2007 AIME II Problems/Problem 3
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Problem
Square 
has side length 
, and points 
and 
are exterior to the square such that 
and 
. Find 
.
Contents |
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://alt1.artofproblemsolving.com/Forum/latexrender/pictures/4/6/1/4613eadf2b658b7f7caec66bb4f15b2a2aa0a118.gif)
, we have 
. Similarly, 
. Since 
, we have 
.
Now, we see that 
. Also, 
. By the Pythagorean Theorem, we have 
. Therefore, 
.
See also
| 2007 AIME II (Problems • Resources) | ||
| Preceded by Problem 2 | Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
