Difference between revisions of "2007 AIME I Problems/Problem 12"
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== Solution == | == Solution == | ||
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− | Call the [[vertex|vertices]] of the new triangle <math>AB'C'</math> (<math>A</math>, the origin, is a vertex of both triangles). <math>B'C'</math> and <math>AB</math> intersect at a single point, <math>D</math>. <math>BC</math> intersect at two points; the one with the higher y-coordinate will be <math>E</math>, and the other <math>F</math>. The intersection of the two triangles is a [[quadrilateral]] <math>ADEF</math>. Notice that we can find this area by subtracting <math>[\triangle ADB'] - [\triangle EFB']</math>. | + | Call the [[vertex|vertices]] of the new triangle <math>\displaystyle AB'C'</math> (<math>A</math>, the origin, is a vertex of both triangles). <math>\displaystyle B'C'</math> and <math>\displaystyle AB</math> intersect at a single point, <math>D</math>. <math>\displaystyle BC</math> intersect at two points; the one with the higher y-coordinate will be <math>E</math>, and the other <math>F</math>. The intersection of the two triangles is a [[quadrilateral]] <math>\displaystyle ADEF</math>. Notice that we can find this area by subtracting <math>[\triangle ADB'] - [\triangle EFB']</math>. |
Since <math>\displaystyle \angle B'AC'</math> and <math>\displaystyle \angle BAC</math> both have measures <math>75^{\circ}</math>, both of their [[complement]]s are <math>15^{\circ}</math>, and <math>\angle DAC' = 90 - 2(15) = 60^{\circ}</math>. We know that <math>C'B'A = 75^{\circ}</math>, and since the angles of a triangle add up to <math>180^{\circ}</math>, we find that <math>ADB' = 180 - 60 - 75 = 45^{\circ}</math>. | Since <math>\displaystyle \angle B'AC'</math> and <math>\displaystyle \angle BAC</math> both have measures <math>75^{\circ}</math>, both of their [[complement]]s are <math>15^{\circ}</math>, and <math>\angle DAC' = 90 - 2(15) = 60^{\circ}</math>. We know that <math>C'B'A = 75^{\circ}</math>, and since the angles of a triangle add up to <math>180^{\circ}</math>, we find that <math>ADB' = 180 - 60 - 75 = 45^{\circ}</math>. | ||
− | So <math>ADB'</math> is a <math>45 - 60 - 75 \triangle</math>. It can be solved by drawing an [[altitude]] splitting the <math>75^{\circ}</math> angle into <math>30^{\circ}</math> and <math>45^{\circ}</math> angles – this forms a <math>\displaystyle 30-60-90</math> [[right triangle]] and a <math>\displaystyle 45-45-90</math> isosceles right triangle. Since we know that <math>DB' = 20</math>, the base of the <math>\displaystyle 30-60-90</math> triangle is <math>10</math>, the height is <math>10\sqrt{3}</math>, and the base of the <math>\displaystyle 45-45-90</math> is <math>10\sqrt{3}</math>. Thus, the total area of <math>[\triangle ADB'] = \frac{1}{2}(10\sqrt{3})(10\sqrt{3} + 10) = 150 + 50\sqrt{3}</math>. | + | So <math>\displaystyle ADB'</math> is a <math>45 - 60 - 75 \triangle</math>. It can be solved by drawing an [[altitude]] splitting the <math>75^{\circ}</math> angle into <math>30^{\circ}</math> and <math>45^{\circ}</math> angles – this forms a <math>\displaystyle 30-60-90</math> [[right triangle]] and a <math>\displaystyle 45-45-90</math> isosceles right triangle. Since we know that <math>\displaystyle DB' = 20</math>, the base of the <math>\displaystyle 30-60-90</math> triangle is <math>10</math>, the height is <math>10\sqrt{3}</math>, and the base of the <math>\displaystyle 45-45-90</math> is <math>10\sqrt{3}</math>. Thus, the total area of <math>[\triangle ADB'] = \frac{1}{2}(10\sqrt{3})(10\sqrt{3} + 10) = 150 + 50\sqrt{3}</math>. |
Now, we need to find <math>[\triangle EFB']</math>, which is a <math>\displaystyle 15-75-90</math> right triangle. We can find its base by subtracting <math>AF</math> from <math>20</math>. <math>\triangle AFB</math> is also a <math>\displaystyle 15-75-90</math> triangle, so we find that <math>AF = 20\sin 75 = 20 \sin (30 + 45) = 20\frac{\sqrt{2} + \sqrt{6}}4 = 5\sqrt{2} + 5\sqrt{6}</math>. <math>FB' = 20 - AF = 20 - 5\sqrt{2} - 5\sqrt{6}</math>. | Now, we need to find <math>[\triangle EFB']</math>, which is a <math>\displaystyle 15-75-90</math> right triangle. We can find its base by subtracting <math>AF</math> from <math>20</math>. <math>\triangle AFB</math> is also a <math>\displaystyle 15-75-90</math> triangle, so we find that <math>AF = 20\sin 75 = 20 \sin (30 + 45) = 20\frac{\sqrt{2} + \sqrt{6}}4 = 5\sqrt{2} + 5\sqrt{6}</math>. <math>FB' = 20 - AF = 20 - 5\sqrt{2} - 5\sqrt{6}</math>. | ||
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To solve <math>[\triangle EFB']</math>, note that <math>\displaystyle [\triangle EFB'] = \frac{1}{2} FB' \cdot EF = \frac{1}{2} FB' \cdot (\tan 75 FB')</math>. Through [[algebra]], we can calculate <math>(FB')^2 \cdot \tan 75</math>: | To solve <math>[\triangle EFB']</math>, note that <math>\displaystyle [\triangle EFB'] = \frac{1}{2} FB' \cdot EF = \frac{1}{2} FB' \cdot (\tan 75 FB')</math>. Through [[algebra]], we can calculate <math>(FB')^2 \cdot \tan 75</math>: | ||
:<math>\displaystyle \frac{1}{2}\tan (30 + 45) \cdot (20 - 5\sqrt{2} - 5\sqrt{6})^2</math> | :<math>\displaystyle \frac{1}{2}\tan (30 + 45) \cdot (20 - 5\sqrt{2} - 5\sqrt{6})^2</math> | ||
− | :<math>\displaystyle = \frac{1}{2} \left(\frac{\frac{1}{\sqrt{3}} + 1}{1 - \frac{1}{\sqrt{3}}}\right) [400 - 100\sqrt{2} - 100\sqrt{6} - 100\sqrt{2} + 50 + 50\sqrt{3} - 100\sqrt{6} + 50\sqrt{3} + 150]</math> | + | :<math>\displaystyle = \frac{1}{2} \left(\frac{\frac{1}{\sqrt{3}} + 1}{1 - \frac{1}{\sqrt{3}}}\right) \left[400 - 100\sqrt{2} - 100\sqrt{6} - 100\sqrt{2} + 50 + 50\sqrt{3} - 100\sqrt{6} + 50\sqrt{3} + 150\right]</math> |
:<math>= \frac{1}{2}(2 + \sqrt{3})[600 - 200\sqrt{2} - 200\sqrt{6} + 100\sqrt{3}]</math> | :<math>= \frac{1}{2}(2 + \sqrt{3})[600 - 200\sqrt{2} - 200\sqrt{6} + 100\sqrt{3}]</math> | ||
:<math>=- 500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} +750</math> | :<math>=- 500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} +750</math> | ||
− | To finish, find <math>[ADEF] = [\triangle ADB'] - [\triangle EFB']</math><math>= (150 + 50\sqrt{3}) - (-500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} + 750)</math><math>=500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600</math>. The solution is <math>\frac{500 + 350 + 300 + 600}2 = \frac{1750}2 = 875</math>. | + | To finish, find <math>[ADEF] = [\triangle ADB'] - [\triangle EFB']</math><math>= \left(150 + 50\sqrt{3}\right) - \left(-500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} + 750\right)</math><math>=500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600</math>. The solution is <math>\frac{500 + 350 + 300 + 600}2 = \frac{1750}2 = 875</math>. |
== See also == | == See also == |
Revision as of 13:37, 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
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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 .
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 |