I will demonstrate to you how to successfully coordinate bash 3 AIME questions on the 2002 AIME (i think its the I)
ok
(paraphrased) Number 10: is a right triangle with hypotenuse . . Find the integer closest to .
First, you must be aware that, although this solution is slightly a coordinate bash, this solution is also a slight trig bash. (all measures are in degrees so i won't include any degree marks)
You will need to know and be familiar with the following trigonometric identities/definitions, which are, by the war, basic.
[trigbash]




These identities will not be proven and the definition is basic, so it needs no citation. (at least in this solution)
NOW!!!! TEH SOLUTION



yeah yeah rationalizing denominators... BUT NOT HERE

(see why we don't rationalize denominators?)

now that's really why we don't rationalize denominators
The slope of
[/trigbash]
the coordinates of points are (respectively) . THe coordinates of D can be found by using the sope of
the equation of the line containing is $ y = - \frac {7}{5}x + 12}$
Therefore, the equatino of the line containing is
Since G is the intersection of , the coordinates of G are
Using Shoestring on triangle AGF tells us that the area of is
Since ,
How beast

(paraphrased) Number 13: In , median has length 18 and has length 27. Also, . If F is the intersection (distinct from C) of the line containing and $ \trangle{ABC}$'s circumcircle, then is equal to if the area is put into simplest radical form. Find m+n.

WARNING: this problem is the messiest one by far

solution: Position points A and B at points and , respectively. Since the median from vertes C has length 27, point D is on the circle defined by the equation
Similarly, point D is on the circle defined by the equation . Since D is the midpoint of , point C must be on the circle resulting from dilation circle A by a factor of 2 with respect to the origin and subtracting the vector to B, or
Since C is on both of these circles, we can solve for point C:



Since the circles intersect at both of these points, i will just pick the point . (This is the coordinate of C)
Since the circumcenter of is on each side's perpendicular bisectors, The circumcenter is on the x axis. We can now form the equation (regarding the circumcenter):
(yeah yeah, i rationalized the root 55, but its supa convenient)
The circumcenter has coordinates
There fore, the square of the circumradius (i will never take the square root) is

Therefore, the equation of the circumcircle is
We want to Find the intersection of the circum circle and the line containing that is distinct from C. The equation of the line containing is
Now we can solve for the intersection (distinct from C):





Multiply by the LCM...


Dividing by the GCF...

Letting simplifies this quadratic:



is simply , where d is the distance from F to , or simply, the absolute value of F's X coordinate, which is
The area is therefore $ \frac12 (24)\left(\frac {2 \sqrt {55}}{3}\right) = 8\sqrt {55} \implies m = 8, n = 55 \implire m + n = \boxed{063}$

(paraphrased) Number 15: Polyhedron ABCDEFG is a hexahedron with square ABCD as its bottom face and triangle EFG as its top face. The other 4 faces are isosceles triangle EDC, isosceles trapezoid AGFB, and congruent quadrilaterals AGED and BFEC. , and the distance from E to the polyhedron's bottom face is 12. If , where this value is in simplest radical form, then find .

Note: This problem is very threatening looking, but it is easier thaan numbers 10, 12, and 13 of the same AIME (in my opinion). Therefore, the solution will be relatively short and there will not be much bashing going on. Note, however, that this solution will require an understanding of the cartesian SPACE (3-d coordinate system). (x axis- left/right y axis- front/back z axis up/down)

Position points at (respectively)
point E has z coordinate 12. Since it is the vertex of the isosceles triangle and , the x coordinate of E is 6. Since it is 14 units away from D, the coordinates of E are
points G and F have x coordinates 3 and 9 respectively. Since face AGED lies in one plane, and we know the coordinates of points A, D, and E, we can ASCERTAIN the coordinates of point G.
Since the plane passes through the y axis, the plane is in the form of . Plugging in E's coordinates tells us that
So G's z coordinate is 6. G's coordinates are in the form Since G is 8 units away from point A, G's coordinates are
Therefore,

p.s., here is my opinionated order of the questions on the AIME in decreasing level of difficulty:

13
12 (i bashed out the first 4 's)
10
7 (you have to be really calm when interpreting this information or else you'll spaz)
14
15 (there is a degree of trickiness, but not much)
9 (can't rush on this one)
8
6 (would have bashed this one out if i hadn't seen this problem from another source)
5 (there is a minor degree of trickiness to this problem)
11(there is an extremely easy trick)
4
3
2
1

The last 2 questions were really not that hard...
btw, i got a 13 on this AIME

here is a pic of my looping the city in like 133.7 seconds (yesm i stalled)