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Problem 5.7.2. Let \( {ABCD} \) be a quadrilateral with perpendicular diagonals. Let \( M \) and \( N \) be the midpoints of \( {AB} \) and \( {AD} \), respectively. Prove that the perpendicular from \( N \) to \( {BC} \) and the perpendicular from \( M \) to \( {CD} \) intersect on the line \( {AC} \) . | Solution. Consider a homothety with center \( A \) and coefficient 2.\n\nThe line \( {AC} \) is fixed, and the perpendiculars from \( N \) to \( {BC} \) and from \( M \) to \( {CD} \) are sent to the altitudes from \( D \) and \( B \) in \( \bigtriangleup {BCD} \), respectively.\n\nHence, these three lines intersect at... | Yes |
Problem 5.7.3. Let \( {ABCD} \) be a cyclic quadrilateral with \( {AC} \cap {BD} = \) \( E \) and \( {AC} \bot {BD} \) . Let \( M \) be the midpoint of the side \( {DC} \) . Prove that \( {ME} \bot {AB} \) . | Solution. Let \( {ME} \cap {AB} = H \) . Set \( \angle {EDM} = x \) . Since \( {EM} \) is a median to the hypotenuse in \( \bigtriangleup {DEC} \), we have \( \angle {DEM} = x \) and \( \angle {BEH} = x \) . Thus, \( \angle {ACD} = {90}^{ \circ } - \angle {EDC} = {90}^{ \circ } - x. \) But \( \angle {HBE} = \angle {ACD... | Yes |
Problem 5.7.4. Let \( {ABCD} \) be a cyclic quadrilateral. Let \( M, N \) , \( P \) and \( Q \) be the midpoints of \( {AB},{BC},{CD} \) and \( {DA} \), respectively. The points \( {H}_{1},{H}_{2},{H}_{3} \) and \( {H}_{4} \) are the feet of the perpendiculars from \( M, N \) , \( P \) and \( Q \) to \( {CD},{DA},{AB} ... | Solution. Midsegments give\n\n\n\nthat \( {PN} \) is parallel and equal to \( {MQ} \) .\n\nTherefore, the quadrilateral \( {PNMQ} \) is a parallelogram. Denote the circumcenter of \( {ABCD} \) by \( O \) .\n\nNow we ... | Yes |
Problem 5.7.5. Let \( {ABCD} \) be a quadrilateral. Denote the feet of the perpendiculars from \( D \) to \( {AB} \) and \( {BC} \) by \( M \) and \( N \), respectively, and the feet of the perpendiculars from \( B \) to \( {CD} \) and \( {DA} \) by \( K \) and \( L \) , respectively. Denote the orthocenters of \( \big... | Solution. Let \( k \) be the cir- cle with diameter \( {BD} \) . Then the hexagon \( {BMLDKN} \) is inscribed in \( k \) . Desargues’ Theorem (Problem 7.1), applied to \( \bigtriangleup {MPL} \) and \( \bigtriangleup {KQN} \), yields that the lines \( {PQ},{KM} \) and \( {LN} \) are concurrent if or only if the points ... | Yes |
Problem 5.7.6. Let \( {ABCD} \) be a cyclic quadrilateral with circumcenter \( O \), and let \( {AC} \cap {BD} = E \) . Let \( F \) be a point such that \( {CF} \bot {CD} \) and \( {FB} \bot {AB} \) . Prove that the points \( E, O \) and \( F \) are collinear. | Solution. We have \( {OB} = {OC} \) and \( \angle {OCB} = \angle {OBC} \) . Also,\n\n\[ \angle {ECO} = \angle {DCO} - \angle {DCA} \]\n\n\[ = {90}^{ \circ } - \angle {DBC} - \angle {DBA} = \angle {CBF}\text{.} \]\n\nAnalogously, we deduce that \( \angle {BCF} = \angle {OBE} \) . This implies \( \angle {EBF} = \) \( \an... | Yes |
Problem 5.7.7. Let \( {ABCD} \) be a cyclic quadrilateral with \( {AC} \cap {BD} = \) \( E \) . Let \( M, N, P \) and \( Q \) be the projections of \( E \) onto the lines \( {AB} \) , \( {BC},{CD} \) and \( {DA} \), respectively. Prove that the quadrilateral \( {MNPQ} \) is circumscribed. | Solution. We have that the quadrilaterals AMEQ and QEPD are cyclic. Hence, \( \angle {MQE} = \angle {MAE} \) and \( \angle {EDP} = \angle {EQP} \) .\n\nBut since the quadrilateral \( {ABCD} \) is also cyclic, we have \( \angle {EDP} = \) \( \angle {EAM} \) and hence \( \angle {MQE} = \angle {PQE} \) . This implies that... | Yes |
Problem 5.7.8. Let \( {ABCD} \) be a convex quadrilateral with \( {AC} \cap \) \( {BD} = E \) and \( {AC} \bot {BD} \) . Let \( M, N, P \) and \( Q \) be the projections of \( E \) onto the lines \( {AB},{BC},{CD} \) and \( {DA} \), respectively. Prove that the quadrilateral \( {MNPQ} \) is cyclic. | Solution. We have that\n\n\n\nthe quadrilaterals AMEQ, QEPD, NEPC and NEMB are cyclic. Therefore,\n\n\[ \angle {MQE} = \angle {MAE} \]\n\n\[ \angle {EDP} = \angle {EQP} \]\n\n\[ \angle {ECP} = \angle {ENP} \]\n\n\[ \... | Yes |
Problem 5.7.9. Let \( {ABCD} \) be a cyclic quadrilateral with perpendicular diagonals, and let \( {AC} \cap {BD} = X \) . Let the feet of the perpendiculars from \( X \) to \( {AB},{BC},{CD} \) and \( {DA} \) be \( E, F, G \) and \( H \), respectively. If the midpoints of \( {AB},{BC},{CD} \) and \( {DA} \) are \( M, ... | Solution. We have \( \angle {FGH} = \angle {XGH} + \angle {XGF} = \angle {XDH} + \angle {XCF} \) and \( \angle {FEH} = \angle {XEH} + \angle {XEF} = \angle {XAH} + \angle {XBF} \) . Hence, \( \angle {FGH} + \) \( \angle {FEH} = {180}^{ \circ } \) and the quadrilateral \( {EFGH} \) is cyclic.\n\n\n\nthe quadrilaterals ANMD and \( {BPQC} \) are cyclic.\n\nThen \( \angle {DAN} = \angle {NMP} \) and \( \angle {PBC} = \angle {PQN} \) .\n\nThe quadrilateral \( {ABCD} \) is cyclic, and th... | Yes |
Problem 6.1.1. Let \( k \) be a circle and let \( {AB} \) be a chord in it. The circle \( {k}_{1} \) touches \( {AB} \) and \( k \) at the points \( E \) and \( C \), respectively. Let \( {CE} \cap k = M \neq C \) . Prove that \( M \) is the midpoint of the arc \( \overset{⏜}{AB} \) which does not contain \( C \) . | Solution. Denote the centers of \( k \) and \( {k}_{1} \) by \( O \) and \( {O}_{1} \) , respectively. Then \( O,{O}_{1} \) and \( C \) are collinear. Note that\n\n\[ \angle {OMC} = \angle {OCM} = \angle {O}_{1}{EC} \]\n\nfrom the isosceles triangles \( {MOC} \) and \( E{O}_{1}C \) .\n\nHence, \( {MO}\parallel E{O}_{1}... | Yes |
Let \( k \) be a circle and let \( {AB} \) be a chord in it. The circle \( {k}_{1} \) touches \( {AB} \) and \( k \) at \( P \) and \( D \), respectively. The circle \( {k}_{2} \) touches \( {AB} \) and \( k \) at \( Q \) and \( C \), respectively. Assume that \( {k}_{1} \) and \( {k}_{2} \) lie in the same half-plane ... | Problem 6.1.1 yields that the points \( D, P \) and \( M \), as well as the points \( C, Q \) and \( M \) are collinear. Observe that \( {CM} \) is the angle bisector of \( \angle {ACB} \). Hence, \( \angle {ACM} = \angle {BCM} = \angle {ABM} = \angle {BAM} \). We obtain that \( \bigtriangleup {BMC} \sim \bigtriangleup... | Yes |
Problem 6.1.3. Let \( {l}_{1} \) and \( {l}_{2} \) be two parallel lines, \( A \in {l}_{1} \) and \( C \in {l}_{2} \) . Let \( {k}_{1} \) and \( {k}_{2} \) be circles touching \( {l}_{1} \) and \( {l}_{2} \) at \( A \) and \( C \), respectively, and let they touch each other externally at \( E \) . Prove that the point... | Solution. Let \( l \) be the com-\n\n\n\nmon internal tangent line of the circles \( {k}_{1} \) and \( {k}_{2} \) . Let \( l \cap {l}_{1} = B \) and \( l \cap {l}_{2} = D \) .\n\nObserve that \( {CD} = {ED} \) , \( {... | Yes |
Let \( {l}_{1} \) and \( {l}_{2} \) be two parallel lines. A circle \( {k}_{1} \) touches \( {l}_{1} \) and \( {l}_{2} \) at the points \( M \) and \( R \), respectively. A circle \( {k}_{2} \) touches \( {l}_{1} \) and \( {k}_{1} \) externally at the points \( A \) and \( D \), respectively. A circle \( {k}_{3} \) tou... | Problem 6.1.3\n\n\n\nyields that the points \( A, C \) and \( N \), as well as the points \( A, D \) and \( R \) are collinear.\n\nNote that \( \angle {ARN} = \) \( \angle {MAR} \) and \( \angle {MAD} = \) \( \angle ... | Yes |
Problem 6.1.5. Let \( k \) and \( {k}_{1} \) be circles that touch each other internally at \( C \) . Let \( {k}_{1} \) be the smaller of the two circles. Let \( D, E \in {k}_{1} \) and \( {DE} \cap k = \{ A, B\} \) . Assume that \( D \) lies between \( A \) and \( E \) . Prove that \( \angle {ACD} = \angle {BCE} \) . | Solution. Consider the homothety \( h \) with center \( C \), such that \( h\left( {k}_{1}\right) = \) \( k \) . Let \( h\left( D\right) = F \) and \( h\left( E\right) = L \) . Then \( h \) sends the line \( {AB} \) to the line \( {FL} \), hence \( {FL}\parallel {AB} \) . The quadrilateral \( {LFAB} \) is a cyclic trap... | Yes |
Problem 6.1.6. Let \( k \) be a circle with diameter \( {AB} \) . A circle \( {k}_{1} \) touches the line \( {AB} \) and the circle \( k \) internally at the points \( L \) and \( C \) , respectively. Prove that \( \angle \left( {{AC},{CL}}\right) = {45}^{ \circ } \) . | Solution. Problem 6.1.1 yields that \( {CL} \) is an angle bisector of \( \angle {ACB} \) . Since \( {AB} \) is a diameter, we have \( \angle {ACB} = {90}^{ \circ } \) . Therefore, \( \angle {ACL} = {45}^{ \circ } \). | Yes |
Problem 6.1.7. The circles \( k \) and \( {k}_{1} \) touch each other internally at the point \( C \) . Let \( {k}_{1} \) be the smaller of the two circles. A circle \( {k}_{2}\left( O\right) \), where \( O \in {k}_{1} \), touches \( k \) . Let \( {k}_{1} \cap {k}_{2} = \{ P, Q\} \) . Let \( {PQ} \cap k = \{ A, B\} \) ... | Solution. Let \( {k}_{2} \cap k = D,{DA} \cap {k}_{2} = X \) and \( {DB} \cap {k}_{2} = Y \) . Then Problem 6.10.11 yields that \( {MX} \) and \( {NY} \) are the common external tangent lines of \( {k}_{1} \) and \( {k}_{2} \) . Denote the intersection point of these tangents by \( Z \) .\n\nNow Problem 4.5.1, applied ... | Yes |
Problem 6.1.8. Let \( {k}_{1} \) and \( {k}_{2} \) be circles that touch another circle \( k \) internally at the points \( A \) and \( B \), respectively. Let \( {k}_{1} \cap {k}_{2} = \{ C, D\} \) . Prove that the intersection point \( L \) of the angle bisectors of \( \angle {CAD} \) and \( \angle {CBD} \) lies on \... | Solution. The angle bisector theorem, applied to \( \bigtriangleup {ACD} \) and \( \bigtriangleup {BCD} \), yields that it suffices to show that \( \frac{AC}{AD} = \frac{BC}{BD} \) . Let \( E \) be the intersection point of the tangent lines to \( k \) at \( A \) and \( B \) . Observe that \( E \) lies on the radical a... | Yes |
Problem 6.1.9. Let \( k\left( {O, r}\right) \) be a circle and let \( A, B \in k \) . A circle \( {k}_{1}\left( {{O}_{1},{r}_{1}}\right) \) touches \( k \) at \( A \) and intersects the segment \( {AB} \) at \( M \) . The circle \( {k}_{2}\left( {{O}_{2},{r}_{2}}\right) \) touches \( k \) at \( B \) and passes through ... | Solution. It is clear that the points \( A,{O}_{2} \) and \( O \), as well as the points \( B,{O}_{1} \) and \( O \) are collinear. Then \[ \angle {MA}{O}_{2} = \angle {BAO} = \angle {ABO} \] \[ = \angle {MB}{O}_{1} = \angle {BM}{O}_{1}\text{.} \] Hence, \( {AO}\parallel M{O}_{1} \) . Analogously, \( {BO}\parallel {\ma... | Yes |
Problem 6.1.10. (Casey’s Theorem.) Let \( k \) be a circle and let \( {k}_{1},{k}_{2} \) , \( {k}_{3} \) and \( {k}_{4} \) touch \( k \) . Let \( {t}_{ij} \) be the length of the common external tangent segment of \( {k}_{i} \) and \( {k}_{j} \) . Let \( {l}_{ij} \) be the length of the common internal tangent segment ... | Solution. We will prove Casey's Theorem only in the case when all circles touch \( k \) internally, and its converse only in the case when \( {t}_{13}{t}_{24} = \) \( {t}_{12}{t}_{34} + {t}_{14}{t}_{23} \), because the proofs of the other cases are analogous. Approaching the problems, let us prove some lemmas:\n\nLemma... | Yes |
Lemma 1. Two circles \( {k}_{1}\left( {{O}_{1};{r}_{1}}\right) \) and \( {k}_{2}\left( {{O}_{2};{r}_{2}}\right) \) are given, and they touch \( k\left( {O;R}\right) \) internally at the points \( A \) and \( B \), respectively. Let the length of their common tangent segment be \( t \) . Then\n\n\[ t = \frac{{AB}\sqrt{\... | Proof. Let one of the com-\n\nmon external tangent lines of \( {k}_{1} \) and \( {k}_{2} \) touch \( {k}_{1} \) at the point \( M \) and \( {k}_{2} \) at the point \( N \) . Let \( \angle {AOB} = \varphi \) .\n\nThe Pythagorean Theorem, applied in the right-angled trapezoid \( {\mathrm{O}}_{1}{\mathrm{O}}_{2}{MN} \), y... | Yes |
Lemma 2. Let \( {k}_{1}\left( {{O}_{1};{r}_{1}}\right) ,{k}_{2}\left( {{O}_{2};{r}_{2}}\right) ,{k}_{3}\left( {{O}_{3};{r}_{3}}\right) \) be circles. Let \( {t}_{ij} \) denote the length of the common external tangent segment of \( {k}_{i} \) and \( {k}_{j} \) . If \( {t}_{12} + {t}_{23} = {t}_{13} \), then there exist... | Proof. Without loss of generality, let \( {k}_{3} \) be the smallest circle (if \( {k}_{2} \) is the smallest one, the proof is analogous). We decrease the radii of the three circles by \( {r}_{3} \), such that \( {k}_{3} \) turns into a point \( - {O}_{3} \), and \( {k}_{1} \) and \( {k}_{2} \) turn into the smaller c... | Yes |
Lemma 3. Let \( \varphi \) be an inversion with center \( O \) and radius \( r \) . It sends the circle \( {k}_{1}\left( {{O}_{1};{r}_{1}}\right) \) to the circle \( {k}_{1}^{\prime }\left( {{O}_{1}^{\prime };{r}_{1}^{\prime }}\right) \) . Let \( {t}_{O{k}_{1}} \) be the length of the tangent segment from the point \( ... | Proof. We will use the notations on the figure. The properties of inversion yield \( {OM} = \frac{{r}^{2}}{{t}_{O{k}_{1}}} \), and the intercept theorem gives the equality \( \frac{OM}{ON} = \frac{M{O}_{1}^{\prime }}{N{O}_{1}} \) . Therefore, \( {r}_{1}^{\prime } = \frac{{r}^{2}{r}_{1}}{{t}_{O{k}_{1}}^{2}} \) . | Yes |
Lemma 4. Let \( \varphi \) be an inversion with center \( O \) and radius \( r \) . It sends the circle \( {k}_{1}\left( {{O}_{1};{r}_{1}}\right) \) to the circle \( {k}_{1}^{\prime }\left( {{O}_{1}^{\prime };{r}_{1}^{\prime }}\right) \), and the circle \( {k}_{2}\left( {{O}_{2};{r}_{2}}\right) \) to the circle \( {k}_... | Proof. Denote \( {O}_{1}{O}_{2} = n \) and \( {O}_{1}^{\prime }{O}_{2}^{\prime } = m \) .\n\nThen (see the figure) we can express \( y \) by using the Pythagorean Theorem. We have\n\n\[ \n{y}^{2} = {m}^{2} - {r}_{1}^{\prime 2} - {r}_{2}^{\prime 2} + 2{r}_{1}^{\prime }{r}_{2}^{\prime } \n\]\n\nWe apply the law of cosine... | Yes |
Problem 6.2.1. Let \( {k}_{1}\left( {O}_{1}\right) \) and \( {k}_{2}\left( {O}_{2}\right) \) be two nonintersecting circles. The tangent segments from \( {O}_{1} \) to \( {k}_{2} - {O}_{1}A \) and \( {O}_{1}B\left( {A, B \in {k}_{2}}\right) \), intersect \( {k}_{1} \) at the points \( M \) and \( N \), respectively. Th... | Solution. The configuration is symmetric with respect to the line \( {O}_{1}{O}_{2} \) . Hence, this line is the perpendicular bisector of both \( {MN} \) and \( {KL} \) . Therefore, it suffices to prove that \( {MK}\parallel {O}_{1}{O}_{2} \) . The quadrilateral \( {O}_{1}{DA}{O}_{2} \) is cyclic because \( \angle {O}... | Yes |
Problem 6.2.2. Let \( k,{k}_{1} \) and \( {k}_{2} \) be circles with radii \( r,{r}_{1} \) and \( {r}_{2} \) , respectively. The circles \( {k}_{1} \) and \( {k}_{2} \) do not intersect each other and they touch \( k \) internally at the points \( A \) and \( B \), respectively. The tangent lines \( {AC} \) and \( {AD}... | Solution. Consider the\n\n\n\nhomothety centered at \( A \) that sends \( {k}_{1} \) to \( k \) . It sends the incircle of the curvilinear triangle \( {ANM} \) to \( {k}_{2} \) and hence \( \frac{{r}^{\prime }}{{r}_{... | Yes |
Let \( k,{k}_{1},{k}_{2} \) and \( {k}_{3} \) be circles. The circles \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) do not intersect each other and they touch \( k \) internally at the points \( A, B \) and \( C \), respectively. The common external tangent lines of each pair from \( {k}_{1},{k}_{2} \) and \( {k}_{3} \), tha... | Problem 6.2.4, applied to the circles \( k,{k}_{1} \) and the incircle of \( \bigtriangleup {DEF} \) , yields that \( A, D \) and the internal homothetic center of \( k \) and the incircle of \( \bigtriangleup {DEF} \) are collinear.\n\nAnalogously, \( {BE} \) and \( {CF} \) also pass through the internal homothetic ce... | Yes |
Problem 6.2.6. Let \( k,{k}_{1},{k}_{2} \) and \( {k}_{3} \) be circles. The circles \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) do not intersect each other, and \( k \) does not contain each of them fully. The common external tangent lines of each pair from \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) , that separate them from... | Solution. Problem 6.2.3, applied to the circles \( k,{k}_{1} \) and the incircle of \( \bigtriangleup {DEF} \) , yields that \( A, D \) and the external homothetic center of \( k \) and the incircle of \( \bigtriangleup {DEF} \) are collinear. Analogously, \( {BE} \) and \( {CF} \) also pass through the external homoth... | Yes |
Problem 6.2.7. Let \( k,{k}_{1},{k}_{2} \) and \( {k}_{3} \) be circles. Prove that the lines \( S\left( {k,{k}_{1}}\right) L\left( {{k}_{2},{k}_{3}}\right), S\left( {k,{k}_{2}}\right) L\left( {{k}_{1},{k}_{3}}\right) \) and \( S\left( {k,{k}_{3}}\right) L\left( {{k}_{1},{k}_{2}}\right) \) are concurrent. | Solution. Desargues' Theorem (Problem 7.1) yields that it suffices to prove that the intersection points \[ S\left( {k,{k}_{1}}\right) S\left( {k,{k}_{2}}\right) \cap L\left( {{k}_{1},{k}_{3}}\right) L\left( {{k}_{2},{k}_{3}}\right) \] \[ S\left( {k,{k}_{2}}\right) S\left( {k,{k}_{3}}\right) \cap L\left( {{k}_{1},{k}_{... | Yes |
Problem 6.2.8. Let \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) be circles. We draw the tangent lines from \( S\left( {{k}_{2},{k}_{3}}\right) \) to \( {k}_{1} \) and from \( S\left( {{k}_{1},{k}_{2}}\right) \) to \( {k}_{3} \) . Prove that the convex quadrilateral, formed by these two pairs of lines is circumscribed. | Solution. Let one of the tangent lines from \( S\left( {{k}_{1},{k}_{2}}\right) \) to \( {k}_{3} \) intersect the tangent lines from \( S\left( {{k}_{2},{k}_{3}}\right) \) to \( {k}_{1} \) at the points \( A \) and \( B \) . We will prove that the circle \( k \), which touches \( {AB}, S\left( {{k}_{2},{k}_{3}}\right) ... | Yes |
Problem 6.2.9. Let \( {k}_{1},{k}_{2},{k}_{3} \) and \( {k}_{4} \) be circles. Prove that the lines \( S\left( {{k}_{1},{k}_{3}}\right) S\left( {{k}_{2},{k}_{4}}\right), S\left( {{k}_{1},{k}_{2}}\right) S\left( {{k}_{3},{k}_{4}}\right) \) and \( L\left( {{k}_{1},{k}_{4}}\right) L\left( {{k}_{2},{k}_{3}}\right) \) are c... | Desargues’ Theorem (Problem 7.1) yields that it suffices to prove that\nthe points\n\n\[ S\left( {{k}_{1},{k}_{3}}\right) S\left( {{k}_{1},{k}_{2}}\right) \cap S\left( {{k}_{3},{k}_{4}}\right) S\left( {{k}_{2},{k}_{4}}\right) \]\n\n\[ S\left( {{k}_{1},{k}_{2}}\right) L\left( {{k}_{1},{k}_{4}}\right) \cap S\left( {{k}_{... | Yes |
Problem 6.2.10. Let \( {ABCD} \) be a square and let \( k \) be a circle that contains it. The circles \( {k}_{1},{k}_{2},{k}_{3} \) and \( {k}_{4} \) are inscribed in the curvilinear triangles, formed by the intersections of the square's lines and the circle \( k \) . Let the circle with respect to the vertex \( A \) ... | Solution. Let \( \omega \) be the incircle of the square \( {ABCD} \) . Monge’s Theorem (see Problems 6.2.3 and 6.2.4), applied to the circles \( k \) , \( \omega \) and \( {k}_{1} \), yields that the points \( A, S\left( {k,{k}_{1}}\right) \) and \( L\left( {k,\omega }\right) \) are collinear, and so the line \( S\lef... | Yes |
Problem 6.2.11. Let \( {ABCDE} \) be a regular pentagon, and let \( k \) be a circle that contains it. The circles \( {k}_{1},{k}_{2},{k}_{3},{k}_{4} \) and \( {k}_{5} \) are inscribed in the curvilinear triangles, formed by the intersections of the pentagon's diagonals and the circle \( k \) . Let the circle with resp... | Solution. Let \( \omega \) be the circle that touches all diagonals of \( {ABCDE} \). Monge's Theorem (see Problems 6.2.3 and 6.2.4), applied to the circles \( k \) , \( \omega \) and \( {k}_{1} \), yields that the points \( A, S\left( {k,{k}_{1}}\right) \) and \( L\left( {k,\omega }\right) \) are collinear, and so the... | Yes |
Problem 6.3.1. Let \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) be circles, such that their disks do not intersect each other. All their common internal tangent lines form the hexagon \( {ABCDEF}. \) Prove that the lines \( {AD},{BE} \) and \( \widetilde{C}F \) are concurrent. | Solution. Let \( {AB} \cap {DE} = Z,{AF} \cap {DC} = X \) and \( {FE} \cap {BC} = Y \) . We apply Desargues’ Theorem (Problem 7.1) to \( \bigtriangleup {FAZ} \) and \( \bigtriangleup {DCY} \) . We have that \( {FA} \cap {DC} = X,{AZ} \cap {CY} = B \) . Also, the lines \( {XB},{FZ} \) and \( {DY} \) are concurrent (see ... | Yes |
Problem 6.3.2. Let \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) be circles, such that their disks do not intersect each other. All their common external tangent lines form the hexagon \( {ABCDEF} \). Let \( {AB} \cap {DE} = Z,{AF} \cap {DC} = X \) and \( {FE} \cap B\bar{C} = Y \). Prove that the lines \( {ZF},{XB} \) and \(... | Solution. Let \( {AF} \cap {DE} = P,{FE} \cap {DC} = Q,{ED} \cap {BC} = R \) , \( {CD} \cap {AB} = S,{BC} \cap {AF} = T \) and \( {BA} \cap {EF} = K \) . Let the centers of \( {k}_{1} \) , \( {k}_{2} \) and \( {k}_{3} \) be \( {I}_{1},{I}_{2} \) and \( {I}_{3} \), and let their radii be \( {r}_{1},{r}_{2} \) and \( {r}... | Yes |
Problem 6.3.3. Let \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) be circles, such that their disks do not intersect each other. The common external homothetic centers of each pair of circles are \( F, D \) and \( B \), as shown in the figure. For each pair of circles, one of their common external tangent lines is drawn - the... | Solution. Let \( {XZ} \cap {BD} = Q,{BF} \cap {YZ} = P \) and \( {FD} \cap {XY} = R \) . Desargues’ Theorem (Problem 7.1), applied to \( \bigtriangleup {BFD} \) and \( \bigtriangleup {ZYX} \), yields that our statement is equivalent to the statement that the points \( P, Q \) and \( R \) are collinear.\n\nWe will prove... | Yes |
Let \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) be circles, such that their disks do not intersect each other. All their common internal tangent lines form the hexagon \( {ABCDEF} \) . Let \( {AB} \cap {DE} = Z,{AF} \cap {DC} = X \) and \( {FE} \cap B\bar{C} = Y \) . Prove that the lines \( {AY},{EX} \) and \( {CZ} \) are ... | We apply Desargues’ Theorem (Problem 7.1) to \( \bigtriangleup {FAZ} \) and \( \bigtriangleup {DCY} \) . We have that \( {FA} \cap {DC} = X,{AZ} \cap {CY} = B \) . In addition, the lines \( {XB},{FZ} \) and \( {DY} \) are concurrent (see Problem 6.3.2). Therefore, the lines \( {ZY},{FD} \) and \( {AC} \) are concurrent... | Yes |
Let \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) be circles, such that their disks do not intersect each other. All their common external tangent lines are drawn. For each pair of circles, let us consider one of their common external tangent lines - the one that leaves all three circles completely in the same half-plane. Th... | Let \( {XZ} \cap {AC} = Q,{BC} \cap {YZ} = P \) and \( {AB} \cap {XY} = R \) . Monge’s Theorem (Problem 6.2.3) yields that the points \( P, Q \) and \( R \) are collinear.\n\nNow Desargues’ Theorem (Problem 7.1), applied to the triangles \( {ABC} \) and \( {XYZ} \), yields that the lines \( {AX},{BY} \) and \( {CZ} \) ... | Yes |
Problem 6.3.8. Let \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) be circles, such that their disks do not intersect each other. Let their centers be \( {I}_{1},{I}_{2} \) and \( {I}_{3} \), and let their radii be \( {r}_{1},{r}_{2} \) and \( {r}_{3} \), respectively. Let \( B \) be the internal homothetic center of \( {k}_{2... | Solution. (See the figure of Problem 6.3.2) We have that the points \( {I}_{1} \) , \( F \) and \( {I}_{2} \), as well as the points \( {I}_{2}, B \) and \( {I}_{3} \), and the points \( {I}_{3}, D \) and \( {I}_{1} \) are collinear. Then\n\n\[ \n\frac{{I}_{2}B}{B{I}_{3}} \cdot \frac{{I}_{3}D}{D{I}_{1}} \cdot \frac{{I}... | Yes |
Problem 6.4.1 Let \( {ABC} \) be a triangle. A circle \( {k}_{1}\left( {O}_{1}\right) \) touches \( {BC} \) and \( {AC} \) at the points \( N \) and \( P \), respectively. The point \( {I}_{1} \) lies on the perpendicular bisector of \( {AB} \) . Denote the midpoint of \( {AB} \) by \( M \) . Prove that the points \( M... | First solution. Since \( {I}_{1} \) belongs to the perpendicular bisector of \( {AB} \), we have \( A{I}_{1} = B{I}_{1} \) . Note that \( {I}_{1}P = {I}_{1}N \) . Hence, \( \bigtriangleup A{I}_{1}P \cong \bigtriangleup B{I}_{1}N \) and the equalities \( {AP} = {BN} \) and \( {CN} = {CP} \) hold. It follows that \( \fra... | Yes |
Problem 6.4.2. Let \( {ABO} \) be a triangle and let \( {AO} = {BO} \) . A circle \( {k}_{1}\left( I\right) \) touches \( {OA} \) and \( {OB} \) at the points \( A \) and \( B \), respectively. A point \( P \) is chosen on the side \( {AB} \) . A line \( l \) passes through \( P \), such that \( {IP} \bot l \) . Let th... | Solution. Observe that \( \angle {IAB} = \angle {IBA} \) . The quadrilaterals \( {PIBD} \) and \( {PICA} \) are cyclic.\n\nHence, \( \angle {PDI} = \angle {PBI} = \angle {IAP} = \angle {ICP} \), which means that \( {IC} = \) \( {ID} \) . Thus, \( \bigtriangleup {CID} \) is an isosceles triangle, and \( {IP} \) is both ... | Yes |
Problem 6.4.3. (The butterfly theorem) Let \( {ABCD} \) be a quadrilateral inscribed in a circle with center \( O \) . The diagonals of \( {ABCD} \) intersect at the point \( E \) . A line perpendicular to \( {EO} \) at \( E \) intersects \( {AD} \) and \( {BC} \) at the points \( P \) and \( Q \), respectively. Prove ... | Solution. Note that \( \angle {DAE} = \angle {CBE} \) and \( \angle {AED} = \angle {BEC} \) . Hence, \( \bigtriangleup {AED} \sim \bigtriangleup {BEC} \) . Denote the midpoints of \( {AD} \) and \( {BC} \) by \( M \) and \( N \) , respectively. Then \( {EM} \) and \( {EN} \) are the corresponding medians of the similar... | Yes |
Problem 6.4.4. Let \( k\\left( O\\right) \) be a circle. The points \( P \) and \( Q \) are external with respect to \( k \) . Denote the orthogonal projection of \( O \) onto \( {PQ} \) by \( H \) . The four tangent lines from \( P \) and \( Q \) to \( k \) form the quadrilateral \( {ABCD} \) . Prove that the line \( ... | Solution. Let the points of tangency of \( {AB},{BC},{CD} \) and \( {DA} \) to \( k \) be \( E, F, M \) and \( N \), respectively. Brianchon’s Theorem, applied to the circumscribed hexagon \( {AEBCMD} \), yields that the lines \( {AC},{BD} \) and \( {EM} \) are concurrent. Let their intersection point be \( L \) .\n\nA... | Yes |
Problem 6.4.5. Let \( {ABC} \) be a triangle. Let \( k\left( O\right) \) be the circumcircle of the triangle and let \( D \) be the reflection of \( C \) with respect to \( O \) . The tangent line to \( k \) at the point \( D \) intersects \( {AB} \) at the point \( E \) . Let \( {OE} \) intersect \( {AC} \) and \( {BC... | Solution. Let \( {F}^{\prime } \in {AC} \) be a point, such that \( {F}^{\prime }D\parallel {BC} \) . Let \( {P}^{\prime } \in {BC} \) be a point, such that \( D{P}^{\prime }\parallel {AC} \) . Then clearly the quadrilateral \( D{F}^{\prime }C{P}^{\prime } \) is a parallelogram and the point \( O \) bisects its diagona... | Yes |
Problem 6.4.6. Let \( k\\left( O\\right) \) be a circle. The circle \( {k}_{1} \) passes through \( O \) . Let \( {k}_{1} \\cap k = \\{ M, N\\} \) . Let \( A \\in {k}_{1} \) and \( P \\in k \) be arbitrary points. Denote one of the intersection points of \( k \) and the reflection of the line \( {PA} \) with respect to... | Solution. In \( \\bigtriangleup {APQ} \), the line \( {OA} \) is a bisector of \( \\angle {PAQ} \) and \( O \) lies on the perpendicular bisector of \( {PQ} \) .\n\nHence, \( O \) lies on the circumcircle of \( \\bigtriangleup {APQ} \) . Observe that the lines \( {MN},{PQ} \) and \( {AO} \) are the radical axes of the ... | Yes |
Problem 6.5.3. Let \( {k}_{1} \) and \( {k}_{2} \) be circles, such that their disks do not intersect each other. The points \( A, C, E \) and \( G \) lie on \( {k}_{1} \) . The points \( B \) , \( D, F \) and \( H \) lie on \( {k}_{2} \) . The lines \( {AB},{GH},{EF} \) and \( {CD} \) are the common external and inter... | Solution. We will use the notations on the figure. Since the powers of \( M \) with respect to \( {k}_{1} \) and \( {k}_{2} \) are equal (due to \( M{A}^{2} = M{B}^{2} \) ), we conclude that \( M \) belongs to the radical axis of \( {k}_{1} \) and \( {k}_{2} \) . Analogously, we see that the points \( N, P \) and \( Q ... | Yes |
Problem 6.5.4. Let \( {k}_{1} \) and \( {k}_{2} \) be circles. Assume that \( {k}_{1} \cap {k}_{2} = \) \( \{ D, E\} \) . Let \( S \) be an arbitrary point on the line \( {DE} \) external for the segment \( {DE} \) . Let the tangent lines from \( S \) to \( {k}_{1} \) and \( {k}_{2} \) be \( {SA}\left( {A \in {k}_{1}}\... | Solution. Observe that the line \( {DE} \) is the radical axis of \( {k}_{1} \) and \( {k}_{2} \), and that \( S \in {DE} \). Hence, the powers of \( S \) with respect to \( {k}_{1} \) and \( {k}_{2} \) are equal. Since these powers are respectively equal to \( A{S}^{2} \) and \( S{B}^{2} \), we conclude that \( A{S}^{... | Yes |
Problem 6.5.5. Let \( k \) be a circle. The circles \( {k}_{1} \) and \( {k}_{2} \) touch \( k \) at the points \( A \) and \( B \), respectively. Let \( S \) be the intersection point of the tangent lines to \( k \) at \( A \) and \( B \) . Prove that \( S \) lies on the radical axis of \( {k}_{1} \) and \( {k}_{2} \)... | Solution. Observe that \( \rho \left( {{k}_{1}, k}\right) \equiv {AS} \) and \( \rho \left( {{k}_{2}, k}\right) \equiv {BS} \) . Since \( S = \rho \left( {{k}_{1}, k}\right) \cap \rho \left( {{k}_{2}, k}\right) \), we conclude that \( S \) is the radical center of the three circles. | No |
Let \( k \) be a circle. The circles \( {k}_{1} \) and \( {k}_{2} \) touch \( k \) internally at the points \( A \) and \( B \), respectively. Denote the external homothetic center of \( {k}_{1} \) and \( {k}_{2} \) by \( C \) . Prove that the points \( A, B \) and \( C \) are collinear. | Solution. Monge's Theorem (Problem 6.2.3), applied to the circles \( {k}_{1},{k}_{2} \) and \( k \) , yields that the external homothetic centers of the pairs \( \left( {{k}_{1},{k}_{2}}\right) ,\left( {{k}_{2}, k}\right) \) and \( \left( {{k}_{1}, k}\right) \) are collinear. These centers coincide with the points \( A... | Yes |
Problem 6.5.7. Let \( {k}_{1} \) and \( {k}_{2} \) be two circles and let \( {k}_{1} \) be the smaller one. Let \( C \) be their external homothetic center. A line passing through \( C \) intersects \( {k}_{1} \) and \( {k}_{2} \) at the points \( E \) and \( D \) . These two points do not correspond to each other unde... | Solution. Let \( M \) be the second intersection point of \( {DE} \) and \( {k}_{2} \) . Let \( Y \) be the intersection of \( {DM} \) and the tangent line to \( {k}_{2} \) at the point \( M \) .\n\nSince \( h\left( E\right) = M \), we have \( {EX}\parallel {MY} \) . Therefore,\n\n\[ \angle {EDX} = \angle {DMY} = \angl... | Yes |
Problem 6.5.8. Let \( {k}_{1} \) and \( {k}_{2} \) be circles and let \( {k}_{1} \cap {k}_{2} = \{ A, B\} \) . The circles \( {k}_{3} \) and \( {k}_{4} \) touch \( {k}_{1} \) and \( {k}_{2} \) externally. Let \( C \) be the external homothetic center of \( {k}_{3} \) and \( {k}_{4} \) . Prove that the points \( A, B \)... | Solution. Let \( {k}_{1} \cap {k}_{3} = M,{k}_{2} \cap {k}_{3} = N,{k}_{2} \cap {k}_{4} = P \) and \( {k}_{1} \cap {k}_{4} = Q \) . Applying Problem 6.2.4, we get that the points \( M, Q \) and \( C \), as well as the points \( N, P \) and \( C \) are collinear.\n\nProblem 6.8.1, applied to all four given circles, yiel... | Yes |
Problem 6.5.9. Let \( {k}_{1} \) and \( {k}_{2} \) be circles and let \( {k}_{1} \cap {k}_{2} = \{ C, D\} \) . The tangent lines to \( {k}_{1} \) at the points \( C \) and \( D \) intersect each other at the point \( A \) . Point \( B \) is defined analogously for the circle \( {k}_{2} \) . Let \( X \in {CD} \) be such... | Solution. The quadrilateral \( {NDMC} \) is harmonic because \( {NM} \cap \) \( t\left( {D,{k}_{1}}\right) \cap t\left( {C,{k}_{1}}\right) = \widetilde{A} \) . Analogously, the quadrilateral \( {DLCK} \) is harmonic.\n\nLet \( t\left( {M,{k}_{1}}\right) \cap t\left( {N,{k}_{1}}\right) = Y \) . Then \( Y \in {CD} \) and... | Yes |
Problem 6.5.10. The points \( A, B, C, D, E \) and \( F \) lie on a circle \( k \) in this order. The circles \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) pass through the pair of points \( \left( {A, F}\right) ,\left( {B, C}\right) \) and \( \left( {D, E}\right) \), respectively. Let \( {k}_{1} \cap {k}_{2} = S \) and let ... | Solution. Let \( {AF} \cap {BC} = X,{BC} \cap {ED} = Y \) and \( {AF} \cap {ED} = Z \) .\n\nDesargues’ Theorem (Problem 7.1), applied to \( \bigtriangleup {XYZ} \) and \( \bigtriangleup {SKL} \) , yields that it suffices to show that the lines \( {XS},{YK} \) and \( {ZL} \) are concurrent.\n\nSince \( {XY} \) and \( {X... | Yes |
Problem 6.6.1. Let \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) be three congruent circles with radii \( R \) . Let \( {k}_{1} \cap {k}_{2} = \{ A, X\} ,{k}_{2} \cap {k}_{3} = \{ B, X\} \) and \( {k}_{3} \cap {k}_{1} = \{ C, X\} \) . Prove that the circumradius of \( \bigtriangleup {ABC} \) equals \( R \) . | Solution. Let \( {CX} \) intersect\n\n\n\n\( {k}_{2} \) for the second time at the point \( M \) . Denote the circumradius of \( \bigtriangleup {ABC} \) by \( {R}_{1} \) . Since the radii of \( {k}_{1},{k}_{2} \) and... | Yes |
Problem 6.6.2. Let \( {k}_{1} \) and \( {k}_{2} \) be two circles with equal radii and let \( {k}_{1} \cap {k}_{2} = \{ E, H\} \) . The points \( C \in {k}_{2} \) and \( D \in {k}_{1} \) are chosen, such that \( C \) and \( D \) lie inside of \( {k}_{1} \) and \( {k}_{2} \), respectively. Let \( {EC} \cap {k}_{1} = \{ ... | Solution. Let \( {AD} \cap {BC} = L \) . Denote the midpoint of \( {EL} \) by \( P \) . Problem 3.8, applied to the quadrilateral ECLD, yields that the points \( K, M \) and \( P \) are collinear.\n\nWe will prove that \( N \) lies on the same line. It suffices to show that \( {MK}\parallel {LH} \) because \( {PN} \) i... | Yes |
Problem 6.6.3. Let \( {k}_{1} \) and \( {k}_{2} \) be two circles with equal radii, and let \( {k}_{1} \cap {k}_{2} = \{ B, D\} \) . A point \( E \in {k}_{1} \) is chosen, such that \( E \) does not lie inside of \( {k}_{2} \) . The lines \( {EB} \) and \( {ED} \) intersect \( {k}_{2} \) for the second time at the poin... | Solution. Let \( H \) be the or-\n\n\n\nthogonal projection of \( E \) onto \( {BD} \) and \( Q \) be the reflection of \( E \) with respect to \( {BD} \) .\n\nIt is clear that \( Q \in {k}_{2} \) because \( {k}_{1} ... | Yes |
Let \( {ABC} \) be a triangle. The point \( D \) lies on the segment \( {AB} \), such that the radii of the incircles of \( \bigtriangleup {ADC} \) and \( \bigtriangleup {DBC} \) are equal. Prove that the \( C \) -excircles of \( \bigtriangleup {ADC} \) and \( \bigtriangleup {BDC} \) have equal radii. | Denote the incircles of \( \bigtriangleup {ADC} \) and \( \bigtriangleup {DBC} \) by \( {k}_{1} \) and \( {k}_{2} \) , respectively. Let \( r \) be the radius of these circles. The common external tangent line of \( {k}_{1} \) and \( {k}_{2} \) (different from \( {AB} \) ) intersects \( {AC} \) and \( {BC} \) at the po... | Yes |
Problem 6.7.1. Let \( {ABC} \) be a triangle with an altitude \( {CH}(H \in \) \( {AB}) \) . A circle \( {k}_{1}\left( L\right) \), where \( L \in {AB} \), touches \( {AC} \) and \( {BC} \) at the points \( M \) and \( N \), respectively. Prove that \( {CH} \) bisects \( \angle {MHN} \) . | Solution. Note that\n\n\n\n\( \bigtriangleup {MLC} \cong \bigtriangleup {NLC} \) . Hence, \( \angle {MLC} = \angle {NLC} \) .\n\nIt is clear that the points \( M, C, N, H \) and \( L \) lie on the circle with diamete... | Yes |
Problem 6.7.2. The points \( A, B, C \) and \( D \) lie on a circle with diameter \( {AB} \) . The tangent lines to the circle at the points \( C \) and \( D \) intersect each other at the point \( F \) . Let \( E = {AC} \cap {BD} \) . Prove that \( {FE} \bot {AB} \) . | Solution. Let \( H = {FE} \cap \n\n\n\n\( {AB} \) .\n\nDenote \( \angle {ABD} = \alpha \) and \( \angle {BAC} = \beta \) . Observe that \( \angle {ADB} = \angle {ACB} = {90}^{ \circ } \) and \( \angle {BAD} = {90}^{ ... | Yes |
Problem 6.7.3. Let \( {ABC} \) be a triangle with \( \angle {ACB} < {90}^{ \circ } \) . The circle \( {k}_{1} \) with diameter \( {AB} \) and center \( M \) intersects \( {AC} \) and \( {BC} \) at the points \( {B}_{1} \) and \( {A}_{1} \), respectively. Denote the intersection point of the tangent lines to \( {k}_{1} ... | Solution. It is easy to see\n\n\n\nthat \( \angle {BM}{A}_{1} = {180}^{ \circ } - {2\beta } \) and \( \angle {AM}{B}_{1} = {180}^{ \circ } - {2\alpha } \) .\n\nHence, \( \angle {A}_{1}M{B}_{1} = {2\alpha } + \) \( {2... | Yes |
Problem 6.7.4. Let \( {ABCD} \) be a quadrilateral with \( \angle {DAB} = \) \( \angle {DCB} = {90}^{ \circ } \) . Let \( X \) and \( Y \) be the orthogonal projections of \( D \) and \( B \) onto \( {AC} \), respectively. Prove that \( {AX} = {CY} \) . | Solution. Denote \( \angle {DCA} = \alpha \) and \( \angle {CAB} = \beta \) . We get \( \angle {BCY} = {90}^{ \circ } - \alpha \) and \( \angle {CBY} = \alpha \) .\n\nAlso, \( \angle {XAD} = {90}^{ \circ } - \beta \) and \( \angle {XDA} = \beta \) .\n\nLet \( R \) be the circumradius of \( {ABCD} \) . Then\n\n\( {AX} =... | Yes |
Problem 6.7.5. Let \( k \) be a circle with diameter \( {AB} \) . The circles \( {k}_{1} \) and \( {k}_{2} \) touch each other externally at the point \( C \) . Moreover, the centers of \( {k}_{1} \) and \( {k}_{2} \) lie on the segment \( {AB} \), and those circles touch \( k \) internally at the points \( A \) and \(... | Solution. Observe that \( \angle {ANC} = \angle {BPC} \) . Now we get the construction of Problem 6.7.4 and the result follows from there. | No |
Problem 6.7.6. Let \( k \) be a circle with diameter \( {AB} \) . Let \( C \in k \) be an arbitrary point. Denote the orthogonal projection of \( C \) onto \( {AB} \) by \( H \) . Let \( {k}_{1}\left( {C,\dot{C}\dot{H}}\right) \) be a circle. If \( k \cap {k}_{1} = \{ M, N\} \), prove that the points \( K \) , \( M \) ... | Solution. Consider the inversion \( I\left( {C,{CH}}\right) \) . Observe that \( I\left( M\right) = M \) and \( I\left( N\right) = N \) . Then \( I\left( k\right) = \) \( {MN} \) .\n\nLet \( {K}_{1} = I\left( K\right) \) . Hence, \( C{K}_{1} = \frac{C{H}^{2}}{CK} \) . Since \( {2CK} = \) \( {CH} \), we have \( C{K}_{1}... | Yes |
Problem 6.7.7. Let \( k\left( I\right) \) be a circle with diameter \( {AB} \) . Let \( D \in k \) be an arbitrary point. Denote the orthogonal projection of \( D \) onto \( {AB} \) by \( C \) . An arbitrary point \( X \) lies on \( {DC} \) . The circles \( {k}_{1}\left( {I}_{1}\right) \) and \( {k}_{2}\left( {I}_{2}\r... | Solution. Let \( {DC} \cap k = \{ D, E\} \) . Applying Problem 6.1.1, we obtain that \( {PX} \) and \( {QX} \) are the bisectors of \( \angle {DPE} \) and \( \angle {DQE} \), respectively. Observe that \( B \) and \( A \) lie on these lines, respectively.\n\nThus, \( \frac{DP}{PE} = \frac{DX}{XE} = \frac{DQ}{QE} \) and... | No |
Problem 6.7.8. The points \( A, C \) and \( B \) lie on a line in this order. Consider the semicircles \( k,{k}_{1} \) and \( {k}_{2} \) with diameters \( {AB},{AC} \) and \( {BC} \) and centers \( O,{O}_{1} \) and \( {O}_{2} \), respectively, lying in the same half-plane with respect to \( {AB} \) . Denote the common ... | Solution. Denote all the points of tangency as shown in the figure. Let the radii of \( k,{k}_{1} \) and \( {k}_{2} \) be \( R,{r}_{1} \) and \( {r}_{2} \), respectively. It is clear that \( \dot{R} = {r}_{1} + {r}_{2} \) . We are going to express \( P{C}^{2} \) in two different ways - by using the right-angled trapezo... | Yes |
Let \( {ABC} \) be a triangle with \( \angle {ACB} = {90}^{ \circ } \). The tangent lines to the circumcircle of \( \bigtriangleup {AB}\bar{C} \) at \( B \) and \( C \) intersect each other at the point \( R \). Let \( P \) and \( N \) be the midpoints of \( {BC} \) and the arc \( \overset{⏜}{AC} \), respectively. Deno... | Denote the midpoint of \( {AB} \) by \( M \). Hence, \( M \) is the center of the circumcircle of \( \bigtriangleup {ABC} \), and the points \( M, P \) and \( R \) are collinear. On the other hand, \( {MN} \) contains the midpoint of \( {AC} \).\n\nHence, \( {MN}\parallel {BC} \) and \( {MP}\parallel {AC} \), which mea... | Yes |
Problem 6.8.1. Let \( {k}_{1},{k}_{2},{k}_{3} \) and \( {k}_{4} \) be circles. Let \( {k}_{1} \) touch \( {k}_{2} \) and \( {k}_{4} \) externally at the points \( A \) and \( D \), respectively. Let \( {k}_{3} \) touch \( {k}_{2} \) and \( {k}_{4} \) externally at the points \( B \) and \( C \), respectively. Prove tha... | Solution. Consider an inversion centered at \( A \) with an arbitrary rá- dius.\n\nNow the statement follows from Problem 6.1.3. | No |
Problem 6.8.2. Let \( {CDEF} \) be a cyclic quadrilateral. Let \( A \) and \( B \) be two internal points, such that the circumcircles of \( \bigtriangleup {DAE} \) and \( \bigtriangleup {CBD} \) intersect for the second time at the point \( N \), which is internal to \( {CDEF} \) . Let the circumcircles of \( \bigtria... | Solution. We consecutively have\n\n\[ \angle {AMB} = {360}^{ \circ } - \angle {AMF} - \angle {BMF} = \angle {AEF} + \angle {BCF} \]\n\n\[ = {180}^{ \circ } - \angle {AED} - \angle {BCD} = \angle {AND} + \angle {BND} - {180}^{ \circ } \]\n\n\[ = {180}^{ \circ } - \angle {ANB}\text{.} \]\n\nTherefore, the quadrilateral A... | Yes |
Problem 6.8.3. Let \( {k}_{1},{k}_{2},{k}_{3} \) and \( {k}_{4} \) be circles. Let \( {k}_{1} \) touch \( {k}_{2} \) and \( {k}_{4} \) externally at the points \( A \) and \( D \), respectively. Let \( {k}_{3} \) touch \( {k}_{2} \) and \( {k}_{4} \) externally at the points \( B \) and \( C \), respectively. The commo... | Solution. Problem 6.8.1 yields that the quadrilateral \( {ABCD} \) is cyclic. Then the perpendicular bisectors of the sides \( {AB},{BC},{CD} \) and \( {DA} \) are concurrent. Also, \( {FA} = {FB} \) because they are tangent segments to \( {k}_{2} \) . Thus, the perpendicular bisector of \( {AB} \) coincides with the a... | Yes |
Problem 6.8.4. Let \( k \) be a circle. The circles \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) touch \( k \) internally at the points \( A, B \) and \( C \), respectively. Let \( {k}_{1} \) touch \( {k}_{2} \) externally at the point \( {C}_{1} \), let \( {k}_{2} \) touch \( {k}_{3} \) externally at the point \( {A}_{1} \... | Solution. This problem is a special case of Problem 6.2.7. | Yes |
Problem 6.8.5. Let \( k,{k}_{1},{k}_{2},{k}_{3},{c}_{1},{c}_{2} \) and \( {c}_{3} \) be circles. Each one of \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) touches each of the other two externally. The circle \( k \) touches these three circles internally. Each of the circles \( {c}_{1},{c}_{2} \) and \( {c}_{3} \) touches tw... | Solution. We apply an inversion centered at the point \( P \) with an arbitrary radius. We will use the prime symbol to denote the images of the objects after the inversion has been applied to them.\n\nThe circles \( k \) and \( {k}_{1} \) map to parallel lines, and the circles \( {c}_{2},{k}_{2},{k}_{3} \) and \( {c}_... | Yes |
Problem 6.8.6. Let \( k,{k}_{1},{k}_{2},{k}_{3},{c}_{1},{c}_{2} \) and \( {c}_{3} \) be circles. Each one of \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) touches each of the other two externally. The circle \( k \) touches these three circles internally. Each of the circles \( {c}_{1},{c}_{2} \) and \( {c}_{3} \) touches tw... | Solution. Problem 6.8.5 yields that the hexagon DYXLKE is cyclic. Then the internal angle bisectors of the hexagon \( {R}_{1}{V}_{1}{T}_{1}{W}_{1}{S}_{1}{U}_{1} \) at the vertices \( {S}_{1},{T}_{1} \) and \( {R}_{1} \) coincide with the perpendicular bisectors of \( {YX} \) , \( {LK} \) and \( {ED} \), respectively. T... | Yes |
Problem 6.8.8. Let \( k,{k}_{1}\left( {O}_{1}\right) ,{k}_{2}\left( {O}_{2}\right) ,{k}_{3}\left( {O}_{3}\right) ,{c}_{1},{c}_{2} \) and \( {c}_{3} \) be circles. Each one of \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) touches each of the other two externally. The circle \( k \) touches these three circles internally. Each... | Solution. We apply the results that we obtained while proving Problem 6.8.7. If \( {L}_{1},{L}_{2} \) and \( {L}_{3} \) are the external homothetic centers of the pairs of circles among \( {k}_{1},{k}_{2} \) and \( {k}_{3} \), then \( {FZ} \cap {O}_{2}{O}_{3} = {L}_{1},{QZ} \cap {O}_{1}{O}_{3} = {L}_{2} \) and \( {FQ} ... | Yes |
Problem 6.8.9. Let \( k,{k}_{1}\left( {O}_{1}\right) ,{k}_{2}\left( {O}_{2}\right) ,{k}_{3}\left( {O}_{3}\right) ,{c}_{1}\left( {I}_{1}\right) ,{c}_{2}\left( {I}_{2}\right) \) and \( {c}_{3}\left( {I}_{3}\right) \) be circles. Each one of \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) touches each of the other two externally.... | Solution. This problem follows from Desargues' Theorem (Problem 7.1) and the fact that the homothetic centers of \( \left( {{c}_{1},{c}_{2}}\right) \) and \( \left( {{k}_{1},{k}_{2}}\right) \), of \( \left( {{c}_{1},{c}_{3}}\right) \) and \( \left( {{k}_{1},{k}_{3}}\right) \), and of \( \left( {{c}_{2},{c}_{3}}\right) ... | Yes |
Problem 6.8.10. Let \( k \) be a circle. The circles \( {k}_{1},{k}_{2},{k}_{3},{k}_{4},{k}_{5} \) and \( {k}_{6} \) touch \( k \) internally. Also, the circle \( {k}_{i} \) touches the circles \( {k}_{i - 1} \) and \( {k}_{i + 1} \) externally \( \left( {{k}_{7} \equiv {k}_{1}}\right) \) . Let \( {k}_{i} \) touch \( k... | Solution. Consider an inversion centered at \( {A}_{1} \) with an arbitrary radius. We will use the prime symbol to denote the images of the objects after the inversion has been applied to them. The problem is now reformulated as follows:\n\nLet \( {k}^{\prime } \) and \( {k}_{1}^{\prime } \) be two parallel lines, and... | Yes |
Problem 6.8.11. Let the points \( {A}_{1},{A}_{2} \) and \( {A}_{3} \) lie on the circle \( k \) . The circle \( {\omega }_{1} \) touches \( k \) internally at the point \( {A}_{1} \) . The circle \( {k}_{2} \) touches \( k \) internally at the point \( {A}_{2} \) and touches \( {\omega }_{1} \) externally. The circle ... | Solution. Problem 10.8 yields\n\n\n\nthat the circle \( {\omega }_{1} \) touches the circle \( {k}_{3} \) externally.\n\nConsider an inversion centered at \( {A}_{3} \) with an arbitrary radius. We will use the prime... | Yes |
Problem 6.8.12. Let the points \( A, B, C, D, E \) and \( F \) lie on the circle \( k \) . The circle \( {k}_{1} \) passes through the points \( E \) and \( F \) . The circle \( {k}_{2} \) passes through the points \( C \) and \( D \), and touches \( {k}_{1} \) at the point \( L \) . The circle \( {k}_{3} \) passes thr... | Solution. Consider an inversion centered at \( E \) . We will use the prime symbol to denote the images of the objects after the inversion has been applied to them. The problem is now reformulated as follows:\n\nLet \( {k}^{\prime },{k}_{1}^{\prime } \) and \( {k}_{4}^{\prime } \) be three lines passing through the poi... | Yes |
Problem 6.8.13. Let \( {k}_{1} \) and \( {k}_{2} \) be circles which touch each other externally at the point \( O \) . The circles \( {k}_{3} \) and \( {k}_{4} \) also touch each other externally at the point \( O \) . Let \( F \) be the second intersection point of \( {k}_{1} \) and \( {k}_{3} \), let \( A \) be the ... | Solution. Consider an inversion centered at \( O \) with an arbitrary radius. The images of \( {k}_{1} \) and \( {k}_{3} \) are two parallel lines, and the images of \( {k}_{2} \) and \( {k}_{4} \) are also two parallel lines.\n\nWe will use the prime symbol to denote the images of the objects after the inversion has b... | Yes |
Problem 6.8.14. Let \( {k}_{1} \) and \( {k}_{2} \) be circles that intersect each other at the points \( X \) and \( Y \) . A chain of circles, each touching the next one \( {\omega }_{1},{\omega }_{2},\ldots \) is constructed, such that each of these circles touches externally one of \( {k}_{1} \) and \( {k}_{2} \), ... | Solution. Let us consider an inversion centered at \( X \) with an arbitrary radius. The images of the circles \( {k}_{1} \) and \( {k}_{2} \) are two intersecting lines. The chain of circles \( {\omega }_{1},{\omega }_{2},\ldots \) maps to a chain of circles that touch the images of \( {k}_{1} \) and \( {k}_{2} \) .\n... | Yes |
Problem 6.8.15. Let \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) be circles that intersect each other at six points. Let the circle \( {k}_{4} \) touch \( {k}_{2} \) and \( {k}_{3} \) internally and \( {k}_{1} \) externally. Let the circle \( {k}_{5} \) touch \( {k}_{1} \) and \( {k}_{3} \) internally and \( {k}_{2} \) exte... | Solution. We will apply Casey’s Theorem (Problem 6.1.10) three times, and its converse once. Let \( {t}_{ij} \) denote the length of the common external tangent segment of \( {k}_{i} \) and \( {k}_{j} \) . Let \( {l}_{ij} \) denote the length of the common internal tangent segment of \( {k}_{i} \) and \( {k}_{j} \) . W... | Yes |
Let \( N \) be one of the intersection points of the circles \( {k}_{1}\left( {O}_{1}\right) \) and \( {k}_{2}\left( {O}_{2}\right) \) . The points \( F \in {k}_{1} \) and \( Q \in {k}_{2} \) are chosen, such that the points \( N, F \) and \( Q \) lie in the same half-plane with respect to \( {O}_{1}{O}_{2} \), and suc... | Let \( {O}_{1}{O}_{2} \cap {FQ} = E \) and \( {EN} \cap {k}_{1} = \{ N, K\} \) . There exists a homothety \( h \) centered at \( E \), such that \( h\left( {k}_{1}\right) = {k}_{2} \) . Hence, \( h\left( F\right) = Q \) and \( h\left( K\right) = N \) . Thus, \( \widehat{KF} = \widehat{NQ} \) and we obtain \( \angle {EQ... | Yes |
Problem 6.9.2. Let \( {k}_{1} \) and \( {k}_{2} \) be two circles that intersect each other. Let \( {EA} \) be their external common tangent line \( \\left( {E \\in {k}_{1}, A \\in {k}_{2}}\\right) \). Let \( F \\in {k}_{1}, B \\in {k}_{2}, C \\in {k}_{2} \) and \( D \\in {k}_{1} \) be points such that \( C \) and \( D... | Solution. Consider \( \\bigtriangleup {ABC} \) that has a median \( {CD} \) and a circumcircle \( {k}_{2} \). Let the tangent line to \( {k}_{2} \) at \( A \) intersect the circumcircle of \( \\bigtriangleup {CDA} \) at the point \( {E}^{\\prime } \). Let \( {k}_{1}^{\\prime } \) be the circle that passes through the p... | Yes |
Problem 6.9.3. Let \( {k}_{1} \) and \( {k}_{2} \) be two circles. Let \( A, D \in {k}_{1} \) and \( B, C \in {k}_{2} \) be points, such that \( {AB} \) and \( {DC} \) are the common external tangent lines of \( {k}_{1} \) and \( {k}_{2} \) . The line \( {AC} \) intersects \( {k}_{1} \) and \( {k}_{2} \) for the second... | Solution. Due to symmetry, we have \( {AB} = {CD} \) . \n\nConsider the power of \( A \) with respect to \( {k}_{2} \) . We get \( A{B}^{2} = {AF}.{AC} \) . Analogously, \( C{D}^{2} = {CE}.{CA} \) . Thus, \( {CE}.{CA} = {AF}.{CA} \) and therefore \( {AE} = {FC} \) . | Yes |
Let \( {AB} \) be a chord in the circle \( k \) . The points \( C \) and \( D \) lie on the tangent lines to \( k \) at \( A \) and \( B \), respectively. Let \( {AC} = {BD} \) . The points \( D \) and \( C \) lie in different half-planes with respect to the line \( {AB} \) . Let \( M = {CD} \cap {AB} \) . Prove that \... | We have \( \angle {AMC} = \)\n\n\n\n\( \angle {BMD} \) and \( \angle {CAM} = {180}^{ \circ } - \) \( \angle {MBD} \) . Therefore, \( \sin \angle {AMC} = \) \( \sin \angle {BMD} \) and \( \sin \angle {CAM} = \) \( \si... | No |
Problem 6.9.5. Let \( {k}_{1} \) and \( {k}_{2} \) be circles such that \( {k}_{1} \cap {k}_{2} = \{ A, B\} \) . Let \( E \in {k}_{1} \) and \( F \in {k}_{2} \) be points such that \( A \) lies inside of \( \bigtriangleup {BEF} \) , and \( {EF} \) is one of the common external tangent lines of \( {k}_{1} \) and \( {k}_... | Solution. Let \( {AB} \cap {EF} = M \) . Consider the power of \( M \) with respect to \( {k}_{1} \) and \( {k}_{2} \) . We have \( M{F}^{2} = {MA}.{MB} = M{E}^{2} \), so \( M \) is the midpoint of \( {EF} \) . The alternate segments theorem yields \( \angle {BEM} = {180}^{ \circ } - \angle {BAE} = \) \( \angle {BAC} =... | Yes |
Problem 6.9.6. Let \( {k}_{1} \) and \( {k}_{2} \) be circles that intersect at the points \( A \) and \( B \) . Let \( E \in {k}_{1} \) and \( F \in {k}_{2} \) be points, such that \( A \) lies inside of \( \bigtriangleup {BEF} \), and \( {EF} \) is one of the common external tangent lines of \( {k}_{1} \) and \( {k}_... | The alternate segments theorem yields\n\n\[ \angle {BEF} = {180}^{ \circ } - \angle {BAE} = \angle {BAC} = \angle {BFC}, \]\n\n\[ \angle {BCF} = \angle {BFE} \Rightarrow \angle {FBC} = \angle {FBE}. \] | Yes |
Problem 6.9.7. Let \( {k}_{1} \) and \( {k}_{2} \) be circles that touch each other externally at the point \( C \) . The points \( A \in {k}_{1} \) and \( B \in {k}_{2} \) are chosen, such that \( {AB} \) is one of the common external tangent lines of \( {k}_{1} \) and \( {k}_{2} \) . Prove that \( \angle {ACB} = {90}... | Solution. Let \( M \) be the intersection point of \( {AB} \) and the common internal tangent line of the circles at \( C \) . Since \( {MC} = {MA} \) and \( {MC} = {MB} \), we obtain that the points \( A, B \) and \( C \) lie on a circle centered at \( M \) . Hence, \( \angle {ACB} = {90}^{ \circ } \). | Yes |
Let \( {k}_{1} \) and \( {k}_{2} \) be circles such that none of them lies inside of the other one. The points \( A, D \in {k}_{1} \) and \( B, D \in {k}_{2} \) are such that \( {AB} \) and \( {CD} \) are the common external tangent lines of \( {k}_{1} \) and \( {k}_{2} \) . Let \( M \) be the midpoint of \( {CD} \) . ... | Solution. Consider the power of of \( M \) with respect to \( {k}_{1} \) and \( {k}_{2} \) . We get \( {MF}.{MB} = M{C}^{2} = M{D}^{2} = {ME}.{MA} \) . Therefore, the quadrilateral \( {AEFB} \) is cyclic. | Yes |
Problem 6.9.9. Let \( {k}_{1}\left( {O}_{1}\right) \) and \( {k}_{2}\left( {O}_{2}\right) \) be circles such that their disks do not intersect each other. The line \( l \) is a common external tangent line of the two circles, and it touches \( {k}_{1} \) and \( {k}_{2} \) at the points \( A \) and \( B \) , respectivel... | Solution. Observe that \( \angle A{O}_{1}D + \angle E{O}_{2}B = {180}^{ \circ } \) . Then \( \angle {O}_{1}{AD} + \) \( \angle {O}_{1}{DA} + \angle {O}_{2}{BE} + \angle {O}_{2}{EB} = {180}^{ \circ } \) . Since \( {O}_{1}D = {O}_{1}A \) and \( {O}_{2}B = {O}_{2}E \), we obtain \( \angle {EB}{O}_{2} + \angle {DA}{O}_{1} ... | Yes |
Problem 6.10.1. Let \( {k}_{1}\left( {O}_{1}\right) \) and \( {k}_{2}\left( {O}_{2}\right) \) be circles that intersect each other at the points \( A \) and \( B \) . Let \( {k}_{3} \) be the circumcircle of \( \bigtriangleup {O}_{1}B{O}_{2} \), and let \( {k}_{3} \) intersect \( {k}_{1} \) and \( {k}_{2} \) at the poi... | Solution. Let \( \angle {DEB} = \alpha \) . Then \( \angle B{O}_{1}D = {180}^{ \circ } - \alpha \) . Hence, \( \angle {BAD} = {90}^{ \circ } + \frac{\alpha }{2} \) . Denote \( \angle {BDE} = \) \( \beta \) . Then \( \angle {BAE} = {90}^{ \circ } + \frac{\beta }{2} \) . Let \( {A}_{1} \) be the incenter of \( \bigtriang... | Yes |
Problem 6.10.2. Let \( {k}_{1}\left( {O}_{1}\right) \) and \( {k}_{2}\left( {O}_{2}\right) \) be circles that intersect each other at the points \( A \) and \( B \) . Let \( {O}_{2}B \cap {k}_{1} = \{ B, D\} \) and \( {O}_{1}B \cap {k}_{2} = \{ B, C\} \) . Prove that the quadrilateral \( {O}_{1}{O}_{2}{CD} \) is cyclic... | Solution. Observe that \( {O}_{1}D = {O}_{1}B \) and \( {O}_{2}B = {O}_{2}C \) . Hence, \[ \angle {O}_{1}D{O}_{2} = \angle {O}_{1}{BD} \] \[ = \angle {O}_{2}{BC} = \angle {O}_{2}C{O}_{1}\text{. } \] Thus, the quadrilateral \( {\mathrm{O}}_{1}{\mathrm{O}}_{2}\mathrm{{CD}} \) is cyclic. | Yes |
Problem 6.10.3. Let \( k\\left( O\\right) \) be a circle and let \( X \) be a point. The points \( A \) and \( B \) lie on \( k \), and the circumcircles of \( \\bigtriangleup {AOX} \) and \( \\bigtriangleup {BOX} \) intersect \( k \) at the points \( C \) and \( D \), respectively. Prove that \( \\angle {AXB} = \) \( ... | Solution. Let \( E \) be a point on the ray \( O{X}^{ \\rightarrow } \) after \( X \) . Hence,\n\n\[ \n\\angle {CXE} = \\angle {OAC} \n\]\n\n\[ \n= \\angle {OCA} = \\angle {OXA}\\text{.} \n\]\n\nAnalogously, we obtain \( \\angle {DXE} = \\angle {OXB} \) .\n\nSumming the above equations, we get \( \\angle {AXB} = \\angl... | Yes |
Problem 6.10.4. Let \( {k}_{1} \) and \( {k}_{2} \) be circles that intersect each other at the points \( A \) and \( B \) . The points \( C, D, E \) and \( F \) are collinear and lie in this order, such that \( E, C \in {k}_{1} \) and \( D, F \in {k}_{2} \) . Prove that \( \angle {CAD} = \angle {FBE} \) . | Solution. We consecu-\ntively have\n\n\[ \angle {FBE} = \angle {FBA} - \angle {EBA} \]\n\n\[ = \angle {FDA} - \angle {ECA} \]\n\n\[ = \angle {CAD}\text{.} \] | Yes |
Problem 6.10.5. Let \( {k}_{1} \) and \( {k}_{2} \) be circles that intersect each other at the points \( A \) and \( B \) . The point \( C \) lies on \( {k}_{1} \) and the line \( {AC} \) intersects \( {k}_{2} \) for the second time at the point \( D \) . The tangent lines to \( {k}_{1} \) and \( {k}_{2} \) at \( C \)... | Solution. Let \( M \) and \( N \) lie on the extensions of \( {EC} \) and \( {ED} \), as shown in the figure.\n\nThe alternate segments theorem yields\n\n\[ \angle {BDN} = \angle {DAB} \]\n\n\[ = {180}^{ \circ } - \angle {CAB} \]\n\n\[ = {180}^{ \circ } - \angle {MCB}\text{.} \] | No |
Let \( \bigtriangleup {ABC} \) be a triangle with circumcircle \( k \) . The tangent lines to \( k \) at \( A \) and \( B \) intersect at the point \( E \) . The circle \( {k}_{1} \) passes through the points \( A \) and \( C \) and touches the line \( {BC} \) . The circle \( {k}_{2} \) passes through the points \( B \... | The alternate seg-ments theorem yields\n\n\[ \angle {ACF} = \angle {CBF} \]\n\nand\n\n\[ \angle {BCF} = \angle {CAF}\text{.} \]\n\nThe statement of the problem is now the same as that of Problem 4.4.6. | No |
Problem 6.10.8. Let \( {k}_{1} \) and \( {k}_{2} \) be circles that touch each other externally at the point \( X \) . The point \( D \) lies outside of both circles. The points \( A \in {k}_{1} \) and \( B \in {k}_{2} \) are such that \( {DA} \) and \( {DB} \) are tangent lines to \( {k}_{1} \) and \( {k}_{2} \), resp... | Solution. Let \( F \) be a point such that \( {FX} \) is the common internal tangent line of \( {k}_{1} \) and \( {k}_{2} \) . \n\nObserve that \( \angle {FXA} = \angle {DAX} \) and \( \angle {DBX} = \angle {BXF} \) . Hence, we consecutively have \n\n\[ \n\angle {ADB} = {360}^{ \circ } - \angle {DBX} - \angle {BXF} - \... | Yes |
Let \( {k}_{1} \) and \( {k}_{2} \) be circles that intersect each other at the points \( A \) and \( B \) . Let \( C \) be an arbitrary point on \( {k}_{1} \) . The tangent line to \( {k}_{1} \) at \( C \) intersects \( {AB} \) at \( M \) . Denote the reflection of \( C \) with respect to the point \( M \) by \( D \) ... | Solution. Let \( {DE} \) and \( {DF} \) be the tangent lines to \( {k}_{2} \), where \( E, F \in {k}_{2} \) . Hence, \( {EF} \) is the polar line of \( D \) with respect to \( {k}_{2} \) . Denote the midpoints of \( {DE} \) and \( {DF} \) by \( N \) and \( P \), respectively. Then \( N \) and \( P \) lie on the radical... | Yes |
Let \( {k}_{1} \) and \( {k}_{2} \) be circles that intersect each other at the points \( A \) and \( B \) . The points \( C \) and \( D \) lie on \( {k}_{1} \) and \( {k}_{2} \), respectively, and \( {CD} \) is the common external tangent line of the circles which is closer to \( B \) . Let \( {AB} \cap {CD} = E \) . ... | Since \( E \) lies on the radical axis of \( {k}_{1} \) and \( {k}_{2} \) , we have that \( {EC} = {ED} \) . Since \( {AE} = {EF} \), we obtain that the quadrilateral \( {CADF} \) is a parallelogram. Thus, \( \angle {CAD} = \angle {CFD} \) . On the other hand, \[ \angle {MAC} = \angle {FCD} \] \[ \angle {DAN} = \angle ... | Yes |
Problem 6.10.11. Let \( k,{k}_{1} \) and \( {k}_{2} \) be circles, such that \( {k}_{1} \) and \( {k}_{2} \) touch \( k \) internally at the points \( A \) and \( B \), respectively. Let \( {k}_{1} \cap {k}_{2} = \) \( \{ X, Y\} \) . The line \( {XY} \) intersects \( k \) at the points \( B \) and \( D \) . Let \( {DA}... | Solution. Consider the inversion \( I\left( {D,\sqrt{{DM}.{DA}}}\right) \) . Since \( D \) lies on the radical axis of \( {k}_{1} \) and \( {k}_{2} \), it follows that \( I\left( {k}_{1}\right) = {k}_{1} \) and \( I\left( {k}_{2}\right) = {k}_{2} \) . On the other hand, \( I\left( M\right) = A \) and \( I\left( N\right... | Yes |
Problem 6.10.12. Let \( {k}_{2}\left( O\right) \) be a circle that is internally tangent to the circle \( {k}_{1} \) at the point \( C \) . The chord \( {EF} \) at \( {k}_{1} \) is tangent to \( {k}_{2} \) at the point \( D \) . The line passing through \( C \), perpendicular to \( {DC} \), intersects the lines through... | Solution. Let \( {HI} \cap {k}_{2} = \{ C, J\} \) and \( {EJ} \cap {FI} = K \) . The segment \( {JD} \) is a diameter of \( {k}_{2} \) and \( {JD} \bot {EF} \) . It suffices to show that \( {IK} = {IF} \) because of Steiner's Theorem (Problem 5.2.1), applied to the trapezoid DFKJ.\n\nThe intercept theorem yields \( \fr... | Yes |
Problem 6.10.13. Let \( {k}_{1} \) and \( {k}_{2} \) be circles that intersect each other at the points \( A \) and \( B \) . The points \( E \) and \( C \) lie on \( {k}_{1} \) and \( {k}_{2} \), respectively, such that the lines \( {EA} \) and \( {CA} \) touch \( {k}_{2} \) and \( {k}_{1} \), respectively. Let \( {BE... | Solution. Consider the cyclic quadrilateral \( {CBDA} \) . We have \( \angle {ADE} = \) \( \angle {ACB} \) . On the other hand, \( \angle {BEA} = \angle {CFA} \) because the quadrilateral \( {BEAF} \) is cyclic. Then \( \bigtriangleup {EAD} \sim \bigtriangleup {FAC} \) . Hence, \( \angle {EAD} = \angle {FAC} \) . Note ... | Yes |
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