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Problem 3.2. Let \( {ABC} \) be a triangle and let \( N \) be its Nagel point. Let \( M \) be its centroid and let \( I \) be its incenter. Prove that the points \( N \) , \( I \) and \( M \) are collinear. | Solution. The Nagel point is defined as the intersection point of the lines that connect each vertex of the triangle and the point of tangency of the excircle and the corresponding side of the triangle (see Problem 2.7). Let the incircle touch the sides \( {BC} \) and \( {AB} \) at the points \( D \) and \( E \) , resp... | Yes |
Problem 3.3. Let \( {ABC} \) be a triangle. Prove that the lines, formed by the first and second Fermat-Torricelli points, and the first and second isodynamic points, intersect at the Lemoine point \( L \) . Also, prove that the circumcenter of \( \bigtriangleup {ABC} \) lies on the line, formed by the isodynamic point... | Solution. First, we will prove that the points \( O, A{p}_{1}, L \) and \( A{p}_{2} \) are collinear. Since \( A{p}_{1}A{p}_{2} \) is the radical axis of the three Apollonius circles, it is enough to show that the powers of \( O \) and \( L \) with respect to two of them are equal. | No |
Problem 4.3.3. yields that the line \( {OC} \) is tangent to the Apollonius circle \( {k}_{c} \) that passes through \( C \) . | Therefore, the power of \( O \) with respect to \( {k}_{c} \) is \( {R}^{2} \) . Analogously, its power with respect to \( {k}_{b} \) and \( {k}_{a} \) is also \( {R}^{2} \) . Problem 4.4.5 yields that if \( {CL} \cap k = {C}_{1} \), then the quadrilateral \( {CB}{C}_{1}A \) is harmonic. Thus, \( \frac{CA}{CB} = \) \( ... | Yes |
Problem 3.5. Let \( {ABC} \) be a triangle. Let \( I \) be its incenter and let \( O \) be its circumcenter. Let \( {Bi} \) be the Bevan point of \( \bigtriangleup {ABC} \) (see Problem 2.20). Prove that the points \( I, O \) and \( {Bi} \) are collinear. | Solution. Let \( B{i}_{1} \) be the symmetric point of \( I \) with respect to \( O \) . Then the projections of the points \( B{i}_{1} \) and \( I \) onto the sides of the triangle are equidistant from the respective midpoints of the sides. Indeed, these midpoints are the projections of the midpoint of the segment \( ... | Yes |
Problem 3.6. Let \( {ABC} \) be a triangle. Let \( I \) be its incenter, let \( G \) be its Gergonne point, and let \( {S}_{1} \) and \( {\\bar{S}}_{2} \) be its first and second Soddy centers, respectively. Prove that the points \( I, G,{S}_{1} \) and \( {S}_{2} \) are collinear. | Solution. Consider an inversion with center \( I \) and radius \( r \) .\n\nIt sends the circles \( {k}_{1}\\left( {A, s - a}\\right) ,{k}_{2}\\left( {B, s - b}\\right) \) and \( {k}_{3}\\left( {C, s - c}\\right) \) to themselves, and it sends the first Soddy circle (with center \( {S}_{1} \) ) to the second Soddy circ... | Yes |
Problem 3.7. Let \( {ABCD} \) be a quadrilateral. Let the feet of the perpendiculars from \( A \) to \( {BC} \) and \( {CD} \) be \( R \) and \( Q \), respectively. Let the feet of the perpendiculars from \( B \) to \( {CD} \) and \( {DA} \) be \( N \) and \( I \), respectively. Let the feet of the perpendiculars from ... | Solution. Consider the circle \( {k}_{1} \) with diameter \( {AC} \) and the circle \( {k}_{2} \) with diameter \( {BD} \) . We have \( {GR}.{GA} = {GB}.{GI} \) because the quadrilateral \( {BRAI} \) is cyclic and we have \( {HM}.{HC} = {HN}.{HB} \) because the quadrilateral \( {BCNM} \) is cyclic. We have \( {LE}.{EC}... | Yes |
Problem 3.8. Let \( {ABCD} \) be a quadrilateral such that \( {AB} \cap {CD} = E \) and \( {AD} \cap {BC} = F \) . Prove that the midpoints \( M, N \) and \( P \) of the segments \( {AC},{BD} \) and \( {EF} \), respectively, are collinear. | Solution. We will use oriented areas.\n\n\[ \left\lbrack {MNP}\right\rbrack = \frac{\left\lbrack {MNE}\right\rbrack + \left\lbrack {MNF}\right\rbrack }{2} \]\n\n\[ = \frac{\frac{\left\lbrack {MBE}\right\rbrack + \left\lbrack {MDE}\right\rbrack }{2} + \frac{\left\lbrack {MBF}\right\rbrack + \left\lbrack {MDF}\right\rbra... | Yes |
Problem 3.9. Let \( {ABCD} \) be a quadrilateral. Let \( J \) and \( I \) be the midpoints of the diagonals \( {AC} \) and \( {BD} \), respectively. Let the perpendicular \( {DG} \) to \( {BC}\left( {G \in {BC}}\right) \) intersect the perpendicular \( {CH} \) to \( {AD}\left( {H \in {AD}}\right) \) at the point \( K \... | Solution. Consider the circle \( {k}_{1} \) with diameter \( {AC} \) and the circle \( {k}_{2} \) with diameter \( {BD} \) . Their centers are \( I \) and \( J \) , respectively. We have that \( {KG}.{KD} = {KC}.{HK} \) because the quadrilateral \( {DCGH} \) is cyclic. Therefore, the point \( K \) lies on the radical a... | Yes |
Problem 3.10. Let \( {ABCD} \) be a quadrilateral such that \( {AB} \cap {DC} = E \) and \( {AD} \cap {BC} = F \) . The circles \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) have \( {AC},{BD} \) and \( {EF} \) as diameters, respectively. Prove that they share a common radical axis. | Solution. Let the altitudes \( {FH},{AN} \) and \( {DI} \) intersect each other at the orthocenter \( K \) of \( \bigtriangleup {AFD} \), and let the altitudes \( {AM},{BJ} \) and \( {EG} \) intersect each other at the orthocenter \( L \) of \( \bigtriangleup {ABE} \) . Then \( {KI}.{KD} = {KH}.{KF} = {KN}.{KA} \) and ... | Yes |
Problem 3.13. Let \( {ABC} \) be a triangle with circumcircle \( k \) . Two arbitrary points \( P \) and \( Q \) are chosen on the arc \( \overset{⏜}{AB} \) which does not contain \( C \) . Points \( M, N \) and \( K \) are chosen on \( {BC},{CA} \) and \( {AB} \), respectively, such that \( \angle \left( {{PM},{BC}}\r... | Solution. Consider the reflections of \( P \) and \( Q \) with respect to the sides of \( \bigtriangleup {ABC} \) . Denote them by \( {P}_{a},{P}_{b},{P}_{c},{Q}_{a},{Q}_{b} \) and \( {\dot{Q}}_{c} \), as shown in the figure. Problem 3.11 yields that they lie on the Simson lines after a homothety with coefficient 2 is ... | Yes |
Problem 3.14. Let \( {ABC} \) be a triangle and let point \( D \) lie on its circumcircle. Prove that the midpoint \( J \) of the segment \( {DH} \) lies on the Simson line of \( \bigtriangleup {ABC} \) and the point \( D \) . | Solution. Let \( M, N \) and \( K \) be the feet of the perpendiculars drawn from \( D \) to the sides \( {BC},{CA} \) and \( {AB} \), respectively. Let \( {D}_{a},{D}_{b} \) and \( {D}_{c} \) be the reflections of \( D \) with respect to the lines \( {BC},{CA} \) and \( {AB} \). Let \( F \) , \( E \) and \( G \) be th... | Yes |
Problem 3.15. Let \( {ABCD} \) be a cyclic quadrilateral. The feet of the perpendiculars from \( A \) to \( {BC} \) and \( {CD} \) are \( E \) and \( F \), respectively. The feet of the perpendiculars from \( B \) to \( {CD} \) and \( {DA} \) are \( I \) and \( J \), respectively. The feet of the perpendiculars from \(... | Solution. We have that \( \angle {JEL} = \angle {BAJ} = \angle {DCL} \) and it follows that \( {JE} \parallel {FI}. \) Analogously, \( {GL} \parallel {KH},{HI} \parallel {JG} \) and \( {KF} \parallel {EL} \) .\n\nWe have \( \frac{KH}{GL} = \frac{KH}{DC} \cdot \frac{DC}{GL} = \frac{\cos \angle \left( {{AB},{DC}}\right) ... | Yes |
Let \( {ABC} \) be a cyclic quadrilateral and let \( X \) be an arbitrary point. The feet of the perpendiculars from \( X \) to \( {AB} \) and \( {CD} \) are \( H \) and \( I \), respectively. The feet of the perpendiculars from \( X \) to \( {BC} \) and \( {DA} \) are \( K \) and \( F \), respectively. The feet of the... | Let \( {AC} \cap {BD} = E \) . Let us move the point \( X \) uniformly along a line through \( E \) . Then \( J \) and \( G \) are moving uniformly along \( {BD} \) and \( {AC} \), and the line \( {JG} \) is moving uniformly, keeping the same direction. Therefore, \( M \) is moving uniformly along a line through \( E \... | Yes |
Problem 4.1.1. Let \( {ABC} \) be a triangle. The circumcircle of \( \bigtriangleup {ABC} \) is \( k \) and its orthocenter is \( H \) . The altitude through \( B \) intersects \( {AC} \) and \( k \) at the points \( {B}_{1} \) and \( {B}_{2} \), respectively. Prove that the points \( H \) and \( {B}_{2} \) are symmetr... | Solution. It is clear that\n\n\[ \angle {B}_{1}A{B}_{2} = \angle {CB}{B}_{2} = \angle {B}_{1}{AH}. \]\n\nThen in \( \bigtriangleup {B}_{2}{AH} \), the segment \( A{B}_{1} \) is both an angle bisector and an altitude. Hence, it is also a median, which means that \( H \) and \( {B}_{2} \) are symmetric with respect to \(... | Yes |
Let \( O \) be the circumcenter of \( \bigtriangleup {ABC} \) with altitudes \( A{A}_{1}, B{B}_{1} \) and \( C{C}_{1} \) . The lines \( C{C}_{1} \) and \( {A}_{1}{B}_{1} \) intersect at the point \( N \) and the lines \( {CO} \) and \( {AB} \) intersect at the point \( E \) . Prove that \( H\dot{M} \parallel {EN} \) , ... | Problem 4.2.1 tells us that the lines \( {HM} \) and \( {CO} \) intersect on the circumcircle at a point that is symmetric to \( H \) with respect to \( M \) . Denote this point by \( D \) . We know that \( \angle {EAC} = \) \( \angle N{A}_{1}C \) and \( \angle {A}_{1}{CN} = \angle {ACE} \) . Hence, the triangles \( {C... | Yes |
Problem 4.1.3. Let \( {ABC} \) be a triangle with circumcircle \( k \) . Let \( D \) be an arbitrary point on the tangent line to \( k \) at \( C \) . The points \( E \) and \( F \) are its projections onto \( {AC} \) and \( {BC} \), respectively. Prove that \( {EF} \bot {AB} \) . | Solution. The quadrilateral\n\n\n\n\( {CEDF} \) is cyclic. If \( \angle {ABC} = \beta \) , then \( \angle {DCE} = \beta \) too. Hence, we have\n\n\[ \angle {CDE} = \angle {CFE} \]\n\n\[ = \angle {BFX} = {90}^{ \circ } ... | Yes |
Let \( {ABC} \) be a triangle. Let \( P \) be an arbitrary point on the smaller arc \( \overset{⏜}{AB} \) of the circumcircle of \( \bigtriangleup {ABC} \) . The projections of \( P \) onto \( {AC} \) and \( {AB} \) are \( X \) and \( Y \), respectively. The points \( M \) and \( N \) are the midpoints of \( {BC} \) an... | Let \( K \) be the projection of \( P \) onto \( {BC} \) . Then the points \( X \) , \( Y \) and \( K \) lie on the Simson line of \( P \) (Problem 3.11). It suffices to prove that the quadrilateral \( {PNMK} \) is cyclic.\n\nWe show that \( \bigtriangleup {PYX} \sim \bigtriangleup {PBC} \) . We have \( \angle {PXY} = ... | Yes |
Problem 4.1.5. Let \( {ABC} \) be a triangle with altitudes \( A{A}_{1}, B{B}_{1} \) and \( C{C}_{1} \) . The points \( M, N, P \) and \( Q \) are the projections of \( {C}_{1} \) onto the lines \( {AC}, A{A}_{1}, B{B}_{1} \) and \( {BC} \), respectively. Prove that the points \( M, N, P \) and \( Q \) are collinear. | Solution. It is clear that \( \angle M{C}_{1}P = {90}^{ \circ } \) . Hence, \( \angle A{C}_{1}M = \angle P{C}_{1}H \) . The quadrilaterals \( A{C}_{1}{NM} \) and \( {HN}{C}_{1}P \) are cyclic. Hence, \( \angle A{C}_{1}M = \) \( \angle {ANM} \) and \( \angle {PNH} = \angle P{C}_{1}H \) . Now the equality \( \angle {ANM}... | Yes |
Problem 4.1.6. Let \( {ABC} \) be a triangle with altitudes \( A{A}_{1}, B{B}_{1} \) and \( C{C}_{1} \) . Denote the reflections of \( {C}_{1} \) with respect to the sides \( {AC} \) and \( {BC} \) by \( M \) and \( N \), respectively. Prove that the points \( M,{B}_{1},{A}_{1} \) and \( N \) are collinear. | Solution. It is clear that \( {C}_{1}M\parallel B{B}_{1} \) . Hence, \( \angle M{C}_{1}{B}_{1} = \angle {C}_{1}{B}_{1}B \) . On the other hand, \( \angle {C}_{1}{B}_{1}B = \angle {C}_{1}A{A}_{1} = \angle {A}_{1}{B}_{1}B \) . Also, \( \angle M{C}_{1}{B}_{1} = \angle {C}_{1}M{B}_{1} \), hence we derive \( \angle M{B}_{1}... | Yes |
Problem 4.1.7. Let \( {ABC} \) be a triangle with altitudes \( A{A}_{1}, B{B}_{1} \) and \( C{C}_{1} \) . The points \( M \) and \( N \) are the projections from \( {C}_{1} \) onto the sides \( {AC} \) and \( {BC} \), respectively. Let \( P = {MN} \cap {B}_{1}{C}_{1} \) . Prove that \( P \) is the midpoint of \( {B}_{1... | Solution. We have \( \angle {ABC} = \) \( \angle N{C}_{1}C = \angle {NMC} \) because \( N{C}_{1}{MC} \) is a cyclic quadrilateral.\n\nMoreover, \( \angle {ABC} = \angle {C}_{1}{B}_{1}A \) . Hence, \( \angle P{B}_{1}M = \angle {PM}{B}_{1} \) . Therefore, \( {PM} = P{B}_{1} \) .\n\nAlso, \( \angle {PM}{C}_{1} = {90}^{ \c... | Yes |
Let \( {ABC} \) be a triangle. Let \( k \) be its circumcircle and let \( H \) be its orthocenter. Let \( A{A}_{1} \) and \( \bar{B}{B}_{1} \) be altitudes in the triangle. Let \( D \) be an arbitrary point on the segment \( {BH} \) . The line \( {AD} \) intersects \( k \) for the second time at the point \( E \) . Let... | Let the line \( {BH} \) intersect \( k \) for the second time at the point \( X \) . We have \( \angle {XA}{B}_{1} = \angle {HB}{A}_{1} = \angle {B}_{1}{AH} \) . Hence, \( {B}_{1}H = {B}_{1}X \) . Also, we have \( \angle {B}_{1}{AD} = \angle {CBE} = \angle {A}_{1}{BF} \) . This implies that \( \bigtriangleup {ADX} \sim... | Yes |
Problem 4.1.9. Let \( {ABC} \) be a triangle with circumcircle \( k \) . The line \( {CM}\left( {M \in {AB}}\right) \) is the angle bisector of \( \angle {ACB} \), and it intersects \( k \) at the point \( N \) . The line through \( M \), perpendicular to \( {BC} \), intersects \( {BC} \) and the smaller arc \( \overse... | Solution. We will use the notations on the figure. We have \( \angle {YMX} = \) \( \angle {BML} = {90}^{ \circ } - \angle {ABC} = {90}^{ \circ } - \angle {AXC} = \angle {YCX} \) . Hence, the quadrilateral \( {YMCX} \) is cyclic. Also, the quadrilateral \( {CZLX} \) is cyclic as well.\n\nHence, \( \angle {MXY} = \angle ... | Yes |
Problem 4.1.10. Let \( A{A}_{1}, B{B}_{1} \) and \( C{C}_{1} \) be the altitudes of a given triangle \( {ABC} \). Let \( P \) be an arbitrary point inside of the triangle. The points \( {C}_{2} \) and \( {C}_{3} \) are the projections of \( P \) onto \( {AB} \) and \( C{C}_{1} \), respectively The points \( {A}_{2} \in... | Solution. First, consider a\n\n\n\nhomothety with center \( P \) and coefficient 2. The image of the line \( {C}_{2}{C}_{3} \) is a line, passing through \( {C}_{1} \) and parallel to \( {C}_{2}{C}_{3} \). Let \( X \) ... | Yes |
Problem 4.1.11. Let \( {ABC} \) be a triangle with altitudes \( A{B}_{1} \) and \( B{A}_{1} \) which intersect at the point \( H \) . The lines \( {A}_{1}{B}_{1} \) and \( {AB} \) intersect at the point \( D \) and \( M \) is the midpoint of \( {AB} \) . Prove that \( {MH} \bot {DC} \) . | Solution. We have that the quadrilateral \( {AB}{B}_{1}{A}_{1} \) is inscribed in a circle with center \( M \) . Consider the pole-polar transformation \( \pi \) with respect to this circle. By one of the polar properties, \( H \in {\pi }_{C} \), hence (by La Hire’s Theorem) \( C \in {\pi }_{H} \) . Also, \( H \in {\pi... | Yes |
Problem 4.1.12. Let \( {ABC} \) be an acute triangle. The point \( H \) is its orthocenter and the point \( M \) is the midpoint of \( {AB} \) . Let \( A{A}_{1} \) and \( B{B}_{1} \) be altitudes in the triangle and let \( {AB} \cap {A}_{1}{B}_{1} = D \) . The line \( {CH} \) intersects the circumcircle of \( \bigtrian... | Solution. Let \( C{C}_{1} \bot {AB}\left( {{C}_{1} \in {AB}}\right) \) . Denote the symmetric point of \( H \) with respect to \( M \) by \( Q \) . Since \( \angle {AQB} = \angle {AHB} = {180}^{ \circ } - \angle {ACB} \), we have that \( Q \) lies on the circumcircle of \( \bigtriangleup {ABC} \) . Also, \( {BH} \bot {... | Yes |
Problem 4.1.13. Let \( {ABC} \) be a triangle with \( \angle {ACB} > {90}^{ \circ } \) . Let \( A{A}_{1}, B{B}_{1} \) and \( C{C}_{1} \) be its altitudes. The point \( M \) is the midpoint of the side \( {AB} \) . Prove that the midpoints of \( A{A}_{1} \) and \( B{B}_{1} \), and the points \( M \) and \( {C}_{1} \) ar... | Solution. Denote the midpoints of \( A{A}_{1}, B{B}_{1},{AC} \) and \( {BC} \) by \( P, Q \) , \( R \) and \( S \), respectively.\n\nThen \( {PR}\parallel C{A}_{1} \) and \( {MR}\parallel {BC} \), as midsegments in \( \bigtriangleup {AC}{A}_{1} \) and \( \bigtriangleup {ABC} \), respectively. Hence, the points \( P, R ... | Yes |
Problem 4.1.14. Let \( {ABC} \) be a triangle with \( \angle {ACB} > {90}^{ \circ } \) and altitudes \( A{A}_{1} \) and \( B{B}_{1} \) . The points \( P \) and \( M \) are the projections of \( {A}_{1} \) onto \( {AC} \) and \( {AB} \), respectively, and \( Q \) and \( N \) are the projections of \( {B}_{1} \) onto \( ... | Solution. The quadrilaterals \( {AB}{B}_{1}{A}_{1},{AMP}{A}_{1} \) and \( {BNQ}{B}_{1} \) are cyclic. We have\n\n\( \angle {B}_{1}{PQ} = \angle {B}_{1}{AQ} = \angle {CAB} \) .\n\nHence, \( {PQ} \parallel {AB} \) .\n\nMoreover, we have\n\n\[ \angle {PM}{A}_{1} = \angle {PA}{A}_{1} = \angle {A}_{1}B{B}_{1} = \angle {B}_{... | Yes |
Problem 4.1.15. Let \( \\bigtriangleup {ABC} \) be a triangle with an altitude \( {CD} \) and circumcenter \( O \) . Let \( M \) be the midpoint of the side \( {AB} \) . Denote the projection of \( A \) onto \( {CO} \) by \( P \) . Prove that \( {DM} = {PM} \) . | Solution. Note that the quadrilaterals \( {APDC} \) and \( {AOMP} \) are cyclic.\n\nWe have \( \\angle {AOP} = \\angle {AMP} \) and \( \\angle {ADP} = \\angle {ACP} \) . Hence, \( \\bigtriangleup {PMD} \\sim \) \( \\bigtriangleup {AOC} \) . But \( \\bigtriangleup {AOC} \) is isosceles, which means that \( {DM} = {PM} \... | Yes |
Problem 4.1.16. Let \( {ABC} \) be a triangle with altitudes \( A{A}_{1} \) and \( B{B}_{1} \) . The circle with diameter \( {AC} \) intersects the line \( B{B}_{1} \) at the points \( P \) and \( M \), so that \( P \) is between \( B \) and \( M \) . The circle with diameter \( {BC} \) intersects \( A{A}_{1} \) at the... | Solution. Note that\n\n\n\n\( \bigtriangleup {BNC} \) is right-angled and has \( N{A}_{1} \) as an altitude. Hence,\n\n\[ C{N}^{2} = C{A}_{1}.{CB}. \]\n\nAnalogously, we derive \( C{P}^{2} = C{B}_{1} \cdot {CA} \) . Bu... | Yes |
Problem 4.1.17. Let \( {ABC} \) be a triangle with an altitude \( {CD} \) . The points \( E \) and \( F \) are the projections of \( D \) onto \( {AC} \) and \( {BC} \), respectively. Prove that the quadrilateral \( {ABFE} \) is cyclic. | Solution. The quadrilateral \( {EDFC} \) is cyclic, hence\n\n\[ \angle {CEF} = \angle {CDF} = \]\n\n\[ = {90}^{ \circ } - \angle {FDB} \]\n\n\[ = \angle {ABF}\text{.} \]\n\nTherefore, the quadrilateral ABFE is cyclic. | Yes |
Problem 4.1.18. Let \( {ABC} \) be a triangle with an altitude \( {CD} \) . The points \( E \) and \( F \) are the projections of \( D \) onto \( {AC} \) and \( {BC} \), respectively. The points \( M \) and \( N \) are the midpoints of \( {AC} \) and \( {BC} \), respectively. Prove that the quadrilateral \( {EFNM} \) i... | Solutoin. Note that Prob- lem 4.1.17 implies \( \angle {BAC} = \) \( \angle {EFC} \) . Since \( {MN} \) is a midsegment in \( \bigtriangleup {ABC} \), we have \( {MN}\parallel {AB} \) , hence \( \angle {NMC} = \angle {BAC} \) . Therefore, \( \angle {NMC} = \angle {EFC} \) which yields that the quadrilateral EFNM is cyc... | No |
Problem 4.1.19. Let \( {ABC} \) be a triangle. The segments \( A{A}_{1} \) and \( B{B}_{1} \) are altitudes, and the bisector of \( \angle {ACB} \) intersects the segment \( {A}_{1}{B}_{1} \) at the point \( L \) . The circumcircle of \( \bigtriangleup A{B}_{1}L \) intersects \( B{B}_{1} \) for the second time at \( X ... | Solution. Let the circumcircle of \( \bigtriangleup B{A}_{1}L \) intersect \( A{A}_{1} \) for the second time at the point \( {Y}^{\prime } \) . We will show that \( Y \equiv {Y}^{\prime } \) . It suffices to prove the equality \( {BX} = A{Y}^{\prime } \) .\n\nWe have \( \angle {Y}^{\prime }{LB} = \angle {Y}^{\prime }{... | Yes |
Problem 4.1.20. Let \( {ABC} \) be a triangle. Its circumcenter is \( O \), its orthocenter is \( H \) and its altitudes are \( A{A}_{1}, B{B}_{1} \) and \( C{C}_{1} \) . The point \( M \) is the projection of \( C \) onto \( {A}_{1}{B}_{1} \) and \( N \) is the reflection of \( C \) with respect to \( {A}_{1}{B}_{1} \... | Solution. It is clear that\n\n\n\n\( \angle {ACM} = \angle {ACN} = \angle {BCH} \) .\n\nBut \( \angle {BCH} = \angle {ACO} \) , hence the points \( C, M, O \) and \( N \) are collinear.\n\nLet \( R \) be the circumradi... | Yes |
Problem 4.1.21. Let \( {ABC} \) be a triangle with altitudes \( A{A}_{1} \) and \( B{B}_{1} \) . A point \( D \) is chosen on the extension of \( A{A}_{1} \) beyond \( {A}_{1} \) . A point \( E \) is chosen on the extension of \( B{B}_{1} \) beyond \( {B}_{1} \), satisfying \( \angle {DCE} = {90}^{ \circ } \) . Let \( ... | Solution. Let \( {B}^{\prime } \in B{B}_{1} \) be a point such that \( {B}^{\prime }H \bot {AH} \) . It suffices to show that \( B \equiv {B}^{\prime } \) . We have that the quadrilaterals \( E{B}_{1}{HC} \) and \( A{B}_{1}H{B}^{\prime } \) are cyclic.\n\nHence, \( \angle {ACH} = \angle {HE}{B}^{\prime } \) and \( \ang... | Yes |
Problem 4.1.22. Let \( {ABC} \) be a triangle. The segments \( A{A}_{1}, B{B}_{1} \) and \( C{C}_{1} \) are its altitudes. The points \( {A}_{2} \) and \( {A}_{3} \) are the projections of \( {A}_{1} \) onto \( {AC} \) and \( {AB} \), respectively. The points \( {B}_{2},{B}_{3},{C}_{2} \) and \( {C}_{3} \) are defined ... | Solution. The quadrilaterals \( {BA}{B}_{1}{A}_{1} \) and \( {A}_{1}{B}_{1}{A}_{2}{B}_{3} \) are cyclic, hence \( \angle {BAC} = \angle {B}_{1}{A}_{1}C = \angle {B}_{3}{A}_{2}C \) . Thus, \( {A}_{3}{B}_{2}{A}_{2}{B}_{3} \) is a trapezoid. Problem 4.1.17 yields that the quadrilaterals \( B{A}_{3}{A}_{2}C \) and \( A{B}_... | Yes |
Problem 4.1.23. Let \( {ABC}\left( {{AC} = {BC}}\right) \) be an isosceles triangle. The segment \( C{C}_{1} \) is an altitude and \( M \) is its midpoint. Let \( P \) be the projection of \( {C}_{1} \) onto \( {BM} \) . Prove that \( \angle {APC} = {90}^{ \circ } \) . | Solution. Note that \( \bigtriangleup B{C}_{1}M \) is right-angled with \( {C}_{1}P \) as its altitude. Then \[ \bigtriangleup {MP}{C}_{1} \sim \bigtriangleup {C}_{1}{PB} \] which implies \[ \frac{M{C}_{1}}{{C}_{1}B} = \frac{MP}{P{C}_{1}} \] But \( M{C}_{1} = {MC} \) and \( {C}_{1}B = A{C}_{1} \), hence \[ \frac{MC}{A{... | Yes |
Let \( {ABC} \) be a triangle with orthocenter \( H \) and let \( M \) be the midpoint of the side \( {AB} \) . Prove that the symmetric point of \( H \) with respect to \( M \) coincides with the diametrically opposite point of \( C \) with respect to the circumcircle of \( \bigtriangleup {ABC} \) . | Let \( O \) be the circumcenter of \( \bigtriangleup {ABC} \) . Let \( N \) be the projection of \( O \) onto \( {AC} \) . We have \( {OM} \bot {AB},{ON} \bot {AC} \) and \( {MN}\parallel {BC} \) . It follows that \( \bigtriangleup {MON} \) and \( \bigtriangleup {CHB} \) are similar with ratio \( \frac{MN}{BC} \) . But... | Yes |
Problem 4.2.2. Let \( {ABC} \) be a triangle with altitudes \( A{A}_{1} \) and \( B{B}_{1} \) intersecting at the point \( H \) . Let \( D \) be the second intersection point of the circumcircles of \( \bigtriangleup {ABC} \) and \( \bigtriangleup {A}_{1}{B}_{1}C \), and let \( M \) be the midpoint of \( {AB} \) . Prov... | Solution. Let \( O \) be the circumcenter of \( {ABC} \) and let \( N \) be the midpoint of \( {CH} \) . It follows that \( N \) is the circumcenter of the cyclic pentagon \( C{B}_{1}H{A}_{1}D \) . Now Problem 4.2.1 implies \( {OM} = {NH} = {CN} = {ND} \), which yields \( {OM} = {NH} \) and \( {OM}\parallel {HN} \) . H... | Yes |
Problem 4.2.3. Let \( {ABC} \) be a triangle with circumcircle \( k \) . Let \( H \) be its orthocenter. Let \( l \) be an arbitrary line through \( H \) . Prove that the reflections of \( l \) with respect to \( {AB},{BC} \) and \( {CA} \) concur at a point on \( k \) . | Solution. Let \( {A}_{1},{B}_{1} \) and \( {C}_{1} \) be the second intersection points of the respective altitudes with the circumcircle of \( \bigtriangleup {ABC} \) . Let \( X, Y \) and \( Z \) be the intersection points of the line through \( H \) with the sides of \( \bigtriangleup {ABC} \), as shown in the figure... | Yes |
Let \( {ABC} \) be a triangle with circumcenter \( O \) . The reflections of \( {AB} \) with respect to the lines \( {AC} \) and \( {BC} \) intersect at the point \( K \) . Prove that the points \( C, O \) and \( K \) are collinear. | Solution. Note that \( C \) is an excenter in \( \bigtriangleup {ABK} \) . Therefore, \( \angle {AOB} = \) \( 2\angle {ACB} = {180}^{ \circ } - \angle {AKB} \) . Hence, the quadrilateral \( {AKBO} \) is cyclic. But we have \( {AO} = {BO} \) . Thus, \( {KO} \) is the angle bisector of \( \angle {AKB} \) . Therefore, the... | Yes |
Problem 4.2.5. Let \( {ABC} \) be a triangle with orthocenter \( H \), and let \( M \) be the midpoint of the side \( {AB} \) . The line through \( H \), perpendicular to \( {HM} \), intersects the sides \( {AC} \) and \( {BC} \) at the points \( D \) and \( E \), respectively. Prove that \( {DH} = {EH} \) . | Solution. Let \( A{A}_{1} \) and \( B{B}_{1} \) be altitudes in \( \bigtriangleup {ABC} \) . Notice that the quadrilateral \( {AB}{A}_{1}{B}_{1} \) is inscribed in a circle \( {k}_{1} \) with center \( M \) and \( H \) is the intersection point of its diagonals. Now we apply the butterfly theorem (Problem 6.4.3) to the... | Yes |
Problem 4.2.6. Let \( {ABC} \) be a triangle with orthocenter \( H \) and circumcenter \( O \) . The point \( M \) lies on the side \( {BC} \) and \( \angle O{C}_{1}M = {90}^{ \circ } \) . Prove that \( \angle {ABC} = \angle {MH}{C}_{1} \) . | Solution. Let \( C{C}_{1} \) be an altitude in \( \bigtriangleup {ABC} \) . We have \( \angle B{C}_{1}C = \angle M{C}_{1}O \) .\n\nHence, \( \angle O{C}_{1}H = \angle B{C}_{1}M \) . Let \( N \) be the reflection of \( M \) with respect to \( C{C}_{1} \) .\n\nNow, the fact that \( O \) and \( H \) are isogonal conjugate... | Yes |
Let \( {ABC} \) be a triangle. The circumcircle of \( \bigtriangleup {ABC} \) is \( k \) and its orthocenter is \( H \) . Consider two lines through \( H \) that are perpendicular to each other. One of them intersects \( {AB},{BC} \) and \( {CA} \) at the points \( F, K \) and \( P \), respectively, and the other one -... | Solution. Let \( {H}_{a},{H}_{b} \) and \( {H}_{c} \) be the reflections of \( H \) with respect to \( {BC},{AC} \) and \( {AB} \), respectively. Since \( \angle B{H}_{a}C = \angle {BHC} = {180}^{ \circ } - \angle {BAC} \) , we get that \( {H}_{a} \) lies on \( k \), and the same is true for \( {H}_{b} \) and \( {H}_{c... | Yes |
Let \( {ABC} \) be a triangle. The circumcircle of \( \bigtriangleup {ABC} \) is \( k \) and its orthocenter is \( H \) . Let \( P \) be an arbitrary point from the inside of \( \bigtriangleup {ABC} \) . The lines \( {AP},{BP} \) and \( {CP} \) intersect \( k \) for the second time at the points \( {A}_{1},{B}_{1} \) a... | We have\n\n\[ \angle {B}_{1}{C}_{1}{A}_{1} = \angle {B}_{1}{C}_{1}C + \angle C{C}_{1}{A}_{1} = \angle {B}_{1}{BC} + \angle {A}_{1}{AC} \]\n\n\[ = \angle P{C}_{2}{A}_{2} + \angle P{C}_{2}{B}_{2} = \angle {B}_{2}{C}_{2}{A}_{2}. \]\n\nAnalogously, \( \angle {B}_{1}{A}_{1}{C}_{1} = \angle {B}_{2}{A}_{2}{C}_{2} \), hence \(... | Yes |
Let \( {ABC} \) be a triangle with orthocenter \( H \) and let \( P \) be an arbitrary point inside of the triangle. The lines \( {AP},{BP} \) and \( {CP} \) intersect the circumcircle \( k \) of \( \bigtriangleup {ABC} \) at the points \( {A}_{1},{B}_{1} \) and \( {C}_{1} \) , respectively. The points \( {A}_{2},{B}_{... | Let \( M, N \) and \( P \) be the midpoints of \( {AB},{BC} \) and \( {CA} \), respectively. Let \( {C}_{3},{A}_{3} \) and \( {B}_{3} \) be the reflections of \( {C}_{2},{A}_{2} \) and \( {B}_{2} \) with respect to \( M, N \) and \( P \), respectively. We have \( \angle A{C}_{2}B = \angle A{C}_{1}B = \angle A{C}_{3}B \... | Yes |
Problem 4.3.1. Let \( {ABC} \) be a triangle and let \( M \) and \( N \) be the feet of the angle bisectors through \( A \) and \( B \), respectively. The point \( P \) is the foot of the external bisector through \( C \) . Prove that the points \( N, M \) and | Solution. The internal bisector theorem yields\n\n\[ \n\frac{BM}{MC} = \frac{AB}{AC} \n\]\n\nand\n\n\[ \n\frac{CN}{NA} = \frac{CB}{AB} \n\]\n\nThe external bisector theorem yields \( \frac{AP}{PB} = \frac{AC}{CB} \) .\n\nTherefore, \( \frac{AP}{PB} \cdot \frac{BM}{MC} \cdot \frac{CN}{NA} = \frac{AC}{CB} \cdot \frac{AB}... | No |
Let \( {ABC} \) be a triangle and let \( {C}_{1} \) and \( {B}_{1} \) be the feet of the bisectors through \( C \) and \( B \), respectively. Let \( O \) be the circumcenter of \( \bigtriangleup {ABC} \) and let \( {I}_{a} \) be the \( A \) -excenter. Prove that \( O{I}_{a} \bot {B}_{1}{C}_{1} \) . | Let \( {X}_{1} \) and \( {X}_{2} \) be the intersection points of the circumcircle \( k \) and the \( A \) -excircle \( {\omega }_{a} \) of \( \bigtriangleup {ABC} \) . Since \( {X}_{1}{X}_{2} \) is the radical axis of these circles, we have \( O{I}_{a} \bot {X}_{1}{X}_{2} \) . Let \( {X}_{1}{X}_{2} \cap {AC} = P \) an... | Yes |
Problem 4.3.3. Let \( {ABC} \) be a triangle \( \\left( {\\angle {ABC} > {90}^{ \\circ }}\\right) \) and let \( {CM} \) and \( {CN} \) be the internal and the external bisectors of \( \\angle {ACB} \), respectively. Prove that the circles \( \\left( {ACB}\\right) \) and \( \\left( {MNC}\\right) \) are orthogonal, i.e. ... | Solution. We have \( \\angle {AMC} = \\beta + \\frac{\\gamma }{2} \) . Note that \( \\angle {AOC} = {360}^{ \\circ } - {2\\beta } \) and \( \\angle {ACO} = \\beta - {90}^{ \\circ } \) . On the other hand, \( \\bigtriangleup {MNC} \) is right-angled with \( \\angle {MCN} = {90}^{ \\circ } \) . Let \( K \) be the midpoin... | Yes |
Let \( {ABC} \) be a triangle with circumcircle \( k \) . The line \( {CL}\left( {L \in {AB}}\right) \) is the angle bisector of \( \angle {ACB} \) . Denote the intersection point of the tangent line to \( k \) at \( C \) and the line \( {AB} \) by \( N \) . Prove that \( {NC} = {NL} \) . | We have \( \angle {CLN} = \angle {CAL} + \angle {ACL} = \angle {BCN} + \angle {BCL} = \) \( \angle {LCN} \), and therefore \( {NC} = {NL} \) . | No |
Problem 4.3.5. Let \( {ABC} \) be a triangle with bisectors \( A{A}_{1}, B{B}_{1} \) and \( C{C}_{1} \) . Let \( P \) be an arbitrary point on the line \( {A}_{1}{B}_{1} \) and let \( X, Y \) and \( Z \) be its projections onto the lines \( {AB},{BC} \) and \( {CA} \), respectively. Prove that the sum of the lengths of... | Solution. We will use the notations on the figure. Without loss of generality, we assume that \( {AC} > {BC} \) and \( {A}_{1} \) is between \( P \) and \( {B}_{1} \) . We will show that \( {PZ} = {PY} + {PX} \) . Note that \( \bigtriangleup {PY}{A}_{1} \sim \bigtriangleup {B}_{1}M{A}_{1} \), which implies \( \frac{PY}... | Yes |
Problem 4.3.6. Let \( {ABC} \) be a triangle with angle bisectors \( A{A}_{1} \) and \( B{B}_{1} \) . Let \( E \) be the intersection point of the line \( {A}_{1}{B}_{1} \) and the circumcircle of \( \bigtriangleup {ABC} \), such that \( E \) and \( A \) lie in the same half-plane with respect to \( {BC} \) . Prove tha... | Solution. Denote the projections of \( E \) onto \( {AB},{BC} \) and \( {CA} \) by \( M \) , \( N \) and \( P \), respectively. Let \( X \in {AC} \) and \( \angle {XEP} = \angle {CEN} \) . We have\n\n\[ \angle {CEN} = {90}^{ \circ } - \angle {ECN} = {90}^{ \circ } - \angle {EAM} = \angle {MEA} \]\n\n\[ \Rightarrow \big... | Yes |
Problem 4.3.7. Let \( {ABC} \) be a triangle with angle bisectors \( {AQ} \) and \( {BP} \) and circumcircle \( k \) . The lines \( {AQ} \) and \( {BP} \) intersect \( k \) at the points \( M \) and \( N \), respectively. Prove that the lines \( {PQ},{MN} \) and the tangent line to \( k \) at \( C \) are concurrent. | Solution. Let \( l \) be the tangent line to \( k \) at \( C \) . We denote \( U = l \cap {MN} \) and it suffices to prove that \( P, Q \) and \( U \) are collinear.\n\nLet \( R = {BN} \cap l,\;{R}_{1} = {AC} \cap {NM},\;S = {AQ} \cap l \) and \( {S}_{1} = {BC} \cap {NM}. \) If we prove that the lines \( {IC}, R{R}_{1}... | Yes |
Let \( {ABCD} \) be a quadrilateral. The rays \( {AB} \) and \( {DC} \) intersect at the point \( E \) and the rays \( {AD} \) and \( {BC} \) intersect at the point \( F \) . The internal angle bisectors of \( \angle {EAF} \) and \( \angle {ECF} \) intersect at \( X \) . The internal angle bisector of \( \angle {ADE} \... | Denote\n\n\[ \n{YD} \cap {ZF} = K \]\n\n\n\n\[ \n{BY} \cap {ZE} = G, \]\n\n\[ \n{CX} \cap {ZE} = H \]\n\n\[ \n{AX} \cap {FZ} = L \]\n\n\[ \n{DY} \cap {AX} = M, \]\n\n\[ \n{CX} \cap {BY} = N\text{.} \]\n\nNow we have th... | Yes |
Problem 4.3.10. Let \( {ABC} \) be a triangle with angle bisector \( {CL}(L \in \) \( {AB}) \) . The circle with diameter \( {AB} \), centered at \( M \), intersects the sides \( {AC} \) and \( {BC} \) for the second time at the points \( {B}_{1} \) and \( {A}_{1} \), respectively. The angle bisector of \( \angle {A}_{... | Solution. Denote the intersection point of the lines \( {MK} \) and \( {A}_{1}{B}_{1} \) by \( N \) . Now in \( \bigtriangleup {A}_{1}M{B}_{1} \) we have that \( M{A}_{1} = M{B}_{1} \) and we have that \( {MK} \) is an angle bisector of \( \angle {A}_{1}M{B}_{1} \) . Hence, the line \( {MK} \) is the perpendicular bise... | Yes |
Problem 4.3.11. Let \( {ABC} \) be a triangle and let \( l \) be its external bisector through \( C \) . The points \( D \) and \( E \) are the projections of \( A \) and \( B \) onto \( l \) . The line \( {CH}\left( {H \in {AB}}\right) \) is the altitude from \( C \) to \( {AB} \) and the point \( F \) is the midpoint... | Solution. Let \( M \) and \( N \) be the midpoints of \( {AC} \) and \( {BC} \) , respectively. Then \( {FM}\parallel {BC} \) and \( {FN}\parallel {AC} \) .\n\nOn the other hand, \( {DM} \) is a median to the hypotenuse in the right-angled \( \bigtriangleup {CDA} \) . Thus, \( \angle {MDE} = \angle {MCD} = \) \( \angle... | Yes |
Problem 4.3.12. Let \( {ABC} \) be a triangle. The points \( M, N, P \) and \( Q \) are the feet of the perpendiculars from \( \bar{C} \) to the internal and external angle bisectors of \( \angle {BAC} \) and \( \angle {ABC} \), as shown in the figure. Prove that the points \( M, N, P \) and \( Q \) are collinear. | Solution. Denote the intersection points of \( {CM},{CN},{CP} \) and \( {CQ} \) , and the line \( {AB} \) by \( R, S, T \) and \( U \), respectively. In \( \bigtriangleup {RAC} \), the segment \( {AM} \) is both an altitude an angle bisector. Thus, \( M \) is the midpoint of \( {RC} \) .\n\nAnalogously, the points \( N... | Yes |
Problem 4.3.13. Let \( {ABC} \) be a triangle with orthocenter \( H \) . The points \( L \) and \( P \) are the projections of \( H \) onto the internal and external angle bisectors of \( \angle {ACB} \), respectively. Prove that the points \( M, L \) and \( P \) are collinear, where \( M \) is the midpoint of \( {AB} ... | Solution. Since \( \angle {PCL} = {90}^{ \circ } \), the quadrilateral LHPC is a rectangle. Let \( N \) be the intersection point of its diagonals. Then \( {LN} = {CN} = {NH} \) and \( \angle {CLN} = \angle {LCN} \) . However, since \( O \) and \( H \) are isogonal conjugates, we have \( \angle {ACO} = \) \( \angle {BC... | Yes |
Problem 4.3.14. Let \( {ABC} \) be a triangle. A line through \( C \) intersects the angle bisector through \( A \) and the circumcircle of \( \bigtriangleup {ABC} \) at the points \( M \) and \( N \), respectively. A circle \( {k}_{1} \) passes through \( A \) and touches \( {CM} \) at the point \( M \) . Let \( {k}_{... | Solution. Let \( {k}_{1} \cap {AC} = \{ A, T\} \) . The quadrilateral \( {APMT} \) is cyclic, hence \( \angle {PMN} = \angle {PAM} = \angle {MAT} = \angle {MPT} \) . Thus, \( {TP}\parallel {CN} \) . Moreover, \( \angle {CNQ} = \angle {CAQ} = \angle {TPQ} \), which means that the points \( N, P \) and \( Q \) are collin... | Yes |
Problem 4.3.15. Let \( \angle {AOC} \) be an angle. The point \( B \) lies on the ray \( {OA} \) and the point \( D \) lies on the ray \( {OC} \) . Moreover, \( {AB} = {CD}, A \) lies between \( O \) and \( B \), and \( C \) lies between \( O \) and \( D \) . The circumcircles of \( \bigtriangleup {ADO} \) and \( \bigt... | Solution. Since the quadrilaterals \( {AKDO} \) and \( {BKCO} \) are cyclic, we have \( \angle {DCK} = \angle {OBK} \) and \( \angle {ODK} = \) \( \angle {KAB} \) . On the other hand, \( {CD} = \) \( {AB} \), thus \( \bigtriangleup {KBA} \cong \bigtriangleup {KCD} \) . Therefore, the distances from \( K \) to the arms ... | Yes |
Problem 4.3.16. Let \( {ABC}\left( {{AC} > {BC}}\right) \) be a triangle, let \( {CL}(L \in \) \( {AB} \) ) be its angle bisector, and let \( O \) be its circumcenter. Denote the circumcenters of \( \bigtriangleup {ALC} \) and \( \bigtriangleup {BLC} \) by \( {O}_{1} \) and \( {O}_{2} \), respectively. Prove that \( O{... | Solution. We have\n\n\[ \angle {ALC} = \beta + \frac{\gamma }{2} \]\n\nwhich gives\n\n\[ \angle A{O}_{1}C = {360}^{ \circ } - {2\beta } - \gamma \]\n\nand \( \angle {O}_{1}{CA} = \beta + \frac{\gamma }{2} - {90}^{ \circ } \) .\n\nThen\n\n\[ \angle {O}_{1}{CO} = \angle {ACO} + \angle {AC}{O}_{1} = \beta + \frac{\gamma }... | Yes |
Problem 4.3.17. Let \( {ABC} \) be a triangle and let \( {CL}\left( {L \in {AB}}\right) \) be its angle bisector. The incircle of \( \bigtriangleup {ALC} \) touches \( {AC},{AL} \) and \( {CL} \) at the points \( M, N \) and \( P \), respectively, and the \( C \) -excircle of \( \bigtriangleup {BCL} \) touches the line... | Solution. Let \( \angle {BLC} = x \) . Denote the semiperimeters of \( \bigtriangleup {BLC} \) and \( \bigtriangleup {ALC} \) by \( {s}_{1} \) and \( {s}_{2} \) , respectively. We have \( {CR} = {s}_{1} \) , \( {QL} = {s}_{1} - {BC},{LP} = {s}_{2} - {AC} \) and \( {PC} = {s}_{2} - {AL} \) . Now we give expressions for ... | Yes |
Problem 4.3.18. Let \( {ABC} \) be a triangle with altitudes \( A{A}^{\prime } \) and \( B{B}^{\prime } \) and incircle \( k \) centered at \( I \) . The lines \( {AI} \) and \( {BI} \) intersect \( {BC} \) and \( {AC} \) at the points \( F \) and \( G \), respectively. The circle \( k \) touches \( {BC} \) and \( {CA}... | Solution. It follows by the cross-ratio properties, that it suffices to prove that \( \frac{{B}^{\prime }E.{CG}}{{EG}.{B}^{\prime }C} = \frac{{A}^{\prime }D.{FC}}{{DF}.C{A}^{\prime }} \) . But the intercept theorem and the angle bisector theorem give\n\n\[ \frac{{B}^{\prime }E}{EG} = \frac{BI}{IG} = \frac{BC}{CG},\frac... | Yes |
Problem 4.3.19. Let \( {ABC} \) be a triangle with angle bisector \( {CL} \) and incenter \( I \) . The perpendicular bisector of the segment \( {CL} \) intersects the angle bisectors of \( \angle {BAC} \) and \( \angle {ABC} \) at the points \( M \) and \( N \), respectively. Prove that the quadrilateral \( {MINC} \) ... | Solution. The points \( M \) and \( N \)\n\n\n\nalways exist because \( {AC} > {AL} \) and \( {BC} > {BL} \) . Since \( M \) lies on both the angle bisector of \( \angle {LAC} \) and the perpendicular bisector of \( {C... | Yes |
Problem 4.3.20. Let \( {ABC} \) be a triangle with incenter \( I \) . The points \( Y \in {AI} \) and \( X \in {BI} \) are chosen such that \( \angle {XCA} = \angle {YCB} \) . Prove that the lines \( {AX},{CI} \) and \( {BY} \) are concurrent. | Solution. By applying Desargues’ Theorem (Problem 7.1) to \( \bigtriangleup {ABC} \) and \( \bigtriangleup {XYI} \), we see that it suffices to show that the points \( {B}_{1} = {AC} \cap {XI} \) , \( {A}_{1} = {BC} \cap {YI} \) and \( Z = {AB} \cap {XY} \) are collinear.\n\nWe will use the notations on the figure. By ... | Yes |
Problem 4.3.21. Let \( {ABC} \) be a triangle with angle bisectors \( {AM} \) and \( {BN} \), intersecting at the point \( I \) . Points \( L \) and \( K \) are chosen on the line \( {AB} \), such that \( {LN} \) and \( {CN} \) are symmetric with respect to \( {BN} \) , and such that \( {CM} \) and \( {KM} \) are symme... | Solution. The symmetry with respect to \( {BN} \) gives \( \angle {BLN} = \angle {BCN} \) , and the symmetry with respect to \( {AM} \) gives \( \angle {AKM} = \angle {ACM} \) . Therefore, \( \bigtriangleup {LDK} \) is isosceles.\n\nSince \( I \) is the \( K \) -excenter of \( \bigtriangleup {BKM} \), we deduce that \(... | Yes |
Problem 4.3.22. Let \( {ABC}\left( {{AC} > {BC}}\right) \) be a triangle with altitudes \( A{A}_{1} \) and \( B{B}_{1} \), intersecting at the point \( H \) . The angle bisectors of \( \angle {HAC} \) and \( \angle {HBC} \) intersect at the point \( L \) . Let \( M \) and \( N \) be the midpoints of \( {AB} \) and \( {... | Solution. Let the point \( O \) be the circumcenter of \( \bigtriangleup {ABC} \) . By Problem 4.2.1, we have that \( {OM} \) is parallel and equal to \( {NC} \), so \( {OMNC} \) is a parallelogram. Thus, \( \angle {OMN} = \angle \bar{O}{CN} = \bar{\beta } - \alpha . \n\nOur goal is to prove that \( \angle {OML} = \ang... | Yes |
Let \( {ABC} \) be a triangle and let \( {C}_{2} \) be the midpoint of \( {AB} \) . The tangent lines to the circumcircle of \( \bigtriangleup {ABC} \) at \( A \) and \( B \) intersect each other at the point \( N \) . Prove that \( \angle {AC}{C}_{2} = \angle {BCN} \) . | The law of sines, applied to \( \bigtriangleup A{C}_{2}C,\bigtriangleup B{C}_{2}C,\bigtriangleup {ANC} \) and \( \bigtriangleup {BNC} \), gives\n\n\[ \frac{\sin \angle {AC}{C}_{2}}{\sin \angle {BC}{C}_{2}} = \frac{A{C}_{2}}{B{C}_{2}} \cdot \frac{\sin \alpha }{\sin \beta } = \frac{\sin \left( {\beta + \gamma }\right) }{... | Yes |
Let \( {ABC} \) be a triangle. The points \( D \) and \( E \) are chosen on the sides \( {AC} \) and \( {BC} \), such that the quadrilateral \( {ABED} \) is cyclic. Let \( M \) be the midpoint of \( {AB} \) and let \( F \) be the intersection point of the tangent lines to the circumcircle of \( \bigtriangleup {DEC} \) ... | Let \( N \) be the midpoint of \( {DE} \) . Problem 4.4.1 yields that \( {CF} \) and \( {CN} \) are isogonal conjugates with respect to the vertex \( C \) in \( \bigtriangleup {EDC} \) . Also, note that \( \bigtriangleup {ABC} \sim \bigtriangleup {EDC} \) . The segments \( {CM} \) and \( {CN} \) are their respective me... | Yes |
Let \( {ABC} \) be a triangle with circumcircle \( k \) . The tangent lines to \( k \) at \( A \) and \( B \) intersect at the point \( D \) . Let \( {CD} \) intersect \( k \) for the second time at the point \( E \) . The projections of \( E \) onto \( {AB},{BC} \) and \( {CA} \) are \( N, P \) and \( M \), respective... | By Problem 3.11, we have\n\n\n\nthat the points \( M, N \) and \( P \) are collinear (they lie on the Simson line). Since \( {PBNE} \) is inscribed in the circle with diameter \( {BE} \) and \( {AENM} \) is inscribed... | Yes |
Let \( {ABC} \) be a triangle. The tangent lines to its circumcircle at the points \( A \) and \( B \) intersect at \( T \) . The line \( {CT} \) intersects the circumcircle of \( \bigtriangleup {ABC} \) for the second time at the point \( D \) . Let \( {CL}(L \in \) \( {AB} \) ) be the angle bisector of \( \angle {ACB... | Problem 4.4.1 yields that \( {CT} \) is a symmedian in \( \bigtriangleup {ABC} \) . Thus, the quadrilateral \( {ADBC} \) is harmonic. Hence,\n\n\[ \n{AC}.{BD} = {AD}.{BC} \Rightarrow \frac{AC}{BC} = \frac{AD}{BD}.\n\]\n\nThe angle bisector theorem applied to \( \bigtriangleup {ABC} \) gives\n\n\[ \n\frac{AC}{BC} = \fra... | Yes |
Let \( {ABC} \) be a triangle. The tangent lines to its circumcircle at the points \( A \) and \( B \) intersect at \( T \) . The line \( {CT} \) intersects the circumcircle of \( \bigtriangleup {ABC} \) for the second time at the point \( D \) . Let \( E \) be the reflection of \( D \) with respect to \( {AB} \) and l... | Let \( N \) be the midpoint of \( {CD} \) . As in Problem 4.4.6 we deduce that \( \angle {CBN} = \angle {DBA} \) . Moreover, the symmetry gives \( \angle {DBA} = \angle {ABE} \) . Hence, the lines \( {BE} \) and \( {BN} \) are isogonal conjugates with respect to the vertex \( B \) in \( \bigtriangleup {ABC} \) . Analog... | Yes |
Let \( {ABC} \) be a triangle. The tangent lines to its circumcircle at the points \( A \) and \( B \) intersect at \( T \) . The line \( {CT} \) intersects the circumcircle of \( \bigtriangleup {ABC} \) for the second time at the point \( D \) . Let \( N \in \) \( {CD} \) be a point such that \( \angle {NBC} = \angle ... | By Problem 4.4.1, we have that \( {CT} \) is a symmedian in \( \bigtriangleup {ABC} \) . Thus, the quadrilateral \( {ADBC} \) is harmonic. On the other hand, we have that \( \angle {ABD} = \angle {ACD} \) . Then \( \angle {NBC} = \angle {DBA} \), and hence the lines \( {BA} \) and \( {BN} \) are isogonal conjugates wit... | Yes |
Problem 4.4.7. Let \( {ABC} \) be a triangle. Let \( E, F \) and \( M \) be the midpoints of \( {AC},{BC} \) and \( {EF} \), respectively. The segment \( {CD} \) is an altitude in \( \bigtriangleup {ABC} \) . Prove that the circumcircles of \( \bigtriangleup {ECF},\bigtriangleup {BDF} \) and \( \bigtriangleup {ADE} \) ... | Solution. Let the circumcircle of \( \bigtriangleup {EFC} \) intersect the segment \( {DM} \) at the point \( L \) . Let \( N \) be the symmetric point to \( L \) with respect to \( M \) . Then ELFN is a parallelogram. We have\n\n\[ {AE} = {EC} = {ED} \]\n\nand\n\n\[ {BF} = {FC} = {FD}\text{.} \]\n\nHence, \( \bigtrian... | Yes |
Problem 4.4.8. Let \( {ABC} \) be a triangle. Let \( M \) be the midpoint of \( {AB} \) and let \( D \) be the intersection point of the tangent lines to the circumcircle of \( \bigtriangleup {ABC} \) at the points \( A \) and \( B \) . The projections of \( D \) onto the lines \( {CA} \) and \( {CB} \) are \( E \) and... | Solution. Let \( {CM} \cap {EF} = H \) . Note that Problem 4.4.1 yields \( \angle {ACM} = \angle {DCB} \) . It is clear that \( {EDFC} \) is a cyclic quadrilateral.\n\nTherefore, \( \angle {ECH} = \angle {DCB} = \angle {DEF} = {90}^{ \circ } - \angle {HEC} \), which means that \( \angle {EHC} \) is right. | Yes |
Problem 4.4.9. Let \( {ABC} \) be a triangle. Let \( N \) be the midpoint of the altitude \( {CD} \) and let \( M \) be the midpoint of \( {AB} \) . Let \( L \) be the Lemoine point of \( \bigtriangleup {ABC} \) . Prove that the points \( N, L \) and \( M \) are collinear. | Solution. By Problem 2.32, we have that the Lemoine point is the center of the second Lemoine circle. Hence, there exists a rectangle PQKE, whose diagonals intersect at \( L \) and such that the points \( P \) and \( Q \) lie on \( {AB} \), and the points \( E \) and \( K \) lie on \( {AC} \) and \( {BC} \), respective... | Yes |
Let \( k \) be a circle. A point \( C \) is chosen outside of \( k \) . Let the tangent lines from \( C \) to \( k \) touch the circle at the points \( A \) and \( B \) . Prove that the incenter of \( \bigtriangleup {ABC} \) lies on \( k \) . | Solution. Let \( I \) be the midpoint of the smaller arc \( \overset{⏜}{AB} \) of \( k \) . Then \( \angle {CAI} = \frac{\overset{⏜}{AI}}{2} \) and \( \angle {BAI} = \frac{\overset{⏜}{BI}}{2} \) . Thus, \( \angle {CAI} = \angle {BAI} \) . Hence, \( {AI} \) is the bisector of \( \angle {BAC} \) . Analogously, \( {BI} \)... | Yes |
Problem 4.5.2. The circles \( {k}_{1} \) and \( {k}_{2} \) touch each other externally at the point \( I \) . Let \( {l}_{1} \) and \( {l}_{2} \) be the common external tangent lines of the two circles. Let the points of tangency of \( {l}_{1} \) and \( {l}_{2} \) to \( {k}_{2} \) be \( A \) and \( B \) , respectively.... | Solution. Observe that \( {l}_{1} \) and \( {l}_{2} \) are symmetric with respect to the line that connects the centers of \( {k}_{1} \) and \( {k}_{2} \), and note that \( I \) lies on that line. Hence, \( I \) is the midpoint of the arc \( \overset{⏜}{CD} \) of \( {k}_{1} \), and it is also the midpoint of the arc \(... | Yes |
Problem 4.5.3. Let \( {ABC} \) be a triangle with incenter \( I \) . The incircle of \( \bigtriangleup {ABC} \) touches \( {AB} \) at \( N \) . The \( C \) -excircle of \( \bigtriangleup {ABC} \) touches \( {AB} \) at \( M \) . Let \( P \) and \( Q \) be the orthogonal projections of \( A \) and \( B \) onto the line \... | Solution. Let \( {I}_{c} \) be the \( C \) -excenter of \( \bigtriangleup {ABC} \) . Then \( C, I \) and \( {I}_{c} \) are collinear.\n\nThe quadrilateral \( {I}_{c}{AIB} \) is cyclic and hence\n\n\[ \angle {I}_{c}{AB} = \angle {BI}{I}_{c} \]\n\nObserve that \( A{I}_{c}{PM} \) is also cyclic, and thus \( \angle {I}_{c}... | Yes |
Problem 4.5.4. Let \( {ABC} \) be a triangle with incircle \( \omega \) and with incenter \( I \). Denote the orthogonal projection of \( B \) onto the line \( {AI} \) by \( K \). Let \( N \) and \( M \) be the points of tangency of \( {AC} \) and \( {BC} \) to \( \omega \), respectively. Prove that the points \( M, N ... | Solution. Using \( \angle {IMB} = \)\n\n\n\n\( \angle {IKB} = {90}^{ \circ } \), we obtain that \( {IBKM} \) is cyclic. Hence,\n\n\[ \angle {KMB} = \angle {KIB} \]\n\n\[ = {90}^{ \circ } - \frac{\angle {ACB}}{2} \]\n... | Yes |
Problem 4.5.5. Let \( {ABC} \) be a triangle with incircle \( \omega \) and with incenter \( I \) . Let \( {AB} \) touch \( \omega \) at the point \( N \) . Let \( K \) and \( M \) be the midpoints of \( {CN} \) and \( {AB} \), respectively. Prove that \( K, I \) and \( M \) are collinear. | Solution. Apply Newton's Theorem (Problem 5.4.6) to the degenerate quadrilateral \( {ANBC} \) , in which \( \angle {ANB} = {180}^{ \circ } \) . The midpoints of its diagonals \( {AB} \) and \( {CN} \), and its incenter \( I \) are collinear. | Yes |
Problem 4.5.6. Let \( {ABC} \) be a triangle with incircle \( k \) and incenter \( I \) . Let \( {AB} \) touch \( k \) at the point \( P \) . Let \( \widetilde{CH}\left( {H \in {AB}}\right) \) be an altitude in \( \bigtriangleup {ABC} \) . Denote the midpoint of \( {CH} \) by \( M \) . Let \( {k}_{c}\left( {I}_{c}\righ... | Solution. Let \( Q \) be the reflection of \( P \) with respect to \( I \) . Let \( h \) be the homothety with center \( C \) that sends \( k \) to \( {k}_{c} \) . Then \( h\left( Q\right) = F \) . Hence, the points \( C, Q \) and \( F \) are collinear and \( \frac{CQ}{CF} = \frac{IQ}{{I}_{c}F} \) .\n\nObserve that \( ... | Yes |
Problem 4.5.7. Let \( {ABC} \) be a triangle. Let its incircle touch the sides \( {BC},{CA} \) and \( {AB} \) at the points \( F, E \) and \( D \), respectively. Let its \( C \) -excircle touch the lines \( {BC},{CA} \) and \( {AB} \) at the points \( Q, P \) and \( M \), respectively. Let \( {MN}\left( {N \in {PQ}}\ri... | Solution. It is enough to prove that\n\n\n\n\[ \frac{\sin \angle {PCN}}{\sin \angle {NCQ}} = \frac{\sin \angle {FCH}}{\sin \angle {HCE}}. \]\n\nWe have that\n\n\[ \angle {DEF} = {90}^{ \circ } - \frac{\beta }{2},\ang... | Yes |
Problem 4.5.8. Let \( {ABC} \) be a triangle with an \( A \) -excircle \( {\omega }_{a}\left( {I}_{a}\right) \) and a \( B \) -excircle \( {\omega }_{b}\left( {I}_{b}\right) \) . These circles touch \( {AB},{BC} \) and \( {AC} \) as shown in the figure. Let \( {PE} \cap {MN} = R \) and \( {MD} \cap {PQ} = S \) . Prove ... | Solution. Observe that \( {I}_{b}, C \) and \( {I}_{a} \) lie on the external bisector of \( \angle {ACB} \) . We will prove that \( \angle {QCS} = {90}^{ \circ } - \frac{\gamma }{2} \), and then it will follow that \( S \in {I}_{a}{I}_{b} \) . Analogously, we will get that \( R \in {I}_{a}{I}_{b} \) .\n\nWe know that ... | Yes |
Let \( {ABC} \) be a triangle with an \( A \) -excircle \( {\omega }_{a} \) and a \( B \) -excircle \( {\omega }_{b} \) . These circles touch \( {AB},{BC} \) and \( {AC} \) as shown in the figure. Let \( {PE} \cap {MD} = F \) . Prove that \( {CF} = r \), where \( r \) is the inradius of \( \bigtriangleup {ABC} \) . | Problem 4.5.10 gives \( {CF} \bot {AB} \) . But we know that \( \angle {BPE} = \) \( \angle {BEP} = \angle {FEC} = \frac{\beta }{2} \), which means that \( \angle {EFC} = {90}^{ \circ } + \frac{\beta }{2} \) .\n\nThe law of sines, applied to \( \bigtriangleup {CEF} \), yields\n\n\[ \n{CF} = \frac{{CE}\sin \frac{\beta }... | Yes |
Problem 4.5.10. Let \( {ABC} \) be a triangle with an \( A \) -excircle \( {\omega }_{a} \) and a \( B \) -excircle \( {\omega }_{b} \) . These circles touch \( {AB},{BC} \) and \( {AC} \) as shown in the figure. Let \( {PE} \cap {MD} = U \) and \( {PQ} \cap {MN} = F \) . Prove that \( {CF} \bot {AB} \) and \( {CU} \bo... | Solution. We have \( \angle {CQF} = {90}^{ \circ } + \frac{\alpha }{2},\angle {CNF} = {90}^{ \circ } + \frac{\beta }{2},{CQ} = s - b \) and \( {CN} = s - a \) . The law of sines, applied to \( \bigtriangleup {CFN} \) and \( \bigtriangleup {CFQ} \), yields \[ \frac{\sin \angle {NFC}}{\sin \angle {CFQ}} = \frac{\left( {s... | Yes |
Problem 4.5.11. Let \( {ABC} \) be a triangle and let its excircles touch \( {AB},{BC} \) and \( {AC} \) as shown in the figure. Prove that the lines \( {PQ},{MN} \) and \( {AB} \) are concurrent. | Solution. Let \( {MN} \cap {AB} = K \) . We know that \( {AM} = s - c,{CM} = s - a \) , \( {CN} = s - b,{NB} = s - c,{PA} = s - b \) and \( {BQ} = s - a \) . By Menelaus’ Theorem, applied to \( \bigtriangleup {ABC} \) and the points \( K, M \) and \( N \), it follows that\n\n\[ \n\frac{BK}{KA} \cdot \frac{AM}{MC} \cdot... | Yes |
Let \( {ABC} \) be a triangle. Let \( K \) be the midpoint of \( {AB} \) . The line \( {CL}\left( {L \in {AB}}\right) \) is the angle bisector of \( \angle {ACB} \) . The \( B \) - excircle of \( \bigtriangleup {ABC} \) touches \( {AC} \) at the point \( N \) . The \( A \) -excircle of \( \bigtriangleup {ABC} \) touche... | We know that \( {CN} = s - a \) and \( {CP} = s - b \) . Define \( \overrightarrow{CA} = \overrightarrow{x} \) and \( \overrightarrow{BC} = \overrightarrow{y} \) . We will express all necessary vectors as functions of \( \overrightarrow{x} \) , \( \overrightarrow{y}, a, b \) and \( c \) .\n\nObserve that \( \overrighta... | Yes |
Problem 4.5.13. Let \( {ABC} \) be a triangle with an \( A \) -excircle \( {\omega }_{a} \) and a \( B \) -excircle \( {\omega }_{b} \) . These circles touch \( {AB},{BC} \) and \( {AC} \) as shown in the figure. Prove that the perpendicular bisectors of \( {AN} \) and \( {BM} \), and the bisector of \( \angle {ACB} \)... | Solution. Let \( L \) be the intersection point of the two given perpendicular bisectors. Then \( {AL} = {LN} \) and \( {BL} = {LM} \) . Because of the excircles, we have that \( {AM} = {BN} \) . Hence, \( \bigtriangleup {LAM} \cong \bigtriangleup {LNB} \) . Thus, the point \( L \) is equidistant from \( {AM} \) and \(... | Yes |
Problem 4.5.14. Let \( {ABC} \) be a triangle with circumcircle \( k \) . Let \( M \) be the midpoint of \( {AB} \), and let \( N \) be the midpoint of \( \widehat{ACB} \) . Let \( {I}_{1} \) and \( {I}_{2} \) be the incenters of \( \bigtriangleup {ACM} \) and \( \bigtriangleup {BCM} \), respectively. Prove that the po... | Solution. Note that\n\n\n\n\[ \n\angle {I}_{1}C{I}_{2} = \angle {MNB} = \frac{\gamma }{2}.\n\]\n\nSo it is enough to prove that\n\n\[ \n\angle {I}_{1}{NM} = \angle {I}_{2}{NB}.\n\]\n\nLet \( {I}_{1}^{\prime } \) be t... | Yes |
Problem 4.5.15. Let \( {ABC} \) be a triangle with median \( {CM} \) . Let \( {k}_{1}\left( {I}_{1}\right) \) and \( {k}_{2}\left( {I}_{2}\right) \) be the incircles of \( \bigtriangleup {ACM} \) and \( \bigtriangleup {BCM} \), respectively. Let \( {k}_{3}\left( {I}_{3}\right) \) and \( {k}_{4}\left( {I}_{4}\right) \) ... | Solution. Let the circles \( {k}_{1},{k}_{2},{k}_{3} \) and \( {k}_{4} \) touch the line \( {AB} \) at the points \( E, P, D \) and \( Q \), respectively. We have\n\n\[ \n\angle {I}_{3}M{I}_{4} = {90}^{ \circ } \Rightarrow \angle {I}_{4}{MQ} = \angle M{I}_{1}E \Rightarrow \bigtriangleup M{I}_{1}E \sim \bigtriangleup {I... | Yes |
Problem 4.5.16. Let \( {ABC} \) be a triangle with incircle \( k \) and incenter \( I \) . Let \( {k}_{1} \) be the circle that passes through \( A \) and \( B \), and touches \( k \) at the point \( T \) . Let \( {CH} \) be an altitude in \( \bigtriangleup {ABC} \), and let \( M \) be the midpoint of \( {CH} \) . Let ... | Solution. (See also Problem 4.5.6) Without loss of generality, let \( {AC} > \) \( {BC} \) . Denote the circle \( \left( {ABC}\right) \) by \( {k}_{2} \) . Let the tangent line to \( k \) at \( T \) intersect \( {AB} \) at \( D \) .\n\nBy the radical axes theorem, applied to the circles \( k,{k}_{1} \) and \( {k}_{2} \... | Yes |
Let \( {ABC} \) be a triangle with altitudes \( A{A}_{1}, B{B}_{1} \) and \( C{C}_{1} \) . Let the incircle of \( \bigtriangleup {ABC} \) touch \( {AB},{BC} \) and \( {CA} \) at \( {C}_{2},{A}_{2} \) and \( {B}_{2} \), respectively. Denote the midpoints of \( A{A}_{1}, B{B}_{1} \) and \( C{C}_{1} \) by \( M, N \) and \... | Problem 4.5.16 yields that the poles of the lines \( M{A}_{2}, N{B}_{2} \) and \( P{C}_{2} \) with respect to the incircle of \( \bigtriangleup {ABC} \) lie on the radical axis of the incircle and the circumcircle of \( \bigtriangleup {ABC} \) . Thus, these three lines are concurrent. (If we use the notations of Proble... | Yes |
Problem 4.5.18. Let \( {ABC} \) be a triangle with incircle \( \omega \left( I\right) \) that touches \( {AB} \) at \( M \) . Let \( L \) be the the reflection of \( M \) with respect to \( I \) . Let \( {AI} \cap {BC} = N,{BI} \cap {AC} = P \), and \( {PN} \cap {IC} = K \) . Let \( H \) be the foot of the perpendicula... | Solution. Menelaus’ Theorem, applied to \( \bigtriangleup {CSI} \) and the line \( {AB} \), to \( \bigtriangleup {AIC} \) and the line \( {PN} \), and to \( \bigtriangleup {ANC} \) and the line \( {BP} \), yields\n\n\[ \n\frac{CS}{SI} \cdot \frac{IA}{AN} \cdot \frac{NB}{BC} = 1 \n\]\n\n\[ \n\frac{AN}{NI} \cdot \frac{IK... | Yes |
Problem 4.5.19. Let \( {ABC} \) be a triangle with incircle \( \omega \) that touches the sides \( {AB},{BC} \) and \( {CA} \) at the points \( M, N \) and \( P \), respectively. A circle \( {k}_{1} \) is constructed such that it touches \( {AB} \) at \( M \) . Let \( L \) be the intersection point of the tangent lines... | Solution. Let \( \angle {PAS} = {2\varphi } \n\n\n\nand \( \angle {NBQ} = {2\psi } \) . Then \( {BN} = \) \( {BM} = {BQ} \) and \( {AP} = {AM} = {AS} \) . Hence, \( \angle {NQL} = {90}^{ \circ } + \psi \) and \( \ang... | Yes |
Problem 4.5.20. Let \( {ABC} \) be a triangle with incircle \( \omega \left( I\right) \) that touches \( {BC} \) and \( {CA} \) at the points \( M \) and \( N \), respectively. Let \( D \) be the projection of \( A \) onto \( {BI} \), and let \( T \) be the projection of \( I \) onto \( {CD} \) . Let \( P \) and \( Q \... | Solution. Without loss of generality, let \( Q \) be, located between \( B \) and \( P \) . The pentagon IMCTN is iscribed in the circle \( {k}_{1} \) with diameter \( {CI} \) . Problem 4.5.4 yields that the points \( D, N \) and \( M \) are collinear.\n\nConsidering the powers of \( D \) with respect to \( {k}_{1} \) ... | Yes |
Problem 4.5.21. Let \( {ABC} \) be a triangle with incircle \( \omega \) that touches \( {AB} \) at \( D \) . Let \( {CD} \cap \omega = E \neq D \) . The circle with center \( B \) and radius \( {BD} \) intersects \( {CD} \) at the point \( F \neq D \) . Let \( {BF} \cap {AE} = K \) . Prove that \( {KF} = {FB} \) . | Solution. Menelaus’ Theorem, applied to \( \bigtriangleup {AKB} \) and the line \( {ED} \) , yields \( \frac{AE}{EK} \cdot \frac{KF}{FB} \cdot \frac{BD}{DA} = 1 \) . | No |
Problem 4.5.23. Let \( {ABC} \) be a triangle with incircle \( \omega \) that touches the sides \( {AB},{BC} \) and \( {CA} \) at the points \( M, F \) and \( E \), respectively. Let \( {CM} \cap \omega = N \neq M \) . Let \( l\parallel {AB} \) be a line through \( C \) . The points \( P \) and \( Q \) are the intersec... | Solution. The parallel lines and the alternate segments theorem give\n\n\[ \angle {ENM} = \angle {EMA} = \angle {EPC} \]\n\nand\n\n\[ \angle {MNF} = \angle {FMB} = \angle {FQC}\text{.} \]\nHence, the quadrilaterals ENCP and FNCQ are cyclic. Therefore,\n\n\[ \angle {ENF} = \angle {ENM} + \angle {MNF} = \angle {AEM} + \a... | Yes |
Problem 4.5.25. Let \( {ABC}\left( {{AC} \geq {BC}}\right) \) be a triangle with incircle \( k \) that touches the sides \( {BC},{CA} \) and \( {AB} \) at the points \( D, E \) and \( F \) , respectively. Let \( L \) and \( K \) be the respective reflections of \( A \) and \( B \) with respect to \( F \) . The lines th... | Solution. Let the secont tangent line from \( N \) to \( k \) touch \( k \) at \( T \), and let it intersect \( {AC} \) at \( {M}^{\prime } \) . We will prove that \( M \equiv {M}^{\prime } \) . \n\nLet \( {K}^{\prime } \in {AB} \) be a point such that \( {M}^{\prime }{K}^{\prime } \bot {AB} \) . We are left to show th... | Yes |
Problem 4.5.26. Let \( {ABC} \) be a triangle with incircle \( \omega \) that touches the sides \( {AB},{BC} \) and \( {CA} \) at the points \( M;N \) and \( P \), respectively. Let \( T \) be the intersection of the segment \( {CI} \) and \( \omega \) . Let \( {MP} \cap {AT} = R \) and \( {MN} \cap {BT} = Q \) . Prove... | Solution. We have \( {CI} \bot {PN} \) . We will show that \( {PN}\parallel {RQ} \) . This is equivalent to proving that \( \frac{PR}{RM} = \frac{NQ}{MQ} \) . Now, \[ \frac{{S}_{\bigtriangleup {ARP}}}{{S}_{\bigtriangleup {AMR}}} = \frac{\sin \angle {RAP}}{\sin \angle {RAM}} = \frac{PR}{RM} \] \[ \frac{{S}_{\bigtriangle... | Yes |
Problem 4.5.27. Let \( {ABC} \) be an isosceles triangle \( \left( {{AC} = {BC}}\right) \) and let \( {AC} > {AB} \) . Its incircle \( \omega \) touches \( {AB} \) and \( {CA} \) at \( M \) and \( N \) , respectively. Let \( P \) be a point of the side \( {BC} \) such that \( {AP} = {AB} \) . Let \( {I}_{1} \) be the i... | Solution. The points \( C,{I}_{1} \) and\n\n\n\n\( M \) lie on the bisector of \( \angle {ACB} \) , which is perpendicular to \( {AB} \), and hence \( \angle {I}_{1}{MA} = {90}^{ \circ } \) . Denote \( \angle {BAC} =... | Yes |
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