HMMT 二月 2010 · 冲刺赛 · 第 10 题
HMMT February 2010 — Guts Round — Problem 10
题目详情
- [ 7 ] Let ABC be a triangle with AB = 8, BC = 15, and AC = 17. Point X is chosen at random on line segment AB . Point Y is chosen at random on line segment BC . Point Z is chosen at random on line segment CA . What is the expected area of triangle XY Z ? y x
解析
- [ 7 ] Let ABC be a triangle with AB = 8, BC = 15, and AC = 17. Point X is chosen at random on line segment AB . Point Y is chosen at random on line segment BC . Point Z is chosen at random on line segment CA . What is the expected area of triangle XY Z ? Answer: 15 Let E ( X ) denote the expected value of X , and let [ S ] denote the area of S . Then E ([ 4 XY Z ]) = E ([ 4 ABC ] − [ 4 XY B ] − [ 4 ZY C ] − [ 4 XBZ ]) = [ 4 ABC ] − E ([ 4 XY B ]) − E ([ 4 ZY C ]) − [ 4 XBZ ]) , 1 1 where the last step follows from linearity of expectation . But [ 4 XY B ] = · BX · BY · sin( B ). 2 1 AB The sin( B ) term is constant, and BX and BY are both independent with expected values and 2 2 BC 1 1 , respectively. Thus E ([ 4 XY B ]) = AB · BC · sin( B ) = [ 4 ABC ]. Similarly, E ([ 4 ZY C ]) = 2 8 4 1 E ([ 4 ZBX ]) = [ 4 ABC ]. 4 1 1 1 1 Then we have E ([ 4 XY Z ]) = (1 − − − )[ 4 ABC ] = [ 4 ABC ] = 15. 4 4 4 4 Note: We can also solve this problem (and the more general case of polygons) by noting that the area of XY Z is linear in the coordinates of X , Y , and Z , so the expected area of XY Z is the same ′ ′ ′ ′ ′ as the area of X Y Z , where X is the expected location of X , Y is the expected location of Y , and ′ Z is the expected location of Z . In our case, this corresponds to the midpoints of the three sides AB , BC , and CA . 1 See http://www-math.mit.edu/~spielman/AdvComplexity/handout.ps for an introduction to linearity of expectation and other important tools in probability Guts Round y x