HMMT 二月 2019 · 冲刺赛 · 第 7 题

7. The smallest even number satisfying this is 3 · 5 · 7 + 1 = 106, so the smallest possible number is 106 / 2 = 53. M . An equiangular hexagon has side lengths 1 , 2 , 3 , 4 , 5 , 6 in some order. Find the nonnegative difference 11 between the largest and the smallest possible area of this hexagon. √ 3 Answer: 2 Extending three sides of the equiangular hexagon gives an equilateral triangle. Thus, if the sides are a, b, c, d, e, f , in order, then a + b + c = a + f + e ⇒ b + c = e + f ⇒ f − c = b − e . By a symmetric argument, we see that d − a = f − c = b − e holds, which means that they must be separated into three groups of two with equal differences. If the grouping is (1 , 2) , (3 , 4) , (5 , 6), then we have 1 , 4 , 5 , 2 , 3 , 6 around the hexagon. If the grouping is (1 , 4) , (2 , 5) , (3 , 6), then we get 1 , 5 , 3 , 4 , 2 , 6 as the other possibility. Finally, we can use our equilateral triangle trick to find the areas. For the first, we get a big triangle of side 1 + 4 + 5 = 10, and must subtract smaller triangles of sides 1 , 5 , 3. This gives √ √ √ √ (100 − 1 − 25 − 9) 3 / 4 = 65 3 / 4. For the other, we get (81 − 1 − 9 − 4) 3 / 4 = 67 3 / 4. The positive √ difference between these is 3 / 2. 3 2 M . Determine the second smallest positive integer n such that n + n + n + 1 is a perfect square. 12 Answer: 7 3 2 2 2 n + n + n + 1 = ( n + 1)( n + 1). Note that gcd( n + 1 , n + 1) = gcd(2 , n + 1) = 1 or 2, and since 2 2 2 2 n + 1 is not a perfect square for n ≥ 1, we must have n + 1 = 2 p and n + 1 = 2 q for some integers p and q . The first equation is a variant of Pell’s equation, which (either by brute-forcing small cases or using the known recurrence) gives solutions ( n, p ) = (1 , 1) , (7 , 5) , . . . . Incidentally, both smallest solutions n = 1 and n = 7 allows an integer solution to the second equation, so n = 7 is the second smallest integer that satisfy the condition. M . Given that A, B are nonzero base-10 digits such that A · AB + B = BB , find AB . 13 Answer: 25 We know A · AB ends in 0. Since neither is 0, they must be 2, 5 in some order. We easily find that A = 2, B = 5 works while the opposite doesn’t, so AB = 25. √ T . Let S, P, A, C, E be (not necessarily distinct) decimal digits where E 6 = 0. Given that N = ESCAP E 1 is a positive integer, find the minimum possible value of N . Answer: 319 5 Since E 6 = 0 , the 6-digit number ESCAP E is at least 10 , so N ≥ 317 . If N were 317 or 318 , the last 2 2 digit of N would not match the first digit of N , which contradicts the condition. However, N = 319 2 will work, since the first and last digit of N are both 1 . T . Let X = b T / 8 c , Y = T − 1 , Z = T − 2. A point P lies inside the triangle ABC such that P A = 2 1 3 4 X, P B = Y, P C = Z . Find the largest possible area of the triangle. Answer: 1344 X = 39 , Y = 33 , Z = 25. Fix some position for P , A , and B , and we shall find the optimal position for C . Letting AB be the base of the triangle, we wish to maximize the height. The legal positions for C are a subset of the circle with center P and radius P C , so the height is maximized when P C is orthogonal to AB . Symmetrically, we deduce that P is the orthocenter of ABC when the area is maximized (moreover, P must be inside the triangle). If ray BP intersects AC at E , then since AEB is similar to P EC , we have √ √ 2 2 2 2 39 − x 25 − x 2 2 2 2 2 2 3 2 = ⇒ x ( x + 33) = (39 − x )(25 − x ) ⇒ 66 x + 3235 x − 950625 = 0 . 33 + x x 2 The LHS factors to ( x − 15)(66 x + 4225 x + 63375), meaning that x = 15 is the only positive solution, giving AE = 36 , BE = 20, and therefore the maximum area of triangle ABC is (33 + 15)(36 + 20) / 2 =

T . How many ways can one tile a 2 × 8 board with 1 × 1 and 2 × 2 tiles? Rotations and reflections of the 3 same configuration are considered distinct. Answer: 34 Let f ( n ) denote the number of ways to fill a 2 × n board. One can fill the leftmost column with two 1 × 1 tiles, leaving f ( n − 1) ways, or one can fill the leftmost two columns with one 2 × 2 tile, leaving f ( n − 2) ways. Therefore, f ( n ) = f ( n − 1) + f ( n − 2) . One can also directly verify that f (0) = f (1) = 1 . th Therefore, f ( n ) = F , where F is the n Fibonacci number. Easy calculation shows that the n +1 n desired quantity is f (8) = F = 34 . 9 2 2 2 2 T . Let S = T . Given real numbers a, b, c such that a + b + c + ( a + b + c ) = S , find the maximum 4 5 possible value of ( a + b )( b + c )( c + a ). Answer: 27 2 2 2 2 2 2 2 Notice that S = 27 = a + b + c + ( a + b + c ) = ( a + b ) + ( b + c ) + ( c + a ) . By AM-GM, S 2 / 3 ≥ (( a + b )( b + c )( c + a )) with equality if and only if a + b = b + c = c + a, i.e. a = b = c. Thus, 3 ( ) 3 / 2 S 3 the maximum possible value is = 27 , achieved at a = b = c = . 3 2 T . A regular tetrahedron has volume 8. What is the volume of the set of all the points in the space ( not 5 necessarily inside the tetrahedron ) that are closer to the center of the tetrahedron than any of the four vertices? Answer: 27 h Let h denote the height of the tetrahedron. The center of the tetrahedron is a distance from each 4 face. Therefore, the perpendicular bisector plane of the segment connecting the center to a vertex lies 3 a distance h away from both the vertex and the center. Symmetrical considerations with the other 8 three vertices will thus show that the desired region is also a regular tetrahedron, with the center of 3 the original tetrahedron a distance h away from each face. Based on the distance from the center to 8 3 h h a face, one can see that the scale factor of this tetrahedron is : = 3 : 2 relative to the original 8 4 ( ) 3 3 tetrahedron, so its volume is 8 · = 27 . 2