Difference between revisions of "2015 AIME I Problems/Problem 7"
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− | ==Problem== | + | ==Problem == |
− | + | 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>. | |
− | |||
− | == | + | <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== | |
− | <math> | + | 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> | ||
− | We | + | 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> | ||
− | This gives that < | + | 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> |
− | and < | + | |
+ | ==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 | 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 == | == See also == |
Latest revision as of 21:02, 9 November 2023
Problem
In the diagram below, is a square. Point
is the midpoint of
. Points
and
lie on
, and
and
lie on
and
, respectively, so that
is a square. Points
and
lie on
, and
and
lie on
and
, respectively, so that
is a square. The area of
is 99. Find the area of
.
Solution 1
Let us find the proportion of the side length of and
. Let the side length of
and the side length of
.
Now, examine . We know
, and triangles
and
are similar to
since they are
triangles. Thus, we can rewrite
in terms of the side length of
.
Now examine . We can express this length in terms of
since
. By using similar triangles as in the first part, we have
Now, it is trivial to see that
Solution 2
We begin by denoting the length
, giving us
and
. Since angles
and
are complementary, we have that
(and similarly the rest of the triangles are
triangles). We let the sidelength of
be
, giving us:
and
Since ,
Solving for in terms of
yields
We now use the given that , implying that
. We also draw the perpendicular from
to
and label the point of intersection
as in the diagram at the top
This gives that
and
Since =
, we get
So our final answer is .
Solution 3
This is a relatively quick solution but a fakesolve. We see that with a ruler, cm and
cm. Thus if
corresponds with an area of
, then
(
's area) would correspond with
- aops5234
See also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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