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> | ||
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label("$M$",M,dir(90)); | label("$M$",M,dir(90)); | ||
label("$N$",N,dir(180)); </asy> | label("$N$",N,dir(180)); </asy> | ||
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==Solution 1== | ==Solution 1== | ||
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label("$L$",L,SE); | label("$L$",L,SE); | ||
label("$M$",M,dir(90)); | label("$M$",M,dir(90)); | ||
− | label("$N$",N,dir(180 | + | label("$N$",N,dir(180)); </asy> |
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<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> | <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 | + | 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== | ==Solution 2== | ||
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<asy> | <asy> | ||
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label("$P$",P,dir(235)); </asy> | label("$P$",P,dir(235)); </asy> | ||
− | This gives that < | + | 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: |
− | and < | + | |
+ | <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>, | ||
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+ | <cmath>2a = \frac{b\sqrt 5}{2} + \frac{b}{\sqrt5}</cmath> | ||
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+ | Solving for <math>b</math> in terms of <math>a</math> yields <cmath>b = \frac{4a\sqrt5}{7}</cmath> | ||
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+ | 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 | ||
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+ | 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> |
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+ | <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> |
− | <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 == |
Latest revision as of 20:02, 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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