HMMT 十一月 2010 · GEN1 赛 · 第 3 题

英文解析

3. [ 3 ] Triangle ABC has AB = 5, BC = 7, and CA = 8. New lines not containing but parallel to AB ,
   BC , and CA are drawn tangent to the incircle of ABC . What is the area of the hexagon formed by the sides of the original triangle and the newly drawn lines?
   √
   Answer: 331
   General Test 5
   D EA
   B CJF
   H G
   ( )
   2 2 2
   5 +8 − 7 − 1 °
   From the law of cosines we compute ] A = cos = 60 . Using brackets to denote the area
   2(5)(8)
   of a region, we find that
   √
   1 °
   [ ABC ] = AB · AC · sin 60 = 10 3.
   The radius of the incircle can be computed by the formula 2
   √
   √
   2[ ABC ]
   20 3
   r = = = 3.
   AB + BC + CA 20
   √ √ √
   2[ ABC ]
   20 3 20 3 6 3
   Now the height from A to BC is = . Then the height from A to DE is − 2 r = .
   BC 7 7 7
   ( )
   √
   6 3 / 72
   √9
   Then [ ADE ] = [ ABC ] = [ ABC ]. Here, we use the fact that 4 ABC and 4 ADE are
   20 3 / 7100
   similar.
   √ √
   2[ ABC ]
   20 3 5 3
   Similarly, we compute that the height from B to CA is = = . Then the height from
   CA 8 2
   ( )
   √
   √ √
   3 / 22
   5 3 3 1
   √
   B to HJ is − 2 r = . Then [ BHJ ] = [ ABC ] = [ ABC ].
   2 2 25
   5 3 / 2
   √
   √
   2[ ABC ]
   20 3
   Finally, we compute that the height from C to AB is = = 4 3. Then the height from C
   5 5
   ( )
   √
   √ √2
   2 3 1
   √
   to F G is 4 3 − 2 r = 2 3. Then [ CF G ] = [ ABC ] = [ ABC ].
   4 34
   Finally we can compute the area of hexagon DEF GHJ . We have
   ( ) ( )
   9 1 1 31
   [ DEF GHJ ] = [ ABC ] − [ ADE ] − [ BHJ ] − [ CF G ] = [ ABC ] 1 − − − = [ ABC ] =
   100 25 4 50
   √ √
   ( )
   31 31
   10 3 = 3.
   50 5