Difference between revisions of "2020 AMC 10B Problems/Problem 21"
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+ | {{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #21]] and [[2020 AMC 12B Problems|2020 AMC 12B #18]]}} | ||
+ | |||
==Problem== | ==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> | <asy> | ||
real x=2sqrt(2); | real x=2sqrt(2); | ||
Line 36: | Line 38: | ||
</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> | ||
− | ==Solution== | + | == 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 | + | 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 its area to be <math>3-2\sqrt{2}</math>. Now, we notice that <math>FIK</math> is also a right isosceles triangle (because <math>\angle EKB=45^\circ</math>) 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 | + | == Solution 2 == |
− | |||
− | |||
− | |||
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>. | 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>. | ||
Line 59: | Line 59: | ||
Solving for <math>y^2</math> gives <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math> ~DrB | Solving for <math>y^2</math> gives <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math> ~DrB | ||
+ | |||
+ | == Solution 3 == | ||
+ | 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{\textbf{(B)}\ 8-4\sqrt{2}}</cmath>. --OGBooger | ||
+ | |||
+ | == Solution 4 (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 | ||
+ | |||
+ | == Solution 5 == | ||
+ | |||
+ | <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 | ||
+ | |||
+ | == Solution 6 == | ||
+ | |||
+ | <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, K, L; | ||
+ | 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); | ||
+ | K = (4-x,4); | ||
+ | L = J + 1.68 * dir(45); | ||
+ | |||
+ | draw(A--B--C--D--A); | ||
+ | draw(H--E); | ||
+ | draw(J--G^^F--I); | ||
+ | draw(H--K,dashed+linewidth(.5)); | ||
+ | draw(L--K,dashed+linewidth(.5)); | ||
+ | draw(rightanglemark(G, J, I), linewidth(.5)); | ||
+ | draw(rightanglemark(F, I, E), linewidth(.5)); | ||
+ | draw(rightanglemark(H, K, L), linewidth(.5)); | ||
+ | draw(rightanglemark(K, L, G), 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); | ||
+ | dot("$K$", K, N); | ||
+ | dot("$L$", L, S); | ||
+ | </asy> | ||
+ | |||
+ | <math>[ABCD] = 4</math>, <math>AB = 2</math>, <math>[AHE] = 1</math>, <math>AH = AE = \sqrt{2}</math>, <math>DH = 2 - \sqrt{2}</math>, <math>JL = HK = \sqrt{2} \cdot DH = 2 \sqrt{2} - 2</math> | ||
+ | |||
+ | Because <math>ABCD</math> is a square and <math>AH = AE</math>, <math>AC</math> is the line of symmetry of pentagon <math>CDHEB</math>. Because <math>[DHJG] = [BFIE]</math>, <math>DHJG</math> is the reflection of <math>BFIE</math> about line <math>AC</math> | ||
+ | |||
+ | Let <math>FI = GJ = x</math>, <math>KL = LG = GJ - LJ = x - 2 \sqrt{2} + 2</math> | ||
+ | |||
+ | <math>[DHK] = \frac{(2 - \sqrt{2})^2}{2} = 3 - 2 \sqrt {2}</math> | ||
+ | |||
+ | <math>[GKL] = \frac{(x - 2 \sqrt{2} + 2)^2}{2} = \frac{x^2}{2} + 2x - 2x \sqrt{2} - 4 \sqrt{2} + 6</math> | ||
+ | |||
+ | <math>[HKJL] = (x - 2 \sqrt{2} + 2) \cdot (2 \sqrt{2} - 2) = 2x \sqrt{2} - 2x + 8 \sqrt{2} -12</math> | ||
+ | |||
+ | <cmath>[DHK] + [GKL] + [HKLJ] = [DHJG]</cmath> | ||
+ | |||
+ | <cmath>3 - 2 \sqrt {2} + \frac{x^2}{2} + 2x - 2x \sqrt{2} - 4 \sqrt{2} + 6 + 2x \sqrt{2} - 2x + 8 \sqrt{2} -12= 1</cmath> | ||
+ | |||
+ | <cmath>\frac{x^2}{2} + 2 \sqrt{2} - 4 = 0</cmath> | ||
+ | |||
+ | <cmath>x^2 = 8 - 4 \sqrt{2}</cmath> | ||
+ | |||
+ | <cmath>FI^2 = \boxed{\textbf{(B)}\ 8-4\sqrt{2}}</cmath> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==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== | ==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}} | {{MAA Notice}} |
Revision as of 11:38, 5 October 2022
- 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 its area to be . Now, we notice that is also a right isosceles triangle (because ) 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
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 3
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 4 (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
Solution 5
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
Solution 6
, , , , ,
Because is a square and , is the line of symmetry of pentagon . Because , is the reflection of about line
Let ,
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.