Difference between revisions of "2007 AIME I Problems/Problem 12"
m (→Solution: + 2nd solution, but I did something wrong somewhere) |
(→Solution 2: got it, complete solution) |
||
Line 24: | Line 24: | ||
Redefine the points in the same manner as the last time (<math>\displaystyle \triangle AB'C'</math>, intersect at <math>D</math>, <math>E</math>, and <math>F</math>). This time, notice that <math>[ADEF] = [\triangle AB'C'] - ([\triangle ADC'] + [\triangle EFB'])</math>. | Redefine the points in the same manner as the last time (<math>\displaystyle \triangle AB'C'</math>, intersect at <math>D</math>, <math>E</math>, and <math>F</math>). This time, notice that <math>[ADEF] = [\triangle AB'C'] - ([\triangle ADC'] + [\triangle EFB'])</math>. | ||
− | The area of <math>[\triangle AB'C'] = [\triangle ABC]</math>. The [[altitude]] of <math>\triangle ABC</math> is clearly <math>10 \tan 75 = 10 \tan (30 + 45)</math>. The [[trigonometric identity|tangent addition rule]] yields <math>10(2 + \sqrt{3})</math> (see above). Thus, <math>\frac | + | The area of <math>[\triangle AB'C'] = [\triangle ABC]</math>. The [[altitude]] of <math>\triangle ABC</math> is clearly <math>10 \tan 75 = 10 \tan (30 + 45)</math>. The [[trigonometric identity|tangent addition rule]] yields <math>10(2 + \sqrt{3})</math> (see above). Thus, <math>\displaystyle [\triangle ABC] = \frac{1}{2} 20 \cdot (20 + 10\sqrt{3}) = 200 + 100\sqrt{3}</math>. |
− | The area of <math>[\triangle ADC']</math> (with a side on the y-axis) can be found by splitting it into two triangles, <math>30-60-90</math> and <math>15-75-90</math> [[right triangle]]s. <math>AC' = AC = \frac{10}{\sin 75}</math>. The [[trigonometric identity|sine | + | The area of <math>[\triangle ADC']</math> (with a side on the y-axis) can be found by splitting it into two triangles, <math>30-60-90</math> and <math>15-75-90</math> [[right triangle]]s. <math>AC' = AC = \frac{10}{\sin 75}</math>. The [[trigonometric identity|sine subtraction rule]] shows that <math>\frac{10}{\sin 15} = \frac{10}{\frac{\sqrt{6} - \sqrt{2}}4} = \frac{40}{\sqrt{6} - \sqrt{2}} = 10(\sqrt{6} + \sqrt{2})</math>. <math>AC'</math>, in terms of the height of <math>\triangle ADC'</math>, is equal to <math>h(\sqrt{3} + \tan 75) = h(\sqrt{3} + 2 + \sqrt{3})</math>. <math>[ADC'] = \frac 12 AC' \cdot h = \frac 12 (10\sqrt{6} + 10\sqrt{2})(\frac{10\sqrt{6} + 10\sqrt{2}}{2\sqrt{3} + 2}</math><math>= \frac{(800 + 400\sqrt{3})}{(2 + \sqrt{3})}\cdot\frac{2 - \sqrt{3}}{2-\sqrt{3}}</math><math>= \frac{400\sqrt{3} + 400}8 = 50\sqrt{3} + 50</math>. |
The area of <math>[\triangle EFB']</math> was found in the previous solution to be <math>- 500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} +750</math>. | The area of <math>[\triangle EFB']</math> was found in the previous solution to be <math>- 500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} +750</math>. |
Revision as of 14:52, 16 March 2007
Problem
In isosceles triangle , is located at the origin and is located at (20,0). 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 .
Solution
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
Solution 1
Call the vertices of the new triangle (, the origin, is a vertex of both triangles). and intersect at a single point, . intersect at two points; the one with the higher y-coordinate will be , and the other . The intersection of the two triangles is a quadrilateral . Notice that we can find this area by subtracting .
Since and both have measures , both of their complements are , and . We know that , and since the angles of a triangle add up to , we find that .
So is a . It can be solved by drawing an altitude splitting the angle into and angles – this forms a right triangle and a isosceles right triangle. Since we know that , the base of the triangle is , the height is , and the base of the is . Thus, the total area of .
Now, we need to find , which is a right triangle. We can find its base by subtracting from . is also a triangle, so we find that . .
To solve , note that . Through algebra, we can calculate :
To finish, find . The solution is .
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 .
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 |