Difference between revisions of "2015 AIME I Problems/Problem 7"
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==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 \~ \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>. | ||
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L=foot(M,H,G); | L=foot(M,H,G); | ||
draw(K--N--M--L); | draw(K--N--M--L); | ||
− | P=foot(E,M,L) | + | P=foot(E,M,L); |
− | label("P",P,E) | + | label("$P$",P,E); |
label("$A$",A,NW); | label("$A$",A,NW); | ||
label("$B$",B,SW); | label("$B$",B,SW); |
Revision as of 19:01, 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 $\triangle CDE \~ \triangle JFC$ (Error compiling LaTeX. ! Missing $ inserted.) (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:
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.