Difference between revisions of "2007 AIME I Problems/Problem 12"
(... this is long ..., will finish in next edit) |
(complete solution (phew)) |
||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | In [[isosceles triangle]] <math>\triangle ABC</math>, <math>A</math> is located at the [[origin]] and <math>B</math> is located at (20,0). Point <math>C</math> is in the [[first quadrant]] with <math>\displaystyle AC = BC</math> and angle <math>BAC = 75^{\circ}</math>. If triangle <math>ABC</math> is rotated counterclockwise about point <math>A</math> until the image of <math>C</math> lies on the positive <math>y</math>-axis, the area of the region common to the original and the rotated triangle is in the form <math>p\sqrt{2} + q\sqrt{3} + r\sqrt{6} + s</math>, where <math>\ | + | In [[isosceles triangle]] <math>\triangle ABC</math>, <math>A</math> is located at the [[origin]] and <math>B</math> is located at (20,0). Point <math>C</math> is in the [[first quadrant]] with <math>\displaystyle AC = BC</math> and angle <math>BAC = 75^{\circ}</math>. If triangle <math>ABC</math> is rotated counterclockwise about point <math>A</math> until the image of <math>C</math> lies on the positive <math>y</math>-axis, the area of the region common to the original and the rotated triangle is in the form <math>p\sqrt{2} + q\sqrt{3} + r\sqrt{6} + s</math>, where <math>\displaystyle p,q,r,s</math> are integers. Find <math>\frac{p-q+r-s}2</math>. |
== Solution == | == Solution == | ||
Line 8: | Line 8: | ||
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>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>. |
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>. | ||
− | To solve <math>[\triangle EFB']</math>, note that <math>[\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>\frac{1}{2}\tan | + | :<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>= \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> | ||
+ | |||
+ | 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>. | ||
== See also == | == See also == | ||
− | {{AIME box|year=2007|n=I|num-b=11|num-a=13}} | + | {{AIME box|year=2007|n=I|num-b=11|num-a=13}} |
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Revision as of 13:33, 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.
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 |