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]] | + | <div style="text-align:center;">[[Image:2007 AIME II-3.png]]</div> |
== Solution == | == Solution == | ||
| − | [[ | + | 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>. |
| − | + | <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= | + | {{AIME box|year=2007|n=II|num-b=2|num-a=4}} |
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
Revision as of 18:03, 29 March 2007
Problem
Square 
has side length 
, and points 
and 
are exterior to the square such that 
and 
. Find 
.
Solution
Extend 
and 
to their points of intersection. Since 
and are both 
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 and the angles are mostly complementary). Thus, we create a square with sides 
.
is the diagonal of the square, with length 
; the answer is 
.
See also
| 2007 AIME II (Problems • Answer Key • Resources) | ||
| 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 | ||
