HMMT 二月 2024 · 冲刺赛 · 第 36 题
HMMT February 2024 — Guts Round — Problem 36
题目详情
- [20] Let ABC be a triangle. The following diagram contains points P , P , . . . , P , which are the following 1 2 7 triangle centers of triangle ABC in some order: • the incenter I ; • the circumcenter O ; • the orthocenter H ; • the symmedian point L , which is the intersections of the reflections of B -median and C -median across angle bisectors of ∠ ABC and ∠ ACB , respectively; • the Gergonne point G , which is the intersection of lines from B and C to the tangency points of the incircle with AC and AB , respectively; • the Nagel point N , which is the intersection of line from B to the tangency point between B -excircle and AC , and line from C to the tangency point between C -excircle and AB ; and • the Kosnita point K , which is the intersection of lines from B and C to the circumcenters of triangles AOC and AOB , respectively. P 2 P 1 P 3 P 4 P 5 P 6 P 7 Note that the triangle ABC is not shown. Compute which triangle centers { I, O, H, L, G, N, K } corre- sponds to P for k ∈ { 1 , 2 , 3 , 4 , 5 , 6 , 7 } . k Your answer should be a seven-character string containing I , O , H , L , G , N , K , or X for blank. For instance, if you think P = H and P = L , you would answer XHXXXLX . If you attempt to identify 2 6 ⌈ ⌉ 5 / 3 n > 0 points and get them all correct, then you will receive ( n − 1) points. Otherwise, you will receive 0 points.
解析
- [20] Let ABC be a triangle. The following diagram contains points P , P , . . . , P , which are the 1 2 7 following triangle centers of triangle ABC in some order: • the incenter I ; • the circumcenter O ; • the orthocenter H ; • the symmedian point L , which is the intersections of the reflections of B -median and C -median across angle bisectors of ∠ ABC and ∠ ACB , respectively; • the Gergonne point G , which is the intersection of lines from B and C to the tangency points of the incircle with AC and AB , respectively; • the Nagel point N , which is the intersection of line from B to the tangency point between B - excircle and AC , and line from C to the tangency point between C -excircle and AB ; and • the Kosnita point K , which is the intersection of lines from B and C to the circumcenters of triangles AOC and AOB , respectively. P 2 P 1 P 3 P P 4 5 P 6 P 7 Note that the triangle ABC is not shown. Compute which triangle centers { I, O, H, L, G, N, K } corresponds to P for k ∈ { 1 , 2 , 3 , 4 , 5 , 6 , 7 } . k Your answer should be a seven-character string containing I , O , H , L , G , N , K , or X for blank. For instance, if you think P = H and P = L , you would answer XHXXXLX . If you attempt to identify 2 6 ⌈ ⌉ 5 / 3 n > 0 points and get them all correct, then you will receive ( n − 1) points. Otherwise, you will receive 0 points. Proposed by: Kevin Zhao, Pitchayut Saengrungkongka Answer: KOLIN GH ′ Solution: Let G be the centroid of triangle ABC . Recall the following. ′ ′ ′ • Points O, G , H lie on Euler’s line of △ ABC with OG : G H = 1 : 2 . ′ ′ ′ • Points I, G , N lie on Nagel’s line of △ ABC with IG : G N = 1 : 2 . Thus, OI ∥ HN with OI : HN = 1 : 2 . Therefore, we can detect parallel lines with ratio 2 : 1 in the figure. The only possible pairs are P P ∥ P P . Therefore, there are two possibilities: ( P , P ) 2 4 7 5 2 7 and ( P , P ) must be ( O, H ) and ( I, N ) in some order. Intuitively, H should be further out, so it’s 4 5 not unreasonable to guess that P = O , P = H , P = I , and P = N . Alternatively, perform the 2 7 4 5 algorithm below with the other case to see if it fails. To identify the remaining points, we recall that the isogonal conjugate of G and N both lie on OI (they are insimilicenter and exsimilicenter of incircle and circumcircle, respectively). Thus, H, G, N, I lie on isogonal conjugate of OI , known as the Feuerbach’s Hyperbola . It’s also known that OI is tangent to this line, and this hyperbola have perpendicular asymptotes. Using all information in the above paragraph, we can eyeball a rectangular hyperbola passing through H, G, N, I and is tangent to OI . It’s then not hard to see that P = G . 6 Finally, we need to distinguish between symmedian and Kosnita points. To do that, recall that Kosnita point is isogonal conjugate of the nine-point center (not hard to show). Thus, H, L, K, O lies on isogonal conjugate of OH , which is the Jerabek’s Hyperbola . One can see that H, L, K, O lies on the same branch. Moreover, they lie on this hyperbola in this order because the isogonal conjugates (in order) are O , centroid, nine-point center, and H , which lies on OH in this order. Using this fact, we can identity P = L and P = K , completing the identification. 5 1 The following is the diagram with the triangle ABC . A P 2 P 1 P 3 P P 4 5 P 6 P 7 B C Here is the Asymptote code that generates the diagram in the problem. import olympiad; import geometry; size(7.5cm); pair A = (0.5,3.2); pair B = (0,0); pair C = (4,0); pair O = circumcenter(triangle(A,B,C)); pair H = orthocentercenter(triangle(A,B,C)); pair L = symmedian(triangle(A,B,C)); pair Ge = gergonne(triangle(A,B,C)); pair I = incenter(triangle(A,B,C)); pair Na = A+B+C - 2I; pair K = extension(A, circumcenter(B,O,C), B, circumcenter(A,O,C)); dot("",Ge , dir (-90)); dot("",O, dir (90)); dot("",H, dir (90)); dot("",L, dir (135)); dot("",I, dir (-45)); dot("",Na , dir (-90)); dot("",K, dir (90));