2007 AIME I Problems/Problem 12
Problem
In isosceles triangle ,
is located at the origin and
is located at
. Point
is in the first quadrant with
and angle
. If triangle
is rotated counterclockwise about point
until the image of
lies on the positive
-axis, the area of the region common to the original and the rotated triangle is in the form
, where
are integers. Find
.
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Solution
Solution 1
Let the new triangle be (
, the origin, is a vertex of both triangles). Let
intersect with
at point
,
intersect with
at
, and
intersect with
at
. The region common to both triangles is the quadrilateral
. Notice that
, where we let
denote area.
![$[\triangle ADB']$](http://latex.artofproblemsolving.com/d/e/8/de89334f50d1b46a5f99afe334fa5470ee37178d.png)
Since and
both have measures
, both of their complements are
, and
. We know that
, so
.
Thus is a
. It can be solved by drawing an altitude splitting the
angle into
and
angles, forming a
right triangle and a
isosceles right triangle. Since we know that
, the base of the
triangle is
, the base of the
is
, and their common height is
. Thus, the total area of
.
![$[\triangle EFB']$](http://latex.artofproblemsolving.com/2/a/b/2ab67efa5b71d9f0a8ad5c7b0c75f2ffb1c443a1.png)
Since is also a
triangle,
![$AF = 20\sin 75 = 20 \sin (30 + 45) = 20\left(\frac{\sqrt{2} + \sqrt{6}}4\right) = 5\sqrt{2} + 5\sqrt{6}$](http://latex.artofproblemsolving.com/5/6/8/568ab424e801654d874c34d645ee2499c0533191.png)
and
![$FB' = AB' - AF = 20 - 5\sqrt{2} - 5\sqrt{6}$](http://latex.artofproblemsolving.com/a/5/f/a5fb198241bedad12f3c64b5f872e483a5f5d630.png)
Since . With some horrendous algebra, we can calculate
To finish,
Hence,
.
Solution 2
Redefine the points in the same manner as the last time (, intersect at
,
, and
). This time, notice that
.
The area of . The altitude of
is clearly
. The tangent addition rule yields
(see above). Thus,
.
The area of (with a side on the y-axis) can be found by splitting it into two triangles,
and
right triangles.
. The sine subtraction rule shows that
.
, in terms of the height of
, is equal to
.
The area of was found in the previous solution to be
.
Therefore,
, and our answer is
.
Solution 3
Call the points of the intersections of the triangles ,
, and
as noted in the diagram (the points are different from those in the diagram for solution 1).
bisects
.
Through HL congruency, we can find that is congruent to
. This divides the region
(which we are trying to solve for) into two congruent triangles and an isosceles right triangle.
![$AE = 20 \cos 15 = 20 \cos (45 - 30) = 20 \cdot \frac{\sqrt{6} + \sqrt{2}}{4} = 5\sqrt{6} + 5\sqrt{2}$](http://latex.artofproblemsolving.com/e/4/4/e4495f61066c301af98aef66f5a310faddc5747c.png)
Since , we find that
.
Now, we need to find . The acute angles of the triangles are
and
. By repeated application of the half-angle formula, we can find that
.
The area of . Thus,
, which eventually simplifies to
.
Adding them together, we find that the solution is
, and the answer is
.
Solution 4
From the given information, calculate the coordinates of all the points (by finding the equations of the lines, equating). Then, use the shoelace method to calculate the area of the intersection.
:
:
: It passes thru
, and has a slope of
. The equation of its line is
.
:
, so it passes thru point
. It has a slope of
. So the equation of its line is
.
Now, we can equate the equations to find the intersections of all the points.
is the intersection of
.
. Therefore,
,
.
is the intersection of
.
. Therefore,
,
.
is the intersection of
.
. Therefore,
,
.
We take these points and tie them together by shoelace, and the answer should come out to be .
See also
2007 AIME I (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 |
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