PUMaC 2011 · 加试 · 第 3 题

(3) The lines AA , BB , CC concur. Then, if any two of these three assertions are valid, then the third one must hold, too. [Hint: We didn’t put the quadrilateral version of Menelaus in the Prerequisites section for nothing! //Btw, this is called the Cevian Nest Theorem ; remember it, it might prove useful!] (b) (2 points). The triangles A B C and ABC are perspective. 0 0 0 Solution . (a) Let the lines AA ”, BB ”, CC ” meet the sides BC , CA , AB of triangle ABC at the points X , Y , Z , respectively. Of course, the lines AA ”, BB ”, CC ” are nothing but the lines AX , BY , CZ . ′ ′ Apply the Menelaus theorem for quadrilaterals to the quadrilateral BCB C with the collinear points X , A , A ”, A on its sides (you are reading right - the point A occurs twice, it is actually the point of intersection of two opposite sides of the quadrilateral!). Then, you obtain ′ ′′ ′ BX CA B A C A · · · = 1 , ′ ′′ ′ XC AB A C AB so that ′′ ′ ′ BX AB A C AB = · · ′ ′ ′′ XC C A B A CA ′ ′′ ′ AB AB A C = · · . ′ ′ ′′ C A CA B A Similarly, ′ ′′ ′ ′ ′′ ′ CY BC BC B A AZ CA CA C B = · · and = · · . ′ ′ ′′ ′ ′ ′′ Y A A B AB C B ZB B C BC A C Hence, ( ) ( ) ′ ′ ′ ′′ ′ ′′ ′ ′′ ′ BX CY AZ AB BC CA A C B A C B · · = · · · · , ′ ′ ′ ′ ′′ ′ ′′ ′ ′′ XC Y A ZB C A A B B C B A C B A C 10 PUMAC 2011 Power Round 11 ′ ′ ′ ′′ ′ ′′ ′ ′′ ′ BX CY AZ AB BC CA A C B A C B i.e. w = uv , where w = · · , u = · · , and v = · · . ′ ′ ′ ′ ′′ ′ ′′ ′ ′′ XC Y A ZB C A A B B C B A C B A C But by Ceva’s theorem assertion (1) holds iff u = 1, (2) holds iff v = 1, and (3) holds iff w = 1, therefore the relation w = uv proves (a). (b) This now follows by Part (a) and Proposition 4. ′ Proposition 6 (a) (2 points) Let l and l be two lines in plane and let P be a ′ point in the same plane but not lying on either l or l . Let A , B , C , D be points on ′ ′ ′ ′ ′ l and let A , B , C , D be the intersections of the lines P A , P B , P C , P D with l , respectively. Then, ′ ′ ′ ′ BA DA B A D A : = : . ′ ′ ′ ′ BC DC B C D C (b) (5 points). The incenter I is the orthocenter of triangle A B C . 0 0 0 Solution . (a) Imitate the proof you (should) have for Lemma 2 from the Pre- requisites section. If your proof is ugly, then try thinking about a new solution using only the Law of Sines! (b) Since A B C is a cevian triangle, the pencil ( B A , B C , B B , B A ) is 1 1 1 0 1 0 0 1 0 harmonic. Thus, by using the collinearity of A , B and C , the pencil ( B C , B A , 0 0 0 1 0 0 B B , B A ) is also harmonic. 0 1 0 Hence, B lies on the polar of A with respect to the incircle, and since A lies 0 0 too it follows that B C is the polar of A with respect to the incircle. 0 0 0 Similarly, C A is the polar of B and A B the polar of C . Therefore, I is 0 0 0 0 0 0 indeed the orthocenter of triangle A B C . 0 0 0 11 PUMAC 2011 Power Round 12 Proposition 7 (7 points). Let R a mobile point on AC and consider S , T the intersections of RA , RC with AB and BC respectively. Then, B lies on ST . 0 0 0 Solution . Let M and N be the intersections of B A and B C with the sides 0 0 0 0 BA and BC , respectively. Since ( C , A ), ( S , T ), ( M , N ) are the traces left by the lines ( B A , B C ), 1 1 1 0 1 0 ( RA , RC ) and ( CA , AC ) on the sides BA and BC , it follows that the cross-ratios 0 0 0 0 ( A , S , C , M ) and ( N , T , A , C ) are equal. 1 1 Hence, the lines AN , ST , C A , M C are concurrent. But { B } = AN ∩ C A ∩ 1 1 0 1 1 M C , thus B lies on ST , which completes the proof of Proposition 7. 0 12 PUMAC 2011 Power Round 13 Proposition 8 (7 points). The triangles ABC and T RS are perspective. Solution 1 . Since AB C is the polar of A with respect to ∠ BAC , the pencil 0 0 0 ( B R , B A , B S , B A ) is harmonic. Thus, by the collinearity of S , B , and T , 0 0 0 0 0 0 it follows that the pencil ( T S , T A , T R , T C ) is harmonic. Hence, the lines AT , BR , CS concur, i.e. the triangles ABC and T RS are perspective. This proves Proposition 8. Solution 2 . Recall that A , E, F , B , D, F , C , D, E are collinear. 0 0 0 We apply Menelaus’s theorem to points A SR on 4 AEF , 0 EA F S AR 0 · · = 1 A F SA RE 0 to points B ST on 4 BDF , 0 F B DT BS 0 · · = 1 B D T B SF 0 13 PUMAC 2011 Power Round 14 to points C RT on 4 CDE , 0 DC ER CT 0 · · = 1 C E RS T D 0 and finally apply Ceva’s theorem to the perspective triangles A B C and DEF : 0 0 0 F A DB EC 0 0 0 · · = 1 . A E B F C D 0 0 0 Multiplying all these together, we get that BS CT AR · · = 1 SA T B RC which is Ceva’s theorem for points T RS in 4 ABC , which shows that the two triangles are perspective, as desired. Proposition 9 (14 points). The triangles A B C and T RS are perspective 0 0 0 and the locus of the perspector is a line tangent to the incircle of triangle ABC . (Hint: everything used so far is part of the hint!) Solution . Let M and N be the intersections of B C and A B with the sidelines 0 0 0 0 BC and AB , respectively. Also, let K , L be the intersections of M N with RS and RT , respectively. First, note that triangles A B C and T RS are perspective by the Cevian Nest 0 0 0 Theorem. Since the triangles ABC and A B C are perspective, the points AA ∩ 0 0 0 0 CC , A M ∩ C N , AM ∩ CN lie on the same line, i.e. the triangles AA M and 0 0 0 0 CC N are perspective; consequently, the lines AC , A C , M N concur at a point, 0 0 0 ′ say O . However the lines AN , ST , M C are concurrent, so the cross-ratios ( B , A , S , M ) ′ ′ ′ ′ and ( B , N , T , C ) are equal. By intersecting the pencil O B , O A , O S , O M with the line RS , it follows that ( B , A , S , M )=( A , R , S , K ). Similarly, by intersecting 0 ′ ′ ′ ′ the pencil O B , O N , O T , O C with the line RT , ( B , N , T , C )=( C , L , T , R ). 0 Therefore, ( A , R , S , K )=( C , L , T , R ), i.e. ( A , R , S , K )=( T , R , C , L ). 0 0 0 0 Hence, the lines A T , SC , KL are concurrent, that is the lines A T , B R , 0 0 0 0 C S , M N are concurrent. From the above result, we deduce that the locus of the 0 perspector of A B C and T RS is the line M N . 0 0 0 By Proposition 4, the lines A D , B E , C F concur on the incircle at Q . Consider 0 0 0 ′ ′ M and N the intersections of the tangent in Q with the sides BA and BC , respec- ′ ′ tively. By Newton’s theorem applied to the circumscribed quadrilateral M N CA , ′ ′ the lines QE , DF , M C , N A are concurrent at B . 0 ′ ′ Hence, M ≡ M and N ≡ N . Therefore, the line M N is tangent to the incircle at Q . This completes the proof of Proposition 9. 14 PUMAC 2011 Power Round 15 3 Historical Fact/Challenge When the point P chosen at the beginning of the excursion is the incenter, the Nagel point or the orthocenter of triangle ABC , the configuration generates a so- called Feuerbach family of circles. More precisely, maintaining the notations from the previous propositions, the circumcircle of triangle T RS always passes through 4 the Feuerbach point of triangle ABC , as R varies on the line AC . So, do try and have fun with it after finishing everything. You will not receive any credit of course, but you definitely have the chance to impress! 4 You should be aware that the incircle and the nine-point circle of a given triangle ABC are tangent to each other and that their tangency point is called the Feuerbach point of triangle ABC ! 15