HMMT 二月 2018 · 几何 · 第 7 题
HMMT February 2018 — Geometry — Problem 7
题目详情
- Triangle ABC has sidelengths AB = 14, AC = 13, and BC = 15. Point D is chosen in the interior of AB and point E is selected uniformly at random from AD . Point F is then defined to be the intersection point of the perpendicular to AB at E and the union of segments AC and BC . Suppose that D is chosen such that the expected value of the length of EF is maximized. Find AD .
解析
- Triangle ABC has sidelengths AB = 14, AC = 13, and BC = 15. Point D is chosen in the interior of AB and point E is selected uniformly at random from AD . Point F is then defined to be the intersection point of the perpendicular to AB at E and the union of segments AC and BC . Suppose that D is chosen such that the expected value of the length of EF is maximized. Find AD . Proposed by: Gabriel Mintzer √ Answer: 70 Let G be the intersection of the altitude to AB at point D with AC ∪ BC . We first note that the [ ADGC ] maximal expected value is obtained when DG = , where [ P ] denotes the area of polygon P . AD Note that if DG were not equal to this value, we could move D either closer or further from A and increase the value of the fraction, which is the expected value of EF . We note that this equality can only occur if D is on the side of the altitude to AB nearest point B . Multiplying both sides of this equation by AD yields AD · DG = [ ADGC ], which can be interpreted as meaning that the area of the rectangle with base AD and height DG must have area equal to that of quadrilateral ADGC . We can now solve this problem with algebra. Let x = BD . We first compute the area of the rectangle with base AD and height DG . We have that AD = AB − BD = 14 − x . By decomposing the 13 − 14 − 15 triangle into a 5 − 12 − 13 triangle and 4 a 9 − 12 − 15 triangle, and using a similarity argument, we find that DG = x . Thus, the area of this 3 4 56 4 2 rectangle is x (14 − x ) = x − x . 3 3 3 We next compute the area of quadrilateral ADGC . We note that [ ADGC ] = [ ABC ] − [ BDG ]. We ( ) 1 4 1 4 2 2 have that [ ABC ] = (12)(14) = 84. We have BD = x and DG = x , so [ BDG ] = ( x ) x = x . 2 3 2 3 3 2 2 Therefore, we have [ ADGC ] = [ ABC ] − [ BDG ] = 84 − x . 3 Equating these two areas, we have 56 4 2 2 2 x − x = 84 − x , 3 3 3 or, simplifying, 2 x − 28 x + 126 = 0 . √ √ Solving yields x = 14 ± 70, but 14 + 70 exceeds AB , so we discard it as an extraneous root. Thus, √ BD = 14 − 70 and √ √ AD = AB − BD = 14 − (14 − 70) = 70 . Remark: if the altitude to point C meets AB at point H , then the general answer to this problem √ is AH · AB . This result can be derived by considering the effects of dilation in the AB direction and dilation in the CH direction then performing dilations such that ∠ C is right and carrying out the calculation described above while considering congruent triangles.