HMMT 二月 2018 · 冲刺赛 · 第 31 题

HMMT February 2018 — Guts Round — Problem 31

[ 17 ] In triangle ABC , AB = 6, BC = 7 and CA = 8. Let D , E , F be the midpoints of sides BC , AC , AB , respectively. Also let O , O , O be the circumcenters of triangles AF D , BDE , and CEF , A B C respectively. Find the area of triangle O O O . A B C

解析

[ 17 ] In triangle ABC , AB = 6, BC = 7 and CA = 8. Let D , E , F be the midpoints of sides BC , AC , AB , respectively. Also let O , O , O be the circumcenters of triangles AF D , BDE , and CEF , A B C respectively. Find the area of triangle O O O . A B C Proposed by: Henrik Boecken Let AB = z, BC = x, CA = y . Let X, Y, Z, O, N be the circumcenter of AEF, BF D, CDE, ABC, DEF respectively. Note that N is the nine-point center of ABC , and X, Y, Z are the midpoints of OA, OB, OC 1 respectively, and thus XY Z is the image of homothety of ABC with center O and ratio , so this tri- 2 y x z angle has side lengths , , . Since N X perpendicularly bisects EF , which is parallel to BC and thus 2 2 2 Y Z , we see that N is the orthocenter of XY Z . Moreover, O lies on Y N and O X is perpendicular 1 1 to XY . To compute the area of O O O , it suffices to compute [ N O O ] + [ N O O ] + [ N O O ]. Note that 1 2 3 1 2 2 3 3 1 O X is parallel to N O , and O Y is parallel to XN , so [ N O O ] = [ N XO ] = [ N XY ]. Similarly the 1 2 2 1 2 2 other two triangles have equal area as [ N Y Z ] and [ N ZX ] respectively, so the desired area is simply the area of [ XY Z ], which is √ √ √ ( x + y + z )( x + y − z )( x − y + z )( − x + y + z ) 1 21 · 9 · 5 · 7 21 15 = = . 4 4 16 16