Live Round Solutions

1.1 [5] The surface area of the regular cylinder is calculated√ as 216π. The equilateral triangles are inscribed in circles of radius 6, so they have length 6 3. We must subtract two times the area of a triangle and add the√ area of√ the interior of the cylinder that now has triangular√ sides. This comes out to −54 3 + 216 3, so the ﬁnal surface area of the ﬁgure is 216π + 162 3, so the answer is 216 + 162 + 3 = 381 . Problem written by Nathan Bergman 1.2 [5] Note that this process just returns the number mod 7. Thus, the average is: 0 + 9 × (1 + 2 + ... + 7) 9(28) 63 = = 64 64 16 Therefore, m + n = 63 + 16 = 79 . Problem written by Kai Zheng 1.3 [5] n5 + n4 + 2n3 + 1 = (n2 + n + 1)(n3 + n − 1) Using Euclid’s algorithm, we see that the two factors above are relatively prime. The least possible number of factors would then be when it is a product of two primes. This happens when n = 3. Thus, the least possible number of factors is 4 . Problem written by Kai Zheng

2.1 [7] Solution 1: We proceed by casework. If the base is painted all one color, there are 6 ways to paint the triangular faces (all orange, all black, 1 orange, 1 black, 2 adjacent orange, 2 opposite orange). If the base has exactly 1 unique square, fix that square in a determined corner, and then all 24 ways of painting the triangles are distinct. If the base has two differently-colored rows, fix the pyramid so both orange squares are on the left, and again, all 24 ways of painting the triangles are distinct. Lastly, if the base is painted like a checkerboard there are 10 ways to paint the triangles- all orange, all black, 1 orange (two ways), 1 black (two ways), 2 adjacent orange surrounding a orange square, 2 adjacent orange surrounding a black square, alternating black/orange (2 ways). The sum of all of these cases is 6 + 6 + 16 + 16 + 16 + 10 = 70 . Solution 2: We use Burnside’s Lemma. There are 28 total configurations. The 90-degree rotations fix 22 configurations (2 choices for the base and for the triangles), and the 180-degree rotation fixes 24 configurations (22 choices for the base and for the triangles). Thus, the 28+22+22+24 number of rotationally indistinguishable pyramids is 4 = 70. Problem written by Jacob Wachspress 2.2 [7] Using linearity of expectation, we sum the probabilities of each student being called before Charlie. For the students before Charlie in alphabetical order, this equals the probability of being in an earlier homeroom or the same homeroom as Charlie. For the students after Charlie in alphabetical order, this equals the probability of being in an earlier homeroom as Charlie. 15 The probability of any student being in the same homeroom as Charlie is 159 , so the desired sum is 1 144 15 1 144 53(174) + 106(144) 53 · + + 106 · = = 77 . 2 159 159 2 159 318 Alternatively, note that on average, 72 people will be in an earlier homeroom than Charlie and 5 of the 15 in her homeroom will be before her. Problem written by Zack Stier

1 2.3 [7] Checking the discriminant, we must have b2 − 8 ≥ 0 to have real roots, so b ≥ 3. If the roots are r and s, we have r2 + s2 = (r + s)2 − 2rs = b2 − 4. Hence, we seek

∞ ∞ X 1 1 X 1 1 = − b2 − 4 4 b − 2 b + 2 b=3 b=3 Using the telescoping property of the sum, we get an answer of

1 + 1 + 1 + 1 25 2 3 4 = . 4 48

The answer is 25 + 48 = 73 .

Calculus 1. [5] f(x + 2) − f(x) has a double root at x = 3, and is quadratic. So we know it’s equal to 6(x − 3)2. This let’s us ﬁnd a = −12 and b = 47. The sum is 35 . Problem written by Alec Leng

Estimation 1. [7] This is equivalent to asking for the number of up-right lattice paths from (0, 0) to (100, 100) that do not go above the diagonal, which is the the 100th Catalan number, or 1 200 101 100 . Using Stirling’s approximation for n!, we have √ 200200 200 1 200 400π · 200 2 M = ≈ √ e = √ ≈ 2189. 101 100 100100 2 101( 200π · e100 ) 101 100π

So M ≈ 2189, and 189 turns out to be an excellent estimate, as the correct answer is ≈ 189.19 . Less precise estimates are possible without Stirling’s approximation. This may rely on exam- x+100 ining the expression x for x ∈ {1, 2,..., 100} or evaluating

Z 100 x + 100 log2 dx 1 x though the latter method requires a decent estimation of ln 2. Problem written by Jacob Wachspress σ(i) 1 Miscellaneous 1. [9] Note = P . Rearranging the whole sum, we ﬁnd that it equals i k|i k

n n X k k k=1 Furthermore, n n n n n X X X n k < k = k k k2 k=1 k=1 k=1 Note n n ∞ 1 X σ(i) X 1 X 1 π2 < < = n i k2 k2 6 i=1 k=1 k=1

2 and n n n n n 1 X σ(i) 1 X π2 1 X π2 lim = lim k = − lim k = n→∞ n i n→∞ n k 6 n→∞ n k 6 i=1 k=1 k=1 π2 π2 implying the least upper bound is . The integer closest to 6 · = π2 is 10 . 6 6 Problem written by Jackson Blitz

4.1 [9] Since P (0) > 0 is composite, it must be at least 4. Therefore, since P is not constant, the sum of the squares of the coefficients of P is at least 17. If the sum of the squares of the coefficients were exactly 17, then P (x) would be equal to xm + 4 or −xm + 4 for some m ≥ 1, but that is impossible because then P (1) would be prime. However, since x2 + x is always even, the polynomial x2 + x + 4 meets the desired conditions. Since the sum of the squares of the coefficients of this polynomial is 18 , this is our answer. Problem written by Zack Stier 4.2 [9] AB | ABCD =⇒ AB | CD =⇒ CD = AB · m. CD | AB =⇒ CD | 100·AB =⇒ 100·AB = CD·n = AB·mn =⇒ mn = 100 =⇒ m | 100. AB and CD both have 2 digits, so m < 10 and can only take on the values 1,2,4,5. For each 10 ≤ AB < 20: 4 possibilities for m with a total of 12 (1,2,4,5). Each AB contributes 412 · AB. For each 20 ≤ AB < 25: 3 possibilities for m with a total of 7 (1,2,4). Each AB contributes 307 · AB. For each 25 ≤ AB < 50: 2 possibilities for m with a total of 3 (1,2). Each AB contributes 203 · AB. For each 50 ≤ AB < 100: 1 possibilities for m with a total of 1 (1). Each AB contributes 101 · AB. Computation gives the total to be 657510 . Problem written by Zack Stier

4 3 2 4.3 [9] We have P2(x) − P1(x) = 8(4x + 9x + 16x + 25x + 36). The four roots in common must 4 3 2 be roots of this polynomial as well. Hence, P1(x) = (4x + 9x + 16x + 25x + 36)(x − a) and 4 3 2 P2(x) = (4x + 9x + 16x + 25x + 36)(x − b). We have 9 − 4a = −311 and 9 − 4b = −279, yielding a = 80, b = 72, and a + b = 152 . Problem written by Zack Stier. Solution written by Sam Mathers.

5.1 [12] It turns out that 2 is the only unhappy number. 1 1 1 The expression can be rewritten as 2 ··· 2 2 (σ(1) + σ(2)) + σ(3) + ··· , i.e., repeatedly taking the replacing the ﬁrst two elements in the list with their arithmetic mean, which we want to be an integer at each step. When N = 2 this clearly fails, since the average of 1 and 2 is a necessary step but this does not yield an integer. However, when N ≥ 3, the permutation σ(1) = 1, σ(2) = 3, σ(3) = 2, σ(n) = n for 4 ≤ n ≤ N gives intermediate values of 2, 2, 3, 4, . . . , n − 1 inductively. Problem written by Zack Stier

3 5.2 [12] Each time the puck hits the side of the rink, the next bounce is the same distance along the arc as the previous bounce. In four bounces, it can hit at 130, 260, 390 (which equals 30), and 160 degrees clockwise of Kapil, as desired. One may verify that it is impossible to hit 30 and 160 degrees with fewer bounces. The total distance traveled is 2(4)(10 sin(130◦/2)) = 80 sin 65◦. The answer is 80 + 65 = 145 . (Note: an alternative interpretation of the problem gives an answer of 60 sin 65◦, so we accepted 125 as well.) 5.3 [12] Let yz = w > 0. Substituting and performing the quadratic formula, 1 x = (−11w ± p(11w)2 − 2(w5 + n)) 2 Note the inside of the radical must be a perfect square k.

−2n + 121w2 − 2w5 = k2

Note that in particular 121w2 − 2w5 > 0. This only holds for w = 1, 2, 3. For w = 1, 2, 3, 121w2 − 2w5 = 119, 420, 621. Now, note for every w there are exactly 2 x that satisfy. Fur- thermore, for every w, there are 2τ(w) pairs (y, z) that satisfy. Thus, the number of solutions is the sum of 4τ(w) over all w for which there exists a single n that satisﬁes. Now, we show that both w = 2, 3 cannot be satisﬁed by the same n. Suppose they could. Then, we would 2 2 have k1 = −2n + 420 and k2 = −2n + 621.

2 2 k2 − k1 = (k2 − k1)(k2 + k1) = 201 = 3 · 67

giving the solutions (k2, k1) = (35, 32), (101, 100). However, both solutions require n < 0. 2 Now, we similarly show that both w = 1, 3 can’t be simultaneously satisﬁed. k1 = −2n + 119 2 and k2 = −2n + 621.

2 2 2 k2 − k1 = (k2 − k1)(k2 + k1) = 502 = 2 · 251

2 2 giving no solutions. Finally, we show w = 1, 2 can’t work. k1 = −2n+119 and k2 = −2n+420.

2 2 k2 − k1 = (k2 − k1)(k2 + k1) = 301 = 7 · 43

giving the solutions (k2, k1) = (25, 18), (151, 150) both requiring n < 0. Thus, no w’s can be simultaneously satisﬁed, implying the maximum amount of solutions is max(4τ(w)) = 4τ(2) = 4τ(3) = 8 . Picking w = 2 arbitrarily, we can pick (n, k) = (10, 200) that satisﬁes. Problem written by Jackson Blitz

Calculus 2. [9] The spheres share both have the z-axis as a line of symmetry. Hence, the volume of the desired region can be found by revolving a cross-section in the plane x = 0 about the 1 z-axis. Since the spheres intersect at z = 2 in that cross-section, the volume equals

Z 1 Z 1 h z3 i1 5π π · 2 y2 dz = π · 2 (1 − 02 − z2) dz = 2π z − = . 1 1 3 1 12 2 2 2

The answer is 5 + 12 = 17 . Problem written by Jacob Wachspress

4 Estimation 2. [12] The answer is 7984 . Observe that a number is stupendous if and only if it can be expressed as pk2 with p prime. n Using the fact that there are about ln n primes less than any integer n, we note that there are about √ n n X k2 ln n k=1 k2 stupendous numbers. We write

√ √ n n n n X 2 X 2 k ≈ k ln n ln n k=1 k2 k=1 k2 √ n n X 1 ≈ ln n k2 k=1 n π2 ≈ · . ln n 6

n This is a slight underestimate of the sum, using the fact that ln k2 ≈ ln n for most values n π2 n of k in the sum. Hence, we seek an integer n such that ln n · 6 ≈ 2018, so ln n ≈ 1200. A good estimate for n is 1200 ln(1200). This is an underestimate, which works to balance out the n fact that we believe that ln n is actually slightly less than 1200. As it turns out, ln(1200) ≈ 7, yielding a very close estimate of 8400. Problem written by Jacob Wachspress

Miscellaneous 2. [15] Solution 1: Deﬁne A(m,n) to be the number of ways to complete a tunnel system that currently has m paths that have not terminated and n ways to descend to a dead end. We seek A(1,1). We have

A(1,1) = A(4,4) 4 4 4 4 A = A + A + A + (4,4) 1 (4,7) 2 (8,10) 3 (12,13) 4 4 4 4 A = A + A + (4,7) 1 (4,10) 2 (8,13) 3 4 4 A = A + (4,10) 1 (4,13) 2 8 8 A = A + (8,10) 1 (4,13) 2

Since A(m,13) = m, we ﬁnd that A(8,10) = 60, A(4,10) = 22, A(4,7) = 140, and A(1,1) = A(4,4) = 969 . Solution 2: Deﬁne T (z) to be the generating function for tunnel systems, enumerated by ongoing paths. A tunnel system consists of a single external node or an internal root node connected to four other tunnel systems. Therefore, we have the generating function equation

T (z) = z + T (z)4.

5 Let f(u) = u − u4, so that we have the equation f(T (z)) = z. Then, since f(0) = 0 and f 0(0) = 1, we can use Lagrange inversion to get that

1 1 16 [z16]T (z) = [u15] . 16 1 − u3

Then, since 16 ∞ 1 X i + 15 = (u15)i, 1 − u3 15 i=0 we have 1 20 [z16]T (z) = = 969. 16 15

Problem written by Matt Tyler

7.1 [15] Say m = |x − y| and n = 1 − 1 = y−x = m =⇒ |xy| = m . Let x > y; then x y xy |xy| n √ −m± m2+ 4m x = y + m, so |y(y + m)| = m . If y(y + m) > 0 then y2 + my − m = 0 so y = n √ n n 2 m± m2+ 4m and x = n . k(x, y)k2 = m2 + 2m (from the distance formula). If y(y + m) < 0 2 √ n √ 2 4m 2 4m 2 m −m± m − n m± m − n 2 2 2m then y + my + n = 0 so y = 2 and x = 2 . k(x, y)k = m − n . 2 2016 q 2016 For k(x, y)k ≤ 2017 < 1 (which comes from k(x, y)k ≤ 2017 ) we need xy < 0 and m = 1. 2 2016 4032 This now reduces to 1 − n ≤ 2017 = 4034 , so n ≤ 4034. We see that n < 4 gives negative discriminants, so 4 ≤ n ≤ 4034. For each of those n values there are 2 possible separated points with x > y (from the ± in the quadratic formula), except for n = 4, where the discriminant is unique and thus there is only one. We thus ﬁnd a ﬁnal answer of 2(1 + 2 · 4030) = 16122 (since (x, y) is separated whenever (y, x) is as well). Problem written by Zack Stier

7.2 [15] Let xn+1 = x1. Let a be the number of times the sign from xi to xi+1 stays the same and b be the number of times it changes. Clearly a + b = n. Furthermore, we see a − b = m. Thus, n + m n − m a = and b = . Note that b must be even as the sign of x must be the same. 2 2 1 lnm Thus, m ≡ n mod 4. Furthermore, since b ≥ 0 n ≥ m. We ﬁnd that for each n there are 4 ordered pairs. Summing from n = 2 to 2017, we get 509544 ordered pairs. Problem written by Jackson Blitz 7.3 [15] Let PE and PF meet the circumcircle at B0 and C0. Then B0 and C0 are the antipodes of B and C, so using Pascal on BACC0PB0 shows that the circumcenter O lies on EF. Since OE is the perpendicular bisector of B0C, we see that AB0CP is an isosceles trapezoid. Hence, AP ⊥ EF. It follows that A and P are symmetric in EF , so the desired quantity is dist(A, EF ). Clearly we have bc dist(A, EF ) = dist(A, BC) − dist(O,BC) = − R cos A 2R

63 which evaluates to 8 . The answer is 63 + 8 = 71 . Problem written by Kapil Chandran

6 8.1 [18] Since there are 40 rows, there are 80 groups of 3 seats. Once a boarder chooses one of these groups, his or her seat is ﬁxed. Hence, the number of ways to board the plane is the number of 240! distinct strings of three A0 s, three A0 s, three A0 s . . . and three A0 s. This is precisely . 1 2 3 80 (3!)80 The numerator has 120 + 60 + 30 + 15 + 7 + 3 + 1 = 236 factors of 2, and the denominator has 80 factors of 2. The answer is 156 . Problem written by Zack Stier

8.2 [18] Let r q √ 1 243 + 3 2187 + 9 19683 ... Then, r 1 1 q √ x = 3 + 2187 + 9 19683 ... 9 27 s r 1 √ = 3 + 3 + 19683 ... 81 r q √ = 3 + 3 + 3 + . . .,

√ 1 2 1 9+9 13 so 9 x = 3 + 9 x. Since x > 0, x = 2 . (The sum converges because the partial sums are increasing and always less than this value.) Hence, the answer is 9 + 9 + 13 + 2 = 33 . Problem written by Matt Tyler

8.3 [18] For any r ∈ C with r < 1, ∞ X r nrn = . (1 − r)2 n=0

ei Letting r = 2 , we have v v u ∞ !2 ∞ !2 u ∞ u X n sin(n) X n cos(n) u X 2 t + = t nrn = . 2n 2n 5 − 4 cos(1) n=0 n=0 n=0

The answer is 22 + 52 + 42 = 45 . Problem written by Matt Tyler

7 Calculus 3. [15] If γ is the counter-clockwise path along the unit circle, then from Cauchy’s coef- ﬁcient formula, we have 1 I f(z) dz = f(0). 2πi γ z

Let z = eiθ, so that dz = ieiθdθ. Then, under this change of variables, the integral from part 1 becomes I Z 2π iθ Z 2π 1 f(z) 1 f(e ) iθ 1 iθ dz = iθ ie dθ = f(e )dθ, 2πi γ z 2πi 0 e 2π 0 so Z 2π f(eiθ)dθ = 2πf(0). 0 Now, we let f(z) = cosh(z), so that

ecos(θ)+i sin(θ) + e− cos(θ)−i sin(θ) < f(eiθ) = < 2 ecos(θ)+i sin(θ) + e− cos(θ)−i sin(θ) = < 2 ecos(θ) cos(sin(θ)) + e− cos(θ) cos(− sin(θ)) = 2 ecos(θ) + e− cos(θ) = cos(sin(θ)) 2 = cosh(cos(θ)) cos(sin(θ)).

Then, from part 2, we have

Z 2π Z 2π cosh(cos(θ)) cos(sin(θ))dθ = < f(eiθ) dθ 0 0 = <(2πf(0)) = <(2π × 1) = 2π.

Hence, 1 Z 2π 2π cosh(cos(θ)) cos(sin(θ))dθ = 3 0 3

and the answer is 2 + 3 = 5 . Problem written by Matt Tyler

Estimation 3. [18] The answer is 12577 . Note that each even integer must be paired with an odd integer in order to get an odd sum, so there are at most 10! such pairs. Also note that slightly more than half of the odd numbers from 1 to 40 are prime. This is still roughly true when we only consider odd numbers near 20. Assuming independence of events (which is wrong) gives 10! a decent underestimate of ≈ 3600. 210 Problem written by Jacob Wachspress

8 Miscellaneous 3. The generating function for this random walk is

30 30 1 1 1 1 X f(x, y) = x−2 + y−1 + xy2 = (x−2 + y−1 + xy2)30 = p(a, b)xayb 3 3 3 3

We need to ﬁnd the E(d2) = E(x2 + y2) = E(x2) + E(y2) by the linearity of expectation. Furthermore note E(x2) + E(y2) = 2E(x2) due to the symmetry in movement in the x and y directions. E(x2) can be found by plugging in the point (x, y) = (1, 1) into the expression

X ∂ ∂f ∂f ∂2f ∂f ∂2f X p(a, b)a2xa−1yb = x = + x = + = p(a, b)a2 ∂x ∂x ∂x ∂x2 ∂x ∂x2 Finding derivatives and plugging in (x, y) = (1, 1)

∂f 130 130 = · 30(x−2 + y−1 + xy2)29(−2x−3 + y2) = · 30(3)29(−1) = −10 ∂x 3 3

∂2f 130 = · 30(29(x−2 + y−1 + xy2)28(−2x−3 + y2)2 + (x−2 + y−1 + xy2)29(6x−4)) ∂x2 3 130 470 = · 30(29(3)28(−1)2 + (3)29(6)) = 3 3 Finally, ∂f ∂2f 470 880 2 (x2) = 2 + = 2 −10 + = E ∂x ∂x2 3 3 which yields 880 + 3 = 883 . Problem written by Jackson Blitz