider omega triangle \( \bigtriangleup {AB\Omega } \) with interior point \( C \) . We first show that \( \overrightarrow{AC} \) intersects ray \( \overrightarrow{B\Omega } \) . Construct \( D \) on \( \overset{\overleftrightarrow{} }{B\Omega } \) such that \( {AD} \bot \overset{\overleftrightarrow{} }{B\Omeg... | No |
Theorem 13.12 [I.16 \( {}_{\Omega } \) : Exterior Angle Theorem for Omega Triangles]. The exterior angle of an omega triangle is greater than the opposite interior angle. | Proof. Let \( \bigtriangleup {AB\Omega } \) be an omega triangle. Consider points \( C \) on line \( \overset{\overleftrightarrow{} }{AB} \) such that \( B \) is between \( A \) and \( C \) . We wish to show that \( \angle {CB\Omega } > \angle {BA\Omega } \) . We will assume that this is not the case and obtain a contr... | No |
Theorem 13.14 [AS \( {}_{\Omega } \) : Angle-Side Congruence for Omega Triangles]. Given omega triangles \( \bigtriangleup {AB\Omega } \) and \( \bigtriangleup {CD\Lambda } \), if \( \angle {AB\Omega } = \angle {CD\Lambda } \) and \( {AB} = {CD} \), then \( \angle {BA\Omega } = \angle {DC\Lambda } \) . | Proof. Let \( \bigtriangleup {AB\Omega } \) and \( \bigtriangleup {CD\Lambda } \) be omega triangles with \( \angle {AB\Omega } = \angle {CD\Lambda } \) and \( {AB} = \) \( {CD} \) . In order to obtain a contradiction, we will assume \( \angle {BA\Omega } \neq \angle {DC\Lambda } \) . WLOG, assume \( \angle {BA\Omega }... | Yes |
Theorem 13.15 [AA \( {}_{\Omega } \) : Angle-Angle Congruence for Omega Triangles]. Given omega triangles \( \bigtriangleup {AB\Omega } \) and \( \bigtriangleup {CD\Lambda } \), if \( \angle {AB\Omega } = \angle {CD\Lambda } \) and \( \angle {BA\Omega } = \angle {DC\Lambda } \), then \( {AB} = {CD} \) . | Proof. Let \( \bigtriangleup {AB\Omega } \) and \( \bigtriangleup {CD\Lambda } \) be omega triangles with \( \angle {AB\Omega } = \angle {CD\Lambda } \) and \( \angle {BA\Omega } \) \( = \angle {DC\Lambda } \) . In order to obtain a contradiction, we will assume \( {AB} \neq {CD} \) . WLOG, assume \( {AB} > {CD} \) . C... | Yes |
Proposition 13.20 [Saccheri's Proposition I]. The summit angles of a Saccheri quadrilateral are equal. | Proof. Assume \( {ABCD} \) is a Saccheri quadrilateral where \( {AC} \bot {AB},{BD} \bot {AB} \) and \( {AC} = {BD} \) . Construct \( {AD} \) and \( {BC} \) . Then by SAS, \( \bigtriangleup {CAB} \cong \bigtriangleup {DBA} \) . Therefore, \( {BC} = \) \( {AD} \) . Triangles \( \bigtriangleup {ACD} \) and \( \bigtriangl... | Yes |
Proposition 13.21 [Saccheri's Proposition II]. The line joining the midpoints of the base and summit of a Saccheri quadrilateral is perpendicular to the base and the summit. | Proof. Assume \( {ABCD} \) is a Saccheri quadrilateral where \( {AC} \bot {AB},{BD} \bot {AB} \) and \( {AC} = {BD} \) . Let \( M \) and \( H \) be the midpoints of \( {AB} \) and \( {CD} \), respectively. Construct \( {MH} \) , \( {CM} \) and \( {DM} \) . By SAS, we have \( \bigtriangleup {CAM} \cong \bigtriangleup {D... | Yes |
Theorem 13.24. The summit angles in a Saccheri quadrilateral are acute. | Proof. Assume \( {ABCD} \) is a Saccheri quadrilateral where \( {AC} \bot {AB},{BD} \bot {AB} \) and \( {AC} = {BD} \) . Let \( E \) be a point on \( \overrightarrow{CD} \) such that \( D \) is between \( C \) and \( E \) . Let \( \ell = \overrightarrow{AB} \) have ideal point \( \Omega \) on the side of \( \overrighta... | Yes |
Theorem 13.27 \( \left\lbrack {\mathrm{{I.32}}}_{\mathrm{H}}\right\rbrack \) . Given any triangle, the sum of the angles is less than two right angles. | Proof. Consider any triangle \( \bigtriangleup {ABC} \) . We start by constructing an associated Saccheri quadrilateral. To do this, bisect \( {AB} \) and \( {AC} \) at \( D \) and \( E \), respectively, and construct the line \( \overrightarrow{DE} \) . Next, construct \( F, G \) and \( H \) on \( \overrightarrow{DE} ... | No |
Corollary 13.29. The angle sum of any quadrilateral is less than four right angles. | Proof. Consider a quadrilateral \( {ABCD} \) . Construct diagonal \( {AC} \) . There are two cases to consider.\n\nCase 1. ABCD is the union of triangles \( \bigtriangleup {ABC} \) and \( \bigtriangleup {ACD} \) .\n\nCase 2. Both \( B \) and \( D \) lie on the same side of \( \overrightarrow{AC} \) . In this case, eith... | No |
Theorem 13.31 \( \left\lbrack {\mathrm{{AAA}}}_{\mathrm{H}}\right\rbrack \) . If the angles of one triangle are equal, respectively, to the angles of another triangle, then the triangles are congruent. | Proof. Consider triangles \( \bigtriangleup {ABC} \) and \( \bigtriangleup {DEF} \), where \( \angle A = \angle D,\angle B = \angle E \) and \( \angle C = \) \( \angle F \) . We wish to show that \( \bigtriangleup {ABC} \cong \bigtriangleup {DEF} \) . If we can show that \( {AB} = {DE} \), then by ASA, we have \( \bigt... | Yes |
Theorem 13.32 [Sensed parallels are nowhere equidistant]. If \( m \) and \( \ell \) are sensed parallels with ideal point \( \Omega \), and \( A \) and \( C \) are distinct points on \( m \), then the distance from A to \( \ell \) does not equal the distance from \( C \) to \( \ell \) . | Actually, we can say a bit more about the relationship between the distances from these points to \( \ell \) . If \( C \) lies on the sensed parallel ray \( \overrightarrow{A\Omega } \), then the distance from \( C \) to \( \ell \) is less than the distance from \( A \) to \( \ell \) . Furthermore, if \( \ell \) and \(... | No |
Theorem 13.35. Suppose that \( \ell \) and \( m \) are ultraparallel with unique common perpendicular \( \overrightarrow{EF} \), where \( E \) lies on \( m \) and \( F \) lies on \( \ell \) . Let \( A \) be any other point on \( m \) . The distance from \( A \) to \( \ell \) is greater than the length of \( {EF} \) . | Proof. Once again, our goal is to construct a Saccheri quadrilateral whose summit lies on \( m \) and whose base lies on \( \ell \) . This time we will use \( A \) as one of the summit vertices and we will construct the quadrilateral with \( E \) as the midpoint of the summit and \( F \) as the midpoint of the base.\n\... | No |
Theorem 13.47. If two polygons are equivalent by finite decomposition, then they have the same area. | Proof. Suppose \( P \) and \( {P}^{\prime } \) are polygons such that \( P \equiv {P}^{\prime } \) . By definition, there are triangulations \( \mathcal{C} = \left\{ {{T}_{1},{T}_{2},\ldots ,{T}_{n}}\right\} \) and \( {\mathcal{C}}^{\prime } = \left\{ {{T}_{1}^{\prime },{T}_{2}^{\prime },\ldots ,{T}_{n}^{\prime }}\righ... | Yes |
Theorem 13.50 [Bolyai’s Theorem for Hyperbolic Triangles]. If two triangles have the same area, then they are equivalent by finite decomposition. That is, if \( \delta \left( {\bigtriangleup {ABC}}\right) = \) \( \delta \left( {\bigtriangleup {DEF}}\right) \), then \( \bigtriangleup {ABC} \equiv \bigtriangleup {DEF} \)... | To prove this theorem, we need the following two lemmas. | No |
Lemma 13.51. Two Saccheri quadrilaterals with congruent summits and congruent summit angles are congruent. | Proof of Lemma. Let \( {ABDC} \) and \( {EFHG} \) be Saccheri quadrilaterals with bases \( {AB} \) and \( {EF} \), respectively. Assume that \( {CD} = {GH} \), and \( \angle C = \angle D = \angle G = \angle H \) . To establish the congruence of these quadrilaterals, we must show that the corresponding sides and bases a... | Yes |
Lemma 13.52. Two triangles with the same defect and one pair of congruent sides are equivalent by finite decomposition. | Proof of Lemma. Let \( \bigtriangleup {ABC} \) and \( \bigtriangleup {DEF} \) be triangles with \( {BC} = {EF} \) and \( \delta \left( {\bigtriangleup {ABC}}\right) \) \( = \delta \left( {\bigtriangleup {DEF}}\right) \) . Construct associated Saccheri quadrilaterals \( {BCHG} \) and \( {EFKJ} \), with summits \( {BC} \... | Yes |
Theorem 14.2. In Four Point geometry, if two distinct lines intersect then they have exactly one point in common. | Proof. Assume lines \( \ell \) and \( m \) meet at more than one point, say \( A \) and \( B \) . Thus, points \( A \) and \( B \) are on more than one line, contradicting Axiom 4P-2. | Yes |
Theorem 14.5. In Four Point geometry, to each line there is exactly one distinct line parallel to it. | Proof. Let \( \ell \) be a line in the geometry. By Axiom 4P-3, \( \ell \) has two points on it, say \( A \) and \( B \) . By Axiom 4P-1 there is a point \( C \) not on \( \ell \) . By Theorem 14.3, \( C \) is on exactly three lines. By Axiom 4P-2, two of these lines intersect \( \ell \) and the third does not intersec... | Yes |
Theorem 14.7. There are at least seven lines and at least seven points in any projective plane. | Proof. Let \( A, B, C \) and \( D \) be the four points guaranteed by Axiom P-3. Axiom P-1 guarantees the existence of six lines: \( {AB},{AC},{AD},{BC},{BD} \) and \( {CD} \) . By Axiom P-3, these lines must be distinct since no three of the points \( A, B, C \) and \( D \) are on the same line. This is illustrated in... | No |
Theorem 14.8. There are at least four distinct lines, no three of which are on the same point. | Proof. By Axiom P-3, we have points \( A, B, C \) and \( D \), and four distinct lines \( {AB},{BC} \) , \( {CD} \) and \( {AD} \), as explained in the previous proof. We will show by contradiction that no three of these lines pass through a common point. Without loss of generality, suppose \( {AB},{BC} \) and \( {CD} ... | Yes |
Theorem 14.9 [Principle of Duality]. In projective geometry, if a statement is true, then its dual is also true. | For example, we could now revisit Theorem 14.7, knowing that we only need to prove the existence of at least seven points, let's say, since duality provides the justification for the existence of at least seven lines. | No |
Theorem 14.10. In a projective plane, every line is on at least three points. | Proof. Let \( \ell \) be a line, and \( A, B, C \) and \( D \) be the points, as given by Axiom P-3.\n\nCase 1. If \( \ell \) is one of the six distinct lines \( {AB},{AC},{AD},{BC},{BD} \) or \( {CD} \), then from the proof of Theorem 14.7, \( \ell \) is on at least three points.\n\nCase 2. On the other hand, suppose ... | Yes |
In a finite projective plane, every line is on the same number of points and every point lies on the same number of lines. Furthermore, the number of points on a line equals the number of lines on a point. | We will show that every line is on the same number of points, and appeal to duality for the other two statements.\n\nOur strategy is to show that, given distinct lines \( \ell \) and \( m \), we can find a one-to-one map from the set of points on \( \ell \) to the set of points on \( m \) . Let \( \ell \) and \( m \) b... | Yes |
Theorem 14.15. In a finite projective plane of order \( n \), there are exactly \( {n}^{2} + n + 1 \) lines. | Proof. Let \( \ell \) be any line of the plane. By definition, there are \( n + 1 \) distinct points on \( \ell \) . Each of these points is on \( n \) distinct lines other than \( \ell \), as demonstrated by Figure 14.18. Axiom P-1 guarantees that a line other than \( \ell \) through point \( {P}_{i} \) is necessarily... | Yes |
Theorem 14.17. There are at least six lines in any affine plane. | Proof. Let points \( A, B, C \) and \( D \) be the four points given by Axiom A-3. By Axiom A-1, there exist lines \( {AB} \) and \( {AC},{AD},{BC},{BD} \) and \( {CD} \) . A pairwise grouping of any of these lines shows that these lines are distinct, otherwise three of these four points are on the same line, violating... | Yes |
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