Difference between revisions of "2020 AMC 10B Problems/Problem 21"
Sugar rush (talk | contribs) (Redirected page to 2020 AMC 12B Problems/Problem 18) (Tag: New redirect) |
MRENTHUSIASM (talk | contribs) m |
||
(One intermediate revision by the same user not shown) | |||
Line 1: | Line 1: | ||
− | # | + | {{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #21]] and [[2020 AMC 12B Problems|2020 AMC 12B #18]]}} |
+ | |||
+ | ==Problem== | ||
+ | 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> | ||
+ | |||
+ | == Solution 1 == | ||
+ | Since the total area is <math>4</math>, the side length of square <math>ABCD</math> is <math>2</math>. We see that since triangle <math>HAE</math> is a right isosceles triangle with area 1, we can determine sides <math>HA</math> and <math>AE</math> both to be <math>\sqrt{2}</math>. Now, consider extending <math>FB</math> and <math>IE</math> until they intersect. Let the point of intersection be <math>K</math>. We note that <math>EBK</math> is also a right isosceles triangle with side <math>2-\sqrt{2}</math> and find it's area to be <math>3-2\sqrt{2}</math>. Now, we notice that <math>FIK</math> is also a right isosceles triangle and find it's area to be <math>\frac{1}{2}</math><math>FI^2</math>. This is also equal to <math>1+3-2\sqrt{2}</math> or <math>4-2\sqrt{2}</math>. Since we are looking for <math>FI^2</math>, we want two times this. That gives <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math>.~TLiu | ||
+ | |||
+ | == Solution 2 (Lucky Measuring) == | ||
+ | Since this is a geometry problem involving sides, and we know that <math>HE</math> is <math>2</math>, we can use our ruler and find the ratio between <math>FI</math> and <math>HE</math>. Measuring(on the booklet), we get that <math>HE</math> is about <math>1.8</math> inches and <math>FI</math> is about <math>1.4</math> inches. Thus, we can then multiply the length of <math>HE</math> by the ratio of <math>\frac{1.4}{1.8},</math> of which we then get <math>FI= \frac{14}{9}.</math> We take the square of that and get <math>\frac{196}{81},</math> and the closest answer to that is <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math>. ~Celloboy (Note that this is just a strategy I happened to use that worked. Do not press your luck with this strategy, for it was a lucky guess) | ||
+ | This cannot work if the problem says not to scale - awu2014 | ||
+ | |||
+ | Note that this will only work if the diagram is to scale, and at the start of the test, they mention that all diagrams are not necessarily to scale (whether or not the problem states that). Therefore, if you are to use this strategy on a problem, you are betting on the fact that this diagram IS to scale, so only use it as a last resort. | ||
+ | |||
+ | == Solution 3 == | ||
+ | Draw the auxiliary line <math>AC</math>. Denote by <math>M</math> the point it intersects with <math>HE</math>, and by <math>N</math> the point it intersects with <math>GF</math>. Last, denote by <math>x</math> the segment <math>FN</math>, and by <math>y</math> the segment <math>FI</math>. We will find two equations for <math>x</math> and <math>y</math>, and then solve for <math>y^2</math>. | ||
+ | |||
+ | Since the overall area of <math>ABCD</math> is <math>4 \;\; \Longrightarrow \;\; AB=2</math>, and <math>AC=2\sqrt{2}</math>. In addition, the area of <math>\bigtriangleup AME = \frac{1}{2} \;\; \Longrightarrow \;\; AM=1</math>. | ||
+ | |||
+ | The two equations for <math>x</math> and <math>y</math> are then: | ||
+ | |||
+ | <math>\bullet</math> Length of <math>AC</math>: <math>1+y+x = 2\sqrt{2} \;\; \Longrightarrow \;\; x = (2\sqrt{2}-1) - y</math> | ||
+ | |||
+ | <math>\bullet</math> Area of CMIF: <math>\frac{1}{2}x^2+xy = \frac{1}{2} \;\; \Longrightarrow \;\; x(x+2y)=1</math>. | ||
+ | |||
+ | Substituting the first into the second, yields | ||
+ | <math>\left[\left(2\sqrt{2}-1\right)-y\right]\cdot \left[\left(2\sqrt{2}-1\right)+y\right]=1</math> | ||
+ | |||
+ | Solving for <math>y^2</math> gives <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math> ~DrB | ||
+ | |||
+ | == Solution 4 == | ||
+ | Plot a point <math>F'</math> such that <math>F'I</math> and <math>AB</math> are parallel and extend line <math>FB</math> to point <math>B'</math> such that <math>FIB'F'</math> forms a square. Extend line <math>AE</math> to meet line <math>F'B'</math> and point <math>E'</math> is the intersection of the two. The area of this square is equivalent to <math>FI^2</math>. We see that the area of square <math>ABCD</math> is <math>4</math>, meaning each side is of length 2. The area of the pentagon <math>EIFF'E'</math> is <math>2</math>. Length <math>AE=\sqrt{2}</math>, thus <math>EB=2-\sqrt{2}</math>. Triangle <math>EB'E'</math> is isosceles, and the area of this triangle is <math>\frac{1}{2}(4-2\sqrt{2})(2-\sqrt{2})=6-4\sqrt{2}</math>. Adding these two areas, we get <cmath>2+6-4\sqrt{2}=8-4\sqrt{2}\rightarrow \boxed{\mathrm{(B)}}</cmath>. --OGBooger | ||
+ | |||
+ | == Solution 5 (HARD Calculation) == | ||
+ | We can easily observe that the area of square <math>ABCD</math> is 4 and its side length is 2 since all four regions that build up the square has area 1. | ||
+ | Extend <math>FI</math> and let the intersection with <math>AB</math> be <math>K</math>. Connect <math>AC</math>, and let the intersection of <math>AC</math> and <math>HE</math> be <math>L</math>. | ||
+ | Notice that since the area of triangle <math>AEH</math> is 1 and <math>AE=AH</math> , <math>AE=AH=\sqrt{2}</math>, therefore <math>BE=HD=2-\sqrt{2}</math>. | ||
+ | Let <math>CG=CF=m</math>, then <math>BF=DG=2-m</math>. | ||
+ | Also notice that <math>KB=2-m</math>, thus <math>KE=KB-BE=2-m-(2-\sqrt{2})=\sqrt{2}-m</math>. | ||
+ | Now use the condition that the area of quadrilateral <math>BFIE</math> is 1, we can set up the following equation: | ||
+ | <math>\frac{1}{2}(2-m)^2-\frac{1}{4}(\sqrt{2}-m)^2=1</math> | ||
+ | We solve the equation and yield <math>m=\frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}</math>. | ||
+ | Now notice that | ||
+ | <math>FI=AC-AL-\frac{m}{\sqrt{2}}=2\sqrt{2}-1-\frac{\sqrt{2}}{2}*\frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}</math> | ||
+ | <math>=2\sqrt{2}-1-\frac{8\sqrt{2}-4-\sqrt{128-64\sqrt2}}{4}</math> | ||
+ | <math>=\frac{\sqrt{128-64\sqrt{2}}}{4}</math>. | ||
+ | Hence <math>FI^2=\frac{128-64\sqrt{2}}{16}=8-4\sqrt{2}</math>. -HarryW | ||
+ | |||
+ | -edit: annabelle0913 | ||
+ | |||
+ | == Solution 6 == | ||
+ | |||
+ | <asy> | ||
+ | real x=2sqrt(2); | ||
+ | real y=2sqrt(16-8sqrt(2))-4+2sqrt(2); | ||
+ | real z=2sqrt(8-4sqrt(2)); | ||
+ | real k= 8-2sqrt(2); | ||
+ | real l= 2sqrt(2)-4; | ||
+ | pair A, B, C, D, E, F, G, H, I, J, L, M, K; | ||
+ | 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); | ||
+ | L = (k,0); | ||
+ | M = F + z * dir(315); | ||
+ | K = (4,l); | ||
+ | |||
+ | draw(A--B--C--D--A); | ||
+ | draw(H--E); | ||
+ | draw(J--G^^F--I); | ||
+ | draw(F--M); | ||
+ | draw(M--L); | ||
+ | draw(E--K,dashed+linewidth(.5)); | ||
+ | draw(K--L,dashed+linewidth(.5)); | ||
+ | draw(B--L); | ||
+ | draw(rightanglemark(G, J, I), linewidth(.5)); | ||
+ | draw(rightanglemark(F, I, E), linewidth(.5)); | ||
+ | draw(rightanglemark(F, M, L), linewidth(.5)); | ||
+ | fill((4,0)--(k,0)--M--(4,y)--cycle, gray); | ||
+ | dot("$A$", A, S); | ||
+ | dot("$C$", C, dir(90)); | ||
+ | dot("$D$", D, dir(90)); | ||
+ | dot("$E$", E, S); | ||
+ | dot("$G$", G, N); | ||
+ | dot("$H$", H, W); | ||
+ | dot("$I$", I, SW); | ||
+ | dot("$J$", J, SW); | ||
+ | dot("$K$", K, S); | ||
+ | dot("$F(G)$", F, E); | ||
+ | dot("$J'$", M, dir(90)); | ||
+ | dot("$H'$", L, S); | ||
+ | dot("$B(D)$", B, S); | ||
+ | |||
+ | |||
+ | </asy> | ||
+ | Easily, we can find that: quadrilateral <math>BFIE</math> and <math>DHJG</math> are congruent with each other, so we can move <math>DHJG</math> to the shaded area (<math>F</math> and <math>G</math>, <math>B</math> and <math>D</math> overlapping) to form a square <math>FIKJ'</math> (<math>DG</math> = <math>FB</math>, <math>CG</math> = <math>FC</math>, <math>{\angle} CGF</math> = <math>{\angle}CFG</math> = <math>45^{\circ}</math> so <math>{\angle} IFJ'= 90^{\circ}</math>). Then we can solve <math>AH</math> = <math>AE</math> = <math>\sqrt{2}</math>, <math>EB</math> = <math>2-\sqrt{2}</math>, <math>EK</math> = <math>2\sqrt{2}-2</math>. | ||
+ | |||
+ | <math>FI^2</math> = <math>area</math> of <math>BFIE</math> <math>+</math> <math>area</math> of <math>FJ'H'B</math> <math>+</math> <math>area</math> of <math>EH'K</math> = <math>1 + 1 + \frac{1}{2}(2\sqrt{2}-2)^2=8-4\sqrt{2}\rightarrow \boxed{\mathrm{(B)}}</math> | ||
+ | |||
+ | --Ryan Zhang @BRS | ||
+ | |||
+ | ==Video Solution 1== | ||
+ | https://www.youtube.com/watch?v=AKJXB07Sat0&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=7 ~ MathEx | ||
+ | |||
+ | ==Video Solution 2 by the Beauty of Math== | ||
+ | https://youtu.be/VZYe3Hu88OA?t=189 | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2020|ab=B|num-b=20|num-a=22}} | ||
+ | {{AMC12 box|year=2020|ab=B|num-b=17|num-a=19}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 11:24, 9 May 2021
- The following problem is from both the 2020 AMC 10B #21 and 2020 AMC 12B #18, so both problems redirect to this page.
Contents
Problem
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 ?
Solution 1
Since the total area is , the side length of square is . We see that since triangle is a right isosceles triangle with area 1, we can determine sides and both to be . Now, consider extending and until they intersect. Let the point of intersection be . We note that is also a right isosceles triangle with side and find it's area to be . Now, we notice that is also a right isosceles triangle and find it's area to be . This is also equal to or . Since we are looking for , we want two times this. That gives .~TLiu
Solution 2 (Lucky Measuring)
Since this is a geometry problem involving sides, and we know that is , we can use our ruler and find the ratio between and . Measuring(on the booklet), we get that is about inches and is about inches. Thus, we can then multiply the length of by the ratio of of which we then get We take the square of that and get and the closest answer to that is . ~Celloboy (Note that this is just a strategy I happened to use that worked. Do not press your luck with this strategy, for it was a lucky guess) This cannot work if the problem says not to scale - awu2014
Note that this will only work if the diagram is to scale, and at the start of the test, they mention that all diagrams are not necessarily to scale (whether or not the problem states that). Therefore, if you are to use this strategy on a problem, you are betting on the fact that this diagram IS to scale, so only use it as a last resort.
Solution 3
Draw the auxiliary line . Denote by the point it intersects with , and by the point it intersects with . Last, denote by the segment , and by the segment . We will find two equations for and , and then solve for .
Since the overall area of is , and . In addition, the area of .
The two equations for and are then:
Length of :
Area of CMIF: .
Substituting the first into the second, yields
Solving for gives ~DrB
Solution 4
Plot a point such that and are parallel and extend line to point such that forms a square. Extend line to meet line and point is the intersection of the two. The area of this square is equivalent to . We see that the area of square is , meaning each side is of length 2. The area of the pentagon is . Length , thus . Triangle is isosceles, and the area of this triangle is . Adding these two areas, we get . --OGBooger
Solution 5 (HARD Calculation)
We can easily observe that the area of square is 4 and its side length is 2 since all four regions that build up the square has area 1. Extend and let the intersection with be . Connect , and let the intersection of and be . Notice that since the area of triangle is 1 and , , therefore . Let , then . Also notice that , thus . Now use the condition that the area of quadrilateral is 1, we can set up the following equation: We solve the equation and yield . Now notice that . Hence . -HarryW
-edit: annabelle0913
Solution 6
Easily, we can find that: quadrilateral and are congruent with each other, so we can move to the shaded area ( and , and overlapping) to form a square ( = , = , = = so ). Then we can solve = = , = , = .
= of of of =
--Ryan Zhang @BRS
Video Solution 1
https://www.youtube.com/watch?v=AKJXB07Sat0&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=7 ~ MathEx
Video Solution 2 by the Beauty of Math
https://youtu.be/VZYe3Hu88OA?t=189
See Also
2020 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 |
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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.