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Problem 4.5.28. Let \( {ABC} \) be a triangle with incenter \( I \) . Let \( {k}_{2}\left( {I}_{2}\right) \) be a circle that touches the segments \( {AC} \) and \( {BC} \). Let \( {k}_{1}\left( {I}_{1}\right) \) be a circle that touches \( {AB} \) at \( G,{AC} \), and \( {k}_{2} \) at \( D \). Let \( {k}_{3}\left( {I}... | Solution. Let \( L \in {AC} \) lie on the common internal tangent line of \( {k}_{1} \) and \( {k}_{2} \). Let \( H \in {BC} \) lie on the common internal tangent line of \( {k}_{3} \) and \( {k}_{2} \). Let \( {HF} \cap {AB} = P,{LD} \cap {AB} = Q \) and \( {HF} \cap {LD} = K \).\n\nThe perpendicular bisectors of \( {... | Yes |
Problem 4.5.30. Let \( {m}_{a},{m}_{b} \) and \( {m}_{c} \) be the Malfatti circles of a triangle \( {ABC} \) with incenter \( I \) . Let \( {m}_{a} \cap {m}_{b} = {T}_{c},{m}_{b} \cap {m}_{c} = {T}_{a} \) and \( {m}_{a} \cap {m}_{c} = {T}_{b} \) . The circles \( {m}_{a} \) and \( {m}_{b} \) touch \( {AB} \) at \( X \)... | Solution. Let \( O \) be the radical center of \( {m}_{a},{m}_{b} \) and \( {m}_{c} \) . | No |
Lemma 1. Let \( {s}_{c} \) be the circle that touches the ray \( {T}_{a}{O}^{ \rightarrow } \) beyond \( O \), the ray \( {T}_{b}{O}^{ \rightarrow } \) beyond \( O \), and the segment \( {AB} \), such that \( {s}_{c} \) and \( C \) lie in the same half-plane, with respect to \( {AB} \). Denote the last point of tangenc... | Proof. The perpendicular bisectors of \( Y{T}_{a}, X{T}_{b} \) and \( {T}_{a}{T}_{b} \) intersect at the center of \( {s}_{c} \) and thus the quadrilateral \( {XY}{T}_{a}{T}_{b} \) is cyclic. Hence, \( {U}_{c} \) is the midpoint of \( {XY} \). The radical axis of \( {m}_{a} \) and \( {m}_{b} \) bisects \( {XY} \), and ... | Yes |
Lemma 2. The second common internal tangent line of \( {s}_{b} \) and \( {s}_{c} \) passes through \( A \) . | Proof. One of the common external tangent lines of \( {s}_{b} \) and \( {m}_{a} \), one of the common external tangent lines of \( {s}_{c} \) and \( {m}_{a} \), and one of the common internal tangent lines of \( {s}_{b} \) and \( {s}_{c} \) are concurrent at the point \( O \) . Problem 5.4.16 yields that the second tan... | Yes |
Lemma 3. The length of the tangent segment from \( {U}_{b} \) to \( {s}_{c} \), and the length of the tangent segment from \( {U}_{c} \) to \( {s}_{b} \) are equal. | Proof. Let \( t\left( {X, k}\right) \) denote the length of the tangent segment from \( X \) to the circle \( k \) . We have\n\n\[ t\left( {{U}_{b},{s}_{c}}\right) - t\left( {{U}_{c},{s}_{b}}\right) = t\left( {O,{s}_{c}}\right) - t\left( {O,{s}_{b}}\right) + {U}_{b}O - {U}_{c}O \]\n\n\[ = t\left( {O,{s}_{c}}\right) - t... | Yes |
Lemma 4. The line \( {AI} \) is the second common internal tangent line of \( {s}_{b} \) and \( {s}_{c} \) . | Proof. Let \( {U}_{b}{U}_{c} \) intersect \( {s}_{b} \) and \( {s}_{c} \) at \( Z \) and \( S \) for the second time, respectively. Denote the centers of \( {s}_{b} \) and \( {s}_{c} \) by \( {I}_{b} \) and \( {I}_{c} \), respectively. Let \( {r}_{b} \) and \( {r}_{c} \) be the radii of those circles, respectively.\n\n... | Yes |
Problem 4.5.32. Let \( {ABC} \) be a triangle. Points \( D \) and \( E \) lie on the segment \( {AB} \) ( \( D \) lies between \( A \) and \( E \) ). Let \( {k}_{1},{k}_{2},{k}_{3} \) and \( {k}_{4} \) be the incircles of \( \bigtriangleup {ADC},\bigtriangleup {AEC},\bigtriangleup {BDC} \) and \( \bigtriangleup {BEC} \... | Solution. Let \( I \) be the incenter of \( \bigtriangleup {ABC} \) . It is clear that \( {I}_{1}{I}_{2} \cap {I}_{3}{I}_{4} = I \) . It suffices to prove that \( {I}_{1}{I}_{4},{I}_{2}{I}_{3} \) and \( {AB} \) are concurrent. By Menelaus’ Theorem, this is equivalent to\n\n\[ \frac{A{I}_{1}}{{I}_{1}I} \cdot \frac{I{I}_... | Yes |
Let \( {ABC} \) be a triangle. Let \( D \) be an arbitrary point on the segment \( {AB} \) . Let \( {k}_{1} \) and \( {k}_{2} \) be the incircles of \( \bigtriangleup {ADC} \) and \( \bigtriangleup {BDC} \) , respectively. Let the common external tangent line (different from \( {AB} \) ) of \( {k}_{1} \) and \( {k}_{2}... | Solution. Denote the points of tangency of \( {k}_{1} \) and \( {k}_{2} \) as shown in the\n\nfigure.\n\nThen we consecutively have\n\n\[ \n{CK} = {CT} - {TK} = {CQ} - {KF} = {CB} - {BQ} - {EF} + {EK} \]\n\n\[ \n= {CB} - {BN} - {MN} + {KH} = {CB} - {BM} + {CH} - {CK} \]\n\n\[ \n= {BC} - {AB} + {AM} + {CP} - {CK} = {BC}... | Yes |
Let \( {ABC} \) be a triangle with altitude \( {CH} \) and with incircle that touches \( {AB} \) at \( M \) . Denote the incenters of \( \bigtriangleup {AHC} \) and \( \bigtriangleup {BHC} \) by \( {I}_{1} \) and \( {I}_{2} \), respectively. A point \( {D}_{1} \) is chosen such that \( M{I}_{2}{D}_{1}{I}_{1} \) is a re... | Problem 4.5.34\n\n\n\nyields that \( \angle {I}_{1}M{I}_{2} = {90}^{ \circ } \) . Also, \( \angle {I}_{2}H{I}_{1} = {90}^{ \circ } \) . Let \( D \in {CH} \) be a point such that \( \angle M{I}_{1}D = \) \( {90}^{ \ci... | Yes |
Problem 4.5.36. Let \( {ABC} \) be a triangle with circumcircle \( k \) . Let the line \( l\parallel {AB} \) intersect \( {AC} \) and \( {BC} \) at the points \( D \) and \( E \), respectively. Let \( {DE} \) intersect \( k \) at the points \( P \) and \( Q \), respectively ( \( D \) lies between \( P \) and \( E \) ).... | Solution. Let \( {k}_{1} \) and \( {k}_{2} \) touch \( l \) at \( K \) and \( L \), respectively.\n\nSawayama’s Lemma (Problem 4.7.13) yields that \( {KM} \) and \( {NL} \) intersect at \( I \) - the incenter of \( \bigtriangleup {PQC} \) .\n\nIt is clear that \( {DK} = {DM} \) and \( {EN} = {EL} \) . Hence,\n\n\[ \ang... | Yes |
Problem 4.5.37. Let \( {ABC} \) be a triangle with circumcircle \( k \) . The midpoints of the smaller arcs \( \widehat{BC} \) and \( \widehat{AC} \) are \( M \) and \( N \), respectively. The lines \( {AM} \) and \( {BN} \) intersect at \( I \) . Let \( {k}_{1}\left( N\right) \) be the circle that touches \( {AC} \), ... | Solution. It is clear that \( {AM} \) and \( {BN} \) are the bisectors of \( \angle {BAC} \) and \( \angle {ABC} \), respectively, and so \( I \) is the incenter of \( \bigtriangleup {ABC} \) .\n\nLet \( l \) be the tangent line to \( k \) at \( C \) .\n\nLet \( l \cap {MN} = R \) . Then \( \angle {RCN} = \angle {CBN} ... | Yes |
Problem 4.5.38. Let \( {ABC} \) be a triangle and let \( l \) be a line passing through \( A \), which does not intersect the segment \( {BC} \) . Let \( {k}_{1}\left( {O}_{3}\right) ,{k}_{2}\left( {O}_{2}\right) \) and \( {k}_{3}\left( {O}_{3}\right) \) be circles that touch the lines \( {AB},{BC},{CA} \) and \( l \),... | Solution. Note that \( \angle B{O}_{3}C = {90}^{ \circ } - \frac{\alpha }{2} \) and\n\n\[ \angle {O}_{1}A{O}_{2} = \alpha + \left( {\angle {O}_{1}{AB} + \angle {O}_{2}{AC}}\right) = \alpha + \left( \frac{{180}^{ \circ } - \alpha }{2}\right) = {90}^{ \circ } + \frac{\alpha }{2}. \]\n\nThen \( \angle {O}_{1}A{O}_{2} + \a... | Yes |
Problem 4.5.39. Given the construction of Problem 4.5.38, prove that the orthocenter of \( \bigtriangleup {A}_{1}{O}_{2}{O}_{3} \) lies on \( {BC} \) . | Solution. We have that the quadrilateral \( {\mathrm{O}}_{1}{\mathrm{{AO}}}_{2}{\mathrm{O}}_{3} \) is cyclic. Let \( H \) be the orthocenter of \( \bigtriangleup {O}_{1}{O}_{2}{O}_{3} \) . If \( H \in {BC} \), then \( {AB} \) and \( {AC} \) are the reflections of \( {BC} \) with respect to two of the sides of \( \bigtr... | Yes |
Let \( {ABCD} \) be a convex quadrilateral. The four excircles of that quadrilateral are constructed (see the figure). Let \( {O}_{1} \) and \( {O}_{2} \) be the centers of the circles that touch the segments \( {AD} \) and \( {BC} \) . The other two circles touch \( {AB} \) and \( {DC} \) at \( L \) and \( \bar{K} \),... | We will use the notations from the figure.\n\nLet \( {PQ} = {NG} = x \) . We have \( {RU} = {VT} \), which means that \( {RS} + {SU} = \) \( {VX} + {XT} \) . Hence, \( {NK} + {PL} = {KG} + {LQ} \) and so \( x - {KG} + {PL} = \) \( {KG} + x - {PL} \) . It follows that \( {KG} = {PL} \) and \( {KN} = {LQ} \) .\n\nWe also... | Yes |
Let \( {ABC} \) be a triangle with altitudes \( {AF},{BD} \) and \( {CE} \) . Let its incircle touch \( {AB} \) and \( \overline{AC} \) at the points \( K \) and \( L \) , respectively. Let \( {I}_{1} \) and \( {I}_{2} \) be the incenters of \( \bigtriangleup {AED} \) and of \( \bigtriangleup {CDF} \) , respectively. P... | We will prove that \( K{I}_{1} = L{I}_{1} = r \), where \( r \) is the inradius of \( \bigtriangleup {ABC} \) . Let the incircle of \( \bigtriangleup {AED} \) touch \( {AB} \) and \( {AC} \) at the points \( M \) and \( N \) , respectively. Note that \( \bigtriangleup {AED} \sim \bigtriangleup {ACB} \) with ratio \( \f... | Yes |
Problem 4.5.42. Let \( {ABC} \) be a triangle, and let \( C{C}_{1} \) and \( C{C}_{2} \) , \( A{A}_{1} \) and \( A{A}_{2}, B{B}_{1} \) and \( B{B}_{2} \) be three pairs of isogonally conjugate lines with respect to \( \bigtriangleup {ABC} \) . Also, we assume that \( {C}_{1} \in {AB},{C}_{2} \in {AB} \) , \( {B}_{1} \i... | Solution. Denote the incenters of \( A{C}_{1}Z{B}_{2} \) and \( B{A}_{1}X{C}_{2} \) by \( {I}_{1} \) and\n\n\( {I}_{2} \), respectively. Let the bisectors of \( \angle {B}_{1}{BC} \) and \( \angle {A}_{2}{AC} \) intersect at \( {I}_{3} \) . Denote \( \angle {CA}{I}_{3} = \angle {I}_{3}A{A}_{2} = \angle {A}_{1}A{I}_{2} ... | Yes |
Problem 4.5.43. Let \( {ABCD} \) be a circumscribed quadrilateral with incircle \( k \) . Let the rays \( {DA} \) and \( {CB} \) intersect at the point \( E \), and let the rays \( {DC} \) and \( {AB} \) intersect at the point \( F \) . Let \( H \) be the projection of \( B \) onto the line \( {EF} \) . Let \( {k}_{1} ... | Solution. Let \( {I}_{1}{I}_{2} \cap {EF} = M \) . Monge’s Theorem (Problem 6.2.3), applied to the circles \( k,{k}_{1} \) and \( {k}_{2} \), yields that \( M \) is the external center of homothety of \( {k}_{1} \) and \( {k}_{2} \) . Note that \( B \) is the internal center of homothety of \( {k}_{1} \) and \( {k}_{2}... | Yes |
Problem 4.6.1. (Trillium theorem) Let \( {ABC} \) be a triangle. Its incenter is \( I \) and its circumcircle is \( k \) . Let \( M \) be the second intersection point of \( {CI} \) and \( k \) . Prove that \( {MA} = {MB} = {MI} \) . | Solution: We have\n\n\n\n\[ \n\angle {AIM} = \angle {IAC} + \angle {ICA} \n\]\n\n\[ \n= \angle {IAB} + \angle {ICB} \n\]\n\n\[ \n= \angle {IAB} + \angle {MAB} \n\]\n\n\[ \n= \angle {MAI}\text{.} \n\]\n\nTherefore, \(... | Yes |
Problem 4.6.2. Let \( {ABC} \) be a triangle with incenter \( I \) . Let \( M \) be the midpoint of \( {AB} \), and let \( L \) be the midpoint of \( \overset{⏜}{ACB} \) . Prove that \( \angle {ILC} = \angle {IMB} \) . | First solution. Let \( {A}_{1} \) and \( {B}_{1} \) be the intersection points of \( {CL} \) and the lines \( {AI} \) and \( {BI} \), respectively. The given conditions imply that \( L \) lies on\nthe external bisector of \( \angle {ACB} \) .\nTherefore, \( {A}_{1} \) and \( {B}_{1} \) are the excenters of \( \bigtrian... | Yes |
For a given triangle \( {ABC} \), prove that \( O{I}^{2} = {R}^{2} - {2Rr} \), where \( O \) and \( I \) are the circumcenter and the incenter, respectively, and \( R \) and \( r \) are the circumradius and the inradius, respectively. | Solution. Let \( M \) be the second intersection point of \( {CI} \) and \( k \), where \( k \) is the circumcircle of \( \bigtriangleup {ABC} \) . Let \( D \) be the diametrically opposite point to \( M \) in \( k \) . Denote the projection of \( I \) onto the line \( {BC} \) by \( N \) . We have that \( \angle {MDB} ... | Yes |
Problem 4.6.4. Let \( {ABC} \) be a triangle with circumcircle \( k \), incircle \( \omega \), and incenter \( I \) . The line through \( I \), perpendicular to \( {CI} \), intersects the line \( {AB} \) at \( M \) . Let \( K \) be the second intersection point of \( k \) and \( {MC} \) . Prove that \( {IK} \bot {MC} \... | Solution. We have \( \angle {CIB} = {90}^{ \circ } + \frac{\angle {CAB}}{2} = {90}^{ \circ } + \angle {IAB} \) and hence \( \angle {MIB} = \angle {BAI} \) .\n\nThus, \( \bigtriangleup {AMI} \sim \bigtriangleup {IMB} \) and \( {MB}.{MA} = M{I}^{2} \) .\n\nConsidering the power of \( M \) with respect to \( k \), we get ... | Yes |
Problem 4.6.5. Let \( {ABC} \) be a triangle with incircle \( \omega \) and circumcircle \( k \) . The circle \( \omega \) touches \( {AB},{BC} \) and \( {CA} \) at the points \( M, N \) and \( P \), respectively. The points \( R, S \) and \( T \) are the midpoints of the smaller arcs \( \overset{⏜}{AB},\overset{⏜}{BC}... | Solution. Let \( O \) be the circumcenter of \( \bigtriangleup {ABC} \) . Let the tangent lines to \( k \) at the points \( R, S \) and \( T \) form the triangle \( {A}_{1}{B}_{1}{C}_{1} \), as shown in the figure. It is clear that there exists a homothety that maps \( \bigtriangleup {ABC} \) to \( \bigtriangleup {A}_{... | Yes |
Problem 4.6.6. Let \( {ABC} \) be a triangle with incircle \( \omega \) and circumcircle \( k \) . The circle \( \omega \) touches \( {BC},{CA} \) and \( {AB} \) at the points \( D, E \) and \( F \), respectively. Let \( {ED} \cap {AB} = K \) . Prove that the midpoint \( M \) of \( {KF} \) lies on the radical axis of \... | Solution. It suffices to prove that \( M{F}^{2} = {MB}.{MA} \) . Applying Menelaus’ Theorem to \( \bigtriangleup {ABC} \) and the points \( E, D \) and \( K \) yields\n\n\[ \frac{AE}{EC} \cdot \frac{CD}{DB} \cdot \frac{BK}{KA} = 1 \]\n\nAfter performing simple manipulations, we get\n\n\[ \frac{BK}{KA} = \frac{BD}{EA} =... | Yes |
Problem 4.7.1. (Verrièr’s Lemma) Let \( {ABC} \) be a triangle and let \( k \) be its circumcircle. Its \( C \) -mixtilinear incircle \( \omega \) touches \( {AC},{BC} \) and \( k \) at the points \( E, D \) and \( F \), respectively. Prove that the incenter of \( \bigtriangleup {ABC} \) is the midpoint of the segment ... | Solution. Let the lines \( {DF} \) and \( {DE} \) intersect \( k \) again at the points \( K \) and \( L \), respectively. Problem 6.1.1 yields that \( L \) is the midpoint of the smaller arc \( \overset{⏜}{AC} \), and \( K \) is the midpoint of the smaller arc \( \overset{⏜}{BC} \). Let \( I = {AK} \cap {BL} \). Then ... | Yes |
Let \( {ABC} \) be a triangle with circumcircle \( k \) . Its \( C \) - mixtilinear incircle \( \omega \) touches \( {AC},{BC} \) and \( k \) at the points \( E, F \) and \( X \) , respectively. The point \( D \) is the midpoint of the arc \( \overset{⏜}{ACB} \) . Prove that the points \( X, D \) and the midpoint of \(... | Problem 4.7.1 gives that the midpoint of \( {EF} \) is the incenter \( I \) of \( \bigtriangleup {ABC} \) . Let the line \( {IX} \) intersect \( k \) again at the point \( {D}_{1} \) . Let \( M \) and \( N \) be the second intersection points of \( {XE} \) and \( {XF} \) with \( k \), respectively. Problem 6.1.1 gives ... | Yes |
Problem 4.7.3. Let \( {ABC} \) be a triangle with circumcircle \( k \) . The \( A \) -mixtilinear incircle \( \omega \) touches \( {AC},{AB} \) and \( k \) at the points \( E, F \) and \( X \) , respectively. The points \( P \) and \( Q \) are the midpoints of the smaller arcs \( \overset{⏜}{AC} \) and \( \overset{⏜}{A... | Solution. Problem 6.1.1 gives that the points \( X, E \) and \( P \), as well as the points \( X, F \) and \( Q \) are collinear. We have that \( \angle {PXC} = \angle {PAC} = \) \( \angle {PCA} \) and hence \( \bigtriangleup {PCE} \sim \bigtriangleup {PXC} \) . We deduce that \( P{A}^{2} = P{C}^{2} = {PE}.{PX} \) . Th... | Yes |
Problem 4.7.4. Let \( {ABC} \) be a triangle with circumcircle \( k \) . Let \( M \) be the midpoint of the smaller arc \( \overset{⏜}{AB} \) of \( k \) . Let the point \( G \) be diametrically opposite to \( C \) with respect to \( k \), and let \( {AG} \cap \bar{C}M = X \) . Let \( {k}_{1} \) be the \( C \) -mixtilin... | Solution. It suffices to show that \( G \) lies on the radical axis of the point \( X \) and the circle \( {k}_{1} \) .\n\nWe will use the notations on the diagram. Since \( \angle {GMC} = \) \( \angle {GAC} = {90}^{ \circ } \), it suffices to prove that \( K \) lies on the radical axis of the point \( X \) and the cir... | Yes |
Let \( {ABC} \) be a triangle with circumcircle \( k \) . The \( C \) -mixtilinear incircle \( {k}_{1} \) touches \( k \) at the point \( M \), and the lines \( {AC} \) and \( {BC} \) at the points \( F \) and \( L \), respectively. Let \( {FL} \cap {CM} = N \) . Prove that \( \angle {ANF} = \angle {BNL} \) . | It suffices to show that \( \bigtriangleup {ANF} \sim \bigtriangleup {BNL} \) . Since \( \angle {CFL} = \) \( \angle {CLF} \), it remains to show that \( \frac{AF}{BL} = \frac{FN}{NL} \) .\n\nWe will use Lemma 1 from the proof of Casey's Theorem (Problem 6.1.10). We have\n\n\[ \n{AF} = {AM}\sqrt{\frac{R - {R}_{1}}{R}},... | Yes |
Problem 4.7.6. Let \( {ABC} \) be a triangle with circumcircle \( k \) . Let \( D \) be an arbitrary point on the smaller arc \( \widehat{AB} \) of \( k \) . The circle \( {k}_{1} \) touches \( k \) internally at the point \( D \) and the line \( {BC} \) at the point \( P \) . The circle \( {k}_{2} \) touches \( k \) i... | Solution. Let \( L \) and \( M \) be the midpoints of the smaller arcs \( \overset{⏜}{AC} \) and \( \overset{⏜}{BC} \) of \( k \), respectively. Problem 6.1.1 implies that the points \( D, H \) and \( L \) , as well as the points \( D, P \) and \( M \) are collinear. Also, the points \( A, I \) and \( M \), as well as ... | Yes |
Let \( {ABC} \) be a triangle with circumcircle \( k \) and \( C \) - mixtilinear incircle \( {k}_{1} \) . Its incircle \( {k}_{2} \) touches \( {AB} \) at the point \( Q \) . Prove that \( \angle {APQ} = \angle {BAC} \) and \( \angle {BPQ} = \angle {ABC} \) . | Note that \( \frac{\sin \angle {APQ}}{\sin \angle {BPQ}} = \frac{AQ}{QB} \cdot \frac{BP}{PA} = \frac{s - a}{s - b} \cdot \frac{\sin \angle {BCP}}{\sin \angle {ACP}} \) . Problem 4.7.11 implies that \[ \frac{\sin \angle {BCP}}{\sin \angle {ACP}} = \frac{s - b}{s - a} \cdot \frac{a}{b} \] \[ \Rightarrow \frac{\sin \angle... | Yes |
Problem 4.7.9. Let \( {ABC} \) be a triangle with circumcircle \( k \) . The \( C \) -mixtilinear circle \( {k}_{1} \) touches \( k \) at the point \( P \) . Let \( Q \) be an arbitrary point on the arc \( \overset{⏜}{AB} \) of \( k \), not containing \( C \) . The points \( {I}_{1} \) and \( {I}_{2} \) are the incente... | Solution. Let \( {k}_{1} \) touch \( {AC} \) and \( {BC} \) at the points \( E \) and \( D \), respectively. The midpoints of the smaller arcs \( \overset{⏜}{AC} \) and \( \overset{⏜}{BC} \) are \( N \) and \( M \), respectively. Let the lines tangent to \( k \) at the points \( C \) and \( P \) intersect at the point ... | Yes |
Problem 4.7.10. Let \( {ABC} \) be a triangle with circumcircle \( k \) and incircle \( \omega \) . The \( C \) -mixtilinear incircle \( {k}_{1} \) touches \( k \) at the point \( K \) . The point \( M \in \widehat{AKB} \) is arbitrary. The tangent lines through \( M \) to \( \omega \) intersect \( {AB} \) at the point... | Solution. Let \( {K}^{\prime } \) be the second intersection point of the circumcircle of \( \bigtriangleup {MEF} \) and \( k \) . We will show that \( {K}^{\prime } \equiv K \) . Problems 4.7.2 and 4.7.1 yield that it suffices to show that \( {K}^{\prime }I \) bisects \( \angle A{K}^{\prime }B \) . Indeed, then \( {K}... | Yes |
Problem 4.7.11. Let \( {ABC} \) be a triangle with circumcircle \( k \) . Its \( C \) -excircle \( {k}_{2} \) touches \( {AB} \) at the point \( F \) . Its \( C \) -mixtilinear incircle \( {k}_{1} \) touches \( k \) at the point \( D \) . Prove that \( \angle {ACF} = \angle {BCD} \) . | Solution. Consider the composition \( \varphi \) of an inversion with center \( C \) and radius \( \sqrt{{CA}.{CB}} \) and a reflection with respect to the angle bisector of \( \angle {ACB} \) .\n\nClearly, \( \varphi \left( A\right) = B \) ,\n\n be a triangle with circumcircle \( k \) . Its incircle \( {k}_{1} \) touches the segment \( {AB} \) at the point \( F \) . Its \( C \) -mixtilinear excircle \( {k}_{2} \) touches \( k \) at the point \( D \) . Prove that \( \angle {ACF} = \angle {BCD} \) . | Solution. As in Problem 4.7.11, the statement follows from the fact that \( {CF} \) is the image of \( {CD} \) under the same composition of inversion and reflection. | No |
Problem 4.7.13. (Sawayama's Lemma) Let \( {ABC} \) be a triangle with circumcircle \( k \) . Let \( D \) be an arbitrary point on \( {AB} \) . The circle \( {k}_{1} \) touches the segments \( {AD} \) and \( {CD} \), as well as \( k \) internally at a point on the arc \( \overset{⏜}{AC} \) . Let \( {k}_{1} \) touch \( {... | Solution. Let the lines \( {CI} \n\n\n\nand \( {AB} \) intersect at \( E \) and let \( {k}_{1} \) touch \( k \) at the point \( K \) .\n\nProblem 6.1.1 yields that the circle \( {k}_{2} \), which touches the segment ... | Yes |
Problem 4.7.14. Let \( {ABC} \) be a triangle with circumcircle \( k \) . The point \( D \) lies on the segment \( {AB} \) . The circle \( {k}_{1} \) touches \( {AB} \) at the point \( F \), touches \( {CD} \) at the point \( M \) and touches \( k \) internally at the point \( K \) on the smaller arc \( \overset{⏜}{AC}... | Solution. Let \( L \) be the second intersection point of \( {CI} \) and \( k \) . Problem 4.6.1 gives that \( {LA} = {LB} = {LI} \) .\n\nProblem 6.1.1 yields that\n\n\n\nthe points \( K, F \) and \( L \) are colline... | Yes |
Problem 4.7.15. (Thébault's Theorem) Let \( {ABC} \) be a triangle with circumcircle \( k \) . The point \( D \) lies on the segment \( {AB} \) . The circle \( {k}_{1} \) touches the segment \( {AD} \) at the point \( K \), the segment \( {CD} \) at the point \( E \) and the circle \( k \) internally at a point on the ... | Solution. By Sawayama’s Lemma (Problem 4.7.13), the points \( K, E \) and \( I \) are collinear, and the points \( M, N \) and \( I \) are collinear.\n\nWe have that \( D{I}_{2} \bot D{I}_{1} \) .  We also have \( D{... | Yes |
Problem 4.7.16 Let \( {ABC} \) be a triangle with circumcircle \( k \) . The point \( D \) lies on the segment \( {AB} \) . The circle \( {k}_{1} \) touches the circle \( k \) internally at a point on the arc \( \overset{⏜}{AC} \), touches the segment \( {AD} \) and touches the segment \( {CD} \) at the point \( F \) .... | Solution. Problem 4.6.1 yields that the midpoint \( T \) of the smaller arc \( \overset{⏜}{AB} \) is the circumcenter of \( \bigtriangleup {ABI} \) . Therefore, it suffices to prove that the point \( T \) lies on the perpendicular bisector of \( {IL} \) . Let \( {YZ} \cap k = \{ P, S\} \) and \( {CD} \cap k = \{ C, R\}... | Yes |
Problem 4.7.17. Let \( {ABCD} \) be a cyclic quadrilateral with circumcircle \( k \) . The circle \( {k}_{1} \) touches the circle \( k \) internally at a point on the smaller arc \( \overset{⏜}{AD} \), touches the segment \( {AC} \) at the point \( F \), and touches the segment \( {BD} \) at the point \( E \) . The ci... | Solution. Clearly, the points \( P \) and \( Q \) lie on the radical axis \( l \) of \( {k}_{1} \) and \( {k}_{2} \) . Therefore, it suffices to prove that \( T \) and \( M \) also lie on \( l \) . Let \( {TG} \) and \( {TH} \) be the tangent lines through \( T \) to \( {k}_{1} \) and \( {k}_{2} \), respectively, so th... | Yes |
Problem 4.7.18. Let \( {CYBED} \) be a cyclic pentagon with circumcircle \( k \) . The circle \( {k}_{1} \) touches the line \( {CB} \) at the point \( N \), touches the line \( {EY} \) and touches the circle \( k \) internally at a point on the smaller arc \( \widehat{CE} \) . The circle \( {k}_{2} \) touches the line... | Solution. Firstly, we will prove the following lemma.\n\nLemma. Using the notations of the problem, let \( {CD} \cap {BE} = A \) . Let \( {I}_{1},{I}_{2} \) and \( I \) be the incenters of \( \bigtriangleup {CBD},\bigtriangleup {CBE} \) and \( \bigtriangleup {CBA} \), respectively. Then the quadrilaterals \( {CX}{I}_{2... | Yes |
Problem 4.7.19. Let \( {ABCD} \) be a cyclic quadrilateral with circumcircle \( k \) . The circle \( {k}_{1} \) touches the lines \( {AC} \) and \( {BD} \), and the circle \( k \) internally at the point \( E \) on the smaller arc \( \overset{⏜}{AD} \) . The circle \( {k}_{2} \) touches the lines \( {AC} \) and \( {BD}... | Solution. Let \( {KX} \cap k = \{ F, Y\} \), so that \( K \) lies between \( X \) and \( F \) . Also, let \( {k}_{1} \cap {k}_{3} = N,{k}_{2} \cap {k}_{3} = S,{XZ} \cap {EK} = Q \) and \( {XS} \cap {KN} = P \) . Problem 6.1.1 yields that \( Q \) is the midpoint of \( \widehat{YF} \) . Moreover, Problem 6.1.2 yields tha... | Yes |
Problem 4.8.1. (Euler Circle) Let \( {ABC} \) be a triangle with orthocenter \( H \) . Prove that the midpoints of its sides, the feet of its altitudes and the midpoints of the segments \( {AH},{BH} \) and \( {CH} \) are concyclic. | Solution. Let the feet of the respective altitudes be \( {A}_{1},{B}_{1} \) and \( {C}_{1} \), and let \( M \) and \( N \) be the midpoints of \( {AB} \) and \( {CH} \), respectively. We have that \( \angle {A}_{1}{C}_{1}{B}_{1} = \) \( {180}^{ \circ } - {2\gamma } \) . Since \( N \) is the circumcenter of the quadrila... | Yes |
Let \( {ABC} \) be a triangle. Let \( k,{k}_{a},{k}_{b} \) and \( {k}_{c} \) be the incircle and the three excircles, respectively. Let \( {A}_{1},{B}_{1} \) and \( {C}_{1} \) be the midpoints of \( {BC},{CA} \) and \( {AB} \), respectively. Prove that the circumcircle \( \omega \) of \( \bigtriangleup {A}_{1}{B}_{1}{C... | Without loss of generality, let \( b \geq a \) . We will show that \( \omega \) is tangent to \( k \) and \( {k}_{c} \) . Let \( k \) touch \( {AB} \) at the point \( {J}_{3} \) and let \( {k}_{c} \) touch \( {AB} \) at the point \( {J}_{c} \) .\n\nClearly, \( {J}_{c}{C}_{1} = {J}_{3}{C}_{1} = \frac{b - a}{2} \) . Let ... | Yes |
Problem 4.8.3. Let \( {ABC} \) be a triangle with incenter \( I \) and circumcenter \( O \) . Let \( P \) be an arbitrary point on the line \( {OI} \) . Denote the feet of the perpendiculars from \( P \) to \( {AB},{BC} \) and \( {CA} \) by \( {C}_{2},{A}_{2} \) and \( {B}_{2} \) , respectively. Prove that the Feuerbac... | Solution. Let the feet of the perpendiculars from \( I \) to \( {AB},{BC} \) and \( {CA} \) be \( {C}_{1},{A}_{1} \) and \( {B}_{1} \), respectively.\n\nThe first and second\n\n\n\nFontene Theorems, applied to the po... | No |
Problem 4.8.7. Let \( {ABC} \) be a triangle. Denote the midpoints of \( {AB},{BC} \) and \( {CA} \) by \( {C}_{1},{A}_{1} \) and \( {B}_{1} \), respectively. Denote the points of tangency of the excircles and the segments \( {AB},{BC} \) and \( {CA} \) by \( {C}_{2},{A}_{2} \) and \( {B}_{2} \), respectively. Let \( {... | Solution. Define the Bevan point \( {Bi} \) as in Problem 2.20. Problem 3.5 yields that \( {Bi} \) lies on the line \( {OI} \), where \( O \) and \( I \) are the circumcenter and the incenter of \( \bigtriangleup {ABC} \), respectively.\n\nBy Problem 4.8.3, we have that \( F \) lies on the circumcircle of \( \bigtriang... | Yes |
Problem 4.8.10. (Conway circle) Let \( {ABC} \) be a triangle. The points \( P \) and \( S \) lie on the line \( {AB} \), such that the points \( P, A, B \) and \( S \) are in this order, \( {AP} = {BC} \) and \( {BS} = {AC} \) . The points \( R \) and \( N \) lie on the line \( {BC} \), such that the points \( R, B, C... | Solution. We have \( {CN} = \) \( {CM} \) and \( {CQ} = {CR} \) . Hence, \( \angle {QRC} = \angle {NMC} \), which means that the quadrilateral QRMN is cyclic. Analogously, the quadrilaterals \( {PQSM} \) and \( {NPRS} \) are cyclic.\n\nAssume the circumcircles of the quadrilaterals \( {QRMN} \) , \( {PQSM} \) and \( {N... | Yes |
Problem 4.8.11. Let \( {ABC} \) be a triangle. The points \( D, E \) and \( F \) are the midpoints of the sides \( {BC},{CA} \) and \( {AB} \), respectively. The point \( G \) is the centroid of \( \bigtriangleup {ABC} \) . Prove that the circumcenters of \( \bigtriangleup {AFG},\bigtriangleup {AEG} \) , \( \bigtriangl... | Solution. Denote the circumcenters by \( H, I, J, K, L \) and \( M \), respectively. We will show that the quadrilateral HIJK is cyclic (the same will hold analogously for HILM and JKLM). Then the desired statement will follow from radical axes, analogously to how we prove the lemma in Problem 5.4.8, but applied to the... | Yes |
Problem 4.8.12. The circles \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) are chosen such that \( {k}_{1} \cap {k}_{2} = \{ R, S\} ,{k}_{2} \cap {k}_{3} = \{ P, Q\} \) and \( {k}_{3} \cap {k}_{1} = \{ M, N\} \) . Denote the radical center of \( {k}_{1},{k}_{2} \) and \( {k}_{3} \) by \( X \) . The points \( A, B \) and \( C ... | Solution. The point \( A \) lies on the radical axis of \( {k}_{1} \) and \( {k}_{3} \) Hence, \( {AG}.{AH} = {AM}.{AN} = {AL}.{AK} \) and thus the quadrilateral \( {GHKL} \) is cyclic. Analogously, the quadrilaterals GHIJ and IJKL are cyclic. The desired statement follows by observing that the radical axes of the circ... | Yes |
Problem 4.8.13. Let \( {ABC} \) be a triangle. A circle \( k \) through \( A \) and \( B \) intersects the sides \( {AC} \) and \( {BC} \) at the points \( E \) and \( D \), respectively. Let \( {AD} \cap {BE} = G \) . Denote the midpoints of \( {AB} \) and \( {DE} \) by \( M \) and \( N \), respectively. Prove that th... | Solution. Denote the reflections of \( G \) with respect to \( M \) and \( N \) by \( K \) and \( L \), respectively. We have that \( {AKBG} \) and \( {EGDL} \) are parallelograms.\n\nThen \( \angle {CEL} = \angle {CAD} = \angle {EBC} \) and \( \frac{EL}{GB} = \frac{GD}{GB} = \frac{AB}{ED} = \frac{CB}{CE} \) .\n\nThere... | Yes |
Problem 4.8.14 Let \( {ABC} \) be a triangle. The points \( N \) and \( M \) are chosen on the sides \( {AC} \) and \( {BC} \), respectively, so that \( {ABMN} \) is cyclic. Let \( {AM} \cap {BN} = D \) and \( {CD} \cap {AB} = P \) . Denote the midpoint of the segment \( {AB} \) by \( Q \) . Prove that the quadrilatera... | Solution. Without loss of generality, let \( {AC} > {BC} \) . Denote \( {AB} \cap {MN} = X \) . Then \( B \) is between \( X \) and \( A \) .\n\nMenelaus’ Theorem, applied to \( \bigtriangleup {ABC} \) and the points \( N, M \) and \( X \), gives\n\n\[ \frac{AX}{BX} = \frac{AN}{NC} \cdot \frac{CM}{MB} \]\n\nCeva's Theo... | Yes |
Problem 4.8.15. Let \( {ABC} \) be a triangle with circumcircle \( k \) . Let \( M \) , \( N \) and \( P \) be the midpoints of \( {AB},{BC} \) and \( {CA} \), respectively. Let \( D \in {AB} \) be a point such that \( {CD} \) is tangent to \( k \) . Denote the circumcircle of \( \bigtriangleup {MNP} \) by \( {k}_{1} \... | Solution. By La Hire’s Theorem, it suffices to show that the point \( D \) lies on the polar line of \( C \) with respect to \( {k}_{1} \) .\n\nLet \( {CX} \) and \( {CY} \) be the tangent lines to \( {k}_{1} \) . We will show that the points \( X, Y \) and \( D \) are collinear. Denote the midpoints of \( {CX},{CY} \)... | Yes |
Problem 4.8.16. Let \( {ABC} \) be a triangle with circumcenter \( O \) . Let \( k \) be a circle through the points \( C \) and \( O \) . Denote \( k \cap {AC} = \{ C, D\} \) and \( k \cap {BC} = \{ C, E\} \) . Prove that the orthocenter of \( \bigtriangleup {DOE} \) lies on \( {AB} \) . | Solution. Let \( F \in {AB} \) be a point such that \( {OF} \bot {DE} \) . We will show that \( F \) is the orthocenter of \( \bigtriangleup {DOE} \) . We have\n\n\[ \angle {FOE} = {90}^{ \circ } + \angle {OED} = {90}^{ \circ } + \angle {OCD} = {90}^{ \circ } + {90}^{ \circ } - \beta \]\n\n\[ = {180}^{ \circ } - \beta ... | Yes |
Problem 4.8.17. Let \( {ABC} \) be a triangle with circumcircle \( k \) . The point \( D \) is the midpoint of \( \overset{⏜}{ACB} \) . An arbitrary point \( M \) is chosen on the side \( {AC} \) . The point \( N \in {BC} \) satisfies \( {AM} = {BN} \) . Prove that the points \( C, D, N \) and \( M \) are concyclic. | Solution. The point \( D \) lies on the perpendicular bisector of the side \( {AB} \) and hence \( {AD} = \) \( {BD} \) . On the other hand, we have \( \angle {CAD} = \angle {CBD} \) and \( {AM} = \) \( {BN} \) . Therefore, we obtain \( \bigtriangleup {MAD} \cong \bigtriangleup {NBD} \), whence we deduce that \( \angle... | Yes |
Problem 4.8.18. Let \( {ABC} \) be a triangle. Let \( X \) be an arbitrary point on \( {AB} \) . The points \( E \) and \( F \) lie on the sides \( {AC} \) and \( {BC} \) and are chosen such that the quadrilaterals \( {AXFC} \) and \( {BCEX} \) are cyclic. Let \( O \) be the circumcenter of \( \bigtriangleup {EFC} \) .... | Solution. Cyclic quadrilaterals give\n\n\[ \angle {EXA} = \angle {ACB} = \angle {FXB}\text{.} \]\n\nOn the other hand, we have \( \angle {EOF} = 2\angle {ACB} \) . Thus, the quadrilateral \( {OEXF} \) is cyclic and finally,\n\n\[ {90}^{ \circ } - \angle {ACB} = \angle {OEF} = \angle {OXF} \Rightarrow \angle {OXB} = {90... | Yes |
Problem 4.8.19. Let \( {ABC} \) be a triangle with circumcircle \( k \) . Let \( M \) and \( N \) be arbitrary points on the side \( {AB} \), so that \( M \) lies between \( A \) and \( N \) . The points \( K \in {AC} \) and \( L \in {BC} \) are chosen such that \( {MK}\parallel {BC} \) and \( {NL}\parallel {AC} \) . T... | Solution. Let \( {PQ} \cap {AB} = R \) . Without loss of generality, we will assume that \( A \) lies between \( B \) and \( R \) . Denote \( {AM} = x,{BN} = y,{AR} = z \) and \( {AB} = c \) .\n\nThe intercept theorem gives that \( \frac{AK}{KC} = \frac{AM}{MB} = \frac{x}{c - x} \) . Analogously, \( \frac{CL}{LB} = \fr... | Yes |
Problem 4.8.20. Let \( {ABCD} \) be a convex quadrilateral, such that \( {AC} = {AD} \) and \( {BC} = {BD} \) . The circumcircle of \( \bigtriangleup {ABC} \) intersects \( {AD} \) and \( {BD} \) at the points \( E \) and \( F \), respectively. Prove that \( \angle {ECD} = \angle {FCD} \) . | Solution. Since \( {AB} \) is the perpendic-\n\n\n\nular bisector of \( {CD} \), we have \( {AB} \bot {CD} \) .\n\nTherefore,\n\n\[ \angle {FCD} = \angle {CFB} - \angle {CDB} \]\n\n\[ = \angle {CAB} - \angle {DCB} \]... | Yes |
Problem 4.8.21. Let \( {ABC} \) be a triangle. Its incircle is \( \omega \) and its incenter is \( I \) . Let \( \omega \) touch \( {AB},{BC} \) and \( {CA} \) at the points \( M, N \) and \( P \) , respectively. Let \( {AB} \cap {PN} = K \) . Prove that \( {CM} \bot {KI} \) . | Solution. Let \( \pi \) be the pole-polar transformation with respect to \( \omega \) . Clearly, \( {NP} \) is the polar line of \( C \) . By La Hire’s Theorem and \( K \in {\pi }_{C} \), we get \( C \in {\pi }_{K} \n\nOn the other hand, we have \( M \in {\pi }_{K} \) . Therefore, \( {\pi }_{K} \equiv {CM} \), which me... | Yes |
Problem 4.8.22. Let \( {ABC} \) be a triangle. Its incircle is \( \omega \) and its incenter is \( I \) . Let \( \omega \) touch \( {AB},{BC} \) and \( \overline{CA} \) at the points \( M, N \) and \( P \), respectively. Let \( {AB} \cap {PN} = K \) and \( {CM} \cap {KN} = L \) . Prove that \( {IL} \bot {KC} \) . | Solution. Consider the pole-polar transformation \( \pi \) with respect to \( \omega \) . Let \( {KQ} \) be the second tangent line to \( \omega \) through \( K \), so that \( Q \in \omega \) . Problem 4.8.21 yields that the points \( C, Q \) and \( M \) lie on \( {\pi }_{K} \) . But since \( L \in {CM} \), it follows ... | Yes |
Problem 4.8.23. Let \( {ABC} \) be a triangle and let \( X \) be a point inside of it. The points \( M, N \) and \( P \) are the feet of the perpendiculars from \( X \) to \( {AB},{BC} \) and \( {CA} \), respectively. Let \( Q \) be the second intersection point of the circumcircle of \( \bigtriangleup {MNP} \) and \( ... | Solution. Let \( Y \) be a point, such that \( Y \in {CQ} \) and \( {XY} \bot {CQ} \) . Then the pentagon \( {PYXNC} \) is inscribed in the circle with diameter \( {CX} \) . Clearly, the quadrilateral \( {QMXY} \) is cyclic. We apply the radical axes theorem (Problem 6.5.2) to the circumcircles of \( {QMNP},{MXYQ} \) a... | Yes |
Problem 4.8.26. Let \( {ABC} \) be a triangle. Let \( X \) be an arbitrary point on \( {AB} \) . Denote the circumcircles of \( \bigtriangleup {AXC} \) and \( \bigtriangleup {BXC} \) by \( {k}_{1} \) and \( {k}_{2} \), respectively. The points \( E \in {AC} \) and \( F \in {BC} \) are chosen such that \( {EX}\parallel ... | Solution. Let the line \( {ME} \) intersect \( {k}_{1} \) for the second time at the point \( {M}^{\prime } \) . Then we have \( \angle A{M}^{\prime }C = \angle {AXC} = \angle {CNB} \) . Moreover, the intercept theorem gives \( \frac{AE}{EC} = \frac{AX}{XB} = \frac{CF}{FB} \) .\n\nWe will now show that \( \bigtriangleu... | Yes |
Problem 4.8.27. Let \( {ABC} \) be a triangle \( \left( {{AC} > {BC}}\right) \) . Let \( D \in {AB} \) , such that \( {DC} \) touches the circumcircle of \( \bigtriangleup {ABC} \) . The point \( {B}^{\prime } \) is the reflection of \( B \) with respect to the line \( {CD} \) . The point \( {C}^{\prime } \) is the ref... | Solution. Denote the reflection of \( B \) with respect to the point \( D \) by \( K \) . Then \( {CB}{C}^{\prime }K \) is a parallelogram. Hence,\n\n\[ \angle {CAK} = \angle {BC}{C}^{\prime } = \angle C{C}^{\prime }K\text{.} \]\n\nThus, the quadrilateral \( A{C}^{\prime }{KC} \) is cyclic. Due to\n\n\[ \angle C{B}^{\p... | No |
Problem 4.8.28 Let \( \\bigtriangleup {ABC} \) be a triangle with \( \\angle {ACB} < {60}^{ \\circ } \) and circumcenter \( O \) . The points \( {A}^{\\prime } \) and \( {B}^{\\prime } \) are the reflections of \( A \) and \( B \) with respect to \( {BC} \) and \( {AC} \), respectively. The point \( {C}_{1} \) is the r... | Solution. Let the perpendicular bisector of \( {B}^{\\prime }C \) intersect \( {AC} \) at the point \( M \) and let the perpendicular bisector of \( {A}^{\\prime }C \) intersect \( {BC} \) at the point \( N \) . Using symmetry, we get that the points \( O, N \) and the midpoint \( L \) of \( {AC} \) are collinear and t... | Yes |
Problem 4.8.29. Let \( {ABC} \) be a triangle with median \( {CM} \) . The symmedian of \( \bigtriangleup {ABC} \) at \( C \) intersects the circumcircle of \( \bigtriangleup {AMC} \) at the point \( N \) . Let \( O,{O}_{1} \) and \( {O}_{2} \) be the circumcenters of \( \bigtriangleup {ABC},\bigtriangleup {AMC} \) and... | Solution. We know that \( O{O}_{1},{O}_{1}{O}_{2} \) and \( O{O}_{2} \) are the perpendicular bisectors of \( {AC},{CN} \) and \( {CB} \), respectively. Let \( {CM} \cap {AN} = X \) . Set \( \angle {ACB} = \gamma ,\angle {ACM} = \varphi ,\angle {NAC} = \delta \) and \( \angle {NBC} = \psi \) . Denote the midpoints of \... | Yes |
Problem 4.8.30. Let \( {ABC} \) be a triangle. The points \( D \) and \( E \) from the segment \( {AB} \) are such that \( \angle {ACD} = \angle {BCE} \) . Let \( F, G, H, I \) be the circumcenters of \( \bigtriangleup {ADC},\bigtriangleup {AEC},\bigtriangleup {BDC} \) and \( \bigtriangleup {BEC} \), respectively. Prov... | Solution. Clearly, the lines \( {FG},{GI},{IH} \) and \( {HF} \) are the perpendicular bisectors of \( {AC},{CE},{CB} \) and \( {CD} \), respectively. We have\n\n\[ \angle {HIG} = \angle \left( {{HI},{IG}}\right) = \angle \left( {{BC},{EC}}\right) = \angle {BCE}, \]\n\n\[ \angle {HFG} = \angle \left( {{HF},{FG}}\right)... | Yes |
Problem 4.8.31. Let \( {ABC} \) be a triangle. Let \( l \) be a line through \( C \) that lies inside of \( \angle {ACB} \) . The lines \( {l}_{1} \) and \( l \) are isogonally conjugate with respect to \( \angle {ACB} \) . The points \( K \) and \( M \) are the projections of \( A \) and \( B \) onto \( {l}_{1} \), an... | Solution. Clearly, we have \( \bigtriangleup {ALC} \sim \bigtriangleup {BMC} \) and \( \bigtriangleup {AKC} \sim \bigtriangleup {BNC} \) . Therefore, \[ \frac{CM}{CL} = \frac{CB}{CA} = \frac{CN}{CK} \] We deduce that \[ \text{CM.CK} = \text{CL.CN.} \] In conclusion, the quadrilateral \( {KMLN} \) is cyclic. | Yes |
Problem 4.8.32. Let \( k \) be a circle with center \( A \) and let \( C \) be a point outside of \( k \) . The lines \( {CD} \) and \( {CE} \) are tangent to \( k \), such that \( D, C \in k \) . The point \( G \) is the projection of \( C \) onto an arbitrary line \( l \) through \( A \), and the point \( H \) is an ... | Solution. First, the pentagon \( {AGDCE} \) is inscribed in the circle with diameter \( {AC} \) . \n\nThen \( \angle {DAG} = \angle {DCG} \) and \( \angle {DFG} = \angle {DHG} \) , which implies that \( \bigtriangleup {DFA} \sim \) \( \bigtriangleup {DHC} \) . \n\nTherefore, \n\n\[ \n{FA} = \frac{{AD} \cdot {HC}}{DC}. ... | Yes |
Let \( k \) be a circle with center \( A \) and let \( C \) be a point outside of \( k \) . The lines \( {CD} \) and \( {CE} \) are tangent to \( k \), such that \( D, E \in k \) . The point \( B \) lies on \( k \), such that the points \( B \) and \( C \) lie in different half-planes with respect to \( {ED} \) . Let \... | First, note that\n\n\n\n\( {DN} = {DC} = {CE} = {EM} \) .\n\nLet \( \angle {EBD} = \gamma \) and \( \angle {ECM} = \varphi \) . Then\n\n\[ \n\angle {DCE} = {180}^{ \circ } - {2\gamma }\n\]\n\n\[ \n\angle {GCH} = {180... | Yes |
Problem 4.8.34. Let \( {ABC} \) be a triangle. Let \( {DH}\parallel {AB} \), where \( D \in {AC} \) and \( H \in {BC} \) . Let \( E \) be an arbitrary point inside of \( \bigtriangleup {DHC} \) . Denote \( G = {EB} \cap {DH} \) and \( F = {AE} \cap {DH} \) . Prove that \( {CE} \) passes through the second intersection ... | Solution. Let the circumcircle of \( \bigtriangleup {DGE} \) intersect \( {AC} \) at the points \( D \) and \( J \) . Let the circumcircle of \( \bigtriangleup {FEH} \) intersect \( {BC} \) at the points \( H \) and \( K \) .\n\nWe have\n\n\[ \angle {ABE} = \angle {DGE} = \angle {EJC} \]\n\nand \( \angle {BAE} = \angle... | Yes |
Problem 4.8.35. Let \( {ABC} \) be a triangle and let \( D \) be an arbitrary point inside of it. The points \( E \) and \( F \) are chosen such that \( {DE}\parallel {AC}, E \in \) \( {AB} \) and \( {DF}\parallel {BC}, F \in {AB} \) . Prove that the line \( {CD} \) passes through the second intersection point \( I \) ... | Solution. Let the circumcircle of \( \bigtriangleup {AFD} \) intersect \( {AC} \) at the points \( A \) and \( G \), and let the circumcircle of \( \bigtriangleup {BED} \) intersect \( {BC} \) at the points \( B \) and \( H \) . We have\n\n\[ \angle {DHC} = \angle {DEB} = \angle {CAB} \]\n\nand \( \angle {DGC} = \angle... | Yes |
Problem 4.8.36. Let \( {ABC} \) be a triangle. The circle \( k \) contains the points \( A \) and \( B \) and intersects \( {BC} \) and \( \breve{C}A \) at the points \( D \) and \( F \) , respectively. Let \( J, K \in k \) and let \( L \) be the second intersection point of the circumcircles of \( \bigtriangleup {FKC}... | Solution. Denote \( N = {AF} \cap {BK}, O = {AJ} \cap {BD} \) and \( X = {FD} \cap {JK} \) . Pascal’s Theorem, applied to the hexagon AJKBDF, gives that the points \( N, O \) and \( X \) are collinear.\n\n\n\nDesargu... | Yes |
Problem 4.8.37. Let \( {ABC} \) be a triangle and let \( J \) be a point inside of it. A circle \( k \) passes through \( A \) and \( B \) and intersects \( {CA} \) and \( {CB} \) for the second time at the points \( E \) and \( D \), respectively. Let \( {JA} \cap k = H \) and \( {JB} \cap k = I \) . The point \( F \)... | Solution. Let the circumcircles of \( \bigtriangleup {DHF} \) and \( \bigtriangleup {EGF} \) intersect for the second time at the point \( {G}^{\prime } \) . We will show that \( G \equiv {G}^{\prime } \) . Denote \( M = {AC} \cap {BJ}, L = A\bar{J} \cap {BC} \) and \( X = {ED} \cap {HI} \) . Pascal’s Theorem, applied ... | Yes |
Problem 4.8.38. Let \( {ABC} \) be a triangle. Let \( D \) and \( E \) be the midpoints of the sides \( {AC} \) and \( {BC} \), respectively. The circles with diameters \( {BD} \) and \( {AE} \) intersect each other at the points \( H \) and \( I \) . Prove that the points \( H, I \) and \( C \) are collinear. | Solution. We will show that if \( F \) and \( G \) are the feet of the altitudes from \( A \) and \( B \) in \( \bigtriangleup {ABC} \) , then the quadrilateral \( {GDEF} \) is cyclic. Since the quadrilateral \( {ABFG} \) is cyclic, we have \( {CD}.{CG} = \frac{{CA}.{CG}}{2} = \frac{{CB}.{CF}}{2} = \) CE.CF.\n\nThe rad... | No |
Problem 4.8.39. Let \( {ABC} \) be a triangle. The points \( D \) and \( E \) are the midpoints of the sides \( {AC} \) and \( {BC} \), respectively. The point \( F \) is the foot of the altitude through \( C \) in \( \bigtriangleup {ABC} \) . The circumcircles of \( \bigtriangleup {DFB} \) and \( \bigtriangleup {AFE} ... | Solution. Let the reflec-\ntion of \( D \) with respect to \( E \) be \( J \) and let the reflection of \( E \) with respect to \( D \) be \( I \) . Clearly, \( J \) coincides with the reflection of \( D \) with respect to the perpendicular bisector of \( {BF} \) . Therefore, \( {DJ} = {AB} \) and the quadrilateral \( ... | Yes |
Let \( {ABC} \) be a triangle. Let \( X \) be an arbitrary point, not lying on any of the sides of the triangle. Prove that the isogonal conjugates of the lines \( {AX},{BX} \) and \( {CX} \) with respect to \( \bigtriangleup {AB}\bar{C} \) are concurrent. | Let \( {X}_{1} \) be the intersection point of the isogonal conjugates of the lines \( {AX} \) and \( {BX} \) with respect to \( \bigtriangleup {ABC} \) . Applying the trigonometric form of Ceva's Theorem to \( \bigtriangleup {ABC} \) and the points \( X \) and \( {X}_{1} \) gives\n\n\[ \frac{\sin \angle {CAX}}{\sin \a... | Yes |
Problem 4.9.2. Let \( {ABC} \) be a triangle. The points \( D \) and \( E \) lie on the side \( {BC} \) and satisfy \( \angle {BAD} = \angle {EAC} \) . The points \( F \) and \( G \) lie on the side \( {AC} \) and satisfy \( \angle {ABG} = \angle {FBC} \) . The points \( H \) and \( I \) are chosen on the line \( {AB} ... | Solution. The law of sines gives\n\n\[ \n\frac{BD}{DC} = \frac{BD}{DA} \cdot \frac{AD}{DC} = \frac{\sin \angle {BAD}}{\sin \beta } \cdot \frac{\sin \gamma }{\sin \angle {CAD}} \]\n\n\[ = \frac{\sin \angle {CAE}}{\sin \gamma } \cdot \frac{\sin \beta }{\sin \angle {BAE}} \cdot \frac{{\sin }^{2}\gamma }{{\sin }^{2}\beta }... | Yes |
Problem 4.9.3. Let \( {ABC} \) be a triangle and let \( X \) be a point inside of it. The line \( {AX} \) intersects \( {BC} \) at the point \( {A}_{1} \) . The points \( {B}_{1} \) and \( {C}_{1} \) are defined analogously. Denote the reflection of \( {A}_{1} \) with respect to the midpoint of \( {BC} \) by \( {A}_{2}... | Solution. Ceva's Theorem,\n\n\n\napplied to \( \bigtriangleup {ABC} \) and the points \( {A}_{1},{B}_{1} \) and \( {C}_{1} \), yields\n\n\[ \n\frac{A{C}_{1}}{{C}_{1}B} \cdot \frac{B{A}_{1}}{{A}_{1}C} \cdot \frac{C{B}... | Yes |
Problem 4.9.4. Let \( {ABC} \) be a triangle. The points \( {A}_{1},{B}_{1} \) and \( {C}_{1} \) are collinear and lie on the lines \( {BC},{CA} \) and \( {AB} \), respectively. Let \( {A}_{2} \) be the reflection of \( {A}_{1} \) with respect to the midpoint of \( {BC} \) . The points \( {B}_{2} \) and \( {C}_{2} \) a... | Solution. Menelaus' Theorem, applied to \( \bigtriangleup {ABC} \) and the points \( {A}_{1},{B}_{1} \) and \( {C}_{1} \), gives\n\n\[ \frac{A{C}_{1}}{{C}_{1}B} \cdot \frac{B{A}_{1}}{{A}_{1}C} \cdot \frac{C{B}_{1}}{{B}_{1}A} = 1. \]\n\nWe have that \( A{C}_{1} = B{C}_{2} \) , \( B{A}_{2} = {A}_{1}C \) and \( C{B}_{2} =... | Yes |
Problem 4.9.5. Let \( {ABC} \) be a triangle. The triangles \( \bigtriangleup {BCD} \) , \( \bigtriangleup {CAE} \) and \( \bigtriangleup {ABF} \) are constructed externally on the sides \( {BC},{CA} \) and \( {AB} \), respectively, such that \( \angle {BCD} = \angle {DBC} = \angle {CAE} = \angle {ACE} = \) \( \angle {... | Solution. This is a special case of Problem 4.9.6, where \( \varphi = \theta = \psi \) . | No |
Problem 4.9.6. Let \( {ABC} \) be a triangle. The triangles \( \bigtriangleup {BCD} \) , \( \bigtriangleup {CAE} \) and \( \bigtriangleup {ABF} \) are constructed externally on the sides \( {BC},{CA} \) and \( {AB} \), respectively, such that \( \angle {BCD} = \angle {DBC} = \varphi ,\angle {CAE} = \angle {ACE} = \thet... | Solution. The trigonometric form of Ceva's Theorem gives that it suffices to show that\n\n\[ \frac{\sin \angle {BAD}}{\sin \angle {DAC}} \cdot \frac{\sin \angle {ACF}}{\sin \angle {FCB}} \cdot \frac{\sin \angle {CBE}}{\sin \angle {EBA}} = 1. \]\n\nWe have\n\n\[ \frac{\sin \angle {BAD}}{\sin \angle {DAC}} = \frac{\sin \... | Yes |
Problem 4.9.7. Let \( {ABC} \) be a triangle. The points \( D, E \) and \( F \) lie inside of \( \bigtriangleup {ABC} \) and satisfy \( \angle {BCD} = \angle {ECA} = \varphi ,\angle {DBC} = \angle {ABF} = \theta \) and \( \angle {CAE} = \angle {BAF} = \psi \) . Prove that the lines \( {AD},{BE} \) and \( {CF} \) are co... | Solution. The solution is analogous to that of Problem 4.9.6. | No |
Problem 4.9.8. Let \( {ABC} \) be a triangle. The points \( D, E \) and \( F \) are such that \( \angle {BCD} = \angle {ECA},\angle {DBC} = \angle {ABF} \) and \( \angle {CAE} = \angle {BAF} \) . Denote \( M = {BF} \cap {CE}, N = {AF} \cap {CD} \) and \( P = {AE} \cap {BD} \) . Prove that the lines \( {MD},{NE} \) and ... | Solution. Desargues’ Theorem (Problem 7.1), applied to \( \bigtriangleup {MNF} \) and \( \bigtriangleup {EDP} \), gives that it suffices to prove that the points \( B = {MF} \cap {PD} \) , \( A = {EP} \cap {CN} \) and \( X = {MN} \cap {ED} \) are collinear.\n\nSince \( \angle {BAF} = \angle {CAP} \) and \( \angle {FBA}... | Yes |
Problem 4.9.9. Let \( {ABC} \) be a triangle. The points \( {A}_{1} \) and \( {A}_{2} \) lie on the side \( {BC} \) and are equidistant from its midpoint \( \left( {A}_{1}\right. \) lies between \( B \) and \( {A}_{2} \) ). The points \( {B}_{1} \) and \( {B}_{2} \) lie on the side \( {AC} \) and are equidistant from i... | Solution. Let \( {AD} \cap {BC} = {D}^{\prime },{BE} \cap {AC} = {E}^{\prime } \) and \( {CF} \cap {AB} = {F}^{\prime } \) . Ceva’s Theorem, applied to \( \bigtriangleup {ABC} \) and the point \( F \), yields\n\n\[ \n\frac{A{F}^{\prime }}{{F}^{\prime }B} = \frac{A{B}_{2}}{{B}_{2}C} \cdot \frac{C{A}_{1}}{{A}_{1}B} \n\]\... | Yes |
Problem 4.9.10. Let \( {ABC} \) be a triangle. The points \( {A}_{1} \) and \( {A}_{2} \) lie on the side \( {BC} \) and are equidistant from its midpoint \( \left( {A}_{1}\right. \) lies between \( B \) and \( {A}_{2} \) ). The points \( {B}_{1} \) and \( {B}_{2} \) lie on the side \( {AC} \) and are equidistant from ... | Solution. Desargues' Theorem (Problem 7.1), applied to \( \bigtriangleup {EDP} \) and \( \bigtriangleup {MNF} \), gives that it suffices to prove that the points \( A = {EP} \cap {FN}, B = \) \( {MF} \cap {PD} \) and \( X = {ED} \cap {MN} \) are collinear. Equivalently, we need \( {ED} \) and \( {MN} \) to intersect \(... | Yes |
Problem 4.9.11. Let \( {ABC} \) be a triangle. The points \( {A}_{1} \) and \( {A}_{2} \) lie on the side \( {BC} \) and are equidistant from its midpoint \( \left( {A}_{1}\right. \) lies between \( B \) and \( {A}_{2} \) ). The points \( {B}_{1} \) and \( {B}_{2} \) lie on the side \( {AC} \) and are equidistant from ... | Solution. The trigonometric form of Ceva's Theorem gives that it suffices to show that\n\n\[ \n\frac{\sin \angle {BAD}}{\sin \angle {DAC}} \cdot \frac{\sin \angle {ACD}}{\sin \angle {DCB}} \cdot \frac{\sin \angle {CBD}}{\sin \angle {DBA}} = 1.\n\]\n\nWe have\n\n\[ \n\frac{\sin \angle {C}_{2}{AD}}{\sin \angle {DA}{B}_{2... | Yes |
Problem 4.9.12. Let \( {ABC} \) be a triangle. The points \( {A}_{1} \) and \( {A}_{2} \) lie on the side \( {BC} \) and are equidistant from its midpoint \( \left( {A}_{1}\right. \) lies between \( B \) and \( {A}_{2} \) ). The points \( {B}_{1} \) and \( {B}_{2} \) lie on the side \( {AC} \) and are equidistant from ... | Solution. Let \( {A}_{2}{C}_{2} \cap {AC} = N,{A}_{1}{C}_{1} \cap {AC} = R,{B}_{1}{A}_{1} \cap {AB} = M \) , \( {A}_{2}{B}_{2} \cap {AB} = Q,{B}_{2}{C}_{2} \cap {BC} = S \) and \( {B}_{1}{C}_{1} \cap {BC} = P \) . Problem 4.9.2 yields that \( {CN} = {AR},{AQ} = {BM} \) and \( {BS} = {CP} \) . Now\n\n\[ \n\frac{\sin \an... | Yes |
Problem 4.9.13. Let \( {ABC} \) be a triangle. The points \( {A}_{1} \) and \( {A}_{2} \) lie on the side \( {BC} \) and are equidistant from its midpoint \( \left( {A}_{1}\right. \) lies between \( B \) and \( {A}_{2} \) ). The points \( {B}_{1} \) and \( {B}_{2} \) lie on the side \( {AC} \) and are equidistant from ... | Solution. We will firstly show that the points \( {A}_{1} \) , \( {A}_{2},{B}_{1},{B}_{2},{C}_{1} \) and \( {C}_{2} \) lie on an ellipse. Let the ellipse through \( {A}_{1},{A}_{2},{B}_{1} \) , \( {B}_{2} \) and \( {C}_{1} \) intersect \( {BC} \) again at the point \( X \) (it is easy to see that this point is well-def... | Yes |
Problem 4.9.14. Let \( {ABC} \) be a triangle. The points \( {A}_{1} \) and \( {A}_{2} \) lie on the side \( {BC} \) and are equidistant from its midpoint \( \left( {A}_{1}\right. \) lies between \( B \) and \( {A}_{2} \) ). The points \( {B}_{1} \) and \( {B}_{2} \) lie on the side \( {AC} \) and are equidistant from ... | Solution. Desargues' Theo-\n\n\n\nrem (Problem 7.1), applied to \( \bigtriangleup {ABC} \) and \( \bigtriangleup {DEF} \), yields that it suffices to prove that the points \( G = {AB} \cap {DE}, H = \) \( {AC} \cap {... | Yes |
Problem 4.9.15. (Ceva's Theorem) Let \( {ABC} \) be a triangle and let \( {A}_{1} \in {BC},{B}_{1} \in {AC} \) and \( {C}_{1} \in {AB} \) be points on its sides. Then the lines \( A{A}_{1}, B{B}_{1} \) and \( C{C}_{1} \) are concurrent if and only if \( \frac{A{C}_{1}}{{C}_{1}B} \cdot \frac{B{A}_{1}}{{A}_{1}C} \cdot \f... | Solution. Necessity. Let the lines \( A{A}_{1}, B{B}_{1} \) and \( C{C}_{1} \) intersect each other at the point \( X \) . Now we have \( \frac{A{C}_{1}}{{C}_{1}B} = \frac{{S}_{\bigtriangleup {AXC}}}{{S}_{\bigtriangleup {BXC}}},\frac{B{A}_{1}}{{A}_{1}C} = \frac{{S}_{\bigtriangleup {ABX}}}{{S}_{\bigtriangleup {AXC}}} \)... | Yes |
Problem 4.9.16. (Carnot's Theorem) Let \( {ABC} \) be a triangle and let \( {A}_{1} \in {BC},{B}_{1} \in {AC} \) and \( {C}_{1} \in {AB} \) be points on its sides. Then the perpendiculars through the points \( {A}_{1},{B}_{1} \) and \( {C}_{1} \) to the sides they lie on are concurrent if and only if \( A{C}_{1}^{2} + ... | Solution. Necessity. The Pythagorean Theorem, applied to the triangles \( \bigtriangleup A{C}_{1}X,\bigtriangleup {A}_{1}{XB},\bigtriangleup {B}_{1}{XC},\bigtriangleup B{C}_{1}X \) , \( \bigtriangleup {A}_{1}{XC},\bigtriangleup {B}_{1}{XC} \), and \( \bigtriangleup {B}_{1}{XA} \), gives:\n\n1) \( A{C}_{1}^{2} = A{X}^{2... | Yes |
Problem 4.9.17. Let \( {ABC} \) be a triangle. The points \( D, E \) and \( F \) are such that \( {BD} = {CD},{CE} = {EA} \) and \( {AF} = {FB} \) . Prove that the perpendiculars from \( A \) to \( {EF} \), from \( B \) to \( {DF} \) and from \( C \) to \( {DE} \) are concurrent. | Solution. Denote the projections of \( A, B \) and \( C \) onto \( {EF},{FD} \) and \( {DE} \) by \( H, I \) and \( J \), respectively.\n\n\n\nThen\n\n\[ \left( {D{J}^{2} - J{E}^{2}}\right) + \left( {E{H}^{2} - H{F}^... | Yes |
Problem 4.9.18. Let \( {ABC} \) be a triangle and let \( D \) be a point inside of it. The points \( F, E \) and \( G \) are chosen on the perpendiculars through \( D \) to the sides \( {AB},{BC} \) and \( {CA} \), respectively. Prove that the perpendiculars from \( A \) to \( {GF} \), from \( B \) to \( {FE} \) and fr... | Solution. Denote the feet of the perpendiculars from \( A, B \) and \( C \) to \( {GF},{FE} \) and \( {GE} \) by \( J, K \) and \( I \), respectively. Denote the feet of the perpendiculars from \( F, E \) and \( G \) to \( {AB} \) , \( {BC} \) and \( {CA} \) by \( N, M \) and \( L \) , respectively.\n\nCarnot’s Theorem... | Yes |
Problem 4.9.19. Let \( {ABC} \) be a triangle and let \( X \) be a point inside of it. The points \( {A}_{1},{B}_{1} \) and \( {C}_{1} \) are the projections of \( X \) onto the sides \( {BC},{CA} \) and \( {AB} \), respectively. Denote the reflection of \( {A}_{1} \) with respect to the midpoint of \( {BC} \) by \( {A... | Solution. Carnot’s Theorem (Problem 4.9.16), applied to \( \bigtriangleup {ABC} \) and the point \( X \), gives \( A{C}_{1}^{2} + B{A}_{1}^{2} + C{B}_{1}^{2} = {C}_{1}{B}^{2} + {A}_{1}{C}^{2} + {B}_{1}{A}^{2} \) . Hence, using \( A{C}_{1} = B{C}_{2} \) , \( B{A}_{2} = {A}_{1}C \) and \( C{B}_{2} = A{B}_{1} \) , we dedu... | Yes |
Problem 4.9.20. Let \( {ABC} \) be a triangle and let \( X \) be a point inside of it. Denote the feet of the perpendiculars from \( X \) to the sides \( {AB},{BC} \) and \( {CA} \) by \( D, E \) and \( F \), respectively. The circumcircle of \( \bigtriangleup {DEF} \) intersects \( {AB},{BC} \) and \( {CA} \) at the p... | Solution. Clearly, the perpendicular bisectors of \( D{D}^{\prime }, E{E}^{\prime } \) and \( F{F}^{\prime } \) intersect each other at the center of the circle \( O \) .\n\nLet \( Y \) be the reflection of \( X \) with respect to \( O \) . The projections of \( O \) onto the sides of the triangle are the midpoints of ... | Yes |
Problem 4.9.21. Let \( {ABC} \) be a triangle and let \( X \) be a point inside of it. The points \( {A}_{1} \in {BC},{B}_{1} \in {AC} \) and \( {C}_{1} \in {AB} \) are chosen, such that the lines \( X{A}_{1}, X{B}_{1} \) and \( X{C}_{1} \) are angle bisectors of \( \angle {BXC},\angle {CXA} \) and \( \angle {AXB} \), ... | Solution. Multiplying the equalities given by the angle bisector theorem, applied to \( \bigtriangleup {ABX},\bigtriangleup {ACX} \) and \( \bigtriangleup {BCX} \), we get\n\n\[ \frac{AX}{BX} \cdot \frac{BX}{CX} \cdot \frac{CX}{AX} = \frac{A{C}_{1}}{{C}_{1}B} \cdot \frac{B{A}_{1}}{{A}_{1}C} \cdot \frac{C{B}_{1}}{{B}_{1... | Yes |
Problem 4.9.22. Let \( {ABC} \) be a triangle. Denote the midpoints of the sides \( {BC},{CA} \) and \( {AB} \) by \( D, E \) and \( F \), respectively. Let \( M, N \) and \( P \) be the incenters of \( \bigtriangleup {AFE},\bigtriangleup {FBD} \) and \( \bigtriangleup {EDC} \), respectively. Prove that the lines \( {D... | Solution. Consider the translation that sends \( A \) to \( F \) . It sends \( \bigtriangleup {AFE} \) to \( \bigtriangleup {FBD} \) and thus it sends \( M \) to \( N \) . Therefore, the segments \( {EM} \) and \( {DN} \) are parallel and equal in length.\n\nThe same holds for the segments \( {PD} \) and \( {MF} \) . H... | Yes |
Problem 4.9.23. Let \( {ABC} \) be a triangle. The points \( F \) and \( K;E \) and \( J;G \) and \( D \) are chosen on the lines \( {BC},{CA} \) and \( {AB} \), respectively, such that the orders on each line are \( F - C - B - K, J - A - C - E \) and \( D - B - A - G \) . Also, \( {DB} = {BC} = {CE},{FC} = {CA} = {AG... | Solution. We have\n\n\[ \n\frac{\sin \angle {BAM}}{\sin \angle {DAC}} = \frac{\sin \angle {DAM}}{\sin \angle {AMD}} \cdot \frac{\sin \angle {AMC}}{\sin \angle {DAC}} = \frac{DM}{AD} \cdot \frac{AC}{MC}.\n\]\n\nMenelaus’ Theorem, applied to \( \bigtriangleup {ADC} \), gives\n\n\[ \n\frac{DM}{MC} = \frac{DB}{BA} \cdot \f... | Yes |
Problem 4.9.24. Let \( {ABC} \) be a triangle. The points \( F \) and \( K;E \) and \( J;G \) and \( D \) are chosen on the lines \( {BC},{CA} \) and \( {AB} \), respectively, such that the orders on each line are \( F - C - B - K, J - A - C - E \) and \( D - B - A - G \) . Also, \( {DB} = {BC} = {CE},{FC} = {CA} = {AG... | Solution. Denote \( X = {BC} \cap {NP}, Y = {AC} \cap {MP} \) and \( Z = {AB} \cap {MN} \) (these points are not shown in the figure). Desargues’ Theorem (Problem 7.1) gives that it suffices to show that the points \( X, Y \) and \( Z \) are collinear.\n\nMenelaus’ Theorem, applied to \( \bigtriangleup {ABC} \) and the... | Yes |
Problem 4.9.25. Let \( {ABCDEF} \) be a hexagon with \( \angle B = \angle E \) , \( \angle C = \angle F \) and \( \angle A = \angle D \) . Prove that if the points \( M, N, P, Q, R \) and \( S \) are the midpoints of \( {AB},{BC},{CD},{DE},{EF} \) and \( {FA} \), respectively, then the lines \( {MQ},{NR} \) and \( {PS}... | Solution. Denote \( {AB} \cap {FE} = X \) and \( {ED} \cap {BC} = Y \) . Using \( \angle {AXF} = \angle {CYD} \), it is easy to see that the quadrilateral \( {XBYE} \) is a parallelogram and thus \( {AB}\parallel {ED} \) . Analogously, \( {BC}\parallel {EF} \) and \( {CD}\parallel {FA} \) . Steiner’s Theorem (Problem 5... | Yes |
Problem 4.9.26. Let \( {AFBDCE} \) be a hexagon, such that \( \angle {FBD} = \) \( \angle {DCE} = \angle {EAF} = \varphi ,{AF} = {FB},{BD} = {DC} \) and \( {CE} = {EA} \) . Prove that the lines \( {AD},{BE} \) and \( {CF} \) are concurrent. | Solution. We will apply the trigonometric form of Ceva's Theorem to \( \bigtriangleup {ABC} \) . We have\n\n\[ \n\frac{\sin \angle {BAD}}{\sin \angle {DAC}} = \frac{\sin \angle {BAD}}{\sin \angle {ABD}} \cdot \frac{\sin \angle {ABD}}{\sin \angle {ACD}} \cdot \frac{\sin \angle {ACD}}{\sin \angle {DAC}}\n\]\n\n\[ \n= \fr... | Yes |
Problem 4.9.27. Let \( {ABC} \) be a triangle and let \( X \) be a point. The projections and reflections of \( X \) with respect to \( {BC},{CA} \) and \( {AB} \) are \( D \) and \( {D}^{\prime }, E \) and \( {E}^{\prime } \), and \( F \) and \( {F}^{\prime } \), respectively. Prove that the circumcircles of \( \bigtr... | Solution. Denote the circumcircles of \( \bigtriangleup A{F}^{\prime }{E}^{\prime },\bigtriangleup B{F}^{\prime }{D}^{\prime },\bigtriangleup C{E}^{\prime }D \) and \( \bigtriangleup {ABC} \) by \( {k}_{1},{k}_{2},{k}_{3} \) and \( k \), respectively. Set \( {k}_{1} \cap {k}_{3} = K \) .\n\nWe will show that \( K \) li... | Yes |
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