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Back-To-School Math Contest Year 2 INSTRUCTIONS 1. This is a twenty-five question multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct. 2. Mark your answer to each problem on the Google form. Only answers properly marked on the answer form will be graded. 3. SCORING: You will receive 6 points for each correct answer, 1.5 points for each blank answer, and 0 points for each incorrect answer, for a maximum score of 150 points. Ties on the contest will be broken first by hardest problem solved, and then by submission time. 4. No aids are permitted other than scratch paper, graph paper, rulers, compass, protractors, and erasers. No calculators, smartwatches, or computing devices are allowed. No problems on the test will require the use of a calculator. 5. Figures are not necessarily drawn to scale. 6. Once you have opened the Google form, you may immediately begin working on the problems. You will have 75 minutes to complete the test. 7. You MAY NOT discuss any aspect of the test, including the problems/solutions and their difficulties, until after the submission period is over. The Math4All committee reserves the right to disqualify the scores from an individual or group of individuals if it is determined that cheating or other suspicious behavior has occurred. The publication, reproduction or communication of the problems or solutions of the Back-To-School Math Contest (BTSMC) Year 2 during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination via copier, telephone, e-mail, World Wide Web or media of any type during this period is a violation of the competition rules. 1. Evaluate: 1 + (1 − 2) + (1 − 2 + 3) + (1 − 2 + 3 − 4) + · · · + (1 − 2 + · · · + 355) (A) − 355 (B) − 177 (C) 177 (D) 178 (E) 355 Proposed by Faizaan Siddique Solution: We evaluate the terms within the parentheses to get 1 − 1 + 2 − 2 + 3 − 3... + 177 − 177 + 178 = (D) 178 2. George the Monkey and Harambe the Gorilla live in a community of monkeys and gorillas. Excluding George, the ratio of monkeys to gorillas is 5 : 4, but excluding Harambe and including George, the ratio of monkeys to gorillas is 4 : 3. What is the total number of animals in their community? (A) 24 (B) 32 (C) 40 (D) 48 (E) 64 Proposed by Aaron Hu Solution: Suppose there are a monkeys and b gorillas in the community. Then 5 a−1 = , b 4 a 4 = , b−1 3 which gives the system of equations 4a − 5b = 4, 3a − 4b = −4. Solving gives a = 36, b = 28, so the answer is 36 + 28 = (E) 64 3. How many ordered pairs of positive integers (x, y) where 1 ≤ x ≤ y exist such that x + y divides 3x + 3y + 18? (A) 14 (B) 16 (C) 17 (D) 18 (E)19 Proposed by Faizaan Siddique Solution: We have x + y divides 3x + 3y + 18 if and only if it divides 3x + 3y + 18 − 3(x + y) = 18., Therefore, we get solutions whenever x+y divides 18. This yields solutions of x+y ∈ {1, 2, 3, 6, 9, 18}. The number of solutions (x, y) for each factor respectively becomes: 0 + 1 + 1 + 3 + 4 + 9 = (D) 18 4. A regular n-sided polygon can form exactly 2023 distinct squares from its vertices. What is the minimum possible value of n? (A) 2023 (B) 4046 (C) 8092 (D) 10115 (E) 12138 Proposed by Faizaan Siddique Solution: The four chosen vertices will form a square when they are equally spaced along the vertices of the n-gon. As given, there are 2023 choices for equally spaced vertices of the polygon. This will happen when n = 2023 ∗ 4 = (C) 8092 , being careful not to count a vertex again if it was on a previously counted square. 5. Suppose 20 pairs of 2 parallel lines lie on a single plane. Find the maximum number of intersection points that can exist in the plane. (A) 95 (B) 190 (C) 380 (D) 570 (E) 760 Proposed by Faizaan Siddique Solution: Notice that if three of the lines intersect at a common point, then the number of intersection points is strictly less than how many there would be if they intersected pairwise at three different places. We also assume that no two pairs of lines are parallel to one another (beyond the 20 specified in the problem), since they would also never intersect. Thus, each line intersects every line other than its parallel pair exactly once, so the answer is 40∗(38) = (E) 760 , halving as to not 2 double count. 6. For how many 4-digit numbers of the form 1ABC is it true that when the first digit is removed, the number that is formed divides the original 4-digit number? (A) 14 (B) 15 (C) 16 (D) 17 (E) 18 Proposed by Milo Stammers Solution: With the first digit removed the number becomes ABC, and this will divide 1ABC if and only if it divides 1ABC − ABC = 1000. Because 1000 = 23 53 , 1000 has (4)(4) = 16 divisors. The only case that would not create a valid ABC is the number 1000 itself, thus we have 16−1 = (B) 15 7. On an odd night, 36 creatures plan to eat dinner at a round table with 36 seats. 12 creatures are ghouls, 16 creatures are ghosts, and 8 creatures are humans. If ghouls cannot sit directly adjacent to ghouls, ghosts cannot sit directly adjacent to humans, and humans cannot sit directly adjacent to ghouls nor themselves, what is the maximum number of creatures that can be seated? (Note empty seats between two creatures means that they are not directly adjacent to each other) (A) 27 (B) 28 (C) 29 (D) 30 (E) 31 Proposed by Faizaan Siddique Solution: Notice that humans cannot sit next to basically anyone, as they cannot sit be next to ghouls or humans, and ghosts cannot sit next to them. It is not difficult to seat all the ghouls and ghosts, as they may all be seating next to one another, but with ghouls placed in between two ghosts. There will be 8 seats left, three of which can be filled by humans, as you must leave a space between them and anything else. Therefore, we can place at most 12 + 16 + 3 = (E) 31 8. Suppose Kane places 4 solid circular pegs with radius 1 on a 10 × 10 grid composed of units squares such that the pegs have centers on lattice points in the grid (part of the peg can go off the grid), no two pegs lie in the same row or column, and the centers of no three pegs are collinear. Kane wraps a tight continuous string of length L around the circumference of the 4 pegs to form an enclosed region. Given that√Kane wants to maximize the area of this enclosure formed by the 4 pegs, if L can be expressed as a b + cπ, where b is square-free, find a + b + c. (A) 88 (B) 90 (C) 93 (D) 95 (E) 99 Proposed by Faizaan Siddique Solution: The area will be maximized when we place the vertices in a tilted square with vertices at (1, 0), (0, 9), (9, 10), (10, 1). The length of string will be in contact with each peg for a quarter arc of each circle, so altogether the length along the pegs is 2π(1) = 2π. The length of string not along corresponds to the lengths of the square formed by the center, which has perimeter √ √ the peg √ 2 2 4 1 + 9 = 4 82. Therefore, the total length is 4 82 + 2π, so our answer is (A) 88 9. Find the sum of all integers 1 < n ≤ 100 such that 2 nn − n2 n+1 is an integer. (A) 2499 (B) 2500 (C) 2550 (D) 2551 (E) 5049 Proposed by Milo Stammers Solution: Because n and n + 1 must be relatively prime, we can factor out n2 from the numerator, and it suffices to find when 2 nn −2 − 1 n+1 is an integer. This is equivalent to nn 2 −2 ≡ 1 mod (n + 1). But n ≡ −1 mod (n + 1) → 2 −2 (−1)n ≡1 mod (n + 1) → n2 − 2 is even so long as −1 ̸≡ 1 mod (n + 1). Therefore, either n is even or 1 (the latter case we don’t consider). Thus the answer becomes 2 + 4 + ... + 100 = 2(1 + 2 + 3... + 50) = 50(51) = (C) 2550 10. Suppose Rob has a standard deck of 52 playing cards. Each second, Rob randomly selects a card from the deck, notes down the card suit, and places it back into the deck randomly. After 52 seconds, Rob obtains the following partial results, as shown below: Card Number 2 3 4 5 Number of Draws 2 1 6 7 Rob then repeats the same process for the next 52 seconds. Based on the above partial data, which of the following statements is true for the second round? (A) Rob should expect to have more than 4 draws of card 2 (B) Rob should expect to have less than 4 draws of card 5 (C) Rob should expect to have less than 7 draws of card 5 (D) Rob should expect to have more than 6 draws of card 4 (E) Rob should expect to have more draws of card 3 than card 5 Proposed by Faizaan Siddique Solution: Here, the expected value of drawing a card type (there are 13 card types (2-9, Ace, King, 1 · 52 = 4 in both the first and second round. The only answer option that Queen, Jack) is equal to 13 satisfies this condition is (C) Remark: This problem is notable because it addresses the Gambler’s Paradox, a belief that if an unlikely event occurs too many times, the likelihood of the unlikely event is lower than other events in subsequent turns (for example, flipping a fair coin 100 times and getting all tails would compel many to assume that the 101th flip is more likely to be heads than tails, which is simply not true. 11. Given that there exists a 4-digit number N below such that N shares exactly 3 digits and a 2-digit N common divisor greater than 25 with some 3-digit number, find the smallest possible value of N . (A) 1285 (B) 2185 (C) 2485 (D) 2850 (E) 3285 Proposed by Faizaan Siddique Solution: Note all answer choices share the digits 2,8, and 5 and are all multiples of 5. Therefore, the mystery 3-digit number must either by 285 or 825, the former of which would yield solutions of 2185 and 2850. The answer is thus (B) 2185 because 285 = 3 · 95 and 2185 = 23 · 95. 12. If r1 and r2 are real numbers such that r1 + r2 = 1, and √ 2 r1 r2 =3 r2 − r1 then find (A) 10 1 (2r1 −1)2 + (B) 20 1 . (2r2 −1)2 (C) 50 (D) 100 (E) 200 Proposed by Faizaan Siddique Solution: We see 4r1 r2 = 9(r2 − r1 )2 = 9((r2 + r1 )2 − 4r1 r2 ) = 9(1 − 4r1 r2 ) =⇒ r1 r2 = Expanding the desired answer gives 9 . 40 (2r2 − 1)2 + (2r1 − 1)2 11 =⇒ 100(4(r12 + r22 ) − 2)) =⇒ 100(4( 20 ) − 2) =⇒ (B) 20 (4r1 r2 − 1)2 . 13. Let ABCD be a square with AB = 4, and let E lie within the square such that AE = BE = CE = E GE IE DE. Suppose F, G, H, I are points that lie outside the square such that FAE = BE = HE = DE = 2. CE If F B and AG intersect at J, GC and BH intersect at K, HD and CI intersect at L, and IA and , where m and n are relatively DF intersect at M , the area of AJBKCLDM can be represented as m n prime positive integers. Find m + n. I H L D C K E M A B J F (A) 19 (B) 31 (C) 59 (D) 67 G (E) 83 Proposed by Faizaan Siddique Solution: It is easy to see that [AJBKCLDM ] = [ABCD] + 4[AJB] by symmetry, and [ABCD] = 16. Therefore, it suffices to find [AJB]. Let P and Q be the feet of the perpendiculars from F and J to line AB respectively. Since BP = 3AP = 3AQ = 3BQ we have by similar triangle F P = 3JQ. In addition, by the ratio we have AP F is an isosceles right triangle, so F P = P A = 2, so JQ = 23 . The answer is 16 + 4 · 2 · 23 = 64 =⇒ (D) 67 3 14. Suppose Bob is rolling a fair 8-sided die labeled with consecutive positive integers. After each roll, Bob records the sum of all the faces on the die excluding the bottom face that is touching the table. On his first roll, he records a sum of 49 while on his second roll, he records a sum of 44. If the , where m and n are relatively prime expected value of his sum on his third roll can be expressed as m n positive integers, find m + n. (A) 81 (B) 93 (C) 95 (D) 103 (E) 115 Proposed by Faizaan Siddique Solution: Let’s establish an upper bound by assuming the largest integer was removed from the set when the sum was 44. If the set is assumed to be {x, x + 1, x + 2, x + 3, x + 4, x + 5, x + 6, x + 7} for positive x, then x + 3(7) ≤ 44 =⇒ x ≤ 3 It suffices to show that x = 3 works but in fact x = 2 and x = 1 do not work. Finally, if expected value of the sum is En , then En = 52 − E0 where E0 is the expected value of the bottom face on a single dice roll. By LoE, this is simply 52 = 13 , so the answer is 8 2 13 91 52 − = =⇒ (B) 93 2 2 15. Let S be the set defined as: S = {a + 1 | − 10 ≤ a, b ≤ 10, b ̸= 0} b for integers a,b. Find the number of two-element subsets of S such that the sum of the subset’s elements is an integer. (A) 253 (B) 484 (C) 652 (D) 2332 (E) 4012 Proposed by Aaron Hu Solution: First, note that the integers in S are −11, −10, . . . , 10, 11, out of which that are 23 = 253 2 1 21 19 two-element subsets. Next, note that there are 22 elements of S with fractional part 2 , (− 2 , − 2 , . . . , 21 ), 2 22 any pair of which add to an integer, so these numbers yield 2 = 231 subsets. Finally, the only case remaining is when a two-element subset is of the form {a + 1b , c − 1b }, where b ≥ 3. There are 21 ∗ 21 = 441 choices for a, c and 8 choices for b, yielding 441 · 8 = 3528 subsets. Then the final answer is 253 + 231 + 3528 = (E) 4012 16. Find the sum of all positive integers n ≤ 100 such that n60 − 1 is divisible by 899 (A) 349 (B) 360 (C) 487 (D) 571 (E) 709 Proposed by Faizaan Siddique Solution: Clearly n60 − 1 = (n30 + 1)(n30 − 1) and 899 = 302 − 12 = 31 · 29 By Euler’s totient n30 ≡ 1 mod 31 for any n, so it suffices to show that n2 ≡ 1 mod 29 for the first condition and n4 ≡ 1 mod 29 for the second. For the first equation, n ≡ 1, 28 mod 29 and these are the only squares that are within reach of a multiple of 29. Therefore the second condition also has numbers n ≡ 1, 28 mod 29 or n2 ≡ 1, 28 mod 29. Since we already covered the first 3 cases, we want to find all n satisfying n2 ≡ −1 mod 29 for which n ≡ 12, 17 mod 29 are the only ones that work. Therefore the answer becomes (1 + 12 + 17 + 28) + 29(4) + (1 + 12 + 17 + 28) + 29(8) + (1 + 12 + 17 + 27) + (29(3) + 1) + (29(3) + 12) =⇒ 29(2)(3) + 29(12) + 29(6) + 13 =⇒ 29(24) + 13 = 720 − 24 + 13 = 709 17. John, Jane, and Dwight are playing the game Triangle Destroyer. First, John starts creating a triangle with sides 90 and 94, and Jane chooses a random integer third side such that the resulting triangle is non-degenerate. Then, Dwight is instructed to choose a random positive real number a < 90. Finally, Dwight subtracts a from every side to produce a new triangle. If the probability that the new , where m and n are relatively prime positive integers, triangle is degenerate can be expressed as m n then find n − m. (Assume negative side lengths constitute a degenerate triangle) (A) 90 (B) 134 (C) 175 (D) 224 (E) 269 Proposed by Faizaan Siddique Solution: Let c be the length of the third side. Note that in order for a triangle with side lengths 90 − a, 94 − a, and c − a to be degenerate for some a, then (90 − a) + (94 − a) ≤ (c − a) =⇒ a ≥ (184 − c) (94 − a) + (c − a) ≤ (90 − a) =⇒ a ≥ 4 + c (90 − a) + (c − a) ≤ (94 − a) =⇒ a ≥ c − 4 By definition 4 < c < 184, and since only 1 of these conditions needs to be satisfied we have a ≥ c − 4 is our target condition. This is now a matter of probability. Since c is an integer, we compute manually, which gives 89(90) 1 89 88 1 1 89 ·2· + + ··· + =⇒ · 2 =⇒ =⇒ 179 − 89 = (A) 90 179 90 90 90 179 90 179 18. If r1 , r2 , r3 are the roots of the equation x3 + 4x2 + 6x − 22, find the value of r12 (A) 50 91 (B) 4 7 (C) 54 91 (D) 1 1 1 + 2 + 2 2 2 + r2 r1 + r3 r2 + r32 8 13 (E) 57 91 Proposed by Bole Ying Solution: From Vieta’s we know that r1 + r2 + r3 = −4, r1 r2 + r1 r3 + r2 r3 = 6, r1 r2 r3 = 22. This 1 1 1 means that r12 + r22 + r32 = (−4)2 − 2(6) = 4. Therefore we can rewrite the sum as 4−r 2 + 4−r 2 + 4−r 2 . 1 2 3 Now, putting this all over one fraction is (4 − r12 )(4 − r22 ) + (4 − r12 )(4 − r32 ) + (4 − r22 )(4 − r32 ) (4 − r12 )(4 − r22 )(4 − r32 ) The denominator seems like a mess, but note that we may first factor it using difference of squares Q3 to get i=1 (2 − ri )(2 + ri ). Note that by the definition of a polynomial, P (x) = x3 + 4x2 + 6x − 22 = (x − r1 )(x − r2 )(x − r3 ). Therefore P (2)P (−2) = (2 − r1 )(2 − r2 )(2 − r3 )(−2 − r1 )(−2 − r2 )(−2 − r3 ) = −(2 − r1 )(2 − r2 )(2 − r3 )(2 + r1 )(2 + r2 )(2 + r3 ) = −(14)(−26) = 364. Now for the numerator, after multiplying it out we get, 3 · 16 − 8(r12 + r22 + r32 ) + (r12 r22 + r12 r32 + r22 r32 ). Note that r12 r22 + r12 r32 + r22 r32 = (r1 r2 + r1 r3 + r2 r3 )2 − 2(r1 r2 r3 )(r1 + r2 + r3 ) so r12 r22 + r12 r32 + r22 r32 = 62 − 2(22)(−4) = 124. We also know that r12 + r22 + r32 = 4 from earlier so our numerator is 48 − 32 + 212 = 228. Therefore our 57 228 answer is 364 = (E) 91 19. In quadrilateral ABCD with positive side lengths, let diagonals AC and BD intersect at P . Suppose a line ℓ passes through P such that ℓ bisects ∠AP B and splits the area of ABCD into two equal halves. If AB and CD subtend arcs of a single circle ω with nonzero area, then which of the following quadrilaterals correctly classifies all possible configurations of ABCD? (A) rhombus (B) rectangle (C) parallelogram (D) isosceles trapezoid (E) kite Proposed by Faizaan Siddique Solution: By the second condition, ABCD is a cyclic quadrilateral. We now show that this is true. Let ℓ intersect AB and CD at E and F respectively, and let X, Y , and Z represent the areas of AP E, AP D, and DP F . We see by similar triangles that if PP B = k, then [EP B] = kX, [BP C] = k 2 Y , and A [CP F ] = kZ. Equating both sides we obtain: X + Y + Z = k(X + kY + Z) =⇒ (1 − k)(X + Z) = (k 2 − 1)Y Since X + Z and Y must be positive, it follows that 1 − k = k 2 − 1 = 0 or k = 1. This means ∠P AB = ∠P BA and ∠P DC = ∠P CD, or P contains the perpendicular bisectors of AB and CD, which implies AB ∥ CD. The only quadrilateral that satisfies these properties is (D) isosceles trapezoid 20. A geometric sequence of positive integers a, b, and c with integral common ratio r is called bogus if the numbers d(a), d(b), and d(c) form an arithmetic sequence, where d(x) is the number of divisors of x. Find the number of values of r where 1 ≤ r ≤ 100 such that there exists at least one bogus sequence with common ratio r. (A) 33 (B) 35 (C) 36 (D) 38 (E) 39 Proposed by Andrew Brahms Solution: Consider the prime factorizations of a, b, and c. For each prime number, p, νp (a), νp (b), νp (c), and therefore νp (a) + 1, νp (b) + 1, νp (c) + 1 form nondecreasing arithmetic sequences, as the common ratio is an integer. d(a), d(b), d(c) is therefore equal to the product of nondecreasing arithmetic sequences (here ”product” refers to the sequence generated by taking the products of corresponding terms of the original sequences). Clearly if the common ratio is 1, prime, or a power of a prime, this will be the product of one increasing arithmetic sequence and a number of constant arithmetic sequences. Otherwise, it is the product of more than one increasing arithmetic sequences. If we have the product of two increasing arithmetic sequences a, a + x, a + 2x, and b, b + y, b + 2y, we get ab, ab + xb + ya + xy, ab + 2xb + 2ya + 4xy. The third term minus the second is clearly 2xy greater than the second minus the first. Therefore the product of two increasing arithmetic sequences will have the property that the difference between the first two terms is less than the difference between the second two. Clearly this also holds true for the product of more than two increasing arithmetic sequences, so the only possibilities are when the common ratio is 1, prime, or a power of a prime. Therefore, we simply count the number of such values less than or equal to 100, which is (C) 36 . 21. In acute triangle ABC, points P , Q, and R lie on AB, BC, and CA respectively such that AP = BQ = CR = 32 . Suppose AQ intersects BR and CP at M and N respectively, and segPB QC RA ments BR and CP intersect at O. Find (A) 1 7 (B) 1 10 (C) 1 16 (D) 1 19 [M N O] . [ABC] (Note [ABC] denotes the area of [ABC]) (E) 1 23 Proposed by Luca Pieleanu Solution: WLOG assume the area [ABC] = 1. First, by symmetry we can see that [AM B] = B [BOC] = [CN A]. Additionally, we see that [AM B] = M · AR · [ABC], so it suffices to find these M R AC 6 aformentioned ratios. We employ mass points, which gives [AM B] = 15 · 52 · 1 = 19 . Therefore 19 6 1 [M N O] = 1 − 3 19 = 19 , giving a final answer of (D) . 22. Let s(n) denote the sum of the digits n and let p(n) denote the product of the digits of n where n is a positive integer. Let N be the sum of all positive integers less than 100 such that s(n)2 + p(n) = n Find the sum of the digits of N . (A) 6 (B) 7 (C) 8 (D) 9 (E) 10 Proposed by Andrew Brahms and Luca Pieleanu Solution: Clearly, no one-digit numbers work. Therefore, we can let n = 10a + b where a and b are digits and a is nonzero. We have s(n)2 + p(n) = n =⇒ =⇒ =⇒ =⇒ (a + b)2 + ab = 10a + b a2 + 2ab + b2 + ab − 10a = b a2 + 3ab − 10a = b − b2 a(a + 3b − 10) = b(1 − b). Now, we proceed by casework. If b = 0, then either a = 0, or a + 0 − 10 = 0 =⇒ a = 10, both of which are impossible. If b = 1, then either a = 0, or a + 3 − 10 = 0 =⇒ a = 7, the former of which is impossible, but the latter of which is possible. If b = 2, then the right-hand side is negative, so the left-hand side must also be negative, which implies that a = 1, 2, or 3. However, plugging them in shows us that none of them work. If b ≥ 3, then the right-hand side is negative, but a ≥ 1 =⇒ a + 3b − 10 ≥ 1 + 9 − 10 = 0 =⇒ a(a + 3b − 10) ≥ 0, so the left-hand side is nonnegative. Hence, there are no other solutions. Therefore, we can conclude that the only positive integer less than 100 that satisfies the given equation is 71, so the answer is 7 + 1 = (C) 8 . 23. In kite ABCD with ∠ABC = ∠ADC = 90o and BC = CD, let point E lie on the extension of AD, where E is closer to D than A. Let F be the unique point on BE such that ∠BF D = ∠BDA. If ED = 8, DA = 10, and BF = 12, then CD2 can be expressed as m , where m and n are relatively n prime positive integers. Find m + n. (A) 411 (B) 485 (C) 676 (D) 711 (E) 785 Proposed by Faizaan Siddique Solution: Draw in circle ω with radius BC centered at C. By the inscribed-angle theorem, F must lie on ω. By PoP on E we obtain EF = 4, and PoP on A gives AB = 10. Applying LoC on ABE gives 7 256 = 100 + 324 − 360 cos(∠BAE) =⇒ cos(∠BAE) = 15 ∠BAD ∠BAE x 2 2 ∠BAE Note that if BC = x, then tan( 2 ) = tan( 2 ) = 10 , so x = 100 tan ( 2 ) Applying double4 angle identity gives tan2 ( ∠BAE ) = 11 , so the answer is 400 =⇒ (A) 411 2 11 24. Suppose a positive integer n is considered squarable if the last two digits of the number n2 also form a perfect square. For example, the numbers 40 and 45 are both squarable since 402 = 1600 and k 452 = 2025. Find the number of positive integers n less than 1000 such that f (n(2 ) ) is squarable for all nonnegative integers k (here f (n) denotes the remainder when n is divided by 100) (A) 259 (B) 279 (C) 289 (D) 299 (E) 319 Proposed by Faizaan Siddique Solution: Note that since only the last two digits of the number n matter, the number of squarable numbers cycles mod 100. Therefore, we consider all n = 10a + b, where n2 = 100a2 + 20ab + b2 ≡ 20ab + b2 mod 100. Note if a ∈ {0, 5}, then the condition is always satisfied. Doing casework on b we obtain a = 4 or a = 9 would work as well. Now, in order for all resultant n2 to cycle, the squares must have starting digits corresponding with a ∈ {0, 4, 5, 9}. These perfect squares are thus 01,04,09, and 49 (i.e. none starting with digit 5 or 9). Taking this one step further, 01 cycles indefinitely and 49 cycles to 01, while 04 halts at 16 and 09 halts at 81. As a result, only n producing 01 or 49 when squared will work. Additionally, all multiples of 10 automatically work (i.e. b = 0). If b = 1, then 01 and 51 produce 01. If b = 3, then 43 and 93 produce 49. If b = 5, then any a will work because the number would always end in 25. If b = 7, then 07 and 57 produce 49. Finally, if b = 9, then 49 and 99 produce 01. For 0 < n < 100, there are 9 + 2 + 2 + 10 + 2 + 2 = 27 solutions. Meanwhile, for each module of numbers 100p ≤ n < 100(p + 1), there are 27 + 1 = 28 solutions (since 100p must count). The answer is thus 27 + 28(9) = (B) 279 25. Suppose triangle ABC has integer side lengths satisfying AB = 10, AC = 17, and BC = 21. Let X be the unique point such that AX = BC, BX = AC, and X and A lie on opposite sides of BC. Let H1 denote the orthocenter of ABX, H2 the orthocenter of BCX, and H3 the orthocenter of ACX. Find the area of H1 H2 H3 . (A) 14 (B) 24 (C) 60 (D) 84 (E) 105 Proposed by Andrew Brahms Solution: The area of the triangle formed by the midpoints of the sides of △ABC is going to be the area of △ABC, because each of the side lengths are multiplied by 12 . 1 4 The area of the triangle formed by the centroids of △ABP , △ACP , and △BCP will be 94 of the area of the triangle formed from the midpoints because it’s a homothety about point P by the factor of 23 , so the area is 19 . Call the circumcenter of △ABC O. Because of the Euler Line, △H1 H2 H3 is thep homothety by a 1 factor of 3 about point O of the triangle of centroids. Therefore the area is 9 · 9 · 24(14)(7)(3) = (D) 84