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HMMT 二月 2019 · 几何 · 第 10 题

HMMT February 2019 — Geometry — Problem 10

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. In triangle ABC , AB = 13 , BC = 14 , CA = 15. Squares ABB A , BCC B , CAA C are constructed 1 2 1 2 1 2 outside the triangle. Squares A A A A , B B B B , C C C C are constructed outside the hexagon 1 2 3 4 1 2 3 4 1 2 3 4 A A B B C C . Squares A B B A , B C C B , C A A C are constructed outside the hexagon 1 2 1 2 1 2 3 4 5 6 3 4 5 6 3 4 5 6 A A B B C C . Find the area of the hexagon A A B B C C . 4 3 4 3 4 3 5 6 5 6 5 6
解析
  1. In triangle ABC , AB = 13 , BC = 14 , CA = 15. Squares ABB A , BCC B , CAA C are constructed 1 2 1 2 1 2 outside the triangle. Squares A A A A , B B B B , C C C C are constructed outside the hexagon 1 2 3 4 1 2 3 4 1 2 3 4 A A B B C C . Squares A B B A , B C C B , C A A C are constructed outside the hexagon 1 2 1 2 1 2 3 4 5 6 3 4 5 6 3 4 5 6 A A B B C C . Find the area of the hexagon A A B B C C . 4 3 4 3 4 3 5 6 5 6 5 6 Proposed by: Yuan Yao Answer: 19444 Solution 1. We can use complex numbers to find synthetic observations. Let A = a, B = b, C = c . Notice that B is 2 ◦ ◦ a rotation by − 90 (counter-clockwise) of C about B , and similarly C is a rotation by 90 of B about C . 1 ◦ Since rotation by 90 corresponds to multiplication by i , we have B = ( c − b ) · ( − i ) + b = b (1 + i ) − ci 2 and C = ( b − c ) · i + c = bi + c (1 − i ). Similarly, we get C = c (1 + i ) − ai , A = ci + a (1 − i ), 1 2 1 A = a (1 + i ) − bi , B = ai + b (1 − i ). Repeating the same trick on B B B B et. al, we get 2 1 1 2 3 4 C = − a + b ( − 1 + i ) + c (3 − i ), C = a ( − 1 − i ) − b + c (3 + i ), A = − b + c ( − 1 + i ) + a (3 − i ), 4 3 4 A = b ( − 1 − i ) − c + a (3 + i ), B = − c + a ( − 1 + i ) + b (3 − i ), B = c ( − 1 − i ) − a + b (3 + i ). Finally, 3 4 3 repeating the same trick on the outermost squares, we get B = − a + b (3 + 5 i ) + c ( − 3 − 3 i ) , C = 6 5 − a + b ( − 3 + 3 i ) + c (3 − 5 i ) , C = − b + c (3 + 5 i ) + a ( − 3 − 3 i ) , A = − b + c ( − 3 + 3 i ) + a (3 − 5 i ) , A = 6 5 6 − c + a (3 + 5 i ) + b ( − 3 − 3 i ) , B = − c + a ( − 3 + 3 i ) + b (3 − 5 i ). 5 From here, we observe the following synthetic observations. S1. B C C B , C A A C , A B B A are trapezoids with bases of lengths BC, 4 BC ; AC, 4 AC ; AB, 4 AB 2 1 4 3 2 1 4 3 2 1 4 3 and heights h , h , h respectively (where h is the length of the altitude from A to BC , and like- a b c a wise for h , h ) b c ∼ S2. If we extend B B and B B to intersect at B , then B B B BB B ∼ B B B with scale = 5 4 6 3 7 7 4 3 1 2 7 5 6 factor 1:5. Likewise when we replace all B ’s with A ’s or C ’s. Proof of S1. Observe C − B = c − b and C − B = 4( c − b ), hence B C ‖ B C and B C = 4 B C . 1 2 4 3 2 1 3 4 3 4 2 1 Furthermore, since translation preserves properties of trapezoids, we can translate B C C B such 2 1 4 3 ′ that B coincides with A . Being a translation of a − B , we see that B maps to B = 2 b − c and 2 2 3 3 ′ C maps to C = − 2 b + 3 c . Both 2 b − c and − 2 b + 3 c lie on the line determined by b and c (since 4 4 ′ ′ − 2 + 3 = 2 − 1 = 1), so the altitude from A to BC is also the altitude from A to B C . Thus h a 3 4 equals the length of the altitude from B to B C , which is the height of the trapezoid B C C B . 2 3 4 2 1 4 3 This proves S1 for B C C B ; the other trapezoids follow similarly. 2 1 4 3 Proof of S2. Notice a translation of − a + 2 b − c maps B to B , B to B , and B to a point 1 4 2 3 ∼ B = − a + 3 b − c . This means B B B BB B . We can also verify that 4 B + B = 5 B and = 8 8 3 4 1 2 8 6 3 4 B + B = 5 B , showing that B B B is a dilation of B B B with scale factor 5. We also get B 8 5 4 8 5 6 8 4 3 8 lies on B B and B B , so B = B . This proves S2 for B B B B , and similar arguments prove the 3 6 5 4 8 7 3 4 5 6 likewise part. Now we are ready to attack the final computation. By S2, [ B B B B ] + [ BB B ] = [ B B B ] = 3 4 5 6 1 2 7 5 6 1 ◦ [ BB B ]. But by the ac sin B formula, [ BB B ] = [ ABC ] (since ∠ B BB = 180 − ∠ ABC ). Hence, 1 2 1 2 1 2 2 [ B B B B ] + [ BB B ] = 25[ ABC ]. Similarly, [ C C C C ] + [ CC C ] = 25[ ABC ] and [ A A A A ] + 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 5 BC [ AA A ] = 25[ ABC ]. Finally, the formula for area of a trapezoid shows [ B C C B ] = · h = 1 2 2 1 4 3 a 2 5[ ABC ], and similarly the other small trapezoids have area 5[ ABC ]. The trapezoids thus contribute area (75 + 3 · 5) = 90[ ABC ]. Finally, ABC contributes area [ ABC ] = 84. By S1, the outside squares have side lengths 4 BC, 4 CA, 4 AB , so the sum of areas of the outside squares 2 2 2 2 2 2 is 16( AB + AC + BC ). Furthermore, a Law of Cosines computation shows A A = AB + AC + 1 2 2 2 2 2 2 2 2 2 · AB · AC · cos ∠ BAC = 2 AB + 2 AC − BC , and similarly B B = 2 AB + 2 BC − AC and 1 2 2 2 2 2 2 2 2 C C = 2 BC + 2 AC − AB . Thus the sum of the areas of A A A A et. al is 3( AB + AC + BC ). 1 1 2 3 4 2 2 2 2 Finally, the small squares have area add up to AB + AC + BC . Aggregating all contributions from trapezoids, squares, and triangle, we get 2 2 2 [ A A B B C C ] = 91[ ABC ] + 20( AB + AC + BC ) = 7644 + 11800 = 19444 . 5 6 5 6 5 6 Solution 2. Let a = BC, b = CA, c = AB . We can prove S1 and S2 using some trigonometry instead. Proof of S1. The altitude from B to B C has length B B sin ∠ BB B = B B sin ∠ BB B = 3 2 1 2 3 2 1 1 2 2 1 BB sin ∠ B BB = AB sin ∠ ABC = h using Law of Sines. Similarly, we find the altitude from 1 1 2 a √ 2 2 2 C to B C equals h , thus proving B C C B is a trapezoid. Using B B = 2 a + 2 c − b from 4 2 1 a 2 1 4 3 1 2 end of Solution 1, we get the length of the projection of B B onto B C is B B cos BB B = 2 3 3 4 2 3 2 1 2 2 2 2 2 2 2 2 2 2 2 (2 a +2 c − b )+ a − c 3 a + c − b 3 a + b − c = , and similarly the projection of C C onto B C has length . 1 4 3 4 2 a 2 a 2 a 2 2 2 2 2 2 3 a + c − b 3 a + b − c It follows that B C = + a + = 4 a , proving S1 for B C C B ; the other cases 3 4 2 1 4 3 2 a 2 a follow similarly. Proof of S2. Define B to be the image of B under the translation taking B B to B B . We claim B 8 1 2 4 3 8 ◦ ∼ lies on B B . Indeed, B B B BB B , so ∠ B B B = ∠ BB B = 180 − ∠ B B C = ∠ B B C . = 3 6 8 4 3 1 2 8 3 4 2 1 3 2 1 2 3 4 ◦ ◦ Thus ∠ B B C = ∠ B B B = 90 . But ∠ B B C = 90 , hence B , B , B are collinear. Similarly we 8 3 4 4 3 2 6 3 4 8 3 6 B B B B 1 7 3 7 4 can prove B B passes through B , so B = B . Finally, = = (using B B = 4 a, B B = 5 4 8 8 7 3 6 4 5 B B B B 5 7 6 7 5 4 c, B B = a, B B = c ) shows B B B ∼ B B B with scale factor 1:5, as desired. The likewise part 7 3 7 4 7 4 3 7 5 6 follows similarly.