2018 MOAA Gunga Bowl: Solutions

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1 08 MOAA Gunga Bowl: Solutions Andover Math Club Set Solutions. We pair up the terms as follows: = (+)+(+0)+(+9)+(4+8)+(5+7)+6 = 5+6 = 66.. By the difference of squares formula, 6 x = (6 x)(6 + x). Thus, Proposed by Justin Chang = = Alternatively, the expression can be computed directly:. Note that = = 6. We can telescope this sum into which results in 0 + =. Set Solutions x(x + ) = x x +. m n = = = 0, 4. We compute! =, 0! =,! =, and 8! = 400. Then ( + )( + 400) = 096. = 6. Proposed by Justin Chang Proposed by Justin Chang 5. We prime factorize 5 = 7. The number n is prime, so it must be one of the numbers,, 7. The roots of x 5x + 6 = (x )(x ) are and, so we know that n must be 7. Proposed by Justin Chang 6. This is equivalent to summing an arithmetic sequence with first term, common difference 5, and last term = 4. Therefore the sum is equal to n(a + a n ) = 7( + 4) = 8.

2 Set Solutions 7. We can apply Heron s formula to triangle ABC. We define a = BC, b = CA, and c = AB. Since its semiperimeter s = a+b+c = 6, we have» [ABC] = s(s a)(s b)(s c) = = 4, which results in 4 + = The key observation is that no move changes the score. If Sebastian performs a move on a token at (x, y), then the score after the move equals the score before the move: x++y + x+y+ = x+y+ = x+y. Therefore, the answer is equal to the original total score, which is just 0+0 =. 9. The second condition is just equivalent to the number being divisible by. The least four or five digit number that is a multiple of is 00, and the greatest is Thus, our answer is 4 Set 4 Solutions = The desired area consists of two triangles with legs of length one and one 45 sector of a circle with radius. Therefore, the desired area is Å ã n = + 8 π( ) = + π 4. Computation reveals that π = 78.. We have two cases here: (a) There is a tree in the middle square. As a result, only the four corner squares can have trees planted in them. There are then ( 4 ) ways to pick three that work. (b) There is no tree in the middle square. As a result, the only ways to pick four squares are to choose all of the corner squares or to choose all of the central edge squares.

3 There are ( 4 ) + = 6 ways in total.. We perform a digit-by-digit extraction as follows. Since n 7 (mod 0), we know that n (mod 0). Let n = 0a +. Then Then 67 n 000a + 900a + 70a a + 7 (mod 00). 70a 40 (mod 00) = a (mod 0). Let a = 0b +. Then n = 0(0b + ) + = 00b +, so Therefore 67 n 0 6 b b b b + 67 (mod 000). 700b 00 (mod 000) = b 6 (mod 0). However, b must be a digit, as otherwise n = 00b + > 000. Thus b = 6, and n = 6. 5 Set 5 Solutions. The most straightforward way to solve this problem is to try to find the explicit values of and Optimistically, we set» = a + b = Ä a + b ä = a a b + 8a b + ab + 9b 4. Then, matching rational and irrational parts, we find that 97 = a 4 +8a b +9b 4 and 56 = 4a b+ab. We look at the second equation because it has less terms: 56 = 4a b + ab 4 = ab(a + b ). Since we want a and b to be positive integers, we find that (a, b) = (, ) satisfies this equation, and in fact it also satisfies the other equation, 97 = a 4 + 8a b + 9b 4! Therefore = +. Similarly, we find that =, so that 4» » = + + = Recall that the measure of an inscribed angle is half that of the central angle. Therefore COE = CBE = B = 68. Also, since AD is a height we know that DAC = 90 C. Thus Now we find DOC = DAC = 80 C = 5. DOE = DOC + COE = = 0. Now draw the perpendicular from O to DE, and call the point of intersection M. Note that OM bisects DOE, so that MOE = 60. This means that ME = OE since OME is a right triangle. Finally, DE = ME = OE, so DE = OE = 4.

4 B D O A M C E 5. We can factor the first term by adding and subtracting 4n ; 4n 4 + = 4n 4 + 4n + 4n = (n + ) (n) = (n + n + )(n n + ) by difference of squares. This number is only prime if one of its factors is equal to. Since 0 < n+ n < +n+n, only n+n can equal. This factor equals when n n = 0 n(n ) = 0, so n = is our only answer. The sum of all possible n is. 6 Set 6 Solutions 6. For the moment, we can ignore the condition p q r, as the equation is symmetric. Since divides the right hand side of the equation, it also divides the left. Thus, pqr =, and since p, q, r are primes, we must have one of them equal to. Without loss of generality, allow p =. Then qr q r + = (q )(r ) = qr = q + r + by Simon s Favorite Factoring Trick. This results in three possible cases: q = and r =, which leads to q =, r =, q = 6 and r =, which leads to q = 7, r =, q = 4 and r =, which leads to q = 5, r = 4, which doesn t work as 4 isn t prime. This results in the two triplets (p, q, r) = (,, ) and (, 7, ), as we must incorporate p q r. Thus the sum of all possible values of p is + = Let x n represent where there are n instances of. We have that, ( (( ) ) ), x n+ = / + /x n. 4

5 Since x 0 = 0+ and x = +, it is safe to assume that x n = n+ by induction. Assuming x n = n, we have x n+ = / + n/ = n +, for all n. In fact, we can show this as desired. We have already resolved the base case of 0. Therefore, x 08 = 09, which results in + 09 = Write as Then which reduces to (08 )( ) = , (mod 07). 7 Set 7 Solutions 9. Since XAY = 45, the arc XY in circle ω must have length 45 = 90, so XO Y = 90. Similarly, XO Y = 90. Therefore, as O X = O Y and O X = O Y, O XY is a triangle and O XY is equilateral. Therefore, if D is the intersection of O O and XY, O O = O D + O D = XY + XY = B A X C D O O Y Now, let C be the foot of the altitude from A onto BO. Then, AC = O O and BC = BO AO = O X O X = Finally, using the Pythagorean Theorem on right triangle ABC gives AB =» AC + AB = (0 + 0 ) + (60 0» ) = , which results in =

6 0. Square the equation to obtain (x ) x = 60 =. Therefore x = = x = = 8. Using our computation skills we find that so the answer is 8. 8 < 8 < 8,. Suppose the bag contains b blue balls after Sam replaces some of them. Then, it has b + 5 red balls and 75 b green balls. The expected number of green balls when selecting 500 balls at random is thus ï ò 500 E (75 b) = E[b]. Note that b ranges over all the values from to 67, each with equal probability (as the number of blue balls fixes the number of red and green ones). Therefore, E[b] = 84, so our answer is = Set 8 Solutions Let s solve the first problem (problem ) first (in terms of the other answers). W X P M Z N Y Consider XY Z. We see that XN is a median of this triangle, and that W Y is also a median of the triangle because the diagonals of a rectangle bisect each other. Thus P is the centroid of XY Z. Using the fact that the medians of a triangle divide the triangle into six regions of equal area, we see that so [MP NY ] = 6 [MP NY ] = [XY Z] = [W XY Z] = [W XY Z], B 5C = BC. We conclude that A = BC, which implies that A = BC. We now turn our attention to the third problem (problem 4). We see that x + y = A implies that x + xy + y = A x xy + y = A 4xy (x y) = A 9 B x y = A 9 B, 6

7 which means that C = x = B, we see that C A 9 B, and therefore C = A 9 B. Plugging in A = 5 BC and letting 6 C = 5 6 BC 9 B 6 = 5x 4x 0 = (4x 9)(x 4). Therefore B C = 9 4 = B = 9 4 C or B C = 4 = B = 4C. The first case gives A = 9 6 C + C = A = 5 C. If we let C = 4k then we get the solution (A, B, C) = (5k, 9k, 4k). The second case gives 4 A = 6 9 C + C = A = 5 C. If we let C = k then we get the solution (A, B, C) = (5k, k, k). Armed with this knowledge, we inspect the second problem (problem ). One thing catches our attention in this knowledge: x, z. This means that C + = C, using the last congruence. Therefore the case (A, B, C) = (5k, k, k) is ruled out by this condition, because C is always divisible by in this case. Now we know that (A, B, C) = (5k, 9k, 4k). Therefore the first congruence reduces to xy 5k 0 (mod 5) and the second congruence reduces to yz 4k 0 (mod 7). Multiplying these congruences by z and x respectively yields and Finally, multiplying the last congruence by y yields xyz 0 (mod 5) xyz 0 (mod 7). xyz (C + )y 0 (mod 9), since we know that 9 y. Therefore xyz 0 (mod 5), and since x, y, z are positive integers we know that xyz 5. To show that equality can hold is trivial; simply let (x, y, z) = (5, 9, 7). This produces the solution B = 5. Then k = 5, and we get the solution (A, B, C) = (75, 5, 40). This is the answer; it can be verified to work by simply plugging the numbers back into each of the problems. Note. We made one huge implicit assumption in the problem, and it was that (A, B, C) = (5k, k, k) implies that A is divisible by 5, B is divisible by, and C is divisible by ; in other words, we assumed that k was an integer. Why can we do this? All we know is that A, B, C are positive integers, nothing about k. However, we do know that k = A A, so we can indeed conclude that k is rational. Then 5 5 = B, so it is possible to now conclude that 5 A. A similar argument proves this for the other cases; in general, this is true whenever all the constants in front of the k are relatively prime when considered together. Therefore, even though and are not relatively prime, we can still conclude that k is an integer because they are relatively prime altogether with 5. This argument works for the case (A, B, C) = (5k, 9k, 4k) as well

8 Set 9 Solutions 5. After making the key observation that a + b = (a + b) + (a b), we write 07 8 = ( ) 8 Thus a + b = 704. = (5 + 5 ) 64 = The side length of the flatbread is equal to. The volume of enclosed syrup is maximized when the side length of the pyramid is maximized. Let this side length be s. Then by packing the faces like so (see image), we can achieve s = = + = 6. It is not hard to see that the volume of the square pyramid in terms of s is given by V = bh = s h = s, so the maximum volume of syrup enclosed is equal to V = s = 5 0 = and thus a + b + c = = 755. Proposed by Andy Xu 7. Reflect P over y = x and y = 0 to get P and P, respectively, and let M and N be the midpoints of P P and P P, respectively. Then P Q + QR + RP = P Q + QR + RP P P = MN, so it suffices to minimize MN. Now let P = (x, y), so that M = ( x+y, x+y ) and N = (x, 0). Then MN = ( x + y ( x + y x) + 0) = (x + y ) = OP. Therefore it suffices to minimize OP. By the triangle inequality, it is clear that OP is minimized when its extension passes through C. We can compute that OC = 5, so OP OC P C = 5 6 = 9, which means that P Q+QR+RP MN = OP 9. It is clear that such a setup is achievable, thus the answer is Ä 9 ä = 7. 8

9 P Q M P C O R N P 0 Set 0 Solutions 8. The main idea of this problem is to subtract off the areas of triangles AEF, BDF, CED from triangle ABC. We have [AEF ] [ABC] = [AEF ] [AEB] [AEB] [ABC] = AF AB AE AC, where [K] is the area of polygon K. Similarly, and [BDF ] [ABC] = BF BA BD BC [CED] [ABC] = CD CB CE CA. Now we just need to compute a few side lengths. Since AE and AF are both tangents to the incircle, we must have AE = AF. Suppose x = AE = AF, y = BF = BD, and z = CD = CE. Then, x + y = AB = 7, y + z = BC =, z + x = CA =. Solving this system gives x = 4, y =, z = 9. Then, This gives = 09. [DEF ] [ABC] = [AEF ] [ABC] [BDF ] [ABC] [CED] [ABC] =

10 B F D I A E C 9. First, note that if Sebastian selects a token at (x, y, z), this token contributes a score of S = x+y+z before replacing the token. After replacing the token, the three tokens together contribute a score of x+y+z+ = S. Therefore it is most beneficial to perform moves on tokens with the highest score, which are the tokens on (x, y, z) with the lowest sum x + y + z. Let s say a token at position (x, y, z) has value x + y + z, so that replacing a token of value k produces three tokens of value k +. Then after the first move (forced), we have three tokens of value. Replacing these three tokens yields nine tokens of value, after four total replacements. Replacing these nine tokens yields 7 tokens of value, after total replacements. Replacing these 7 tokens yields 8 tokens of value 4, after 40 total replacements. Now we have only 60 replacements left, so we can replace 60 of the 8 tokens of value 4, which yields tokens of value 4 and 80 tokens of value 5. Finally, note that the score of a token given its value k is equal to, so the final score of this set of tokens is k Then + 6 = = We approach this problem by recursion. Note that the digits reduce into 0,,, (mod 5). We let a n be the number of n digit numbers with the given digits, and with a digit sum equivalent to 0 (mod 5), and we define b n, c n, d n, e n analogously for,,, 4 (mod 5). Clearly, a = b = c = d = and e = 0. Then, for every new digit added, a n = c n + d n + e n + a n, as we can add a digit of any residue modulo 5 except 4. Now, we can set s n = a n + b n + c n + d n + e n, so Similarly, a n = s n b n. b n = s n c n, c n = s n d n, d n = s n e n, e n = s n a n. 0

11 Now, we can plug these formulas into the right hand side of a n = s n b n, yielding a n = s n b n = s n (s n c n ) = s n s n + s n d n = s n s n + s n s n 4 + e n 4 = s n s n + s n s n 4 + s n 5 a n 5 It suffices to compute s n. However, this is just the total number of numbers with digits as, 5, 7, or 8 (as the digit sum can be anything), so s n = 4 n, as there are four choices for each digit. Thus, a 6 = = 46 = Note. Here, we will present an elegant alternate method, utilizing complex numbers. After using the same definitions of a n, b n, c n, d n, e n as above, we let ζ = cos π 5 + i sin π 5 = e iπ 5 be the first fifth root of unity. Then, notice that a n ζ 0 + b n ζ + c n ζ + d n ζ + e n ζ 4 = ζ x+x+ +xn. x,x,...,x n {,5,7,8} The idea here is that the right hand side takes in all inputs of n digit integers with digits, 5, 7, 8, and the root of unity ζ takes the sum modulo 5, outputting the left hand side. Now, the right hand side is equal to (ζ + ζ 5 + ζ 7 + ζ 8 ) n = ( + ζ + ζ + ζ ) n, and since 4 i=0 ζi = 0, this is equal to ( ζ 4 ) n. If we take n = 6, we obtain so We proceed with a lemma. a 6 ζ 0 + b 6 ζ + c 6 ζ + d 6 ζ + e 6 ζ 4 = ζ 4 = ζ 4, () a 6 ζ 0 + b 6 ζ + c 6 ζ + d 6 ζ + (e 6 )ζ 4 = 0 Lemma. For any prime p and integers a 0, a, a p, if then a 0 = a = = a p. a 0 ζ 0 + a ζ + + a p ζ p = 0, Proof. Just consider the polynomials P (X) = a 0 + a X + a X + + a p X p and Q(X) = + X + X + + X p. It is pretty well known that Q(X) is irreducible over Q. In fact, Q(X + ) = (X + )p p Ç å p = X i, X p i which is irreducible by Eisenstein s Criterion on the prime p, as p ( p i) for all 0 < i < p. However, Q and P share the root ζ, so Q must divide P. Thus, a 0 = a = = a p. i=0 We apply this lemma on (), giving for some constant k. Then, so a 6 = 46 5 = 89. a 6 = b 6 = c 6 = d 6 = e 6 = k 5k + = a 6 + b 6 + c 6 + d 6 + e 6 = s 6 = 4 6,

12 Set Solutions. We break the sum into i=6 T i + These fractions decompose into i=6 ( = S i i=6 ) ( + i(i + ) i=6 i(i + ) = i i + and i(i + ) = i i +. ). i(i + ) This allows us to telescope the sums into Å ã Å = 6 ã + Å ã = , which results in = We reflect Will s barnhouse B(4, 5) over the river at y = to a point B, and over the strip mall at y = x to a point B. The idea now is that if Will s house is at H, he reaches the river at point X, and the strip mall at point Y, then the distance he travels is HX + XB + BY + Y H = HX + XB + B Y + Y H. This distance is clearly minimized when H lies on B B. Now it suffices to find the points B and B, and we can intersect lines B B and y = x. B K Y H B X B Clearly, B = (4, ), since we reflect B over a horizontal line. As for B, we will determine the foot of the perpendicular from B onto y = x, which we will call K. Since BK is perpendicular to y = x, its slope is. Therefore, BK is defined by y 5 = (x 4).

13 Intersecting this line with y = x, we can solve the system to get We can solve for y, so x = x x = K = x = Ç = , å. 4 Since K is the midpoint of BB, we can determine B to be Ç 5 4 B =, 4 å + 5. Now, we need to find the equation for B B. Its slope is so its defined by (4 + 5)/ ( ) (5 4)/ 4 = , y + = (x 4). It remains to intersect this line with y = x. Solving, Therefore ( )x + ( ) = (7 + 4 )x 4(7 + 4 ). x = ( 09 9 )x = which results in = ( )(09 9 ) = 09 (9 = ) 68. Take the equation n = (m + ) m for some integer m. Then n = m + m + 4n = m + m + (n + )(n ) = (m + )., Because n + and n are relatively prime, we know that one of these terms must be a perfect square and the other one must be three times a perfect square. However, we cannot have n be three times a square, because then n + would be a perfect square congruent to (mod ), which is impossible. Therefore n is a perfect square; say n = a. Now we proceed to the second equation. Set n + 87 = k, and plug in n = a to get a + 88 = k 88 = (k + a)(k a).

14 Since maximizing n is the same as maximizing a, we need only to maximize the difference between the two factors k + a and k a. We quickly see that 88 and are impossible, as they do not produce integer values for k and a. However, 44 and do work, yielding k = 7 and a = 7. Then n = 7 n = 504 n = 5, so the maximum possible value of n is 5. Plugging it back in shows that it indeed works, yielding 5 = and = 7. Set Solutions 4. conjecture has some great information about the Collatz conjecture if you re interested. In case you were wondering, the Collatz sequence of , the answer, has a length of 986 steps! 5. We claim that ex(n, K i ) = i i n r + Ç å r, where r is the remainder when n is divided by i. In fact, this known as Turan s Theorem in Extremal Graph Theory. Note that equality can hold; take a complete multipartite graph of i groups, r of which with q + elements, and i r of which with q elements, where n = q(i ) + r. This is called a Turan Graph. Now, we will prove Turan s using the following lemma. Lemma (Zarankiewicz s Lemma). If G is a graph of degree n without a subgraph K i, then it contains a vertex with degree of at most ö i i nù. Proof. Let N(v) denote the neighborhood of a vertex v; that is, the set of all vertices that are connected to v by an edge. For the sake of contradiction, assume otherwise. Consider an arbitrary vertex v. It must satisfy õ û i N(v ) > i n > 0, so there exists v N(v ). Then, points that are neighbors to both: N(v ) N(v ) = N(v ) + N(v ) N(v ) N(v ) Å õ ûã i + i n n > 0, so there exists a v N(v ) N(v ). We continue this for the rest of the vertices, and since k j= Å õ ûã i N(v j ) k + i n (k )n > 0 for all k, we set j = i so that we can choose a vertex v i i j= N(v j ), but then v, v,..., v i form a complete graph on i vertices, which violates the assumption. 4

15 Now we will use this lemma to prove Turan s Theorem. We induct on n. Suppose the statement is true for n = k. We will show it is true for n = k +. Since the graph doesn t contain a K i, applying Zarankiewicz s Lemma on the graph means we can choose a vertex x such that deg(x) i i (k + ). If we remove this vertex and all edges attached to it, we get a graph of k vertices, and by our inductive hypothesis, the maximal number of edges in this graph is precisely ex(k, K i ) = i Ç å i k rk rk +, where r k is the remainder when k is divided by i. Then, after adding back that vertex, it suffices to confirm that Ç å i i k rk õ û rk i + + (k + ) = i i i (k + Ç å ) r r +, where r is the remainder when k + is divided by i, which is in fact true. This completes the inductive step. The Turan Graph suffices for every base case. Then, the problem reduces to computing 08 i= i i 08 (r(08, i )) + where r(08, i ) is the remainder when 08 is divided by i. Ç å r(08, i ) = 40908, 6. No solution for this problem. We are unaware of any strategy that would net you the greatest expected number of points, but if you find one, we d be happy to hear it! and Sebastian Zhu 5