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]]}}
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#redirect [[2020 AMC 12B Problems/Problem 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>
 
 
 
== Solutions ==
 
=== 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 ===
 
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)
 
 
 
=== 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
 
 
 
=== Video Solution ===
 
https://www.youtube.com/watch?v=AKJXB07Sat0&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=7 ~ MathEx
 
 
 
==Video Solution 2==
 
Solution starts at 3:09 timestamp in description
 
https://youtu.be/VZYe3Hu88OA
 
 
 
~IceMatrix
 
 
 
==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}}
 
{{MAA Notice}}
 

Revision as of 17:57, 9 December 2020

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