2005 AIME II Problems/Problem 12
Problem
Square has center
and
are on
with
and
between
and
and
Given that
where
and
are positive integers and
is not divisible by the square of any prime, find
Solution
Solution 1 (trignometry)
![[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), G=(4.5,9), O=(4.5,4.5); draw(A--B--C--D--A);draw(E--O--F);draw(G--O); dot(A^^B^^C^^D^^E^^F^^G^^O); label("\(A\)",A,(-1,1));label("\(B\)",B,(1,1));label("\(C\)",C,(1,-1));label("\(D\)",D,(-1,-1)); label("\(E\)",E,(0,1));label("\(F\)",F,(1,1));label("\(G\)",G,(-1,1));label("\(O\)",O,(1,-1)); label("\(x\)",E/2+G/2,(0,1));label("\(y\)",G/2+F/2,(0,1)); label("\(450\)",(O+G)/2,(-1,1)); [/asy]](http://latex.artofproblemsolving.com/f/f/6/ff61d20895ed22a782e7a4ab5be1c93452a6eed0.png)
Let be the foot of the perpendicular from
to
. Denote
and
, and
(since
and
). Then
, and
.
By the tangent addition rule , we see that
Since
, this simplifies to
. We know that
, so we can substitute this to find that
.
Substituting again, we know have
. This is a quadratic with roots
. Since
, use the smaller root,
.
Now, . The answer is
.
Solution 2 (synthetic)
![[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), O=(4.5,4.5), G=O+(E-O)*dir(-90), J=O+(F-O)*dir(-90); draw(A--B--C--D--A);draw(E--O--F);draw(G--O--J);draw(F--G,linetype("4 4")); dot(A^^B^^C^^D^^E^^F^^G^^J^^O); label("\(A\)",A,(-1,1));label("\(B\)",B,(1,1));label("\(C\)",C,(1,-1));label("\(D\)",D,(-1,-1)); label("\(E\)",E,(0,1));label("\(F\)",F,(1,1));label("\(G\)",G,(1,0));label("\(J\)",J,(1,0));label("\(O\)",O,(1,-1)); label("\(x\)",(B+F)/2,(0,1)); label("\(400\)",(E+F)/2,(0,1)); label("\(900\)",(C+D)/2,(0,-1)); [/asy]](http://latex.artofproblemsolving.com/d/a/d/dadd3767e5670a0ab2633d7a3e97591ad885f8ae.png)
Label , so
. Rotate
about
until
lies on
. Now we know that
therefore
also since
is the center of the square. Label the new triangle that we created
. Now we know that rotation preserves angles and side lengths, so
and
. Draw
and
. Notice that
since rotations preserve the same angles so
too. By SAS we know that
so
. Now we have a right
with legs
and
and hypotenuse
. By the Pythagorean Theorem,
and applying the quadratic formula we get that
. Since
we take the positive root, and our answer is
.
See also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.