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Problem 6.10.14. Let \( {k}_{1} \) and \( {k}_{2} \) be circles that intersect each other at the points \( C \) and \( D \) . A line passing through \( C \) intersects \( {k}_{1} \) and \( {k}_{2} \) at the points \( A \) and \( B \), respectively. Denote the midpoints of the arcs \( \overset{⏜}{AD} \) and \( \overset{... | Solution. Let \( P \) and \( Q \) be the midpoints of \( {AD} \) and \( {DB} \), respectively. Then \( \angle {APK} = \angle {ADB} = \angle {KQB} \) and \( \angle {MPK} = \angle {KQN} \) . Since\n\n\[ \angle {MDP} = \frac{\angle {ACD}}{2} = \frac{\angle {DNB}}{2} = \angle {DNQ}, \]\n\nwe have \( \bigtriangleup {MPD} \s... | Yes |
Problem 6.10.15. Let the circles \( {k}_{1} \) and \( {k}_{2}\left( O\right) \) intersect each other at the points \( A \) and \( B \) . Denote the reflection of \( B \) with respect to \( O \) by \( C \) . Let \( {BC} \cap {k}_{1} = \{ B, F\} \) . The point \( G \in {k}_{1} \) is such that \( {BG} = {BF} \) . Let \( {... | Solution. Let \( {GA} \) intersect \( {BC} \) at \( D \) . We will prove that \( \left( {C, D, B, F}\right) = 1 \) and the desired result will follow.\n\nNote that \( B \) is the midpoint of the arc \( \overset{⏜}{GAF} \) of \( {k}_{1} \) , and \( \angle {IAB} = {90}^{ \circ } \) . Hence, \( \angle {FAI} = \angle {GAI}... | Yes |
Problem 6.10.16. Let \( k\\left( O\\right) \) be a circle with a chord \( {AB} \) . Let \( D \) be an arbitrary point on \( k \) . The circumcircle of \( \\bigtriangleup {ADO} \) intersects \( {AB} \) for the second time at the point \( E \) . Prove that \( {DE} = {BE} \) . | Solution. We have\n\n\n\n\\[ \n\\angle {OEB} = \\angle {ODA} \n\\]\n\n\\[ \n= \\angle {OAD} = \\angle {OED}\\text{.} \n\\]\n\nAlso,\n\n\\[ \n\\angle {OBE} = \\angle {OAE} = \\angle {ODE}\\text{.} \n\\]\n\nHence, \( \... | Yes |
Problem 6.10.17. Let \( {ABCD} \) be a convex quadrilateral, such that \( {BA} \cap {CD} = E \) . Denote the midpoints of \( {AB},{BC},{CD},{DA} \) and \( {DB} \) by \( H, G, F, I \) and \( J \), respectively. Prove that the circumcircles of \( \bigtriangleup {HJG} \) , \( \bigtriangleup {IJF} \) and \( \bigtriangleup ... | Solution. Let \( k\left( {HJG}\right) \cap k\left( {IJF}\right) = \{ J, K\} \) . Then we consecutively have\n\n\[ \angle {HKF} = \angle {HKJ} + \angle {JKF} = \angle {HGJ} + {180}^{ \circ } - \angle {JIF} \]\n\n\[ = \angle {JGB} - \angle {HGB} + {180}^{ \circ } - \angle {JID} + \angle {FID} \]\n\n\[ = {180}^{ \circ } +... | Yes |
Problem 6.10.18. Let \( \bigtriangleup {ABC} \cong \bigtriangleup A{B}^{\prime }{C}^{\prime } \) . Let \( {B}^{\prime }{C}^{\prime } \cap {BC} = D \) and let \( O \) be the circumcenter of \( \bigtriangleup D{B}^{\prime }C \) . Prove that \( {AO} \bot B{C}^{\prime } \) . | Solution. Let \( E \) be the or-\n\n\n\nthogonal projection of \( A \) onto the line \( {BC} \), and let \( H \) be the reflection of \( B \) with respect to the point \( E \) . We will prove that\n\n\[ \n{C}^{\prime... | Yes |
Problem 6.10.19. Let \( k\left( A\right) \) be a circle and let the line \( l \) touch it at the point \( B \) . A line parallel to \( l \) intersects \( k \) at the points \( C \) and \( D \) . The tangent lines to \( k \) at \( B \) and \( D \) intersect at \( E \) . Let \( {CE} \cap k = \{ C, F\} \) . Denote the mid... | Solution. Let the line through \( F \), parallel to \( {CD} \) , intersect \( k \) for the second time at the point \( {F}^{\prime } \) .\n\nWe are going to project the quadrilateral \( B{F}^{\prime }{CD} \) onto the line \( {BE} \) with respect to the point \( F \) . The point \( {F}^{\prime } \) maps to the point of ... | Yes |
Problem 6.10.20. Let \( k \) and \( {k}_{1} \) be circles, such that \( {k}_{1} \) lies inside of \( k \) . The circles \( {k}_{2} \) and \( {k}_{3} \) touch \( k \) internally and \( {k}_{1} \) externally. The circles \( {k}_{4} \) and \( {k}_{5} \) touch \( k \) internally and \( {k}_{1} \) externally. Denote the int... | Solution. Consider an inversion \( I \) with center \( F \) and an arbitrary radius. We will use the prime symbol to denote the images of the objects after the inversion has been applied to them. Then \( I\left( {k}_{2}\right) = {k}_{2}^{\prime } \) and \( I\left( {k}_{3}\right) = \) \( {k}_{3}^{\prime } \), where \( {... | Yes |
Problem 6.10.21. Let \( {AB} \) and \( {CD} \) be two chords in the circle \( k \) that intersect each other at the point \( X \) . Points \( E \) and \( F \) are chosen on the arcs \( \overset{⏜}{AD} \) and \( \overset{⏜}{BC} \), respectively. The circumcircles of \( \bigtriangleup {FCX} \) and \( \bigtriangleup {AEX}... | Solution. Consider an inversion \( I\left( X\right) \) with an arbitrary radius.\n\nThe circle \( \left( {ADBC}\right) \) transforms into a circle \( {k}^{\prime } \) passing through \( X \) . The other points that lie on \( k \) map on \( {k}^{\prime } \) . Let \( I\left( A\right) = {A}_{1}, I\left( B\right) = {B}_{1}... | Yes |
Let \( {ABCD} \) be a cyclic quadrilateral, inscribed in the circle \( k \) . The line \( l \) forms equal angles with the pair of lines \( \left( {{AB},{CD}}\right) \) while intersecting them. Let us denote the circle that touches the pair \( \left( {{AB},{CD}}\right) \) at their points of intersection with \( l \) by... | Angle chasing gives that \( l \) forms equal angles with the pair \( \left( {{AC},{BD}}\right) \) . Therefore, the circle \( {k}_{1} \) touches the line \( {AC} \) at its point of intersection with \( l \) . First, we will prove the following two lemmas: | No |
Lemma 1. Two circles \( {k}_{1}\left( {{O}_{1}, r}\right) \) and \( {k}_{2}\left( {{O}_{2}, r}\right) \) are given. If \( P \) is a point and \( l \) is the radical axis of the two circles, then\n\n\[ \mid \operatorname{pow}\left( {P,{k}_{1}}\right) - \operatorname{pow}\left( {P,{k}_{2}}\right) \mid = 2{O}_{1}{O}_{2}.\... | Proof. Without loss of gener-\n\nality, let \( \operatorname{pow}\left( {P,{k}_{1}}\right) \leq \operatorname{pow}\left( {P,{k}_{2}}\right) \) .\n\nLet \( {O}_{1}{O}_{2} \cap l = M \) and let \( H \) be the foot of the perpendicular from \( P \) to \( {O}_{1}{O}_{2} \) . Let \( {O}_{1}H = x \) , \( {HM} = y \) and \( M... | Yes |
Lemma 2. Let \( {k}_{1}\left( {O}_{1}\right) \) and \( {k}_{2}\left( {O}_{2}\right) \) be circles. The locus of points \( P \), such that \( \frac{\operatorname{pow}\left( {P,{k}_{1}}\right) }{\operatorname{pow}\left( {P,{k}_{2}}\right) } = \) const, is a circle and the three circles have a common radical axis. | Proof. Let \( k\left( O\right) \) is a circle, such that \( k,{k}_{1} \) and \( {k}_{2} \) have a common radical axis \( l \) . Let \( P \in k \) . We have\n\n\[ \frac{\operatorname{pow}\left( {P,{k}_{1}}\right) }{\operatorname{pow}\left( {P,{k}_{2}}\right) } = \frac{\left| \operatorname{pow}\left( P,{k}_{1}\right) - \... | Yes |
Problem 6.10.24. Let \( {ABCDEF} \) be a cyclic hexagon. Prove that the lines \( {AD},{BE} \) and \( {CF} \) are concurrent if and only if the equality DE.FA.BC = EF.AB.CD holds. | Solution. Let \( {AD},{BE} \) an\n\n\n\n\( {CF} \) be concurrent.\n\nThe trigonometric form of Ceva's Theorem, applied to \( \bigtriangleup {ACE} \) and the lines \( {AD},{CF} \) and \( {EB} \), and a few application... | Yes |
Problem 6.10.25. Let \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) be three circles. Let \( {k}_{1} \cap {k}_{2} = \) \( \{ A, D\} ,{k}_{2} \cap {k}_{3} = \{ C, F\} \) and \( {k}_{3} \cap {k}_{1} = \{ B, E\} \) (see the figure). Prove that \( {AB}.{CD}.{EF} = {BC}.{DE}.{FA} \) . | Solution. The radical axes theorem (Problem 6.5.1) yields that the lines \( {AD},{BE} \) and \( {CF} \) are concurrent.\n\nApplying the trigonometric form of Ceva’s Theorem to \( \bigtriangleup {BDF} \) and the lines \( {DA},{BE} \) and \( {FC} \), we get\n\n\[ \frac{\sin \angle {FDA}}{\sin \angle {ADB}} \cdot \frac{\s... | Yes |
Problem 7.3. (Pappus’ Theorem) The points \( A, B \) and \( C \) lie on a line (in this order). The points \( D, E \) and \( F \) lie on another line (in this order). Let \( {AE} \cap {BD} = P,{AF} \cap {CD} = N \) and \( {BF} \cap {CE} = M \) . Prove that the points \( P, N \) and \( M \) are collinear. | Solution. We will consider the statement in the case when the two lines intersect. A good exercise for the reader is to solve the problem in the case when they are parallel.\n\nLet the lines intersect at the point \( O \) and let \( k \) be a circle with center \( O \) . Consider a pole-polar transformation \( \pi \) w... | No |
Let \( O \) be a point. Two distinct rays with origin \( O \) are given. Let \( {X}_{1},{X}_{2} \) and \( {X}_{3} \) be points lying (in this order) on one of the rays, and let \( {X}_{4},{X}_{5} \) and \( {X}_{6} \) be points lying (in this order) on the other one. Let \( {X}_{2}{X}_{6} \cap {X}_{3}{X}_{5} = {X}_{7} \... | Let \( {X}_{9} = {X}_{1}{X}_{5} \cap {X}_{2}{X}_{4} \) . Pappus’ Theorem (Problem 7.3), applied to the collinear triples of points \( {X}_{1},{X}_{2},{X}_{3} \) and \( {X}_{4},{X}_{5},{X}_{6} \), gives that the points \( {X}_{7},{X}_{8} \) and \( {X}_{9} \) are collinear. So \( {X}_{9} \) is the intersection point of t... | Yes |
Problem 7.5. Let \( A \) be a point. Two distinct rays with origin \( A \) are given. The points \( E \) and \( B \) lie on the first one \( \left( {E\text{lies between}A\text{and}B}\right) \) and the points \( D \) and \( C \) lie on the second one ( \( D \) lies between \( A \) and \( C \) ). Let \( F \in {BD}, G \in... | Solution. Let \( {EF} \cap {DG} = I \) . Applying Pappus’ Theorem to \( {BG} \cap {CF} \) , \( {BE} \cap {CD} \) and \( {EF} \cap {DG} \), yields that the points \( A, H \) and \( I \) are collinear and the desired statement follows. | Yes |
Problem 7.6. Let \( {DJEB} \) be a quadrilateral. Let \( {BD} \cap {EJ} = A \) , \( {BE} \cap {DJ} = C \) and \( {BJ} \cap {DE} = K \) . The line \( l \) intersects the lines \( {AE} \) , \( {AB},{CB} \) and \( {CD} \) at the points \( H, G, I \) and \( F \), respectively. The lines \( {KH},{KG},{KI} \) and \( {KF} \) ... | Solution. We will show that the points \( L, M \) and \( N \) are collinear (the proof is analogous for the other triples). Menelaus' Theorem, applied to \( \bigtriangleup {AJD} \), yields that it suffices to prove that \( \frac{AL}{LJ} \cdot \frac{JN}{ND} \cdot \frac{DM}{MA} = 1 \) . The same theorem, applied to \( \b... | Yes |
Problem 7.7. Let \( {ABC} \) be a triangle. The points \( D \) and \( E \) lie on the sides \( {CA} \) and \( {CB} \), respectively. Let \( F \) be a point and let \( {DE} \cap {CF} = X \) , \( {FD} \cap {AB} = H,{FE} \cap {AB} = G,{BX} \cap {FD} = K,{AX} \cap {EG} = L \) , \( {XH} \cap {BC} = J \) and \( {XG} \cap {AC... | Solution. Let us consider the problem in the projective plane. We send \( X \) to the point of infinity. then the lines \( {AL} \) , \( {DE},{HJ},{GI},{CF} \) and \( {BK} \) become parallel. We will prove that the points \( L, J \) and \( K \) are collinear (the collinearity of the points \( L, I \) and \( K \) follows... | Yes |
Problem 7.8. Let \( {ALFD} \) be a quadrilateral will non-parallel sides. Let \( G \) be a point that does not lie on the line \( {AF} \) . Let \( {AD} \cap {FL} = B \) and \( {AL} \cap {FD} = C \) . The lines \( {GB},{GD},{GL} \) and \( {GC} \) intersect \( {AF} \) at the points \( I, M, K \) and \( J \), respectively... | Solution. Let us consider the problem in the projective plane, sending the point \( G \) to the point of infinity (in a direction different from that of the line \( {AF} \) ). Then the lines \( {KL},{DM},{BI} \) and \( {CJ} \) become parallel. Let \( {KD} \cap {LM} = S \) . Menelaus’ Theorem, applied to \( \bigtriangle... | Yes |
Problem 7.9. Let \( {BEFD} \) be a convex quadrilateral. Let \( {BD} \cap {EF} = \) \( A \) and \( {BE} \cap {DF} = C \) . The point \( G \) is arbitrary. The lines \( {GD},{GB},{GF} \) and \( {GE} \) intersect \( {AC} \) at the points \( H, I, J \) and \( K \), respectively. Prove that the lines \( {KD},{JB},{IF} \) a... | Solution. Let us consider the problem in the projective plane, sending the point \( G \) to the point of infinity. Then the lines \( {DH},{BI},{FJ} \) and \( {EK} \) become parallel. Let \( {HE} \cap {FI} = X \) . We will prove that the points \( D \) , \( X \) and \( K \) are collinear (the collinearity of the points ... | Yes |
Problem 7.10. Let \( l \) be a line and \( P \notin l \) be a point. The points \( A, B, C \) and \( D \) lie (in this order) on \( l \) . The collinear points \( {A}_{1},{B}_{1},{C}_{1} \) and \( {D}_{1} \) lie on the lines \( {PA},{PB},{PC} \) and \( {PD} \), respectively. Prove that \( \left( {A, B, C, D}\right) = \... | Solution. Let \( \angle {APB} = \alpha ,\angle {BPC} = \beta \) and \( \angle {CPD} = \gamma \) . Then\n\n\[ \left( {A, B, C, D}\right) = \frac{AB}{BC} \cdot \frac{CD}{DA} = \frac{{S}_{\bigtriangleup {ABP}}}{{S}_{\bigtriangleup {BCP}}} \cdot \frac{{S}_{\bigtriangleup {CDP}}}{{S}_{\bigtriangleup {ADP}}} \]\n\n\[ = \frac... | Yes |
Let \( {ABC} \) be a triangle. The points \( D, E \) and \( G \) lie on the sides \( {BC},{CA} \) and \( {AB} \), respectively, such that \( {AD} \cap {BE} \cap {CG} = F \) . The point \( J \) lies inside of \( \bigtriangleup {DEG} \) . Let \( {GJ} \cap {DE} = I,{EJ} \cap {DG} = K \) and \( {DJ} \cap {EG} = H \) . Prov... | The trigonometric form of Ceva’s Theorem, applied to \( \bigtriangleup {ABC} \) , yields that it suffices to prove that\n\n\[ \frac{\sin \angle {EAH}}{\sin \angle {HAG}} \cdot \frac{\sin \angle {GBK}}{\sin \angle {KBD}} \cdot \frac{\sin \angle {DCI}}{\sin \angle {ICE}} = 1. \]\n\n\[ \text{But }\frac{\sin \angle {EAH}}{... | Yes |
Problem 8.1. In a regular 12-gon, prove that the lines \( {A}_{1}{A}_{6},{A}_{3}{A}_{7} \) and \( {A}_{5}{A}_{10} \) are concurrent. | Solution. It suffices to prove\n\n\n\nthat\n\n\[ \frac{{A}_{1}{A}_{3}}{{A}_{3}{A}_{5}} \cdot \frac{{A}_{5}{A}_{6}}{{A}_{6}{A}_{7}} \cdot \frac{{A}_{7}{A}_{10}}{{A}_{10}{A}_{1}} = 1 \]\n\nBut it is easy to see that \(... | Yes |
In a regular 14-gon, prove that the lines \( {A}_{1}{A}_{7},{A}_{5}{A}_{9} \) and \( {A}_{6}{A}_{11} \) are concurrent. | It suffices to prove that\n\n\[ \frac{{A}_{1}{A}_{5}}{{A}_{5}{A}_{6}} \cdot \frac{{A}_{6}{A}_{7}}{{A}_{7}{A}_{9}} \cdot \frac{{A}_{9}{A}_{11}}{{A}_{11}{A}_{1}} = 1 \]\n\nBut \( {A}_{1}{A}_{5} = {A}_{1}{A}_{11},{A}_{5}{A}_{6} = \) \( {A}_{6}{A}_{7} \) and \( {A}_{7}{A}_{9} = {A}_{11}{A}_{9} \) and the desired equality f... | Yes |
Problem 8.3. In a regular 18-gon, prove that the lines \( {A}_{1}{A}_{9},{A}_{7}{A}_{11} \) and \( {A}_{8}{A}_{13} \) are concurrent. | Solution. It suffices to prove that \[ \frac{{A}_{1}{A}_{7}}{{A}_{7}{A}_{8}} \cdot \frac{{A}_{8}{A}_{9}}{{A}_{9}{A}_{11}} \cdot \frac{{A}_{11}{A}_{13}}{{A}_{13}{A}_{1}} = 1 \] But note that \( {A}_{1}{A}_{7} = \) \( {A}_{1}{A}_{13},{A}_{7}{A}_{8} = {A}_{8}{A}_{9} \) and \( {A}_{9}{A}_{11} = {A}_{11}{A}_{13} \) . The de... | Yes |
Problem 8.4. In a regular 18-gon, prove that the lines \( {A}_{1}{A}_{9},{A}_{4}{A}_{11} \) and \( {A}_{8}{A}_{17} \) are concurrent. | Solution. It suffices to prove that\n\n\[ \frac{{A}_{1}{A}_{4}}{{A}_{4}{A}_{8}} \cdot \frac{{A}_{8}{A}_{9}}{{A}_{9}{A}_{11}} \cdot \frac{{A}_{11}{A}_{17}}{{A}_{17}{A}_{1}} = 1. \]\n\nThis is equivalent to\n\n\[ \frac{\sin \frac{3\pi }{18}}{\sin \frac{4\pi }{18}} \cdot \frac{\sin \frac{\pi }{18}}{\sin \frac{2\pi }{18}} ... | Yes |
In a regular 18-gon, prove that the lines \( {A}_{1}{A}_{8},{A}_{3}{A}_{9} \) and \( {A}_{7}{A}_{16} \) are concurrent. | It suffices to prove\n\n\n\nthat\n\n\[ \frac{{A}_{1}{A}_{3}}{{A}_{3}{A}_{7}} \cdot \frac{{A}_{7}{A}_{8}}{{A}_{8}{A}_{9}} \cdot \frac{{A}_{9}{A}_{16}}{{A}_{16}{A}_{1}} = 1 \]\n\nBut \( {A}_{7}{A}_{8} = {A}_{8}{A}_{9} ... | Yes |
Problem 8.6. In a regular 18-gon, prove that the lines \( {A}_{1}{A}_{7},{A}_{3}{A}_{8} \) and \( {A}_{6}{A}_{15} \) are concurrent. | Solution. It suffices to prove\n\n\n\nthat\n\n\[ \frac{{A}_{1}{A}_{3}}{{A}_{3}{A}_{6}} \cdot \frac{{A}_{6}{A}_{7}}{{A}_{7}{A}_{8}} \cdot \frac{{A}_{8}{A}_{15}}{{A}_{15}{A}_{1}} = 1 \]\n\nBut \( {A}_{6}{A}_{7} = {A}_{... | Yes |
In a regular 18-gon, prove that the lines \( {A}_{6}{A}_{10},{A}_{7}{A}_{11} \) and \( {A}_{8}{A}_{14} \) are concurrent. | It suffices to prove\n\n\n\nthat\n\n\[ \frac{{A}_{6}{A}_{7}}{{A}_{7}{A}_{8}} \cdot \frac{{A}_{8}{A}_{10}}{{A}_{10}{A}_{11}} \cdot \frac{{A}_{11}{A}_{14}}{{A}_{14}{A}_{6}} = 1 \]\n\nBut \( {A}_{6}{A}_{7} = {A}_{7}{A}_... | Yes |
Problem 8.8. In a regular 18-gon, prove that the lines \( {A}_{4}{A}_{9},{A}_{7}{A}_{10} \) and \( {A}_{8}{A}_{12} \) are concurrent. | Solution. It suffices to prove\n\n\n\nthat\n\n\[ \frac{{A}_{4}{A}_{7}}{{A}_{7}{A}_{8}} \cdot \frac{{A}_{8}{A}_{9}}{{A}_{9}{A}_{10}} \cdot \frac{{A}_{10}{A}_{12}}{{A}_{12}{A}_{4}} = 1 \]\n\nBut \( {A}_{8}{A}_{9} = {A}... | Yes |
In a regular 24-gon, prove that the lines \( {A}_{1}{A}_{13},{A}_{6}{A}_{14} \) , \( {A}_{10}{A}_{16} \) and \( {A}_{12}{A}_{20} \) are concurrent. | Note that \( {A}_{1}{A}_{13} \) is the common perpendicular bisector of \( {\bar{A}}_{12}{\bar{A}}_{14} \) and \( {A}_{6}{A}_{20} \), which gives that the lines \( {A}_{1}{A}_{13},{A}_{6}{A}_{14} \) and \( {A}_{12}{A}_{20} \) are concurrent. | No |
In a regular 30-gon, prove that the lines \( {A}_{4}{A}_{17},{A}_{8}{A}_{18} \) , \( {A}_{12}{A}_{20},{A}_{13}{A}_{21},{A}_{15}{A}_{25} \) and \( {A}_{16}{A}_{29} \) are concurrent. | Solution. We divide the solution into several steps.\n\n1.) We will prove that the lines \( {A}_{4}{A}_{17},{A}_{8}{A}_{18} \) and \( {A}_{16}{A}_{29} \) are concurrent.\n\nIt suffices to prove that \( \frac{{A}_{4}{A}_{8}}{{A}_{8}{A}_{16}} \cdot \frac{{A}_{16}{A}_{17}}{{A}_{17}{A}_{18}} \cdot \frac{{A}_{18}{A}_{29}}{{... | Yes |
Problem 8.11. Let \( {A}_{1}{A}_{2}{A}_{3}\ldots {A}_{18} \) be a regular 18-gon with center \( O \) . Let \( A, B, C \) and \( D \) be the midpoints of \( {A}_{15}{A}_{17}, O{A}_{18}, O{A}_{2} \) and \( {A}_{3}{A}_{5} \) , respectively. Prove that the points \( A, B, C \) and \( D \) are collinear. | Solution. We will prove that the points \( B, C \) and \( D \) are collinear (the fact that \( A \) lies on this line follows analogously).\n\nLet \( E \) be the midpoint of \( O{A}_{3} \) . Then \( {BC} \) is a midsegment in \( \bigtriangleup O{A}_{2}{A}_{18} \) , \( {CE} - \) in \( \bigtriangleup O{A}_{2}{A}_{3} \) a... | Yes |
Let \( {ABCDEF} \) be a regular hexagon. The points \( M \) and \( N \) are the midpoints of \( {AB} \) and \( {DF} \), respectively. Prove that \( \bigtriangleup {MCN} \) is equilateral. | Solution. It is clear that \( \bigtriangleup {ACE} \) is equilateral. We have\n\n\[ 2\overrightarrow{MC} = \overrightarrow{AC} + \overrightarrow{BC}\text{. }\n\]\n\n\( \mathrm{A} + {60}^{ \circ } \) rotation of all vectors in the above equality gives\n\n\[ 2\overrightarrow{a} = \overrightarrow{AE} + \overrightarrow{AF}... | Yes |
Let \( {ABCDE} \) and \( A{B}_{1}{C}_{1}{D}_{1}{E}_{1} \) be regular pentagons with the same orientation and a common vertex \( A \) . Prove that the lines \( B{B}_{1}, C{C}_{1}, D{D}_{1} \) and \( E{E}_{1} \) are concurrent. | Solution. Let \( X \) be the second intersection point of the circumcircles of the two pentagons. We will prove that \( X \) is the intersection point of the lines \( B{B}_{1}, C{C}_{1}, D{D}_{1} \) and \( E{E}_{1} \) . We have\n\n\[ \angle {DXA} = \angle {DCA} = {72}^{ \circ }\text{.} \]\n\nAlso,\n\n\[ \angle {D}_{1}{... | Yes |
Let \( {ABC} \) be an equilateral triangle. The point \( D \) lies inside of the triangle. The symmetric points of \( D \) with respect to the sides \( {AB},{BC} \) and \( {AC} \) are \( {C}^{\prime },{A}^{\prime } \) and \( {B}^{\prime } \), respectively. Prove that the lines \( A{A}^{\prime } \) , \( B{B}^{\prime } \... | Note that\n\n\[ \angle {AB}{A}^{\prime } + \angle {CB}{C}^{\prime } = 2\angle {ABC} + \angle {CB}{A}^{\prime } + \angle {AB}{C}^{\prime } \]\n\n\[ = 2\angle {ABC} + \angle {CBD} + \angle {ABD} = 3\angle {ABC} = {180}^{ \circ }\text{. } \]\n\nTherefore, \( \sin \angle {AB}{A}^{\prime } = \sin \angle {CB}{C}^{\prime } \)... | Yes |
Problem 8.1.2. Let \( {ABC} \) be an equilateral triangle. The point \( D \) lies outside of the triangle. The symmetric points of \( D \) with respect to the lines \( {AB},{BC} \) and \( {AC} \) are \( {C}^{\prime },{A}^{\prime } \) and \( {B}^{\prime } \), respectively. Prove that the circumcircles of \( \bigtriangle... | Solution. It suffices to show that the three circles are coaxial (have a common radical axis). Clearly, \( D \) has equal powers with respect to each of them. If we prove that the circumcenter \( O \) of \( \bigtriangleup {ABC} \) has the same property, then the circles will either have a common radical axis (which wil... | Yes |
Problem 8.1.3. Let \( {ABC} \) be an equilateral triangle. The point \( D \) lies on the smaller arc \( \overset{⏜}{AB} \) of its circumcircle. Prove that \( {AD} + {BD} = {CD} \) . | Solution. Let the sidelength of \( \bigtriangleup {ABC} \) be \( a \) . The quadrilateral \( {ACBD} \) is cyclic. By Ptolemy's Theorem (Problem 5.5.6) we get that \[ {AD}.{BC} + {BD}.{AC} = {CD}.{AB} \] \[ \Leftrightarrow a.{AD} + a.{BD} = a.{CD} \] \[ \Leftrightarrow {AD} + {BD} = {CD}\text{.} \] | Yes |
Problem 8.1.4. Let \( {ABC} \) be an equilateral triangle with circumcircle \( k \) . The point \( P \) lies on the smaller arc \( \overset{⏜}{AB} \) of \( k \) . The circle \( {k}_{1} \) touches \( k \) externally at the point \( P \) . The lines \( {AX},{BY} \) and \( {CZ} \) touch \( {k}_{1} \) at the points \( X, Y... | Solution. Let the sidelength of \( \bigtriangleup {ABC} \) be a. Casey’s Theorem (Problem 6.1.10), applied to \( A,{k}_{1}, B \) and \( C \) (points are circles with zero radius), yields\n\n\[ \n{AX}.{BC} + {BY}.{CA} = {CZ}.{AB} \]\n\n\[ \n\Leftrightarrow a.{AX} + a.{BY} = a.{CZ} \]\n\n\[ \n\Leftrightarrow {AX} + {BY} ... | Yes |
Problem 8.1.5. Let \( {ABC} \) and \( {A}_{1}{B}_{1}{C}_{1} \) be two equilateral triangles with a common center \( O \) and different orientations. Prove that the lines \( A{A}_{1}, B{B}_{1} \) and \( C{C}_{1} \) are concurrent. | Solution. A \( {120}^{ \circ } \) rotation centered at \( O \) gives\n\n\[ \angle {AO}{B}_{1} = \angle {BO}{A}_{1} = \angle {CO}{C}_{1}. \]\n\nSince \( {AO} = {BO} = {CO} \) and \( {A}_{1}O = {B}_{1}O = {C}_{1}O \), we get that\n\n\[ \bigtriangleup A{B}_{1}O \cong \bigtriangleup B{A}_{1}O \cong \bigtriangleup C{C}_{1}O... | Yes |
Problem 8.1.6. Let \( {ABC} \) be an equilateral triangle. The points \( N \) and \( M \) trisect the side \( {BC} \) ( \( N \) is between \( C \) and \( M \) ). The points \( E \) and \( F \) trisect the outside arc \( \overset{⏜}{BC} \) of the circle with diameter \( {BC} \) ( \( E \) lies between \( C \) and \( F \)... | Solution. We will only prove that the points \( A, M \) and \( F \) are collinear (the proof for the other triple is analogous).\n\nLet \( {AF} \cap {BC} = {M}^{\prime } \) . We aim to prove that \( M \equiv {M}^{\prime } \) . If \( D \) is the midpoint of \( {BC} \) (i.e. the center of the circle), then \( \overset{⏜}... | Yes |
Problem 8.1.7. Let \( {ABC} \) be an equilateral triangle. Let \( E \) be a point such that \( A \) is the circumcenter of \( \bigtriangleup {BEC} \) . Let the circumradius of \( \bigtriangleup {BEC} \) be \( R \) . The circumcircle of \( \bigtriangleup {ABE} \) intersects the segment \( {CE} \) at the point \( F \) . ... | Solution. Since \( A \) is the circumcenter of \( \bigtriangleup {BEC} \), we get\n\n\[ \angle {BCF} = \frac{\angle {BAE}}{2} = \frac{\angle {BFE}}{2}. \]\n\nTherefore,\n\n\[ \angle {BCF} = \angle {CBF} \Rightarrow {CF} = {FB}\text{.} \]\n\nWe also have that\n\n\[ \angle {BEF} = \angle {BEC} = \frac{\angle {BAC}}{2} = ... | Yes |
Let \( {ABC} \) be an equilateral triangle. The point \( D \) lies inside of the triangle. The feet of the perpendiculars from \( D \) to the sides \( {BC},{AC} \) and \( {AB} \) are \( {A}_{1},{B}_{1} \) and \( {C}_{1} \), respectively. Prove that \( A{C}_{1} + B{A}_{1} + C{B}_{1} = {C}_{1}B + {A}_{1}C + {B}_{1}A. \) | Denote\n\n\[ \n{AB} = {BC} = {CA} = x, \n\]\n\n\[ \nB{C}_{1} = c \n\]\n\n\[ \nC{A}_{1} = a \n\]\n\n\[ \nA{B}_{1} = b\text{.} \n\]\n\nCarnot's Theorem (Problem 4.9.16) yields\n\n\[ \n{a}^{2} + {b}^{2} + {c}^{2} = {\left( x - a\right) }^{2} + {\left( x - b\right) }^{2} + {\left( x - c\right) }^{2} \n\]\n\n\[ \n\Rightarro... | Yes |
Problem 8.1.10. Let \( {ABO},{CDO} \) and \( {EFO} \) be congruent equilateral triangles, as shown in the figure. The circumcircles of \( \bigtriangleup {FCO} \) and \( \bigtriangleup {EOB} \) intersect at the points \( O \) and \( P \) . The circumcircles of \( \bigtriangleup {FCO} \) and \( \bigtriangleup {AOD} \) in... | Solution. Let the circumcenters of \( \bigtriangleup {BOE},\bigtriangleup {AOD} \) and \( \bigtriangleup {COF} \) be \( {O}_{1},{O}_{2} \) and \( {O}_{3} \), respectively. Consider the symmetry with respect to the line \( {\mathrm{O}}_{2}{\mathrm{O}}_{3} \) . It is clear that \( N \rightarrow O \) . Let \( D \rightarro... | Yes |
Problem 8.1.11. Let \( {ABC} \) be an equilateral triangle. The equilateral triangles \( \bigtriangleup {CED},\bigtriangleup {AFL} \) and \( \bigtriangleup {BQP} \) have the same orientation as \( \bigtriangleup {ABC} \) . Let \( M, K \) and \( N \) be the midpoints of \( {DF},{EP} \) and \( {LQ} \) , respectively. Pro... | Solution. We will use vector rotation. We have\n\n\[ \overrightarrow{MK} = \frac{\overrightarrow{DE} + \overrightarrow{FP}}{2} \]\n\n\[ = \frac{\overrightarrow{DE} + \overrightarrow{FA} + \overrightarrow{AB} + \overrightarrow{BP}}{2}\text{. } \]\n\nWe rotate each vector in this equality by \( + {60}^{ \circ } \), so it... | Yes |
Problem 8.1.12. Two perpendicular lines \( x \) and \( y \) intersect each other at the point \( O \) . Let \( k \) be a circle with center \( I \) that passes through \( O \) . The points \( A \in x \) and \( B \in y \) are chosen, such that the points \( I \) and \( A \) lie in the same half-plane with respect to the... | Solution. We will use the notations from the figure. Denote \( \angle {AOI} = \alpha \) , \( \angle {OCD} = \varphi \) and \( \angle {BAO} = \psi \) . The triangles \( \bigtriangleup {COD} \) and \( \bigtriangleup {AOB} \) are right-angled and hence \( {CY} = {YD} = {YO} \) and \( {AX} = {XB} = {XO} \) . Therefore,\n\n... | Yes |
Problem 8.1.13. Let \( {ABC} \) be an equilateral triangle. The point \( D \) lies on the smaller arc \( \overset{⏜}{AC} \) of the circumcircle of \( \bigtriangleup {ABC} \), and the point \( E \) bisects the smaller arc \( \overset{⏜}{AB} \) . Let \( P \) be the foot of the perpendicular from \( E \) to \( {BD} \), an... | Solution. If \( \angle {DBC} = \alpha \), then\n\n\n\n\( \angle {PEB} = \alpha \) . Let \( R \) be the circumradius of \( \bigtriangleup {ABC} \) . The law of sines yields that\n\n\[ \n{CF} = \frac{CD}{2} = \frac{{2R... | Yes |
Problem 8.1.14. Let \( {ABC} \) be an equilateral triangle and let \( l \) be an arbitrary line through the vertex \( C \) . Let \( P \) and \( Q \) be the feet of the perpendiculars from \( A \) and \( B \) to \( l \), respectively. If \( S \) is the midpoint of \( {AB} \), prove that \( \bigtriangleup {PQS} \) is equ... | Solution. Let \( M \) and \( N \) be the midpoints of \( {BC} \) and \( {AC} \), respectively. Considering the mid-segments and the fact that \( \bigtriangleup {BQC} \) and \( \bigtriangleup {APC} \) are right-angled, we get that \( {MQ} = {MB} = {MC} = \) \( {MS} = {SN} = {AN} = {NC} = {NP} \) . Moreover, \[ \angle {C... | Yes |
Problem 8.1.15. Let \( {ABC} \) be a triangle. The equilateral triangles \( {AB}{C}_{1},{A}_{1}{BC} \) and \( A{B}_{1}C \) are constructed externally to its sides. These three triangles have circumcenters \( {O}_{3},{O}_{2} \) and \( {O}_{1} \), respectively. Prove that \( \bigtriangleup {O}_{1}{O}_{2}{O}_{3} \) is equ... | Solution. Note that \( \angle {\mathrm{O}}_{3}{\mathrm{{BO}}}_{2} = \) \( \angle {AB}{A}_{1} \) . Also, \( \bigtriangleup {AB}{O}_{3} \sim \bigtriangleup B{A}_{1}{O}_{2} \) . Hence, \( \frac{B{O}_{3}}{B{O}_{2}} = \frac{BA}{B{A}_{1}} \) . It follows that \( \bigtriangleup {AB}{A}_{1} \sim \bigtriangleup {O}_{3}B{O}_{2} ... | Yes |
Problem 8.1.16. Let \( {ABC} \) be a triangle. The equilateral triangles \( {ABF},{ACD} \) and \( {BCE} \) are constructed externally. Let \( M \) and \( N \) be the midpoints of \( {DF} \) and \( {EF} \), respectively. Prove that \( \bigtriangleup {MNC} \) is equilateral. | Solution. Let \( {M}^{\prime } \) be the point such that \( \bigtriangleup {NC}{M}^{\prime } \) is equilateral and has the same orientation as \( \bigtriangleup {NCM} \) . Then we have\n\n\[ \overrightarrow{CN} = \frac{1}{2}\left( {\overrightarrow{CE} + \overrightarrow{CF}}\right) = \frac{\overrightarrow{CE} + \overrig... | Yes |
Problem 8.1.17. Let \( {ABC} \) be a triangle. The equilateral triangles \( {BC}{A}_{1} \) and \( A{B}_{1}C \) are constructed externally to its sides. Let \( R, S \) and \( M \) be the midpoints of \( {B}_{1}C,{A}_{1}C \) and \( {AB} \), respectively. Prove that \( \bigtriangleup {MSR} \) is equilateral. | Solution. Let \( P \) and \( Q \) be the midpoints of \( {BC} \) and \( {AC} \) , respectively. Considering the midsegments, we get \( {MQ}\parallel {BC} \) and \( {MP}\parallel {AC} \), which implies that \( \angle {MQC} = \angle {MPC} \) . On the other hand, \( {RQ} \) and \( {SP} \) are mid-segments in \( \bigtriang... | Yes |
Problem 8.1.20. Let \( {ABCD} \) be a square. The equilateral triangles \( {ABP} \) and \( {ADQ} \) are constructed internally. Prove that \( \bigtriangleup {CPQ} \) is equilateral. | Solution. We have that \( \angle {PAD} = \)\n\n\n\n\( {30}^{ \circ },{AP} = {AD},\angle {PBC} = {30}^{ \circ } \) and \( {BP} = {BC} \) . It follows that \( \bigtriangleup {PAD} \cong \bigtriangleup {PBC} \) and henc... | Yes |
Problem 8.1.21. Let \( {ABCD} \) be a square with center \( O \) . The equilateral triangle \( {ABE} \) is constructed internally. Let \( M \) and \( N \) be the midpoints of \( E\bar{D} \) and \( {EC} \), respectively. Prove that \( \bigtriangleup {OMN} \) is equilateral. | Solution. We have that\n\n\n\n\( {AE} = {BE} = {AB} = {CD} \) . On the other hand, \( {OM} \) is a midsegment in \( \bigtriangleup {BED} \) . Thus, \( {OM} = \frac{BE}{2} \) .\n\nAlso, \( {ON} \) is a midsegment in \... | Yes |
Problem 9.1. Let \( {ABC} \) be a triangle. The equilateral triangles \( {AB}{C}_{1},{A}_{1}{BC} \) and \( A{B}_{1}C \) are constructed externally. These three triangles have circumcenters \( {O}_{1},{O}_{2} \) and \( {O}_{3} \), respectively. Prove that the lines \( A{O}_{2} \) , \( B{O}_{3} \) and \( C{O}_{1} \) are ... | Solution. It is easy to see\n\n\n\nthat\n\n\[ \angle {O}_{1}{AB} = \angle {O}_{1}{BA} = {30}^{ \circ }, \]\n\n\[ \angle {O}_{2}{BC} = \angle {O}_{2}{CB} = {30}^{ \circ }, \]\n\n\[ \angle {O}_{3}{CA} = \angle {O}_{3}{... | Yes |
Problem 9.3. Let \( {ABC} \) be a triangle. The equilateral triangles \( {AB}{C}_{1},{A}_{1}{BC} \) and \( A{B}_{1}C \) are constructed externally. Prove that \( A{A}_{1} = \) \( B{B}_{1} = C{C}_{1} \) . | Solution. Note that \( {AB} = \) \( B{C}_{1},{BC} = B{A}_{1} \) and \( \angle {AB}{A}_{1} = \) \( \angle {C}_{1}{BC} \) . These equalities imply that \[ \bigtriangleup {AB}{A}_{1} \cong \bigtriangleup {C}_{1}{BC}. \] Hence, \( C{C}_{1} = A{A}_{1} \) . Analogously, we get that \( A{A}_{1} = B{B}_{1} \) . | Yes |
Problem 9.4. Let \( {ABCD} \) be a parallelogram. The equilateral triangles \( {ABF},{BCP},{CDE} \) and \( {DAK} \) are constructed externally. Let \( O \) be the center of \( \bigtriangleup {DAK} \), and let \( M \) and \( N \) be the midpoints of \( {DE} \) and \( {AF} \) , respectively. Prove that \( {OP} \bot {MN} ... | Solution. It suffices to prove that \( \overrightarrow{NM} \cdot \overrightarrow{PO} = 0 \) . Denote \( \angle {BAD} = \alpha \) and \( {AB} = a,{AD} = b \) . We consecutively have\n\n\[ \overrightarrow{NM}.\overrightarrow{PO} = \left( {\overrightarrow{NA} + \overrightarrow{AD} + \overrightarrow{DM}}\right) \left( {\ov... | Yes |
Problem 9.5. Let \( {ABC} \) be a triangle. The squares \( {ACDE} \) and \( {BCFP} \) are constructed externally. Prove that the squares with diagonals \( {AB} \) and \( {DF} \) have a common vertex. | Solution. Let \( X \) be a point such that \( \angle {AXB} = {90}^{ \circ } \) and \( {AX} = {BX} \) . Let \( C \) and \( X \) lie in the same half-plane with respect to \( {AB} \) . It suffices to prove that \( {DX} = {FX} \) and \( \angle {DX}\dot{F} = {90}^{ \circ } \) . Note that \( \angle {XBF} = \angle {XBC} + {4... | Yes |
Problem 9.6. Let \( {ABC} \) be a triangle. The squares \( {ACFK} \) and \( {BCDE} \) are constructed externally. Let \( P \) and \( Q \) be the centers of \( {ACFK} \) and \( {BCDE} \), respectively. The points \( M \) and \( N \) are the midpoints of \( {AB} \) and \( {FD} \), respectively. Prove that the quadrilater... | Solution. We have that \( {CF} = {CA},{CB} = {CD} \) and \( \angle {ACD} = {90}^{ \circ } + \) \( \angle {ACB} = \angle {BCF} \), which implies that \( \bigtriangleup {BCF} \cong \bigtriangleup {DCA} \) . Therefore, the angle between the lines \( {AD} \) and \( {BF} \) is equal to the angle between the lines \( {AC} \)... | Yes |
Problem 9.7. Let \( {ABC} \) be a triangle. The squares \( {ACMN} \) and \( {BCQP} \) are constructed externally. Let \( S \) be the midpoint of \( {AB} \) . Prove that \( {CS} \bot {QM} \) . | Solution. We have that \( \angle {MCQ} = {180}^{ \circ } - \gamma \)\n\nLet \( D \) be the symmetric point of \( C \) with respect to the point \( S \) . Then \( {ACBD} \) is a parallelogram. Thus, \( {BD} = {AC} = \) \( {MC} \) and \( \angle {CBD} = {180}^{ \circ } - \gamma = \) \( \angle {MCQ} \) .\n\nWe deduce that ... | Yes |
Problem 9.8. Let \( {ABC} \) be a triangle. The squares \( {BCED} \) and \( {ACFP} \) are constructed externally. Prove that the lines \( {AD},{BP} \) and the altitude from \( C \) to \( {AB} \) are concurrent. | Solution. This problem is a special case of Problem 4.12.7 \( \left( {\varphi = {45}^{ \circ }}\right) \). | Yes |
Problem 9.9. Let \( {ABC} \) be a triangle. The squares \( {ABMN},{BPQC} \) and \( {ACRS} \) are constructed externally. Let \( {O}_{1} \) be the center of \( {ABMN} \) . Prove that the lines \( C{O}_{1},{BR},{AQ} \) and \( {PS} \) are concurrent. | Solution. Let \( {AQ} \cap {BR} = X \) . Then \( \bigtriangleup {BCR} \cong \bigtriangleup {QCA} \) due to the equalities \( \angle {BCR} = \angle {QCA},{BC} = {QC} \) and \( {AC} = {CR} \) . We deduce that \( \angle {CAQ} = \angle {CRB} \), which implies that the quadrilateral \( {CRAX} \) is cyclic. Thus, the pentago... | Yes |
Problem 9.10. Let \( {ABC} \) be a triangle. The squares \( {BCDK} \) and \( {ACEF} \) are constructed externally. Prove that the perpendiculars from \( K \) to \( {AC} \), from \( F \) to \( {BC} \) and from \( C \) to \( {AB} \) are concurrent. | Solution. Denote the foot of the perpendicular from \( K \) to the line \( {AC} \) by \( M \), and the feet of the perpendicular from \( F \) to the line \( {BC} \) by \( P \) . Let \( {KM} \cap {FP} = Q \) and \( {QC} \cap {AB} = H \) . We will prove that \( {CH} \) is an altitude in \( \bigtriangleup {ABC} \) . It su... | Yes |
Problem 9.11. Let \( {ABDC},{AKFE} \) and \( {FNQP} \) be squares with the same orientation. The midpoint of \( {BP} \) is \( M \) . Prove that \( {CN} \bot {KM} \) . | Solution. We will use vector rotation. We have\n\n\[ \overrightarrow{KM} = \frac{1}{2}\left( {\overrightarrow{KB} + \overrightarrow{KP}}\right) = \frac{1}{2}\left( {\overrightarrow{KA} + \overrightarrow{AB} + \overrightarrow{KF} + \overrightarrow{FP}}\right) .\n\]\n\n\( \mathrm{A} + {90}^{ \circ } \) rotation of each v... | Yes |
Problem 9.13. Let \( {OAPE},{OBLD} \) and \( {OCKF} \) be squares with a common vertex \( O \) and the same orientation. Let their centers be \( {O}_{1},{O}_{2} \) and \( {O}_{3} \), respectively. Denote the midpoints of \( {EB},{CD} \) and \( {AF} \) by \( Q \) , \( N \) and \( M \), respectively. Prove that the lines... | Solution. We will use complex numbers. Denote the center of the complex plane by \( O \) . As usual, the complex number of each point will be denoted as the small letter of its name. Let \( a, b \) and \( c \) be arbitrary. We have that \( d = {bi}, f = {ci} \) and \( e = {ai} \) . Moreover, \( {O}_{1} \) is the midpoi... | Yes |
Problem 9.14. Let \( {ABCD} \) be a parallelogram. The squares \( {ABNM} \) , \( {BPQC},{DCRS} \) and \( {ADTU} \) are constructed externally. Their centers are \( {O}_{1},{O}_{2},{O}_{3} \) and \( {O}_{4} \), respectively. Prove that the quadrilateral \( {O}_{1}{O}_{2}{O}_{3}{O}_{4} \) is a square. | Solution. The quadrilateral \( {ABCD} \) is a parallelogram, and so the squares \( {ABNM} \) and \( {CDSR} \) are congruent.\n\nAngle chasing gives that \( \angle {O}_{1}B{O}_{2} = {270}^{ \circ } - \angle {ABC} = \) \( {90}^{ \circ } + \angle {BCD} = \angle {O}_{2}C{O}_{3} \).\n\nMoreover, \( B{O}_{1} = C{O}_{3} \) an... | Yes |
Let \( {ABCD} \) be a quadrilateral. The squares \( {ABEF} \) , \( {BCKL},{CDMN} \) and \( {ADPQ} \) are constructed externally. Their centers are \( {O}_{1},{O}_{2},{O}_{3} \) and \( {O}_{4} \), respectively. Prove that \( {O}_{1}{O}_{3} \bot {O}_{2}{O}_{4} \) and that \( {O}_{1}{O}_{3} = {O}_{2}{O}_{4} \) | Solution. Firstly, we will prove that \( \overrightarrow{KC} + \overrightarrow{NC} = \overrightarrow{AQ} + \overrightarrow{AF} \) . Consider the equality \( \overrightarrow{BC} + \overrightarrow{CD} + \overrightarrow{DA} + \overrightarrow{AB} = \overrightarrow{0} \) . A \( - {90}^{ \circ } \) rotation of each vector in... | Yes |
Problem 9.16. Let \( {ABCD} \) and \( {A}_{1}{B}_{1}{C}_{1}{D}_{1} \) be squares with the same orientation. Let the midpoints of \( A{A}_{1}, B{B}_{1}, C{C}_{1} \) and \( D{D}_{1} \) be \( {A}_{2},{B}_{2},{C}_{2} \) and \( {D}_{2} \), respectively. Prove that the quadrilateral \( {A}_{2}{B}_{2}{C}_{2}{D}_{2} \) is a sq... | Solution. We will apply Problem 4.12.9 twice. Applying it to \( \bigtriangleup {ABD} \) and \( \bigtriangleup {A}_{1}{B}_{1}{D}_{1} \) gives that \( \bigtriangleup {A}_{2}{B}_{2}{D}_{2} \) is isosceles and right-angled with \( \angle {B}_{2}{A}_{2}{D}_{2} = {90}^{ \circ } \) . Also, applying it to \( \bigtriangleup {CB... | Yes |
Problem 9.17. Let \( {ABCD} \) be a convex quadrilateral, such that \( {AC} \cap {BD} = E \) and \( {AC} \bot {BD} \) . The squares \( {ADMN} \) and \( {BCQP} \) are constructed externally. Their centers are \( {O}_{1} \) and \( {O}_{2} \), respectively. Prove that the points \( {O}_{1},{O}_{2} \) and \( E \) are colli... | Solution. Note that \( \angle {AED} + \angle A{O}_{1}D = {90}^{ \circ } + {90}^{ \circ } = {180}^{ \circ } \), which means that the quadrilateral \( {AED}{O}_{1} \) is cyclic. Analogously, the quadrilateral \( B{O}_{2}{CE} \) is also cyclic.\n\nNow \( \angle {O}_{1}E{O}_{2} = \angle {O}_{1}{EA} + \angle {AEB} + \angle ... | Yes |
Let \( {ABC} \) be a triangle. The squares \( {ABNM},{BPQC} \) and \( {ACRS} \) are constructed externally. Their centers are \( {O}_{1},{O}_{2} \) and \( {O}_{3} \) , respectively. Prove that the lines \( A{O}_{2}, B{O}_{3} \) and \( C{O}_{1} \) are concurrent. | Since \( {O}_{1} \) , \( {\mathrm{O}}_{2} \) and \( {\mathrm{O}}_{3} \) are the centers of \( {ABNM},{BPQC} \) and \( {ACRS} \), respectively, the measure of each of the marked angles in the figure equals \( {45}^{ \circ } \) . Now Problem 4.9.5 implies that the lines \( A{O}_{2} \) , \( B{O}_{3} \) and \( C{O}_{1} \) ... | Yes |
Problem 9.19. Let \( {ABC} \) be a triangle. The rectangles \( {ABLK} \) , \( {ACFE} \) and \( {BCDQ} \) are constructed externally to its sides. Prove that the perpendicular bisectors of the segments \( {EK},{LQ} \) and \( {FD} \) are concurrent. | Solution. Denote the midpoints of \( {EK},{LQ} \) and \( {FD} \) by \( M, N \) and \( P \), respectively. Also, let \( {O}_{1},{O}_{2} \) and \( {O}_{3} \) be the circumcenters of \( \bigtriangleup {EKA} \) , \( \bigtriangleup {LQB} \) and \( \bigtriangleup {FDC} \), respectively. We have that \( {O}_{2}N \bot {LQ},{O}... | Yes |
Let \( {ABC} \) be a triangle such that \( \angle {ACB} < {90}^{ \circ } \) . The rectangles \( {ACDE} \) and \( {BCFP} \) are constructed externally, such that \( {CD} = {CB} \) and \( {CF} = {CA} \) . Prove that the lines \( {AF},{BD} \) and \( {EP} \) are concurrent. | Let \( {AF} \cap {BD} = Q \) . It suffices to prove that the points \( E, Q \) and \( P \) are collinear. Denote \( \angle {ACB} = \gamma \) .\n\nThen \( \angle {BCD} = \angle {ACF} = {90}^{ \circ } + \gamma ,{AC} = {CF} \) and \( {BC} = {CD} \) . Therefore,\n\n\[ \angle {CBQ} = \angle {CDQ} = \angle {CFQ} = \angle {CA... | Yes |
Let \( {ABC} \) be a triangle. The isosceles triangles \( {BCF} \) \( \left( {{BF} = {CF}}\right) \) and \( {AEC}\left( {{AE} = {CE}}\right) \) are constructed externally, such that \( \angle {ACE} = \angle {BCF} \) . The point \( D \) lies inside of \( \bigtriangleup {ABC} \), such that \( {AD} = {BD} \) and \( \angle... | The problem conditions imply that \( \bigtriangleup {ABD} \sim \bigtriangleup {BCF} \sim \) \( \bigtriangleup {ACE} \) . Then \( \frac{AE}{AD} = \frac{AC}{AB} \) .\n\nOn the other hand, we have \( \angle {DAE} = \angle {BAC} \), which implies that \( \bigtriangleup {ABC} \sim \bigtriangleup {ADE} \) .\n\n![56360f5e-030... | Yes |
Let \( {ABC} \) be a triangle. The points \( D, G \in {AC} \) , \( E, H \in {BC} \) and \( F, I \in {AB} \) are such that \( \widetilde{DI} \parallel {BC},\widetilde{IH} \parallel {AC},{HG} \parallel {AB} \) , \( {GF}\parallel {BC} \) and \( {FE}\parallel {AC} \) . Prove that \( {DE}\parallel {AB} \) . | Applying the intercept theorem, we get\n\n\[ \frac{CD}{DA} = \frac{BI}{IA} = \frac{B \cdot H}{HC} \]\n\n\[ = \frac{AG}{GC} = \frac{AF}{FB} = \frac{CE}{EB} \]\n\nBy the same theorem, we conclude that \( {DE}\parallel {AB} \) . | Yes |
Problem 10.2. Let \( {ABC} \) be a triangle. The points \( D, G \in {AB} \) , \( E, H \in {BC} \) and \( F, K \in {AC} \) are such that the quadrilaterals \( {ACED} \) , \( {BEFA},{BCFG},{ACHG} \) and \( {ABHK} \) are cyclic. Prove that the quadrilateral \( {BCKD} \) is cyclic. | Solution. Let the reflection of \( A \) with respect to \( {BC} \) be \( {A}^{\prime } \) . Let the reflection of \( B \) with respect to \( {A}^{\prime }C \) be \( {B}^{\prime } \) . Let the reflection of \( C \) with respect to \( {A}^{\prime }{B}^{\prime } \) be \( {C}^{\prime } \) . Let the reflection of \( {A}^{\p... | Yes |
Problem 10.3. Let \( {ABC} \) be a triangle. The points \( D, G \in {BC} \) , \( E, H \in {AB} \) and \( F, I \in {AC} \) are chosen, such that \( {AI} = {AH},{BH} = {BG} \) , \( {CG} = {CF},{AF} = {AE} \) and \( {BE} = {BD} \) . Prove that \( {CI} = {CD} \) . | Solution. Set \( {CI} = x \) . Then \( {AH} = {AI} = b - x,{BG} = {BH} = c - {AH} = \) \( c - b + x,{CF} = {CG} = a - {BG} = a - c + b - x,{AE} = {AF} = b - {CF} = c - a + x, \) \( {BD} = {BE} = c - {AE} = a - x \) and finally \( {CD} = x \) . \n\nTherefore, \( {CI} = x = {CD} \) . | Yes |
Problem 10.4. Let \( {ABC} \) be a triangle. The point \( D \) lies on the ray \( C{A}^{ \rightarrow } \) beyond \( A \) . Two arbitrary lines through \( D \) intersect \( {AB} \) at the points \( E \) and \( I \), and \( {BC} \) at the points \( F \) and \( K \), respectively. The point \( H \) lies on the ray \( A{B}... | Solution. We will apply Menelaus’ Theorem to \( \bigtriangleup {ABC} \) a few times. We have \[ \frac{CD}{DA} = \frac{{EB} \cdot {FC}}{{AE} \cdot {BF}} = \frac{{IB} \cdot {KC}}{{AI} \cdot {BK}} \] Also, \[ \frac{BK}{KC} = \frac{BH}{HA} \cdot \frac{AL}{LC} \] and \[ \frac{BF}{FC} = \frac{BH}{HA} \cdot \frac{AG}{GC} \] T... | Yes |
Let \( {ABC} \) be a triangle. The points \( X \) lies inside of \( \bigtriangleup {ABC} \) . The points \( D, E \) and \( F \) are the midpoints of \( {BC},{AC} \) and \( {AB} \), respectively. The point \( H \) lies on the ray \( X{F}^{ \rightarrow } \) beyond \( F \) . Let \( {HB} \cap {XD} = I,{IC} \cap {XE} = J,{J... | Menelaus’ Theorem, applied to \( \bigtriangleup {KXJ} \) and the line \( {MAH} \) , implies that it suffices to prove that\n\n\[ \frac{XM}{MJ} \cdot \frac{JA}{AK} \cdot \frac{KH}{HX} = 1 \]\n\nThe same theorem, applied to \( \bigtriangleup {JIX} \) and the line \( {MCL} \), yields \( \frac{XM}{MJ} = \frac{XL}{LI} \cdot... | Yes |
Problem 10.6. Let \( {m}^{ \rightarrow },{n}^{ \rightarrow } \) and \( {p}^{ \rightarrow } \) be three rays with a common origin \( X \) . Let \( A \in m \) and \( B \in n \) . The line symmetric to \( {AB} \) with respect to \( n \) intersects \( p \) at the point \( C \) . The line symmetric to \( {BC} \) with respec... | Solution. Let the line symmetric to \( {AB} \) with respect to \( m \) intersect \( p \) at the point \( {F}^{\prime } \) . It suffices to show that \( {F}^{\prime } \equiv F \) .\n\nRecall that a point on an angle bisector is equidistant from the arms of the angle. We have\n\n\[ \operatorname{dist}\left( {X, A{F}^{\pr... | Yes |
Problem 10.7. Let \( {m}^{ \rightarrow },{n}^{ \rightarrow } \) and \( {p}^{ \rightarrow } \) be three rays with a common origin \( X \) . Let \( A \in m \) and \( B \in n \) . The point \( C \in p \) satisfies \( \angle {BAX} = \) \( \angle {BCX} \), the point \( D \in m \) satisfies \( \angle {CBX} = \angle {CDX} \),... | Solution. Let \( {F}^{\prime } \in p \) be such that \( \angle {XBA} = \angle X{F}^{\prime }A \) .\n\nIt suffices to show that the points \( F \) and \( {F}^{\prime } \) coincide. We have\n\n\[ {180}^{ \circ } = {540}^{ \circ } - \angle {AXB} - \angle {BXC} - \angle {CXA} \]\n\n\[ = \angle X{F}^{\prime }A + \angle {XA}... | Yes |
Problem 10.8. Let the points \( {A}_{1},{A}_{2},{B}_{1},{B}_{2},{C}_{1} \) and \( {C}_{2} \) lie on a circle \( k \) in this order. Let \( {k}_{1} \) be a circle through \( {A}_{1} \) and \( {A}_{2} \) . The circles \( {k}_{2} \) , \( {k}_{3},{k}_{4},{k}_{5} \) and \( {k}_{6} \) are constructed as follows: \( {\bar{k}}... | Solution. Consider an inversion centered at \( {A}_{1} \) with an arbitrary radius. Now the problem can be reformulated as follows:\n\nLet \( l,{l}_{1} \) and \( {l}_{2} \) be three rays with a common origin \( {A}_{2} \) . The points \( {B}_{1},{B}_{2},{C}_{1} \) and \( {C}_{2} \) lie on \( l \) . The points \( L \) a... | Yes |
Problem 10.9. Let \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) be three circles intersecting at the points \( G, H \) and \( I \), as shown in the figure. The point \( O \) lies on the smaller arc \( \overset{⏜}{GH} \) of \( {k}_{1} \) . The point \( N \in {k}_{2} \) is such that the quadrilateral \( {HONI} \) is cyclic, as... | Solution. Consider an inversion centered at \( H \) with an arbitrary radius. The problem is now equivalent to the following one:\n\nThe points \( {M}^{\prime },{K}^{\prime } \) and \( {I}^{\prime } \) lie on a line in this order, and the points \( {L}^{\prime },{O}^{\prime } \) and \( {G}^{\prime } \) lie on another l... | Yes |
Problem 10.10. Let \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) be circles that intersect each other at the points \( X, Y, Z, M, L \) and \( K \), as shown in the figure. Let \( A \in {k}_{1} \) . The line \( {AM} \) intersects \( {k}_{2} \) for the second time at the point \( B \) . The line \( {BL} \) intersects \( {k}_{... | Solution. We have that \( \angle {DAK} = \angle {KZC} = {180}^{ \circ } - \angle {CEK} \), which implies that \( {EC} \parallel {AD} \) . Moreover, \( \angle {FBL} = \angle {EXL} = {180}^{ \circ } - \angle {ECL} \) and thus \( {BF}\parallel {EC} \) . Hence, \( {AD}\parallel {BF} \) . Now we have \( \angle {FYM} = \angl... | Yes |
Let \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) be three circles. The three radical axes are drawn and the point \( A \) is chosen on one of them (say on \( l \) - the radical axis of \( {k}_{1} \) and \( {k}_{2} \) ). One of the tangent lines to \( {k}_{1} \) that passes through \( A \) is drawn - the one that does not in... | We will use the notations on the figure. We have\n\n\[ {CB} + {ND} = {CH} + {HB} + {NX} + {DX} = {CM} + {BK} + {NK} + {DM} = {CD} + {BN}. \]\n\n\n\nTherefore, the quadrilateral \( {CBND} \) is circumscribed around a ... | Yes |
Problem 10.12. Let \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) be three circles, such that their disks do not intersect each other. Let the external homothetic centers of the three pairs of circles be \( A, B \) and \( C \), as shown in the figure. A line passing through \( B \) intersects \( {k}_{1} \) and \( {k}_{2} \) a... | Solution. We will solve the problem by applying Problem 6.5.7 six times and the analog of Problem 10.11 once (considering the other external common tangent lines of the circles). The proof of this analog is identical to that of Problem 10.11, so we will not describe it here.\n\n![56360f5e-0308-4b19-bbc8-48cfc211c0a8_46... | No |
Problem 10.15. Let \( {ABC} \) be a triangle. The circles \( {k}_{1},{k}_{2},{k}_{3},{k}_{4},{k}_{5} \) and \( {k}_{6} \) are internal for \( \bigtriangleup {ABC} \) and are defined as follows. The circle \( {k}_{1} \) is inscribed in \( \angle {BAC} \) . The circle \( {k}_{2} \) is inscribed in \( \angle {ABC} \) and ... | Solution. Let \( {k}_{7} \) be the circle inside of \( \bigtriangleup {ABC} \), which is inscribed in \( \angle {BAC} \) and is externally tangent to \( {k}_{6} \) . It suffices to show that \( {k}_{1} \equiv {k}_{7} \) . Let \( l, m, n,{l}^{\prime },{m}^{\prime },{n}^{\prime } \) and \( {l}^{\prime \prime } \) be the ... | Yes |
Lemma 3. In the setting of Problem 6.10.22, define the circle \( {k}_{3} \) as the one that touches the pair of lines \( \left( {{AD},{BC}}\right) \) at their points of intersection with \( l \) (you can prove that \( l \) forms equal angles with this pair by angle chasing). Prove that \( {k}_{3},{k}_{2},{k}_{1} \) and... | Proof. Analogous to the one of 6.10.22. | No |
Lemma 4. The point \( A \) lies on the circle \( k \) . Let \( {AB} \) and \( {AC} \) be lines that touch \( {\omega }_{1} \) and \( {\omega }_{2} \) - circles inside of \( k \), which are coaxial (have a common radical axis) with \( k\left( {B \in k\text{and}C \in k}\right) \) . Then if we vary \( A \) along \( k \), ... | Proof. Define the points \( {A}^{\prime },{B}^{\prime } \) and \( {C}^{\prime } \) as the points \( A, B \) and \( C \) for a different placement of \( A \) on \( k \) . Applying Lemma 3 to the quadrilateral \( A{A}^{\prime }B{B}^{\prime } \) gives that \( A{A}^{\prime } \) and \( B{B}^{\prime } \) are tangent to a cir... | Yes |
Theorem 1 If \( \left( {ABC}\right) ,\left( {ACD}\right) \) and \( \left( {ABD}\right) \), then \( \left( {BCD}\right) \) and therefore, (ABCD). | Proof By hypothesis and the theorem of Section 1.6, (ACD) implies \( 0 < {CD} = {AD} - {AC} < {AD} \leq \alpha \) and \( \overrightarrow{CD} = \overrightarrow{AB} \) so that \( A, B, C \), and \( D \) are collinear. We also obtain directly from (ABD)\n\n\[{BD} = {AD} - {AB}\]\n\nSubstituting \( {AC} + {CD} \) for \( {A... | Yes |
Theorem 2 If the triangle inequality is valid, then (ABC) and (ACD) imply (ABCD). | Proof In view of Theorem 1, it suffices to prove (ABD). As in the proof of Theorem 1, points \( A, B, C \), and \( D \) are collinear. By the triangle inequality,\n\n\[ \n{AD} \leq {AB} + {BD} \n\]\n\nIf we can prove that \( {AD} \geq {AB} + {BD} \), then equality would follow and \( \left( {ABD}\right) \) would result... | Yes |
Theorem 4 Suppose that \( \left( {PAQ}\right) \) and \( l = \overrightarrow{AP} \cup \overrightarrow{AQ} \), with \( {PQ} < \alpha \) . For any two interior points \( B \in \overrightarrow{AP} \) and \( C \in \overrightarrow{AQ} \), then\n\n(a) if \( {AB} + {AC} \leq \alpha ,\left( {BAC}\right) \) holds, or\n\n(b) if \... | Proof By Theorem 3, \( \overrightarrow{AB} = \overrightarrow{AP} \) and \( \overrightarrow{AC} = \overrightarrow{AQ} \) . It follows that neither \( \left( {ABC}\right) \) nor \( \left( {ACB}\right) \) can hold, for otherwise, \( \overrightarrow{AB} = \overrightarrow{AC} \) and \( \overrightarrow{AP} = \overrightarrow{... | Yes |
Theorem 3 Let \( B \) and \( F \) lie on opposite sides of a line \( l \) and let \( A \) and \( G \) be any two distinct points on \( l \) such that \( {AG} < \alpha \) . Then, any segment \( \overline{GB} \) and any ray \( \overrightarrow{AF} \) are disjoint sets (have no points in common). (See Figure 2.9.) | Proof Suppose \( B \in {\mathbf{H}}_{1} \) and \( F \in {\mathbf{H}}_{2} \) where \( {\mathbf{H}}_{1} \) and \( {\mathbf{H}}_{2} \) are the half planes determined by \( l \) . By Theorem 1, the interior points of any segment having endpoints \( B \) and \( G \) lie in \( {\mathbf{H}}_{1} \), and by Theorem 2 those of a... | Yes |
Theorem 3 If rays \( u \) and \( v \) are concurrent with ray \( h \) and lie on the same side of the line containing \( h \), then either ( \( {hvu} \) ) or ( \( {huv} \) ). | Proof Let \( \mathbf{H} \) be the half plane determined by \( h \) that contains \( u \) and \( v \) (except for the endpoint) let \( {\mathbf{H}}_{1} \) and \( {\mathbf{H}}_{2} \) be those determined by the line \( u \cup {u}^{\prime } \), with \( h \) in \( {\mathbf{H}}_{1} \) and \( {h}^{\prime } \) in \( {\mathbf{H... | Yes |
Theorem 7 For any three distinct, concurrent rays \( h, k \), and \( u \) , either \( \left( {huk}\right) ,\left( {hku}\right) ,\left( {khu}\right) \), or \( \left( {h{u}^{\prime }k}\right) \), where \( {u}^{\prime } \) is the opposite ray of \( u \) . | [Hint: Show by applications of the plane separation postulate and definition of angle interior that a point \( W \) on ray \( u \) lies in the interior of either \( \angle {hk},\angle h{k}^{\prime },\angle {h}^{\prime }{k}^{\prime } \), or \( \angle {h}^{\prime }k \), then apply Axiom 10 and the result of Problem 11.] | No |
Theorem 1 Every line passing through \( A \) also passes through \( {A}^{ * } \) . | Proof Suppose \( {AB} = \alpha \) . If \( l \) is a line passing through \( A \), by Theorem 1 \( l \) passes through both \( {A}^{ * } \) and \( B \), with \( A{A}^{ * } = {AB} = \alpha \) . Hence \( B = {A}^{ * } \) . \( ▱ \) | No |
There exists a line perpendicular to a given line and passing through a given point not on the given line. That perpendicular is unique unless \( \alpha < \infty \) and the distance from the given point to every point on the given line is \( \alpha /2 \) . | Consider \( P \notin l \) and choose any point \( Q \in l \), as shown in Figure 3.23. By hypothesis, \( {PQ} < \alpha \) since otherwise line \( l \) passes through \( P \) .\n\n\n\nFigure 3.23 Perpendiculars of / p... | Yes |
Theorem 3 The distance from any point \( A \) to any line \( l \) is the minimum value of \( {AX} \) for all \( X \) on \( l \) . | The proof of this theorem awaits the section on inequalities in triangles, and the result that the hypotenuse of a sufficiently small right triangle is greater than either leg (Section 3.7). | No |
Theorem 1 If the sides of a right triangle \( {ABC} \) with right angle at \( C \) are each less than \( \alpha /2 \), the angles at \( A \) and \( B \) are acute, and the hypotenuse \( {AB} \) is the longest side. | 1. If \( {BC} \) and \( {AC} \) are both of length \( < \alpha /2 \), locate \( D \) such that \( \left( {CAD}\right) \) and \( {CD} = \alpha /2 \) ; the cevian inequality (1) Section 3.6 then implies \( {AB} < \alpha /2 \), and Theorem 1 shows that the angles at \( A \) and \( B \) are acute and \( {AB} \) is the long... | No |
Theorem 2 Let \( \\bigtriangleup {ABC} \) be any right triangle with right angle at \( C \) . Then the following is true:\n\n(a) If \( {AC},{BC} < \\alpha /2 \), then \( {AB} < \\alpha /2,{AB} \) is the longest side, and the angles at \( A \) and \( B \) are both acute.\n\n(b) If \( {AC} = \\alpha /2 \) and \( {BC} \\l... | ## Proof\n\n1. Suppose that in \( \\bigtriangleup {ABC} \) and \( \\bigtriangleup {XYZ} \), we have \( {AB} = {XY},{AC} = {XZ} \) , but \( \\angle A > \\angle X \) . Then, we want to prove \( {BC} > {YZ} \) . As shown in Figure 3.35, construct ray \( {AD} \) between rays \( {AB} \) and \( {AC} \) such that \( \\angle {... | Yes |
Theorem 1 The angle bisectors of a triangle with sides \( < \alpha /2 \) are concurrent in a point \( I \) that lies in the interior of the triangle and is equidistant from the sides. | Proof Consider \( \bigtriangleup {ABC} \) and its angle bisectors \( \overline{AX}\overline{BY} \) and \( \overline{CZ} \) (Figure 3.45). It follows by the crossbar principle that since these bisectors are interior to the angles they bisect, any two bisectors meet at an interior point of each bisector and at an interio... | Yes |
Theorem 2 The perpendicular bisectors of the sides of any triangle are concurrent at a point \( O \) that is equidistant from the vertices, provided any two of them intersect. | (Hint for proof Use a pattern analogous to that of Theorem 1, to be left as Problem 13.) | No |
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