Difference between revisions of "2007 AIME II Problems/Problem 3"

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[[Square]] <math>ABCD</math> has side length <math>13</math>, and [[point]]s <math>E</math> and <math>F</math> are exterior to the square such that <math>BE=DF=5</math> and <math>AE=CF=12</math>. Find <math>\displaystyle EF^{2}</math>.
 
[[Square]] <math>ABCD</math> has side length <math>13</math>, and [[point]]s <math>E</math> and <math>F</math> are exterior to the square such that <math>BE=DF=5</math> and <math>AE=CF=12</math>. Find <math>\displaystyle EF^{2}</math>.
  
[[Image:2007 AIME II-3.png]]
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<div style="text-align:center;">[[Image:2007 AIME II-3.png]]</div>
  
 
== Solution ==
 
== Solution ==
[[Image:2007 AIME II-3b.PNG]]
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Extend <math>\overline{AE}, \overline{DF}</math> and <math>\overline{BE}, \overline{CF}</math> to their points of intersection. Since <math>\triangle ABE \cong \triangle CDF</math> and are both <math>5-12-13</math> [[right triangle]]s, we can come to the conclusion that the two new triangles are also congruent to these two (use [[ASA]], as we know all the sides are <math>13</math> and the angles are mostly complementary). Thus, we create a [[square]] with sides <math>5 + 12 = 17</math>.
  
Extend <math>\overline{AE}, \overline{DF}</math> and <math>\overline{BE}, \overline{CF}</math> to their points of intersection. Since <math>\triangle ABE \cong \triangle CDF</math> and are both <math>5-12-13</math> [[right triangle]]s, we can come to the conclusion that the two new triangles are also congruent to these two (use [[ASA]], as we know all the sides are <math>13</math> and the angles are mostly complementary). Thus, we create a [[square]] with sides <math>5 + 12 = 17</math>.
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<div style='text-align:center;'>[[Image:2007 AIME II-3b.PNG]]</div>
  
 
<math>\overline{EF}</math> is the diagonal of the square, with length <math>17\sqrt{2}</math>; the answer is <math>EF^2 = (17\sqrt{2})^2 = 578</math>.
 
<math>\overline{EF}</math> is the diagonal of the square, with length <math>17\sqrt{2}</math>; the answer is <math>EF^2 = (17\sqrt{2})^2 = 578</math>.
  
 
== See also ==
 
== See also ==
{{AIME box|year=2007|n=II|num-b=1|num-a=3}}
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{{AIME box|year=2007|n=II|num-b=2|num-a=4}}
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 19:03, 29 March 2007

Problem

Square $ABCD$ has side length $13$, and points $E$ and $F$ are exterior to the square such that $BE=DF=5$ and $AE=CF=12$. Find $\displaystyle EF^{2}$.

2007 AIME II-3.png

Solution

Extend $\overline{AE}, \overline{DF}$ and $\overline{BE}, \overline{CF}$ to their points of intersection. Since $\triangle ABE \cong \triangle CDF$ and are both $5-12-13$ right triangles, we can come to the conclusion that the two new triangles are also congruent to these two (use ASA, as we know all the sides are $13$ and the angles are mostly complementary). Thus, we create a square with sides $5 + 12 = 17$.

2007 AIME II-3b.PNG

$\overline{EF}$ is the diagonal of the square, with length $17\sqrt{2}$; the answer is $EF^2 = (17\sqrt{2})^2 = 578$.

See also

2007 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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