HMMT 二月 2010 · 几何 · 第 8 题

HMMT February 2010 — Geometry — Problem 8

[ 6 ] Let O be the point (0 , 0). Let A, B, C be three points in the plane such that AO = 15, BO = 15, and CO = 7, and such that the area of triangle ABC is maximal. What is the length of the shortest side of ABC ?

解析

[ 6 ] Let O be the point (0 , 0). Let A, B, C be three points in the plane such that AO = 15, BO = 15, and CO = 7, and such that the area of triangle ABC is maximal. What is the length of the shortest side of ABC ? Answer: 20 We claim that O should be the orthocenter of the triangle ABC . If O is not on an altitude of 4 ABC , suppose (without loss of generality) that AO is not perpendicular to BC . We can Geometry Subject Test rotate A around O , leaving B and C fixed, to make AO perpendicular to BC , which strictly increases the area. Therefore, if [ ABC ] is maximal then 4 ABC is an isosceles triangle with orthocenter O and base AB . C E D O A B F Let F be the foot of the perpendicular from C to AB . Since ∠ F OA and ∠ COE are vertical, ∠ F AO = AF CF OF +7 2 2 ∠ OCE . Then 4 F AO is similar to 4 F CB , so we have = = , so AF = OF + 7 · OF . OF BF AF 2 2 2 Since AF = 225 − OF , 2 · OF + 7 · OF − 225 = 0, so OF = 9. Then AF = 12, so AB = 24 and BC = 20. Thus, the length of the shortest side of 4 ABC is 20.