Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2020 AMC 10B Problems/Problem 21"

m (→‎Solution 5 (Basically Same as Solution 3))
m
 
(2 intermediate revisions by 2 users not shown)
Line 42: Line 42:
  
 
== Solution 1 ==
 
== 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 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.(This is a misplaced problem) That gives <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math>.~TLiu
+
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 ==
Line 131: Line 131:
 
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>.  
 
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=\text{area of} \: BFIE+\text{area of} \:FJ'H'B+\text{area of} \:EH'K = 1 + 1 + \frac{1}{2}(2\sqrt{2}-2)^2=8-4\sqrt{2}\rightarrow \boxed{\mathrm{(B)}}</math>
+
<math>FI^2=\text{area of} \: BFIE+\text{area of} \:FJ'H'B+\text{area of} \:EH'K \\= 1 + 1 + \frac{1}{2}(2\sqrt{2}-2)^2=8-4\sqrt{2}\rightarrow \boxed{\mathrm{(B)}}</math>
  
 
--Ryan Zhang @BRS
 
--Ryan Zhang @BRS
Line 211: Line 211:
 
~Dhillonr25
 
~Dhillonr25
  
 +
== Solution 8 ==
 +
 +
We extend <math>\overline{FB}</math> and <math>\overline{IE}</math> to meet at a point <math>X</math>. Since <math>\angle AEI = \angle BEX</math> and <math>FX</math> is parallel to <math>DA</math>, we know that <math>\triangle{BEX} \sim \triangle{AEH}</math>, and because <math>\angle BXE = \angle IXF</math> and <math>\angle XBE = \angle XIF</math>, we can conclude that <math>\triangle{BEX} \sim \triangle{AEH} \sim \triangle{IFX}</math>.
 +
 +
Now, because <math>\triangle{AEH}</math> is isosceles, right, and has an area of 1, we can conclude that <math>AE = AH = \sqrt{2}</math> and that <math>BE = 2-\sqrt{2}</math>. Armed with this knowledge, and setting <math>IF = a</math> and the area of <math>\triangle{BEX} = b</math>, we can use similarity to say that
 +
<cmath>(\frac{a}{2-\sqrt{2}})^2 = \frac{1+b}{b}</cmath>
 +
Since we know the side lengths of <math>\triangle{BEX}</math> due to the fact that it is also an isosceles right triangle, we know that the area is <math>\frac{(2-\sqrt{2})^2}{2}</math>.
 +
Simplifying further and plugging in values, we have
 +
<cmath>\frac{a^2}{(2-\sqrt{2})^2} = 1 + \frac{2}{(2-\sqrt{2})^2)}</cmath>
 +
Multiplying by <math>(2-\sqrt{2})^2</math> on both sides, we get
 +
<cmath>a^2 = (2-\sqrt{2})^2 + 2 = \boxed{\textbf{(B)}\ 8-4\sqrt{2}}</cmath>
 +
~yingkai_0_
 
==Video Solution (HOW TO THINK CREATIVELY!!!)==
 
==Video Solution (HOW TO THINK CREATIVELY!!!)==
 
https://youtu.be/oRvHHywcw4w
 
https://youtu.be/oRvHHywcw4w

Latest revision as of 16:52, 3 November 2024

The following problem is from both the 2020 AMC 10B #21 and 2020 AMC 12B #18, so both problems redirect to this page.

Problem

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 1

Since the total area is $4$, the side length of square $ABCD$ is $2$. We see that since triangle $HAE$ is a right isosceles triangle with area 1, we can determine sides $HA$ and $AE$ both to be $\sqrt{2}$. Now, consider extending $FB$ and $IE$ until they intersect. Let the point of intersection be $K$. We note that $EBK$ is also a right isosceles triangle with side $2-\sqrt{2}$ and find its area to be $3-2\sqrt{2}$. Now, we notice that $FIK$ is also a right isosceles triangle (because $\angle EKB=45^\circ$) and find it's area to be $\frac{1}{2}$$FI^2$. This is also equal to $1+3-2\sqrt{2}$ or $4-2\sqrt{2}$. Since we are looking for $FI^2$, we want two times this. That gives $\boxed{\textbf{(B)}\ 8-4\sqrt{2}}$.~TLiu

Solution 2

Draw the auxiliary line $AC$. Denote by $M$ the point it intersects with $HE$, and by $N$ the point it intersects with $GF$. Last, denote by $x$ the segment $FN$, and by $y$ the segment $FI$. We will find two equations for $x$ and $y$, and then solve for $y^2$.

Since the overall area of $ABCD$ is $4 \;\; \Longrightarrow \;\; AB=2$, and $AC=2\sqrt{2}$. In addition, the area of $\bigtriangleup AME = \frac{1}{2} \;\; \Longrightarrow \;\; AM=1$.

The two equations for $x$ and $y$ are then:

$\bullet$ Length of $AC$: $1+y+x = 2\sqrt{2} \;\; \Longrightarrow \;\; x = (2\sqrt{2}-1) - y$

$\bullet$ Area of CMIF: $\frac{1}{2}x^2+xy = \frac{1}{2} \;\; \Longrightarrow \;\; x(x+2y)=1$.

Substituting the first into the second, yields $\left[\left(2\sqrt{2}-1\right)-y\right]\cdot \left[\left(2\sqrt{2}-1\right)+y\right]=1$

Solving for $y^2$ gives $\boxed{\textbf{(B)}\ 8-4\sqrt{2}}$ ~DrB

Solution 3

Plot a point $F'$ such that $F'I$ and $AB$ are parallel and extend line $FB$ to point $B'$ such that $FIB'F'$ forms a square. Extend line $AE$ to meet line $F'B'$ and point $E'$ is the intersection of the two. The area of this square is equivalent to $FI^2$. We see that the area of square $ABCD$ is $4$, meaning each side is of length 2. The area of the pentagon $EIFF'E'$ is $2$. Length $AE=\sqrt{2}$, thus $EB=2-\sqrt{2}$. Triangle $EB'E'$ is isosceles, and the area of this triangle is $\frac{1}{2}(4-2\sqrt{2})(2-\sqrt{2})=6-4\sqrt{2}$. Adding these two areas, we get \[2+6-4\sqrt{2}=8-4\sqrt{2}\rightarrow \boxed{\textbf{(B)}\ 8-4\sqrt{2}}\]. --OGBooger

Solution 4 (HARD Calculation)

We can easily observe that the area of square $ABCD$ is 4 and its side length is 2 since all four regions that build up the square has area 1. Extend $FI$ and let the intersection with $AB$ be $K$. Connect $AC$, and let the intersection of $AC$ and $HE$ be $L$. Notice that since the area of triangle $AEH$ is 1 and $AE=AH$ , $AE=AH=\sqrt{2}$, therefore $BE=HD=2-\sqrt{2}$. Let $CG=CF=m$, then $BF=DG=2-m$. Also notice that $KB=2-m$, thus $KE=KB-BE=2-m-(2-\sqrt{2})=\sqrt{2}-m$. Now use the condition that the area of quadrilateral $BFIE$ is 1, we can set up the following equation: $\frac{1}{2}(2-m)^2-\frac{1}{4}(\sqrt{2}-m)^2=1$ We solve the equation and yield $m=\frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}$. Now notice that $FI=AC-AL-\frac{m}{\sqrt{2}}=2\sqrt{2}-1-\frac{\sqrt{2}}{2} \cdot \frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}$ $=2\sqrt{2}-1-\frac{8\sqrt{2}-4-\sqrt{128-64\sqrt2}}{4}$ $=\frac{\sqrt{128-64\sqrt{2}}}{4}$. Hence $FI^2=\frac{128-64\sqrt{2}}{16}=8-4\sqrt{2}$. -HarryW

Solution 5 (Basically Same as Solution 3)

[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 $BFIE$ and $DHJG$ are congruent with each other, so we can move $DHJG$ to the shaded area ($F$ and $G$, $B$ and $D$ overlapping) to form a square $FIKJ'$ ($DG$ = $FB$, $CG$ = $FC$, ${\angle} CGF$ = ${\angle}CFG$ = $45^{\circ}$ so ${\angle} IFJ'= 90^{\circ}$). Then we can solve $AH$ = $AE$ = $\sqrt{2}$, $EB$ = $2-\sqrt{2}$, $EK$ = $2\sqrt{2}-2$.

$FI^2=\text{area of} \: BFIE+\text{area of} \:FJ'H'B+\text{area of} \:EH'K \\= 1 + 1 + \frac{1}{2}(2\sqrt{2}-2)^2=8-4\sqrt{2}\rightarrow \boxed{\mathrm{(B)}}$

--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]

$[ABCD] = 4$, $AB = 2$, $[AHE] = 1$, $AH = AE = \sqrt{2}$, $DH = 2 - \sqrt{2}$, $JL = HK = \sqrt{2} \cdot DH = 2 \sqrt{2} - 2$

Because $ABCD$ is a square and $AH = AE$, $AC$ is the line of symmetry of pentagon $CDHEB$. Because $[DHJG] = [BFIE]$, $DHJG$ is the reflection of $BFIE$ about line $AC$

Let $FI = GJ = x$, $KL = LG = GJ - LJ = x - 2 \sqrt{2} + 2$

$[DHK] = \frac{(2 - \sqrt{2})^2}{2} = 3 - 2 \sqrt {2}$

$[GKL] = \frac{(x - 2 \sqrt{2} + 2)^2}{2} = \frac{x^2}{2} + 2x - 2x \sqrt{2} - 4 \sqrt{2} + 6$

$[HKJL] = (x - 2 \sqrt{2} + 2) \cdot (2 \sqrt{2} - 2) = 2x \sqrt{2} - 2x + 8 \sqrt{2} -12$

\[[DHK] + [GKL] + [HKLJ] = [DHJG]\]

\[3 - 2 \sqrt {2} + \frac{x^2}{2} + 2x - 2x \sqrt{2} - 4 \sqrt{2} + 6 + 2x \sqrt{2} - 2x + 8 \sqrt{2} -12= 1\]

\[\frac{x^2}{2} + 2 \sqrt{2} - 4 = 0\]

\[x^2 = 8 - 4 \sqrt{2}\]

\[FI^2 = \boxed{\textbf{(B)}\ 8-4\sqrt{2}}\]

~isabelchen

Solution 7 (Easy to See)

Note that the side length of $ABCD$ is 2 and thus the diagonal is of length $2\sqrt{2}$. However, the height to side $HE$ in triangle $HAE$ is 1, implying that $CM = 2\sqrt{2}-1$ where $M$ is the midpoint of $JI$. From here suppose that $N$ is the midpoint of $\overline{FG}$ and let $x = NC$, which means $FG=2x$. The area of the pentagon is then \[[FIJG]+[GCF]=GF \cdot FI + x^2 = (2x)(2\sqrt{2}-1-x)+x^2=1\] Solving this quadratic for $x$ yields $x=2\sqrt{2}-1 \pm \sqrt{8-4\sqrt{2}}$ (technically the smaller value is the correct one but it doesn’t matter for our purposes). We can then calculate $FI^2 = (2\sqrt{2} -1 -x)^2 = \boxed{\textbf{(B) } 8-4\sqrt{2}}$.

~Dhillonr25

Solution 8

We extend $\overline{FB}$ and $\overline{IE}$ to meet at a point $X$. Since $\angle AEI = \angle BEX$ and $FX$ is parallel to $DA$, we know that $\triangle{BEX} \sim \triangle{AEH}$, and because $\angle BXE = \angle IXF$ and $\angle XBE = \angle XIF$, we can conclude that $\triangle{BEX} \sim \triangle{AEH} \sim \triangle{IFX}$.

Now, because $\triangle{AEH}$ is isosceles, right, and has an area of 1, we can conclude that $AE = AH = \sqrt{2}$ and that $BE = 2-\sqrt{2}$. Armed with this knowledge, and setting $IF = a$ and the area of $\triangle{BEX} = b$, we can use similarity to say that \[(\frac{a}{2-\sqrt{2}})^2 = \frac{1+b}{b}\] Since we know the side lengths of $\triangle{BEX}$ due to the fact that it is also an isosceles right triangle, we know that the area is $\frac{(2-\sqrt{2})^2}{2}$. Simplifying further and plugging in values, we have \[\frac{a^2}{(2-\sqrt{2})^2} = 1 + \frac{2}{(2-\sqrt{2})^2)}\] Multiplying by $(2-\sqrt{2})^2$ on both sides, we get \[a^2 = (2-\sqrt{2})^2 + 2 = \boxed{\textbf{(B)}\ 8-4\sqrt{2}}\] ~yingkai_0_

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/oRvHHywcw4w

~Education, the Study of Everything

Video Solution by MathEx

https://www.youtube.com/watch?v=AKJXB07Sat0

Video Solution by TheBeautyOfMath

https://youtu.be/VZYe3Hu88OA?t=189

Really Good Vid Explanation

https://www.youtube.com/watch?v=AUndgrOH8U8&ab_channel=ReachTheStars

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_21&oldid=231413"