HMMT 十一月 2010 · GEN1 赛 · 第 3 题
HMMT November 2010 — GEN1 Round — Problem 3
题目详情
- [ 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?
解析
- [ 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? √ 31 Answer: 3 5 General Test A D E F J B C 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. 2 The radius of the incircle can be computed by the formula √ √ 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 ( ) √ 2 6 3 / 7 9 √ Then [ ADE ] = [ ABC ] = [ ABC ]. Here, we use the fact that 4 ABC and 4 ADE are 100 20 3 / 7 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 ( ) √ √ √ 2 3 / 2 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 4 3 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