Difference between revisions of "2007 AIME II Problems/Problem 3"
Jerry122805 (talk | contribs) (→Solution 7 (Trig Bash)) |
Jerry122805 (talk | contribs) (→Solution 7 (Trig Bash)) |
||
Line 94: | Line 94: | ||
First note that <math>\angle BAO = 45^{\circ}</math> since <math>O</math> is the center of square <math>ABCD.</math> Also note that <math>\angle EAB = \arccos{\left(\frac{12}{13}\right)}</math> or <math>\arcsin{\left(\frac{5}{13}\right)}.</math> Finally, we know that <math>AO =\frac{13\sqrt{2}}{2}.</math> Now we apply laws of cosines on <math>\bigtriangleup AEO.</math> | First note that <math>\angle BAO = 45^{\circ}</math> since <math>O</math> is the center of square <math>ABCD.</math> Also note that <math>\angle EAB = \arccos{\left(\frac{12}{13}\right)}</math> or <math>\arcsin{\left(\frac{5}{13}\right)}.</math> Finally, we know that <math>AO =\frac{13\sqrt{2}}{2}.</math> Now we apply laws of cosines on <math>\bigtriangleup AEO.</math> | ||
− | We have <math>EO^2 = 12^2 + (\frac{13\sqrt{2}}{2})^2 - 2 \cdot 12 \cdot \left(\frac{13\sqrt{2}}{2}\right) \cdot \cos{\angle EAO}.</math> We know that <math>\angle EAO = 45^{\circ} + \arccos{\left(\frac{12}{13}\right)}.</math> Thus we have <math>\cos{\angle EAO} = \cos\left(45^{\circ} + \arccos{\left(\frac{12}{13}\right)}\right)</math> which applying the cosine sum identity yields <math>\cos{45^{\circ}}\cos{\arccos\frac{12}{13}} - \sin{45^{\circ}}\sin\arcsin{\frac{5}{12}} = \ | + | We have <math>EO^2 = 12^2 + (\frac{13\sqrt{2}}{2})^2 - 2 \cdot 12 \cdot \left(\frac{13\sqrt{2}}{2}\right) \cdot \cos{\angle EAO}.</math> We know that <math>\angle EAO = 45^{\circ} + \arccos{\left(\frac{12}{13}\right)}.</math> Thus we have <math>\cos{\angle EAO} = \cos\left(45^{\circ} + \arccos{\left(\frac{12}{13}\right)}\right)</math> which applying the cosine sum identity yields <math>\cos{45^{\circ}}\cos{\arccos\frac{12}{13}} - \sin{45^{\circ}}\sin\arcsin{\frac{5}{12}} = \displaystyle\frac{12\sqrt{2}}{26} -\frac{5\sqrt{2}}{26} =\frac{7\sqrt{2}}{26}.</math> |
− | Note that we are looking for <math>4EO^2</math> so we multiply <math>EO^2 = 12^2 + (\frac{13\sqrt{2}}{2})^2 - 2 \cdot 12 \cdot \left(\frac{13\sqrt{2}}{2}\right) \cdot \cos{\angle EAO}</math> by <math>4</math> obtaining <math>4EO^2 = 576 + 338 - 8 \cdot | + | Note that we are looking for <math>4EO^2</math> so we multiply <math>EO^2 = 12^2 + (\frac{13\sqrt{2}}{2})^2 - 2 \cdot 12 \cdot \left(\frac{13\sqrt{2}}{2}\right) \cdot \cos{\angle EAO}</math> by <math>4</math> obtaining <math>4EO^2 = 576 + 338 - 8 \cdot\frac{13\sqrt{2}}{2} \cdot 12 \cdot\frac{7\sqrt{2}}{26} = 576 + 338 - 4 \cdot 12 \cdot 7 = \boxed{578}.</math> |
== See also == | == See also == |
Revision as of 20:46, 3 August 2020
Problem
Square has side length , and points and are exterior to the square such that and . Find .
Contents
Solution
Solution 1
Let , so that . By the diagonal, .
The sum of the squares of the sides of a parallelogram is the sum of the squares of the diagonals.
Solution 2
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 3
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 4
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
Solution 5 (Ptolemy's Theorem)
Drawing , it clearly passes through the center of . Letting this point be , we note that and are congruent cyclic quadrilaterals, and that Now, from Ptolemy's, . Since , the answer is
Solution 6
Coordinate bash
Solution 7 (Trig Bash)
We first See that the whole figure is symmetrical and reflections across the center that we will denote as bring each half of the figure to the other half. Thus we consider a single part of the figure, namely
First note that since is the center of square Also note that or Finally, we know that Now we apply laws of cosines on
We have We know that Thus we have which applying the cosine sum identity yields
Note that we are looking for so we multiply by obtaining
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.