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Difference between revisions of "2020 AMC 12B Problems"

(→‎Problem 21)
(→‎Problem 21)
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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>?
 
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]
+
<asy>
 
real x=2sqrt(2);
 
real x=2sqrt(2);
 
real y=2sqrt(16-8sqrt(2))-4+2sqrt(2);
 
real y=2sqrt(16-8sqrt(2))-4+2sqrt(2);
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draw(rightanglemark(F, I, E), linewidth(.5));
 
draw(rightanglemark(F, I, E), linewidth(.5));
  
−
dot("<math>A</math>", A, S);
+
dot("$A$", A, S);
−
dot("<math>B</math>", B, S);
+
dot("$B$", B, S);
−
dot("<math>C</math>", C, dir(90));
+
dot("$C$", C, dir(90));
−
dot("<math>D</math>", D, dir(90));
+
dot("$D$", D, dir(90));
−
dot("<math>E</math>", E, S);
+
dot("$E$", E, S);
−
dot("<math>F</math>", F, dir(0));
+
dot("$F$", F, dir(0));
−
dot("<math>G</math>", G, N);
+
dot("$G$", G, N);
−
dot("<math>H</math>", H, W);
+
dot("$H$", H, W);
−
dot("<math>I</math>", I, SW);
+
dot("$I$", I, SW);
−
dot("<math>J</math>", J, SW);
+
dot("$J$", J, SW);
  
−
[/asy]
+
</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>
 
<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>
  

Revision as of 15:06, 7 February 2020

2020 AMC 12B (Answer Key)
Printable versions: Wiki • AoPS Resources • PDF

Instructions

  1. This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.
  2. You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer.
  3. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the test if before 2006. No problems on the test will require the use of a calculator).
  4. Figures are not necessarily drawn to scale.
  5. You will have 75 minutes working time to complete the test.
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

Problem 1

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 2

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 3

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 4

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 5

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 6

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 7

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 8

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 9

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 10

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 11

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 12

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 13

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 14

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 15

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 16

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 17

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 18

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 19

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 20

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 21

In square $ABCD$, points $E$ and $H$ lie on $\overline{AB}$ and $\overline{DA}$, respectively, so that $AE=AH.$ Points $F$ and $G$ lie on $\overline{BC}$ and $\overline{CD}$, respectively, and points $I$ and $J$ lie on $\overline{EH}$ so that $\overline{FI} \perp \overline{EH}$ and $\overline{GJ} \perp \overline{EH}$. See the figure below. Triangle $AEH$, quadrilateral $BFIE$, quadrilateral $DHJG$, and pentagon $FCGJI$ each has area $1.$ What is $FI^2$? [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] $\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2$

Solution

Problem 22

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 23

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 24

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

Problem 25

These problems will not be available until the 2020 AMC 12B contest is released on Wednesday, February 5, 2020.

Solution

See also

2020 AMC 12B (Problems • Answer Key • Resources)
Preceded by
2020 AMC 12A Problems 		Followed by
2021 AMC 12A 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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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