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2023 LMT Fall

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2023 LMT Fall Guts Round p16-p27 - Lexington Mathematical Tournament

Part 6
p16. Let p(x)p(x) and q(x)q(x) be polynomials with integer coefficients satisfying p(1)=q(1)p(1) = q(1). Find the greatest integer nn such that p(2023)−q(2023)n\frac{p(2023)-q(2023)}{n} is an integer no matter what p(x)p(x) and q(x)q(x) are.
p17. Find all ordered pairs of integers (m,n)(m,n) that satisfy n3+m3+231=n2m2+nm.n^3 +m^3 +231 = n^2m^2 +nm.
p18. Ben rolls the frustum-shaped piece of candy (shown below) in such a way that the lateral area is always in contact with the table. He rolls the candy until it returns to its original position and orientation. Given that AB=4AB = 4 and BD=CD=3BD =CD = 3, find the length of the path traced by AA.
Part 7
p19. In their science class, Adam, Chris, Eddie and Sam are independently and randomly assigned an integer grade between 7070 and 7979 inclusive. Given that they each have a distinct grade, what is the expected value of the maximum grade among their four grades?
p20. Let ABCDABCD be a regular tetrahedron with side length 22. Let point EE be the foot of the perpendicular from DD to the plane containing △ABC\vartriangle ABC. There exist two distinct spheres ω1\omega_1 and ω2\omega_2, centered at points O1O_1 and O2O_2 respectively, such that both O1O_1 and O2O_2 lie on DE→\overrightarrow{DE} and both spheres are tangent to all four of the planes ABCABC, BCDBCD, CDACDA, and DABDAB. Find the sum of the volumes of ω1\omega_1 and ω2\omega_2.
p21. Evaluate ∑i=0∞∑j=0∞∑k=0∞1(i+j+k+1)2i+j+k+1.\sum^{\infty}_{i=0}\sum^{\infty}_{j=0}\sum^{\infty}_{k=0} \frac{1}{(i + j +k +1)2^{i+j+k+1}}.
Part 8
p22. In △ABC\vartriangle ABC, let IAI_A, IBI_B , and ICI_C denote the AA, BB, and CC-excenters, respectively. Given that AB=15AB = 15, BC=14BC = 14 and CA=13C A = 13, find [IAIBIC][ABC]\frac{[I_A I_B I_C ]}{[ABC]} .
p23. The polynomial x+2x2+3x3+4x4+5x5+6x6+5x7+4x8+3x9+2x10+x11x +2x^2 +3x^3 +4x^4 +5x^5 +6x^6 +5x^7 +4x^8 +3x^9 +2x^{10} +x^{11} has distinct complex roots z1,z2,...,znz_1, z_2, ..., z_n. Find ∑k=1n∣R(z2n))∣+∣I(z2n)∣,\sum^n_{k=1} |R(z^2n))|+|I(z^2n)|, where R(z)R(z) and I(z)I(z) indicate the real and imaginary parts of zz, respectively. Express your answer in simplest radical form.
p24. Given that sin⁡33o+2sin⁡161o⋅sin⁡38o=sin⁡no\sin 33^o +2\sin 161^o \cdot \sin 38^o = \sin n^o , compute the least positive integer value of nn.
Part 9
p25. Submit a prime between 22 and 20232023, inclusive. If you don’t, or if you submit the same number as another team’s submission, you will receive 00 points. Otherwise, your score will be min⁡(30,⌊4⋅ln(x)⌋)\min \left(30, \lfloor 4 \cdot ln(x) \rfloor \right), where xx is the positive difference between your submission and the closest valid submission made by another team.
p26. Sam, Derek, Jacob, andMuztaba are eating a very large pizza with 20232023 slices. Due to dietary preferences, Sam will only eat an even number of slices, Derek will only eat a multiple of 33 slices, Jacob will only eat a multiple of 55 slices, andMuztaba will only eat a multiple of 77 slices. How many ways are there for Sam, Derek, Jacob, andMuztaba to eat the pizza, given that all slices are identical and order of slices eaten is irrelevant? If your answer is AA and the correct answer is CC, the number of points you receive will be: irrelevant? If your answer is AA and the correct answer is CC, the number of points you receive will be: max⁡(0,⌊30(1−2∣A−C∣C)⌋)\max \left( 0, \left\lfloor 30 \left( 1-2\sqrt{\frac{|A-C|}{C}}\right)\right\rfloor \right)
p27. Let Ω(k) \Omega_(k) denote the number of perfect square divisors of kk. Compute ∑k=110000Ω(k).\sum^{10000}_{k=1} \Omega_(k). If your answer is AA and the correct answer is CC, the number of points you recieve will be max⁡(0,⌊30(1−4∣A−C∣C)⌋)\max \left( 0, \left\lfloor 30 \left( 1-4\sqrt{\frac{|A-C|}{C}}\right)\right\rfloor \right)
PS. You should use hide for answers. Rounds 1-5 have been posted [url=https://artofproblemsolving.com/community/c3h3267911p30056982]here. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here.

2023 LMT Fall Guts Round p1-p15- Lexington Mathematical Tournament

Part 1
p1. Calculate (4!−5!+25+26)⋅12!7!+(1−3)(4!−24).(4!-5!+2^5 +2^6) \cdot \frac{12!}{7!}+(1-3)(4!-2^4).
p2. The expression 9!+10!+11!\sqrt{9!+10!+11!} can be expressed as aba\sqrt{b} for positive integers aa and bb, where bb is squarefree. Find aa.
p3. For real numbers aa and bb, f(x)=ax10−bx4+6x+10f(x) = ax^{10}-bx^4+6x +10 for all real xx. Given that f(42)=11f(42) = 11, find f(−42)f (-42).
Part 2
p4. How many positive integers less than or equal to 20232023 are divisible by 2020, 2323, or both?
p5. Larry the ant crawls along the surface of a cylinder with height 4848 and base radius 14π\frac{14}{\pi} . He starts at point AA and crawls to point BB, traveling the shortest distance possible. What is the maximum this distance could be?
p6. For a given positive integer nn, Ben knows that ⌊20x⌋=n\lfloor 20x \rfloor = n, where xx is real. With that information, Ben determines that there are 33 distinct possible values for ⌊23x⌋\lfloor 23x \rfloor. Find the least possible value of nn.
Part 3
p7. Let ABCABC be a triangle with area 11. Points DD, EE, and FF lie in the interior of △ABC\vartriangle ABC in such a way that DD is the midpoint of AEAE, EE is the midpoint of BFBF, and FF is the midpoint of CDCD. Compute the area of DEFDEF.
p8. Edwin and Amelia decide to settle an argument by running a race against each other. The starting line is at a given vertex of a regular octahedron and the finish line is at the opposite vertex. Edwin has the ability to run straight through the octahedron, while Amelia must stay on the surface of the octahedron. Given that they tie, what is the ratio of Edwin’s speed to Amelia’s speed?
p9. Jxu is rolling a fair three-sided die with faces labeled 00, 11, and 22. He keeps going until he rolls a 11, immediately followed by a 22. What is the expected number of rolls Jxu makes?
Part 4
p10. For real numbers xx and yy, x+xy=10x +x y = 10 and y+xy=6y +x y = 6. Find the sum of all possible values of xy\frac{x}{y}.
p11. Derek is thinking of an odd two-digit integer nn. He tells Aidan that nn is a perfect power and the product of the digits of nn is also a perfect power. Find the sum of all possible values of nn.
p12. Let a three-digit positive integer N=abc‾N = \overline{abc} (in base 1010) be stretchable with respect to mm if NN is divisible by mm, and when NN‘s middle digit is duplicated an arbitrary number of times, it‘s still divisible by mm. How many three-digit positive integers are stretchable with respect to 1111? (For example, 432432 is stretchable with respect to 66 because 433...32433...32 is divisible by 66 for any positive integer number of 33s.)
Part 5
p13. How many trailing zeroes are in the base-20232023 expansion of 2023!2023! ?
p14. The three-digit positive integer k=abc‾k = \overline{abc} (in base 1010, with a nonzero) satisfies abc‾=c2ab−1\overline{abc} = c^{2ab-1}. Find the sum of all possible kk.
p15. For any positive integer kk, let aka_k be defined as the greatest nonnegative real number such that in an infinite grid of unit squares, no circle with radius less than or equal to aka_k can partially cover at least kk distinct unit squares. (A circle partially covers a unit square only if their intersection has positive area.) Find the sumof all positive integers n≤12n \le 12 such that an≠an+1a_n \ne a_{n+1}.
PS. You should use hide for answers. Rounds 6-9 have been posted [url=https://artofproblemsolving.com/community/c3h3267915p30057005]here. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here.
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2023 Fall LMT Team #15 area of PQYX , (BCM), (BCN)

In triangle ABCABC with AB=26AB = 26, BC=28BC = 28, and CA=30C A = 30, let MM be the midpoint of ABAB and let NN be the midpoint of CAC A. The circumcircle of triangle BCMBCM intersects ACAC at X≠CX\ne C, and the circumcircle of triangle BCNBCN intersects ABAB at Y≠BY\ne B. Lines MXMX and NYNY intersect BCBC at PP and QQ, respectively. The area of quadrilateral PQYXPQY X can be expressed as pq\frac{p}{q} for positive integers pp and qq such that gcd(p,q)=1(p,q) = 1. Find qq.

2023 Fall LMT Speed p15 , 7^2 x11 divides n^3 -n^2

Find the least positive integer nn greater than 11 such that n3−n2n^3 -n^2 is divisible by 72×117^2 \times 11.
Proposed by Jacob Xu
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2023 Fall LMT Team #12 excircle

In triangle ABCABC with AB=7AB = 7, AC=8AC = 8, and BC=9BC = 9, the AA-excircle is tangent to BCBC at point DD and also tangent to lines ABAB and ACAC at points and FF, respectively. Find [DEF][DEF]. (The AA-excircle is the circle tangent to segment BCBC and the extensions of rays ABAB and ACAC. Also, [XYZ][XY Z] denotes the area of triangle XYZXY Z.)

2023 Fall LMT Speed p12 - flipping a weighted coin

Sam and Jonathan play a game where they take turns flipping a weighted coin, and the game ends when one of them wins. The coin has a 89\frac89 chance of landing heads and a 19\frac19 chance of landing tails. Sam wins when he flips heads, and Jonathan wins when he flips tails. Find the probability that Samwins, given that he takes the first turn.
Proposed by Samuel Tsui
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2023 Fall LMT Team #10 distinct cells in a 4x4 grid

Aidan and Andrew independently select distinct cells in a 44 by 44 grid, as well as a direction (either up, down, left, or right), both at random. Every second, each of them will travel 11 cell in their chosen direction. Find the probability that Aidan and Andrew willmeet (be in the same cell at the same time) before either one of them hits an edge of the grid. (If Aidan and Andrew cross paths by switching cells, it doesn’t count as meeting.)

2023 Fall LMT Speed p10 sqaure and quad. in x-y plane

A square has vertices (0,10)(0,10), (0,0)(0, 0), (10,0)(10, 0), and (10,10)(10,10) on the x−yx-y coordinate plane. A second quadrilateral is constructed with vertices (0,10)(0,10), (0,0)(0, 0), (10,0)(10, 0), and (15,15)(15,15). Find the positive difference between the areas of the original square and the second quadrilateral.
Proposed byWilliam Hua
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2023 Fall LMT Team #6 standard 6 sided die

Jeff rolls a standard 66 sided die repeatedly until he rolls either all of the prime numbers possible at least once, or all the of even numbers possible at least once. Find the probability that his last roll is a 22.

2023 Fall Speed p6

Blue rolls a fair nn-sided die that has sides its numbered with the integers from 11 to nn, and then he flips a coin. Blue knows that the coin is weighted to land heads either 13\dfrac{1}{3} or 23\dfrac{2}{3} of the time. Given that the probability of both rolling a 77 and flipping heads is 115\dfrac{1}{15}, find nn.
Proposed by Jacob Xu
Solution. 10\boxed{10} The chance of getting any given number is 1n\dfrac{1}{n} , so the probability of getting 77 and heads is either 1n⋅13=13n\dfrac{1}{n} \cdot \dfrac{1}{3}=\dfrac{1}{3n} or 1n⋅23=23n\dfrac{1}{n} \cdot \dfrac{2}{3}=\dfrac{2}{3n}. We get that either n=5n = 5 or n=10n = 10, but since rolling a 77 is possible, only n=10n = \boxed{10} is a solution.
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2023 Fall LMT Team #5 regular hexagon

In regular hexagon ABCDEFABCDEF with side length 22, let PP, QQ, RR, and SS be the feet of the altitudes from AA to BCBC, EFEF, CFCF, and BEBE, respectively. Find the area of quadrilateral PQRSPQRS.

2023 Fall Speed p5

Let aa and bb be two-digit positive integers. Find the greatest possible value of a+ba+b, given that the greatest common factor of aa and bb is 66.
Proposed by Jacob Xu
Solution. 186\boxed{186} We can write our two numbers as 6x6x and 6y6y. Notice that xx and yy must be relatively prime. Since 6x6x and 6y6y are two digit numbers, we just need to check values of xx and yy from 22 through 1616 such that xx and yy are relatively prime. We maximize the sum when x=15x = 15 and y=16y = 16, since consecutive numbers are always relatively prime. So the sum is 6⋅(15+16)=1866 \cdot (15+16) = \boxed{186}.
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2023 Fall LMT Team #3 pickles and a fair coin

Adamand Topher are playing a game in which each of them starts with 22 pickles. Each turn, they flip a fair coin: if it lands heads, Topher takes 11 pickle from Adam; if it lands tails, Adam takes 22 pickles from Topher. (If Topher has only 11 pickle left, Adam will just take it.) What’s the probability that Topher will have all 44 pickles before Adam does?

2023 Fall Speed p3

Sam Wang decides to evaluate an expression of the form x+2⋅2+yx +2 \cdot 2+ y. However, he unfortunately reads each ’plus’ as a ’times’ and reads each ’times’ as a ’plus’. Surprisingly, he still gets the problem correct. Find x+yx + y.
Proposed by Edwin Zhao
Solution. 4\boxed{4} We have x+2∗2+y=x⋅2+2⋅yx+2*2+y=x \cdot 2+2 \cdot y. When simplifying, we have x+y+4=2x+2yx+y+4=2x+2y, and x+y=4x+y=4.
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2023 Fall Theme p5C

In equilateral triangle ABCABC, AB=2AB=2 and MM is the midpoint of ABAB. A laser is shot from MM in a certain direction, and whenever it collides with a side of ABCABC it will reflect off the side such that the acute angle formed by the incident ray and the side is equal to the acute angle formed by the reflected ray and the side. Once the laser coincides with a vertex, it stops. Find the sum of the smallest three possible integer distances that the laser could have traveled.
Proposed by Jerry Xu
Solution. 21\boxed{21} Whenever the laser hits a side of the triangle, reflect the laser's path over that side so that the path of the laser forms a straight line. We want the path of the laser to coincide with a vertex of one of the reflected triangles. Thus, we can restate the problem as follows:
Tessellate the plane with equilateral triangles of side length 33. Consider one of these equilateral triangles ABCABC with MM being the midpoint of AB=2AB=2. Find the sum of the three minimum integer distances from MM to any vertex in the plane.
[asy] import geometry; size(8cm); pair A = (0,sqrt(3)); pair B = (-1,0); pair C = (1,0); pair M = (0,0); for (int i = -1; i <= 2; ++i) { draw((i-3,i*sqrt(3))--(-i+3,i*sqrt(3))); draw(((i-1)*2,-sqrt(3))--(i+1,(2-i)*sqrt(3))); draw((-i-1,(2-i)*sqrt(3))--((1-i)*2,-sqrt(3))); } draw(A--B--C--A, red); dot(M); label("AA",A+(0,0.25),N); label("BB",B-(0.25,0),SW); label("CC",C+(0.25,0),SE); label("MM",M,S); [/asy] It is trivial to see that the vertical distance between MM and a given vertex is n3n\sqrt{3} for n∈N0n \in \mathbb{N}^{0}. If nn is even, the horizontal distance between OO and a given vertex is 1+2m1+2m for m∈N0m \in \mathbb{N}^{0}. If nn is odd, the horizontal distance is 2m2m for m∈N0m \in \mathbb{N}^{0}. We consider two separate cases:
1.1. nn is even. We thus want to find l∈Nl \in \mathbb{N} such that (n3)2+(1+2m)2=l2.\left(n\sqrt{3}\right)^2+(1+2m)^2=l^2.Make the substitution 1+2m=k1+2m=k to get that 3n2+k2=l2.3n^2+k^2=l^2.Notice that these equations form a family of generalized Pell equations y2−3x2=Ny^2-3x^2=N with N=k2N=k^2. We can find some set of roots to these equations using the multiplicative principle: we will use this idea to find three small ll values, and that gives us an upper bound on what the three ll values can be. From there, a simple bash of lower ll values to see if solutions to each generalized Pell equation not given by the multiplicative principle exist finishes this case.
By the multiplicative principle some set of solutions (xn,yn)(x_n,y_n) to the above equation with sufficiently small xnx_n follow the formulaxn3+yn=(x03+y0)(un3+vn),x_n\sqrt{3}+y_n=\left(x_0\sqrt{3}+y_0\right)\left(u_n\sqrt{3}+v_n\right),where (x0,y0)\left(x_0,y_0\right) is a solution to the generalized Pell equation and (un,vn)\left(u_n,v_n\right) are solutions to the Pell equation y2−3x2=1y^2-3x^2=1. Remember that the solutions to this last Pell equation satisfyun3+vn=(u03+v0)ku_n\sqrt{3}+v_n=\left(u_0\sqrt{3}+v_0\right)^kwhere the trivial positive integer solution (u0,v0)=(1,2)\left(u_0, v_0\right)=(1,2)(this can easily be found by inspection or by taking the convergents of the continued fraction expansion of 3\sqrt{3}).
We thus get that(u1,v1)=(4,7),(u2,v2)=(15,26),(u2,v2)=(56,97)…\left(u_1,v_1\right)=(4,7),\left(u_2,v_2\right)=(15,26),\left(u_2,v_2\right)=(56,97)\dots(also don't forget that (u,v)=(0,1)(u,v)=(0,1) is another solution).
From here, note that kk must be odd since k=1+2mk=1+2m for m∈N0m \in \mathbb{N}^{0}. For k=1k=1, the smallest three solutions to the Pell equation with nn even are \begin{align*} (x,y)&=(0,1),(4,7),(56,97) \\ \longrightarrow (n,m,l)&=(0,0,1),(4,0,7),(56,0,97) \end{align*}Our current smallest three values of ll are thus 1,7,971,7,97. A quick check confirms that all of these solutions are not extraneous (extraneous solutions appear when the path taken by the laser prematurely hits a vertex).
For k=3k=3, using the multiplicative principle we get two new smaller solutions \begin{align*} (x,y)&=(0,3),(12,21) \\ \longrightarrow (n,m,l)&=(0,1,3),(12,1,21) \end{align*}However, note that (n,m,l)=(0,1,3)(n,m,l)=(0,1,3) is extraneous since is equivalent to the path that is traced out by the solution (n,m,l)=(0,0,1)(n,m,l)=(0,0,1) found previously and will thus hit a vertex prematurely. Thus, our new three smallest values of ll are 1,7,211,7,21.
For k≥5k \ge 5, it is evident that there are no more smaller integral values of ll that can be found using the multiplicative principle: the solution set (n,m,l)=(0,k−12,k)(n,m,l)=\left(0,\dfrac{k-1}{2},k\right) is always extraneous for k>1k > 1 since it is equivalent to the path traced out by (0,0,1)(0,0,1) as described above, and any other solutions will give larger values of ll.
Thus, we now only need to consider solutions to each generalized Pell equation not found by the multiplicative principle. A quick bash shows that l=3,5,9,11l=3,5,9,11 gives no solutions for any odd kk and even nn, however n=13n=13 gives k=11k=11 and n=4n=4, a non-extraneous solution smaller than one of the three we currently have. Thus, our new three smallest ll values are 1,7,131,7,13.
22. nn is odd. We thus want to find l∈Nl \in \mathbb{N} such that (n3)2+(2m)2=l2.\left(n\sqrt{3}\right)^2+(2m)^2=l^2.Make the substitution 2m=k2m=k to get that 3n2+k2=l2.3n^2+k^2=l^2.This is once again a family of generalized Pell equations with N=k2N=k^2, however this time we must have kk even instead of kk odd. However, note that there are no solutions to this family of Pell equation with nn odd: k2≡0 (mod 4)k^2 \equiv 0 \text{ (mod }4) since kk is even, and 3n2≡3 (mod 4)3n^2 \equiv 3 \text{ (mod }4) since nn is odd, however 0+3≡3 (mod 4)0+3 \equiv 3 \text{ (mod }4) is not a possible quadratic residue mod 44. Thus, this case gives no solutions.
Our final answer is thus 1+7+13=211+7+13=\boxed{21}.
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2023 Fall Theme p4C

The equation of line ℓ1\ell_1 is 24x−7y=31924x-7y = 319 and the equation of line ℓ2\ell_2 is 12x−5y=12512x-5y = 125. Let aa be the number of positive integer values nn less than 20232023 such that for both ℓ1\ell_1 and ℓ2\ell_2 there exists a lattice point on that line that is a distance of nn from the point (20,23)(20,23). Determine aa.
Proposed by Christopher Cheng
Solution. 6\boxed{6} Note that (20,23)(20,23) is the intersection of the lines ℓ1\ell_1 and ℓ2\ell_2. Thus, we only care about lattice points on the the two lines that are an integer distance away from (20,23)(20,23). Notice that 77 and 2424 are part of the Pythagorean triple (7,24,25)(7,24,25) and 55 and 1212 are part of the Pythagorean triple (5,12,13)(5,12,13). Thus, points on ℓ1\ell_1 only satisfy the conditions when nn is divisible by 2525 and points on ℓ2\ell_2 only satisfy the conditions when nn is divisible by 1313. Therefore, aa is just the number of positive integers less than 20232023 that are divisible by both 2525 and 1313. The LCM of 2525 and 1313 is 325325, so the answer is 6\boxed{6}.
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2023 Fall Theme p5B

Bamal, Halvan, and Zuca are playing The Game. To start, they‘re placed at random distinct vertices on regular hexagon ABCDEFABCDEF. Two or more players collide when they‘re on the same vertex. When this happens, all the colliding players lose and the game ends. Every second, Bamal and Halvan teleport to a random vertex adjacent to their current position (each with probability 12\dfrac{1}{2}), and Zuca teleports to a random vertex adjacent to his current position, or to the vertex directly opposite him (each with probability 13\dfrac{1}{3}). What is the probability that when The Game ends Zuca hasn‘t lost?
Proposed by Edwin Zhao
Solution. 2990\boxed{\dfrac{29}{90}} Color the vertices alternating black and white. By a parity argument if someone is on a different color than the other two they will always win. Zuca will be on opposite parity from the others with probability 310\dfrac{3}{10}. They will all be on the same parity with probability 110\dfrac{1}{10}.
At this point there are 2⋅2⋅32 \cdot 2 \cdot 3 possible moves. 33 of these will lead to the same arrangement, so we disregard those. The other 99 moves are all equally likely to end the game. Examining these, we see that Zuca will win in exactly 22 cases (when Bamal and Halvan collide and Zuca goes to a neighboring vertex). Combining all of this, the answer is 310+29⋅110=2990\dfrac{3}{10}+\dfrac{2}{9} \cdot \dfrac{1}{10}=\boxed{\dfrac{29}{90}}
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2023 Fall Theme p3B

Evin and Jerry are playing a game with a pile of marbles. On each players' turn, they can remove 22, 33, 77, or 88 marbles. If they can’t make a move, because there's 00 or 11 marble left, they lose the game. Given that Evin goes first and both players play optimally, for how many values of nn from 11 to 14341434 does Evin lose the game?
Proposed by Evin Liang
Solution. 573\boxed{573} Observe that no matter how many marbles a one of them removes, the next player can always remove marbles such that the total number of marbles removed is 1010. Thus, when the number of marbles is a multiple of 1010, the first player loses the game. We analyse this game based on the number of marbles modulo 1010:
If the number of marbles is 00 modulo 1010, the first player loses the game
If the number of marbles is 22, 33, 77, or 88 modulo 1010, the first player wins the game by moving to 00 modulo 10
If the number of marbles is 55 modulo 1010, the first player loses the game because every move leads to 22, 33, 77, or 88 modulo 1010
In summary, the first player loses if it is 00 mod 5, and wins if it is 22 or 33 mod 55. Now we solve the remaining cases by induction. The first player loses when it is 11 modulo 55 and wins when it is 44 modulo 55. The base case is when there is 11 marble, where the first player loses because there is no move. When it is 44 modulo 55, then the first player can always remove 33 marbles and win by the inductive hypothesis. When it is 11 modulo 55, every move results in 33 or 44 modulo 55, which allows the other player to win by the inductive hypothesis. Thus, Evin loses the game if n is 00 or 11 modulo 55. There are 573\boxed{573} such values of nn from 11 to 14341434.
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2023 Fall Theme p5A

Paul Revere is currently at (x0,y0)\left(x_0, y_0\right) in the Cartesian plane, which is inside a triangle-shaped ship with vertices at (−725,2425),(−45,35)\left(-\dfrac{7}{25},\dfrac{24}{25}\right),\left(-\dfrac{4}{5},\dfrac{3}{5}\right), and (45,−35)\left(\dfrac{4}{5},-\dfrac{3}{5}\right). Revere has a tea crate in his hands, and there is a second tea crate at (0,0)(0,0). He must walk to a point on the boundary of the ship to dump the tea, then walk back to pick up the tea crate at the origin. He notices he can take 3 distinct paths to walk the shortest possible distance. Find the ordered pair (x0,y0)(x_0, y_0).
Proposed by Derek Zhao
Solution. (−725,625)\left(-\dfrac{7}{25},\dfrac{6}{25}\right) Let LL, MM, and NN be the midpoints of BCBC, ACAC, and ABAB, respectively. Let points DD, EE, and FF be the reflections of O=(0,0)O = (0,0) over BCBC, ACAC, and ABAB, respectively. Notice since MN∥BCMN \parallel BC, BC∥EFBC \parallel EF. Therefore, OO is the orthocenter of DEFDEF. Notice that (KMN)(KMN) is the nine-point circle of ABCABC because it passes through the midpoints and also the nine-point circle of DEFDEF because it passes through the midpoints of the segments connecting a vertex to the orthocenter. Since OO is both the circumcenter of ABCABC and the orthocenter of DEFDEF and the triangles are 180∘180^\circ rotations of each other, Revere is at the orthocenter of ABCABC. The answer results from adding the vectors OA+OB+OCOA +OB +OC, which gives the orthocenter of a triangle.
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2023 Fall Theme p4A

Let Revolution(x)=x3+Ux2+Sx+A(x) = x^3 +Ux^2 +Sx + A, where UU, SS, and AA are all integers and U+S+A+1=1773U +S + A +1 = 1773. Given that Revolution has exactly two distinct nonzero integer roots GG and BB, find the minimum value of ∣GB∣|GB|.
Proposed by Jacob Xu
Solution. 392\boxed{392} Notice that U+S+A+1U + S + A + 1 is just Revolution(1)(1) so Revolution(1)=1773(1) = 1773. Since GG and BB are integer roots we write Revolution(X)=(X−G)2(X−B)(X) = (X-G)^2(X-B) without loss of generality. So Revolution(1)=(1−G)2(1−B)=1773(1) = (1-G)^2(1-B) = 1773. 17731773 can be factored as 32⋅19732 \cdot 197, so to minimize ∣GB∣|GB| we set 1−G=31-G = 3 and 1−B=1971-B = 197. We get that G=−2G = -2 and B=−196B = -196 so ∣GB∣=392|GB| = \boxed{392}.
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2023 Fall Theme p3A

A rectangular tea bag PARTPART has a logo in its interior at the point YY . The distances from YY to PTPT and PAPA are 1212 and 99 respectively, and triangles △PYT\triangle PYT and △AYR\triangle AYR have areas 8484 and 4242 respectively. Find the perimeter of pentagon PARTYPARTY.
Proposed by Muztaba Syed
Solution. 78\boxed{78} Using the area and the height in △PYT\triangle PYT, we see that PT=14PT = 14, and thus AR=14AR = 14, meaning the height from YY to ARAR is 66. This means PA=TR=18PA = TR = 18. By the Pythagorean Theorem PY=122+92=15PY=\sqrt{12^2+9^2} = 15 and YT=122+52=13YT =\sqrt{12^2 +5^2} = 13. Combining all of these gives us an answer of 18+14+18+13+15=7818+14+18+13+15 = \boxed{78}.
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