Difference between revisions of "2020 AMC 10B Problems"
Kevinmathz (talk | contribs) |
|||
Line 123: | Line 123: | ||
==Problem 21== | ==Problem 21== | ||
− | + | In square <math>ABCD</math>, points <math>E</math> and <math>H</math> lie on <math>\overline{AB}</math> and <math>\overline{DA}</math>, respectively, so that <math>AE=AH.</math> Points <math>F</math> and <math>G</math> lie on <math>\overline{BC}</math> and <math>\overline{CD}</math>, respectively, and points <math>I</math> and <math>J</math> lie on <math>\overline{EH}</math> so that <math>\overline{FI} \perp \overline{EH}</math> and <math>\overline{GJ} \perp \overline{EH}</math>. See the figure below. Triangle <math>AEH</math>, quadrilateral <math>BFIE</math>, quadrilateral <math>DHJG</math>, and pentagon <math>FCGJI</math> each has area <math>1.</math> What is <math>FI^2</math>? | |
+ | <asy> | ||
+ | real x=2sqrt(2); | ||
+ | real y=2sqrt(16-8sqrt(2))-4+2sqrt(2); | ||
+ | real z=2sqrt(8-4sqrt(2)); | ||
+ | pair A, B, C, D, E, F, G, H, I, J; | ||
+ | A = (0,0); | ||
+ | B = (4,0); | ||
+ | C = (4,4); | ||
+ | D = (0,4); | ||
+ | E = (x,0); | ||
+ | F = (4,y); | ||
+ | G = (y,4); | ||
+ | H = (0,x); | ||
+ | I = F + z * dir(225); | ||
+ | J = G + z * dir(225); | ||
+ | |||
+ | draw(A--B--C--D--A); | ||
+ | draw(H--E); | ||
+ | draw(J--G^^F--I); | ||
+ | draw(rightanglemark(G, J, I), linewidth(.5)); | ||
+ | draw(rightanglemark(F, I, E), linewidth(.5)); | ||
+ | |||
+ | dot("$A$", A, S); | ||
+ | dot("$B$", B, S); | ||
+ | dot("$C$", C, dir(90)); | ||
+ | dot("$D$", D, dir(90)); | ||
+ | dot("$E$", E, S); | ||
+ | dot("$F$", F, dir(0)); | ||
+ | dot("$G$", G, N); | ||
+ | dot("$H$", H, W); | ||
+ | dot("$I$", I, SW); | ||
+ | dot("$J$", J, SW); | ||
+ | |||
+ | </asy> | ||
+ | <math>\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2</math> | ||
[[2020 AMC 10B Problems/Problem 21|Solution]] | [[2020 AMC 10B Problems/Problem 21|Solution]] | ||
Line 129: | Line 164: | ||
==Problem 22== | ==Problem 22== | ||
− | + | What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>? | |
+ | |||
+ | <math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math> | ||
[[2020 AMC 10B Problems/Problem 22|Solution]] | [[2020 AMC 10B Problems/Problem 22|Solution]] | ||
Line 135: | Line 172: | ||
==Problem 23== | ==Problem 23== | ||
− | + | Square <math>ABCD</math> in the coordinate plane has vertices at the points <math>A(1,1), B(-1,1), C(-1,-1),</math> and <math>D(1,-1).</math> Consider the following four transformations: | |
+ | <math>L,</math> a rotation of <math>90^{\circ}</math> counterclockwise around the origin; | ||
+ | <math>R,</math> a rotation of <math>90^{\circ}</math> clockwise around the origin; | ||
+ | <math>H,</math> a reflection across the <math>x</math>-axis; and | ||
+ | <math>V,</math> a reflection across the <math>y</math>-axis. | ||
+ | Each of these transformations maps the squares onto itself, but the positions of the labeled vertices will change. For example, applying <math>R</math> and then <math>V</math> would send the vertex <math>A</math> at <math>(1,1)</math> to <math>(-1,-1)</math> and would send the vertex <math>B</math> at <math>(-1,1)</math> to itself. How many sequences of <math>20</math> transformations chosen from <math>\{L, R, H, V\}</math> will send all of the labeled vertices back to their original positions? (For example, <math>R, R, V, H</math> is one sequence of <math>4</math> transformations that will send the vertices back to their original positions.) | ||
+ | |||
+ | <math>\textbf{(A)}\ 2^{37} \qquad\textbf{(B)}\ 3\cdot 2^{36} \qquad\textbf{(C)}\ 2^{38} \qquad\textbf{(D)}\ 3\cdot 2^{37} \qquad\textbf{(E)}\ 2^{39}</math> | ||
[[2020 AMC 10B Problems/Problem 23|Solution]] | [[2020 AMC 10B Problems/Problem 23|Solution]] | ||
==Problem 24== | ==Problem 24== | ||
− | + | How many positive integers <math>n</math> satisfy<cmath>\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?</cmath>(Recall that <math>\lfloor x\rfloor</math> is the greatest integer not exceeding <math>x</math>.) | |
+ | |||
+ | <math>\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32</math> | ||
[[2020 AMC 10B Problems/Problem 24|Solution]] | [[2020 AMC 10B Problems/Problem 24|Solution]] | ||
Line 147: | Line 193: | ||
==Problem 25== | ==Problem 25== | ||
− | + | Let <math>D(n)</math> denote the number of ways of writing the positive integer <math>n</math> as a product<cmath>n = f_1\cdot f_2\cdots f_k,</cmath>where <math>k\ge1</math>, the <math>f_i</math> are integers strictly greater than <math>1</math>, and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number <math>6</math> can be written as <math>6</math>, <math>2\cdot 3</math>, and <math>3\cdot2</math>, so <math>D(6) = 3</math>. What is <math>D(96)</math>? | |
+ | |||
+ | <math>\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184</math> | ||
[[2020 AMC 10B Problems/Problem 25|Solution]] | [[2020 AMC 10B Problems/Problem 25|Solution]] |
Revision as of 15:10, 7 February 2020
2020 AMC 10B (Answer Key) Printable versions: • AoPS Resources • PDF | ||
Instructions
| ||
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 |
Contents
[hide]- 1 Problem 1
- 2 Problem 2
- 3 Problem 3
- 4 Problem 4
- 5 Problem 5
- 6 Problem 6
- 7 Problem 7
- 8 Problem 8
- 9 Problem 9
- 10 Problem 10
- 11 Problem 11
- 12 Problem 12
- 13 Problem 13
- 14 Problem 14
- 15 Problem 15
- 16 Problem 16
- 17 Problem 17
- 18 Problem 18
- 19 Problem 19
- 20 Problem 20
- 21 Problem 21
- 22 Problem 22
- 23 Problem 23
- 24 Problem 24
- 25 Problem 25
- 26 See also
Problem 1
These problems will not be available until the 2020 AMC 10B contest is released on Wednesday, February 5, 2020.
Problem 2
These problems will not be available until the 2020 AMC 10B contest is released on Wednesday, February 5, 2020.
Problem 3
These problems will not be available until the 2020 AMC 10B contest is released on Wednesday, February 5, 2020.
Problem 4
These problems will not be available until the 2020 AMC 10B contest is released on Wednesday, February 5, 2020.
Problem 5
These problems will not be available until the 2020 AMC 10B contest is released on Wednesday, February 5, 2020.
Problem 6
These problems will not be available until the 2020 AMC 10B contest is released on Wednesday, February 5, 2020.
Problem 7
These problems will not be available until the 2020 AMC 10B contest is released on Wednesday, February 5, 2020.
Problem 8
These problems will not be available until the 2020 AMC 10B contest is released on Wednesday, February 5, 2020.
Problem 9
These problems will not be available until the 2020 AMC 10B contest is released on Wednesday, February 5, 2020.
Problem 10
These problems will not be available until the 2020 AMC 10B contest is released on Wednesday, February 5, 2020.
Problem 11
These problems will not be available until the 2020 AMC 10B contest is released on Wednesday, February 5, 2020.
Problem 12
These problems will not be available until the 2020 AMC 10B contest is released on Wednesday, February 5, 2020.
Problem 13
These problems will not be available until the 2020 AMC 10B contest is released on Wednesday, February 5, 2020.
Problem 14
These problems will not be available until the 2020 AMC 10B contest is released on Wednesday, February 5, 2020.
Problem 15
These problems will not be available until the 2020 AMC 10B contest is released on Wednesday, February 5, 2020.
Problem 16
These problems will not be available until the 2020 AMC 10B contest is released on Wednesday, February 5, 2020.
Problem 17
These problems will not be available until the 2020 AMC 10B contest is released on Wednesday, February 5, 2020.
Problem 18
These problems will not be available until the 2020 AMC 10B contest is released on Wednesday, February 5, 2020.
Problem 19
These problems will not be available until the 2020 AMC 10B contest is released on Wednesday, February 5, 2020.
Problem 20
These problems will not be available until the 2020 AMC 10B contest is released on Wednesday, February 5, 2020.
Problem 21
In square , points and lie on and , respectively, so that Points and lie on and , respectively, and points and lie on so that and . See the figure below. Triangle , quadrilateral , quadrilateral , and pentagon each has area What is ?
Problem 22
What is the remainder when is divided by ?
Problem 23
Square in the coordinate plane has vertices at the points and Consider the following four transformations: a rotation of counterclockwise around the origin; a rotation of clockwise around the origin; a reflection across the -axis; and a reflection across the -axis.
Each of these transformations maps the squares onto itself, but the positions of the labeled vertices will change. For example, applying and then would send the vertex at to and would send the vertex at to itself. How many sequences of transformations chosen from will send all of the labeled vertices back to their original positions? (For example, is one sequence of transformations that will send the vertices back to their original positions.)
Problem 24
How many positive integers satisfy(Recall that is the greatest integer not exceeding .)
Problem 25
Let denote the number of ways of writing the positive integer as a productwhere , the are integers strictly greater than , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number can be written as , , and , so . What is ?
See also
2020 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by 2020 AMC 10A Problems |
Followed by 2021 AMC 10A Problems | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.