Difference between revisions of "2007 AIME II Problems/Problem 3"
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== Solution == | == Solution == | ||
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| + | == '''Solution 1.''' == | ||
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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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<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>. | ||
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| + | == '''Solution 2.''' == | ||
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| + | A slightly more analytic/brute-force approach: | ||
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| + | Drop perpendiculars from <math>E</math> and <math>F</math> to <math>I</math> and <math>J</math>, respectively; construct right triangle <math>EKF</math> with right angle at K and <math>EK || BC</math>. Since <math>2[CDF]=DF*CF=CD*JF</math>, we have <math>JF=5\times12/13 = \frac{60}{13}</math>. Similarly, <math>EI=\frac{60}{13}</math>. Since <math>\triangle DJF \sim \triangle DFC</math>, we have <math>DJ=\frac{5JF}{12}=\frac{25}{13}</math>. | ||
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| + | Now, we see that <math>FK=DC-(DJ+IB)=DC-2DJ=13-\frac{50}{13}=\frac{119}{13}</math>. Also, <math>EK=BC+(JF+IE)=BC+2JF=13+\frac{120}{13}=\frac{289}{13}</math>. By the Pythagorean Theorem, we have <math>EF=\sqrt{\left(\frac{289}{13}\right)^2+\left(\frac{119}{13} \right)^2}=\frac{\sqrt{(17^2)(17^2+7^2)}}{13}=\frac{17\sqrt{338}}{13}=\frac{17(13\sqrt{2})}{13}=17\sqrt{2}</math>. Therefore, <math>EF^2=(17\sqrt{2})^2=578</math>. | ||
== See also == | == See also == | ||
Revision as of 18:27, 31 March 2007
Problem
Square 
has side length 
, and points 
and 
are exterior to the square such that 
and 
. Find 
.
Solution
Solution 1.
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 
.
Solution 2.
A slightly more analytic/brute-force approach:
Drop perpendiculars from and to 
and 
, respectively; construct right triangle 
with right angle at K and 
. Since ![$2[CDF]=DF*CF=CD*JF$](http://latex.artofproblemsolving.com/7/4/e/74e74fabd72a3055573cdae880ca96865e528909.png)
, we have 
. Similarly, 
. Since 
, we have 
.
Now, we see that 
. Also, 
. By the Pythagorean Theorem, we have 
. Therefore, 
.
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 | ||
