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> | <asy> | ||
Line 37: | Line 37: | ||
label("$N$",N,dir(180)); </asy> | label("$N$",N,dir(180)); </asy> | ||
− | + | ==Solution 1== | |
− | ==Solution== | + | 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}.\Box</math> | ||
+ | |||
+ | ==Solution 2== | ||
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: | 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: | ||
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So our final answer is <math>(7\sqrt{11})^2 = \boxed{539}</math> | 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 == |
Revision as of 17:34, 21 November 2020
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 :
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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