Difference between revisions of "2007 AIME II Problems/Problem 3"
m (fmt a bit) |
|||
Line 4: | Line 4: | ||
<div style="text-align:center;">[[Image:2007 AIME II-3.png]]</div> | <div style="text-align:center;">[[Image:2007 AIME II-3.png]]</div> | ||
+ | __TOC__ | ||
== Solution == | == Solution == | ||
− | + | === Solution 1 === | |
− | == | ||
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>. | ||
Line 16: | Line 16: | ||
− | == | + | === Solution 2 === |
− | |||
− | |||
A slightly more analytic/brute-force approach: | A slightly more analytic/brute-force approach: | ||
Line 25: | Line 23: | ||
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>. | 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>. | ||
− | 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>. | + | 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}</math><math>=\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 09:41, 4 April 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 , 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 |