Difference between revisions of "2007 AIME II Problems/Problem 3"
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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\frac{13\sqrt{2}}{2} \cdot \cos{\angle EAO}.</math> We know that <math>\angle EAO = 45 + \arccos{\left\frac{12}{13}\right)}</math> | + | We have <math>EO^2 = 12^2 + (\frac{13\sqrt{2}}{2})^2 - 2 \cdot 12 \cdot\frac{13\sqrt{2}}{2} \cdot \cos{\angle EAO}.</math> We know that <math>\angle EAO = 45 + \arccos{\left(\frac{12}{13}\right)}</math> |
== See also == | == See also == |
Revision as of 20:36, 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
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.