MathDB

2020 BMT Fall

Part of BMT Problems

Subcontests

(32)
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2020 BMT Team 25

Submit an integer between 11 and 5050, inclusive. You will receive a score as follows: If some number is submitted exactly once: If EE is your number, AA is the closest number to EE which received exactly one submission, and MM is the largest unique submission, you will receive 25M(A−∣E−A∣)\frac{25}{M} (A - |E - A|) points, rounded to the nearest integer. If no number was submitted exactly once: If EE is your number, KK is the number of people who submitted EE, and MM is the number of people who submitted the most commonly submitted number, then you will receive 25(M−K)M\frac{25(M-K)}{M} points, rounded to the nearest integer.

2020 BMT Individual 25

Let f:R+→R+f : R^+ \to R^+ be a function such that for all x,y∈R+x, y \in R^+, f(x)f(y)=f(xy)+f(xy)f(x)f(y) = f(xy) + f\left( \frac{x}{y}\right), where R+R^+ represents the positive real numbers. Given that f(2)=3f(2) = 3, compute the last two digits of f(222020)f(2^{2^{2020}}). .
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2020 BMT Team 22

Suppose that x,yx, y, and zz are positive real numbers satisfying {x2+xy+y2=64y2+yz+z2=49z2+zx+x2=57\begin{cases} x^2 + xy + y^2 = 64 \\ y^2 + yz + z^2 = 49 \\ z^2 + zx + x^2 = 57 \end{cases} Then xyz3\sqrt[3]{xyz} can be expressed as m/nm/n , where mm and nn are relatively prime positive integers. Compute m+nm + n.

2020 BMT Individual 22

Three lights are placed horizontally on a line on the ceiling. All the lights are initially off. Every second, Neil picks one of the three lights uniformly at random to switch: if it is off, he switches it on; if it is on, he switches it off. When a light is switched, any lights directly to the left or right of that light also get turned on (if they were off) or off (if they were on). The expected number of lights that are on after Neil has flipped switches three times can be expressed in the form m/nm/ n , where mm and nn are relatively prime positive integers. Compute m+nm + n.
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2020 BMT Team 15

Consider a random string ss of 10202010^{2020} base-ten digits (there can be leading zeroes). We say a substring s′s' (which has no leading zeroes) is self-locating if s′s' appears in ss at index s′s' where the string is indexed at 1 1. For example the substring 1111 in the string “122352242411122352242411” is selflocating since the 1111th digit is 1 1 and the 1212th digit is 1 1. Let the expected number of self-locating substrings in s be GG. Compute ⌊G⌋\lfloor G \rfloor.

2020 BMT Individual 15

The graph of the degree 20212021 polynomial P(x)P(x), which has real coefficients and leading coefficient 1, meets the x-axis at the points (1,0)(1, 0), (2,0)(2, 0), (3,0)(3, 0) , ...... , (2020,0)(2020, 0) and nowhere else. The mean of all possible values of P(2021)P(2021) can be written in the form a!/ba!/b, where aa and bb are positive integers and aa is as small as possible. Compute a+ba + b.
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AI x IX wanted, 6-8-10 triangle, incenter circumcircle 2020 BMT Team 21

Let △ABC\vartriangle ABC be a right triangle with legs AB=6AB = 6 and AC=8AC = 8. Let II be the incenter of △ABC\vartriangle ABC and XX be the other intersection of AIAI with the circumcircle of △ABC\vartriangle ABC. Find AI‾⋅IX‾\overline{AI} \cdot \overline{IX}.

2020 BMT Individual 21

Let PP be the probability that the product of 20202020 real numbers chosen independently and uniformly at random from the interval [−1,2][-1, 2] is positive. The value of 2P−12P - 1 can be written in the form (mn)b\left(\frac{m}{n}\right)^b , where m,nm, n and bb are positive integers such that mm and nn are relatively prime and bb is as large as possible. Compute m+n+bm + n + b.
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2020 BMT Team 19

John is flipping his favorite bottle, which currently contains 1010 ounces of water. However, his bottle is broken from excessive flipping, so after he performs a flip, one ounce of water leaks out of his bottle. When his bottle contains k ounces of water, he has a 1k+1\frac{1}{k+1} probability of landing it on its bottom. What is the expected number of number of flips it takes for John’s bottle to land on its bottom ?

circular pillar in a large circular room 2020 BMT Individual 19

Alice is standing on the circumference of a large circular room of radius 1010. There is a circular pillar in the center of the room of radius 55 that blocks Alice’s view. The total area in the room Alice can see can be expressed in the form mπn+pq\frac{m\pi}{n} +p\sqrt{q}, where mm and nn are relatively prime positive integers and pp and qq are integers such that qq is square-free. Compute m+n+p+qm + n + p + q. (Note that the pillar is not included in the total area of the room.) https://cdn.artofproblemsolving.com/attachments/5/1/26e8aa6d12d9dd85bd5b284b6176870c7d11b1.png
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2020 BMT Team 13

Compute the expected sum of elements in a subset of {1,2,3,...,2020}\{1, 2, 3, . . . , 2020\} (including the empty set) chosen uniformly at random.

areas in a regular-hexagon-shaped 2020 BMT Individual 13

Sheila is making a regular-hexagon-shaped sign with side length 1 1. Let ABCDEFABCDEF be the regular hexagon, and let R,S,TR, S,T and U be the midpoints of FAFA, BCBC, CDCD and EFEF, respectively. Sheila splits the hexagon into four regions of equal width: trapezoids ABSRABSR, RSCFRSCF , FCTUFCTU, and UTDEUTDE. She then paints the middle two regions gold. The fraction of the total hexagon that is gold can be written in the form m/nm/n , where m and n are relatively prime positive integers. Compute m+nm + n. https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYS9lLzIwOTVmZmViZjU3OTMzZmRlMzFmMjM1ZWRmM2RkODMyMTA0ZjNlLnBuZw==&rn=MjAyMCBCTVQgSW5kaXZpZHVhbCAxMy5wbmc=
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2020 BMT Team 23

Let 0<θ<2π0 < \theta < 2\pi be a real number for which cos⁡(θ)+cos⁡(2θ)+cos⁡(3θ)+...+cos⁡(2020θ)=0\cos (\theta) + \cos (2\theta) + \cos (3\theta) + ...+ \cos (2020\theta) = 0 and θ=πn\theta =\frac{\pi}{n} for some positive integer nn. Compute the sum of the possible values of n≤2020n \le 2020.

AP/AQ wanted, diameter, circumcenter 2020 BMT Individual 23

Circle Γ\Gamma has radius 1010, center OO, and diameter ABAB. Point CC lies on Γ\Gamma such that AC=12AC = 12. Let PP be the circumcenter of △AOC\vartriangle AOC. Line APAP intersects Γ\Gamma at QQ, where QQ is different from AA. Then the value of APAQ\frac{AP}{AQ} can be expressed in the form mn\frac{m}{n}, where m and n are relatively prime positive integers. Compute m+nm + n.
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