Difference between revisions of "2015 AIME I Problems/Problem 7"
(→Problem) |
|||
Line 41: | Line 41: | ||
==Solution== | ==Solution== | ||
− | 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 | + | 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 \similar \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 \cdot FC = \sqrt5 \cdot FJ/2 = \frac{b\sqrt 5}{2}</math> and <math>BJ = \frac{1}{\sqrt5} \cdot HJ = \frac{b}{\sqrt5}</math>. | <math>JC = \sqrt5 \cdot FC = \sqrt5 \cdot FJ/2 = \frac{b\sqrt 5}{2}</math> and <math>BJ = \frac{1}{\sqrt5} \cdot HJ = \frac{b}{\sqrt5}</math>. | ||
Line 51: | Line 51: | ||
Solving for <math>b</math> in terms of <math>a</math> yields <math>b = \frac{4a\sqrt5}{7}</math>. | Solving for <math>b</math> in terms of <math>a</math> yields <math>b = \frac{4a\sqrt5}{7}</math>. | ||
− | 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 E to ML and label the point of intersection P | + | 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 E to ML and label the point of intersection P: |
+ | |||
+ | <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); | ||
+ | P=foot(E,M,L) | ||
+ | label("P",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)); </asy> | ||
This gives that <math>AM = 2 \cdot AN = 2 \cdot \frac{3\sqrt{11}}{\sqrt5}</math> | This gives that <math>AM = 2 \cdot AN = 2 \cdot \frac{3\sqrt{11}}{\sqrt5}</math> |
Revision as of 19:00, 21 March 2015
Problem
7. 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
We begin by denoting the length , giving us and . Since angles and are complimentary, we have that $\triangle CDE \similar \triangle JFC$ (Error compiling LaTeX. ! Undefined control sequence.) (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 E to ML and label the point of intersection P:
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); P=foot(E,M,L) label("P",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)); (Error compiling LaTeX. label("P",P,E) ^ 7a023ad7e27f5ed1c7813ddb2ed5dce918789de8.asy: 26.1: syntax error error: could not load module '7a023ad7e27f5ed1c7813ddb2ed5dce918789de8.asy')
This gives that and
Since = , we get
So our final answer is
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.