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Difference between revisions of "2015 AIME I Problems/Problem 7"
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==Problem==
+
==Problem ==
−
7. In the diagram below, <math>ABCD</math> is a square. Point <math>E</math> is the midpoint of <math>\overline{AD}</math>. Points <math>F</math> and <math>G</math> lie on <math>\overline{CE}</math>, and <math>H</math> and <math>J</math> lie on <math>\overline{AB}</math> and <math>\overline{BC}</math>, respectively, so that <math>FGHJ</math> is a square. Points <math>K</math> and <math>L</math> lie on <math>\overline{GH}</math>, and <math>M</math> and <math>N</math> lie on <math>\overline{AD}</math> and <math>\overline{AB}</math>, respectively, so that <math>KLMN</math> is a square. The area of <math>KLMN</math> is 99. Find the area of <math>FGHJ</math>.
+
In the diagram below, <math>ABCD</math> is a square. Point <math>E</math> is the midpoint of <math>\overline{AD}</math>. Points <math>F</math> and <math>G</math> lie on <math>\overline{CE}</math>, and <math>H</math> and <math>J</math> lie on <math>\overline{AB}</math> and <math>\overline{BC}</math>, respectively, so that <math>FGHJ</math> is a square. Points <math>K</math> and <math>L</math> lie on <math>\overline{GH}</math>, and <math>M</math> and <math>N</math> lie on <math>\overline{AD}</math> and <math>\overline{AB}</math>, respectively, so that <math>KLMN</math> is a square. The area of <math>KLMN</math> is 99. Find the area of <math>FGHJ</math>.
−
-------------
 
  
−
==Problem==
+
<asy>
 +
pair A,B,C,D,E,F,G,H,J,K,L,M,N;
 +
B=(0,0);
 +
real m=7*sqrt(55)/5;
 +
J=(m,0);
 +
C=(7*m/2,0);
 +
A=(0,7*m/2);
 +
D=(7*m/2,7*m/2);
 +
E=(A+D)/2;
 +
H=(0,2m);
 +
N=(0,2m+3*sqrt(55)/2);
 +
G=foot(H,E,C);
 +
F=foot(J,E,C);
 +
draw(A--B--C--D--cycle);
 +
draw(C--E);
 +
draw(G--H--J--F);
 +
pair X=foot(N,E,C);
 +
M=extension(N,X,A,D);
 +
K=foot(N,H,G);
 +
L=foot(M,H,G);
 +
draw(K--N--M--L);
 +
label("$A$",A,NW);
 +
label("$B$",B,SW);
 +
label("$C$",C,SE);
 +
label("$D$",D,NE);
 +
label("$E$",E,dir(90));
 +
label("$F$",F,NE);
 +
label("$G$",G,NE);
 +
label("$H$",H,W);
 +
label("$J$",J,S);
 +
label("$K$",K,SE);
 +
label("$L$",L,SE);
 +
label("$M$",M,dir(90));
 +
label("$N$",N,dir(180)); </asy>
  
−
We begin by denoting the length <math>ED</math> <math>a</math>, giving us <math>DC = 2a</math> and <math>EC = a\sqrt5</math>. Since angles <math>\angle DCE</math> and <math>\angle FCJ</math> are complimentary, we have that <math>\triangle CDE ~ \triangle JFC</math> (and similarly the rest of the triangles are <math>1-2-\sqrt5</math> triangles). We let the sidelength of <math>FGHJ</math> be <math>b</math>, giving us <math>JC = \sqrt5 * FC = \sqrt5 * FJ/2 = \frac{b\sqrt 5}{2}</math> and <math>BJ = \frac{1}{\sqrt5} * HJ = \frac{a}{\sqrt5}</math>.
+
==Solution 1==
 +
 
 +
Let us find the proportion of the side length of <math>KLMN</math> and <math>FJGH</math>. Let the side length of <math>KLMN=y</math> and the side length of <math>FJGH=x</math>.
 +
 
 +
<asy>
 +
pair A,B,C,D,E,F,G,H,J,K,L,M,N,P;
 +
B=(0,0);
 +
real m=7*sqrt(55)/5;
 +
J=(m,0);
 +
C=(7*m/2,0);
 +
A=(0,7*m/2);
 +
D=(7*m/2,7*m/2);
 +
E=(A+D)/2;
 +
H=(0,2m);
 +
N=(0,2m+3*sqrt(55)/2);
 +
G=foot(H,E,C);
 +
F=foot(J,E,C);
 +
draw(A--B--C--D--cycle);
 +
draw(C--E);
 +
draw(G--H--J--F);
 +
pair X=foot(N,E,C);
 +
M=extension(N,X,A,D);
 +
K=foot(N,H,G);
 +
L=foot(M,H,G);
 +
draw(K--N--M--L);
 +
label("$A$",A,NW);
 +
label("$B$",B,SW);
 +
label("$C$",C,SE);
 +
label("$D$",D,NE);
 +
label("$E$",E,dir(90));
 +
label("$F$",F,NE);
 +
label("$G$",G,NE);
 +
label("$H$",H,W);
 +
label("$J$",J,S);
 +
label("$K$",K,SE);
 +
label("$L$",L,SE);
 +
label("$M$",M,dir(90));
 +
label("$N$",N,dir(180)); </asy>
 +
 
 +
 
 +
Now, examine <math>BC</math>. We know <math>BC=BJ+JC</math>, and triangles <math>\Delta BHJ</math> and <math>\Delta JFC</math> are similar to <math>\Delta EDC</math> since they are <math>1-2-\sqrt{5}</math> triangles. Thus, we can rewrite <math>BC</math> in terms of the side length of <math>FJGH</math>.
 +
<cmath>BJ=\frac{1}{\sqrt{5}}HJ=\frac{x}{\sqrt{5}}=\frac{x\sqrt{5}}{5}, JC=\frac{\sqrt{5}}{2}JF=\frac{x\sqrt{5}}{2}\Rightarrow BC=\frac{7x\sqrt{5}}{10}</cmath>
 +
 
 +
Now examine <math>AB</math>. We can express this length in terms of <math>x,y</math> since <math>AB=AN+NH+HB</math>. By using similar triangles as in the first part, we have
 +
<cmath>AB=\frac{1}{\sqrt{5}}y+\frac{\sqrt{5}}{2}y+\frac{2}{\sqrt{5}}x</cmath>
 +
<cmath>AB=BC\Rightarrow \frac{7y\sqrt{5}}{10}+\frac{2x\sqrt{5}}{5}=\frac{7x\sqrt{5}}{10}\Rightarrow \frac{7y\sqrt{5}}{10}=\frac{3x\sqrt{5}}{10}\Rightarrow 7y=3x</cmath>
 +
 
 +
Now, it is trivial to see that <math>[FJGH]=\left(\frac{x}{y}\right)^2[KLMN]=\left(\frac{7}{3}\right)^2\cdot 99=\boxed{539}.</math>
 +
 
 +
==Solution 2==
 +
 
 +
<asy>
 +
pair A,B,C,D,E,F,G,H,J,K,L,M,N,P;
 +
B=(0,0);
 +
real m=7*sqrt(55)/5;
 +
J=(m,0);
 +
C=(7*m/2,0);
 +
A=(0,7*m/2);
 +
D=(7*m/2,7*m/2);
 +
E=(A+D)/2;
 +
H=(0,2m);
 +
N=(0,2m+3*sqrt(55)/2);
 +
G=foot(H,E,C);
 +
F=foot(J,E,C);
 +
draw(A--B--C--D--cycle);
 +
draw(C--E);
 +
draw(G--H--J--F);
 +
pair X=foot(N,E,C);
 +
M=extension(N,X,A,D);
 +
K=foot(N,H,G);
 +
L=foot(M,H,G);
 +
draw(K--N--M--L);
 +
P=foot(E,M,L);
 +
draw(P--E);
 +
label("$A$",A,NW);
 +
label("$B$",B,SW);
 +
label("$C$",C,SE);
 +
label("$D$",D,NE);
 +
label("$E$",E,dir(90));
 +
label("$F$",F,NE);
 +
label("$G$",G,NE);
 +
label("$H$",H,W);
 +
label("$J$",J,S);
 +
label("$K$",K,SE);
 +
label("$L$",L,SE);
 +
label("$M$",M,dir(90));
 +
label("$N$",N,dir(180));
 +
label("$P$",P,dir(235)); </asy>
 +
 
 +
We begin by denoting the length <math>ED</math> <math>a</math>, giving us <math>DC = 2a</math> and <math>EC = a\sqrt5</math>. Since angles <math>\angle DCE</math> and <math>\angle FCJ</math> are complementary, we have that <math>\triangle CDE \sim \triangle JFC</math> (and similarly the rest of the triangles are <math>1-2-\sqrt5</math> triangles). We let the sidelength of <math>FGHJ</math> be <math>b</math>, giving us:
 +
 
 +
<cmath>JC = \sqrt5 \cdot FC = \sqrt5 \cdot FJ/2 = \frac{b\sqrt 5}{2}</cmath>
 +
and
 +
<cmath>BJ = \frac{1}{\sqrt5} \cdot HJ = \frac{b}{\sqrt5}</cmath>
 +
 
 +
Since <math>BC = CJ + BJ</math>,
 +
 
 +
<cmath>2a = \frac{b\sqrt 5}{2} + \frac{b}{\sqrt5}</cmath>
 +
 
 +
Solving for <math>b</math> in terms of <math>a</math> yields <cmath>b = \frac{4a\sqrt5}{7}</cmath>
 +
 
 +
We now use the given that <math>[KLMN] = 99</math>, implying that <math>KL = LM = MN = NK = 3\sqrt{11}</math>. We also draw the perpendicular from <math>E</math> to <math>ML</math> and label the point of intersection <math>P</math> as in the diagram at the top
 +
 
 +
This gives that <cmath>AM = 2 \cdot AN = 2 \cdot \frac{3\sqrt{11}}{\sqrt5}</cmath>
 +
and <cmath>ME = \sqrt5 \cdot MP = \sqrt5 \cdot \frac{EP}{2} = \sqrt5 \cdot \frac{LG}{2} = \sqrt5 \cdot \frac{HG - HK - KL}{2} = \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}}{2}</cmath>
 +
 
 +
Since <math>AE</math> = <math>AM + ME</math>, we get
 +
 
 +
<cmath>2 \cdot \frac{3\sqrt{11}}{\sqrt5} + \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}}{2} = a</cmath>
 +
 
 +
<cmath>\Rightarrow 12\sqrt{11} + 5(\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}) = 2\sqrt5a</cmath>
 +
 
 +
<cmath>\Rightarrow \frac{-21}{2}\sqrt{11} + \frac{20a\sqrt5}{7} = 2\sqrt5a</cmath>
 +
 
 +
<cmath>\Rightarrow -21\sqrt{11} = 2\sqrt5a\frac{14 - 20}{7}</cmath>
 +
 
 +
<cmath>\Rightarrow \frac{49\sqrt{11}}{4} = \sqrt5a</cmath>
 +
 
 +
<cmath>\Rightarrow 7\sqrt{11} = \frac{4a\sqrt{5}}{7}</cmath>
 +
 
 +
So our final answer is <math>(7\sqrt{11})^2 = \boxed{539}</math>.
 +
 
 +
==Solution 3==
 +
This is a relatively quick solution but a fakesolve. We see that with a ruler, <math>KL = \frac{3}{2}</math> cm and <math>HG = \frac{7}{2}</math> cm. Thus if <math>KL</math> corresponds with an area of <math>99</math>, then <math>HG</math> (<math>FGHJ</math>'s area) would correspond with <math>99*(\frac{7}{3})^2 = \boxed{539}</math> - aops5234
 +
 
 +
== See also ==
 +
{{AIME box|year=2015|n=I|num-b=6|num-a=8}}
 +
{{MAA Notice}}
 +
[[Category:Introductory Geometry Problems]]

Latest revision as of 20:02, 9 November 2023

Problem

In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$. Points $F$ and $G$ lie on $\overline{CE}$, and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$, respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$, and $M$ and $N$ lie on $\overline{AD}$ and $\overline{AB}$, respectively, so that $KLMN$ is a square. The area of $KLMN$ is 99. Find the area of $FGHJ$.

[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,dir(90)); label("$F$",F,NE); label("$G$",G,NE); label("$H$",H,W); label("$J$",J,S); label("$K$",K,SE); label("$L$",L,SE); label("$M$",M,dir(90)); label("$N$",N,dir(180)); [/asy]

Solution 1

Let us find the proportion of the side length of $KLMN$ and $FJGH$. Let the side length of $KLMN=y$ and the side length of $FJGH=x$.

[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N,P; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,dir(90)); label("$F$",F,NE); label("$G$",G,NE); label("$H$",H,W); label("$J$",J,S); label("$K$",K,SE); label("$L$",L,SE); label("$M$",M,dir(90)); label("$N$",N,dir(180)); [/asy]


Now, examine $BC$. We know $BC=BJ+JC$, and triangles $\Delta BHJ$ and $\Delta JFC$ are similar to $\Delta EDC$ since they are $1-2-\sqrt{5}$ triangles. Thus, we can rewrite $BC$ in terms of the side length of $FJGH$. \[BJ=\frac{1}{\sqrt{5}}HJ=\frac{x}{\sqrt{5}}=\frac{x\sqrt{5}}{5}, JC=\frac{\sqrt{5}}{2}JF=\frac{x\sqrt{5}}{2}\Rightarrow BC=\frac{7x\sqrt{5}}{10}\]

Now examine $AB$. We can express this length in terms of $x,y$ since $AB=AN+NH+HB$. By using similar triangles as in the first part, we have \[AB=\frac{1}{\sqrt{5}}y+\frac{\sqrt{5}}{2}y+\frac{2}{\sqrt{5}}x\] \[AB=BC\Rightarrow \frac{7y\sqrt{5}}{10}+\frac{2x\sqrt{5}}{5}=\frac{7x\sqrt{5}}{10}\Rightarrow \frac{7y\sqrt{5}}{10}=\frac{3x\sqrt{5}}{10}\Rightarrow 7y=3x\]

Now, it is trivial to see that $[FJGH]=\left(\frac{x}{y}\right)^2[KLMN]=\left(\frac{7}{3}\right)^2\cdot 99=\boxed{539}.$

Solution 2

[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N,P; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); P=foot(E,M,L); draw(P--E); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,dir(90)); label("$F$",F,NE); label("$G$",G,NE); label("$H$",H,W); label("$J$",J,S); label("$K$",K,SE); label("$L$",L,SE); label("$M$",M,dir(90)); label("$N$",N,dir(180)); label("$P$",P,dir(235)); [/asy]

We begin by denoting the length $ED$ $a$, giving us $DC = 2a$ and $EC = a\sqrt5$. Since angles $\angle DCE$ and $\angle FCJ$ are complementary, we have that $\triangle CDE \sim \triangle JFC$ (and similarly the rest of the triangles are $1-2-\sqrt5$ triangles). We let the sidelength of $FGHJ$ be $b$, giving us:

\[JC = \sqrt5 \cdot FC = \sqrt5 \cdot FJ/2 = \frac{b\sqrt 5}{2}\] and \[BJ = \frac{1}{\sqrt5} \cdot HJ = \frac{b}{\sqrt5}\]

Since $BC = CJ + BJ$,

\[2a = \frac{b\sqrt 5}{2} + \frac{b}{\sqrt5}\]

Solving for $b$ in terms of $a$ yields \[b = \frac{4a\sqrt5}{7}\]

We now use the given that $[KLMN] = 99$, implying that $KL = LM = MN = NK = 3\sqrt{11}$. We also draw the perpendicular from $E$ to $ML$ and label the point of intersection $P$ as in the diagram at the top

This gives that \[AM = 2 \cdot AN = 2 \cdot \frac{3\sqrt{11}}{\sqrt5}\] and \[ME = \sqrt5 \cdot MP = \sqrt5 \cdot \frac{EP}{2} = \sqrt5 \cdot \frac{LG}{2} = \sqrt5 \cdot \frac{HG - HK - KL}{2} = \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}}{2}\]

Since $AE$ = $AM + ME$, we get

\[2 \cdot \frac{3\sqrt{11}}{\sqrt5} + \sqrt{5} \cdot \frac{\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}}{2} = a\]

\[\Rightarrow 12\sqrt{11} + 5(\frac{4a\sqrt5}{7} - \frac{9\sqrt{11}}{2}) = 2\sqrt5a\]

\[\Rightarrow \frac{-21}{2}\sqrt{11} + \frac{20a\sqrt5}{7} = 2\sqrt5a\]

\[\Rightarrow -21\sqrt{11} = 2\sqrt5a\frac{14 - 20}{7}\]

\[\Rightarrow \frac{49\sqrt{11}}{4} = \sqrt5a\]

\[\Rightarrow 7\sqrt{11} = \frac{4a\sqrt{5}}{7}\]

So our final answer is $(7\sqrt{11})^2 = \boxed{539}$.

Solution 3

This is a relatively quick solution but a fakesolve. We see that with a ruler, $KL = \frac{3}{2}$ cm and $HG = \frac{7}{2}$ cm. Thus if $KL$ corresponds with an area of $99$, then $HG$ ($FGHJ$'s area) would correspond with $99*(\frac{7}{3})^2 = \boxed{539}$ - aops5234

See also

2015 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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All AIME Problems and Solutions

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