Difference between revisions of "2020 AMC 10B Problems/Problem 21"
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==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 38: | 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.(This is a misplaced problem) 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 62: | Line 60: | ||
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 | + | == 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. | 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>. | 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>. | 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= | + | Let <math>CG=CF=m</math>, then <math>BF=DG=2-m</math>. |
− | Also notice that <math>KB | + | 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: | 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> | <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>. | We solve the equation and yield <math>m=\frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}</math>. | ||
Now notice that | Now notice that | ||
− | <math>FI=AC-AL=2\sqrt{2}-1-\frac{\sqrt{2}}{2} | + | <math>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}</math> |
<math>=2\sqrt{2}-1-\frac{8\sqrt{2}-4-\sqrt{128-64\sqrt2}}{4}</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>. | <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 | Hence <math>FI^2=\frac{128-64\sqrt{2}}{16}=8-4\sqrt{2}</math>. -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 <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> | ||
+ | |||
+ | --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] | ||
+ | |||
+ | == Solution 7 (Easy to See) == | ||
+ | |||
+ | Note that the side length of <math>ABCD</math> is 2 and thus the diagonal is of length <math>2\sqrt{2}</math>. However, the height to side <math>HE</math> in triangle <math>HAE</math> is 1, implying that <math>CM = 2\sqrt{2}-1</math> where <math>M</math> is the midpoint of <math>JI</math>. From here suppose that <math>N</math> is the midpoint of <math>\overline{FG}</math> and let <math>x = NC</math>, which means <math>FG=2x</math>. The area of the pentagon is then | ||
+ | <cmath>[FIJG]+[GCF]=GF \cdot FI + x^2 = (2x)(2\sqrt{2}-1-x)+x^2=1</cmath> | ||
+ | Solving this quadratic for <math>x</math> yields <math>x=2\sqrt{2}-1 \pm \sqrt{8-4\sqrt{2}}</math> (technically the smaller value is the correct one but it doesn’t matter for our purposes). We can then calculate <math>FI^2 = (2\sqrt{2} -1 -x)^2 = \boxed{\textbf{(B) } 8-4\sqrt{2}}</math>. | ||
+ | |||
+ | ~Dhillonr25 | ||
+ | |||
+ | ==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== | ==See Also== | ||
− | |||
{{AMC10 box|year=2020|ab=B|num-b=20|num-a=22}} | {{AMC10 box|year=2020|ab=B|num-b=20|num-a=22}} | ||
{{AMC12 box|year=2020|ab=B|num-b=17|num-a=19}} | {{AMC12 box|year=2020|ab=B|num-b=17|num-a=19}} | ||
+ | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 23:33, 27 March 2024
- The following problem is from both the 2020 AMC 10B #21 and 2020 AMC 12B #18, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4 (HARD Calculation)
- 6 Solution 5 (Basically Same as Solution 3)
- 7 Solution 6
- 8 Solution 7 (Easy to See)
- 9 Video Solution (HOW TO THINK CREATIVELY!!!)
- 10 Video Solution by MathEx
- 11 Video Solution by TheBeautyOfMath
- 12 Really Good Vid Explanation
- 13 See Also
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.(This is a misplaced problem) 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 (Basically Same as Solution 3)
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
=
=
,
=
,
=
.
--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 ,
Solution 7 (Easy to See)
Note that the side length of is 2 and thus the diagonal is of length
. However, the height to side
in triangle
is 1, implying that
where
is the midpoint of
. From here suppose that
is the midpoint of
and let
, which means
. The area of the pentagon is then
Solving this quadratic for
yields
(technically the smaller value is the correct one but it doesn’t matter for our purposes). We can then calculate
.
~Dhillonr25
Video Solution (HOW TO THINK CREATIVELY!!!)
~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.