Difference between revisions of "2007 AIME II Problems/Problem 3"
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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}</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>. | 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>. | ||
+ | |||
+ | ===Solution 3=== | ||
+ | Based on the symmetry, we know that <math>F</math> is a reflection of <math>E</math> across the center of the square, which we will denote as <math>O</math>. Since <math>\angle BEA</math> and <math>\angle AOB</math> are right, we can conclude that figure <math>AOBE</math> is a cyclic quadrilateral. Pythagorean Theorem yields that <math>BO=AO=\frac{13\sqrt{2}}{2}</math>. Now, using Ptolemy's Theorem, we get that | ||
+ | <cmath>AO\cdot BE + BO\cdot AE = AB\cdot AO</cmath> | ||
+ | <cmath>\frac{13\sqrt{2}}{2}\cdot 5+\frac{13\sqrt{2}}{2}\cdot 12 = 13\cdot OE</cmath> | ||
+ | <cmath>OE=\frac{17\sqrt{2}}{2}</cmath> | ||
+ | Now, since we stated in the first step that <math>F</math> is a reflection of <math>E</math> across <math>O</math>, we can say that <math>EF=2EO=17\sqrt{2}</math>. This gives that <cmath>EF^2=(17\sqrt{2})^2=578</cmath> AWD with this bash solution | ||
== See also == | == See also == |
Revision as of 20:40, 21 January 2017
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, .
Solution 3
Based on the symmetry, we know that is a reflection of across the center of the square, which we will denote as . Since and are right, we can conclude that figure is a cyclic quadrilateral. Pythagorean Theorem yields that . Now, using Ptolemy's Theorem, we get that Now, since we stated in the first step that is a reflection of across , we can say that . This gives that AWD with this bash solution
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.