Difference between revisions of "2007 AIME I Problems/Problem 12"
m (fix typos regarding parentheses) |
(reformat) |
||
Line 1: | Line 1: | ||
+ | __NOTOC__ | ||
== 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> | + | 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>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>p,q,r,s</math> are integers. Find <math>\frac{p-q+r-s}2</math>. |
− | + | [[Image:AIME I 2007-12.png]] | |
− | |||
− | |||
− | |||
− | |||
− | |||
== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
− | Call the [[vertex|vertices]] of the new triangle <math> | + | 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>. |
− | Since <math> | + | Since <math>\angle B'AC'</math> and <math>\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> | + | 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>30-60-90</math> [[right triangle]] and a <math>45-45-90</math> isosceles right triangle. Since we know that <math>DB' = 20</math>, the base of the <math>30-60-90</math> triangle is <math>10</math>, the height is <math>10\sqrt{3}</math>, and the base of the <math>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> | + | Now, we need to find <math>[\triangle EFB']</math>, which is a <math>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>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> | + | 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>: |
− | :<math> | + | :<math>\frac{1}{2}\tan (30 + 45) \cdot (20 - 5\sqrt{2} - 5\sqrt{6})^2</math> |
− | :<math> | + | :<math>= \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']< | + | To finish, find <math>[ADEF] = [\triangle ADB'] - [\triangle EFB']<cmath>= \left(150 + 50\sqrt{3}\right) - \left(-500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} + 750\right)</cmath>=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>. |
=== Solution 2 === | === Solution 2 === | ||
− | Redefine the points in the same manner as the last time (<math> | + | Redefine the points in the same manner as the last time (<math>\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> | + | 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>[\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 15}</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})\left(\frac{10\sqrt{6} + 10\sqrt{2}}{2\sqrt{3} + 2}\right)< | + | 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 15}</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})\left(\frac{10\sqrt{6} + 10\sqrt{2}}{2\sqrt{3} + 2}\right)<cmath>= \frac{(800 + 400\sqrt{3})}{(2 + \sqrt{3})}\cdot\frac{2 - \sqrt{3}}{2-\sqrt{3}}</cmath>= \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>. | ||
Line 40: | Line 36: | ||
[[Image:AIME I 2007-12b.png|left]] | [[Image:AIME I 2007-12b.png|left]] | ||
− | Call the points of the intersections of the triangles as <math>D</math>, <math>E</math>, and <math>F</math> as noted in the above diagram (the points are different from those in the diagram for solution 1). <math>\overline{AD}</math> [[bisect]]s <math> | + | Call the points of the intersections of the triangles as <math>D</math>, <math>E</math>, and <math>F</math> as noted in the above diagram (the points are different from those in the diagram for solution 1). <math>\overline{AD}</math> [[bisect]]s <math>\angle EDE'</math>. Through [[HL]] congruency, we can find that <math>\triangle AED</math> is [[congruent]] to <math>\triangle AE'D</math>. This divides the region <math>AEDF</math> (which we are trying to solve for) into two congruent triangles and an [[isosceles triangle|isosceles]] [[right triangle]]. |
<math>AE = 20 \cos 15 = 20 \cos (45 - 30) = 20 \cdot \frac{\sqrt{6} + \sqrt{2}}{4} = 5\sqrt{6} + 5\sqrt{2}</math>. Since <math>FE' = AE' = AE</math>, we find that <math>[AE'F] = \frac 12 (5\sqrt{6} + 5\sqrt{2})^2 = 100 + 50\sqrt{3}</math>. | <math>AE = 20 \cos 15 = 20 \cos (45 - 30) = 20 \cdot \frac{\sqrt{6} + \sqrt{2}}{4} = 5\sqrt{6} + 5\sqrt{2}</math>. Since <math>FE' = AE' = AE</math>, we find that <math>[AE'F] = \frac 12 (5\sqrt{6} + 5\sqrt{2})^2 = 100 + 50\sqrt{3}</math>. | ||
− | Now, we need to find <math> | + | Now, we need to find <math>[AED] = [AE'D]</math>. The acute angles of the triangles are <math>\frac{15}{2},\ 90 - \frac{15}{2}</math>. By repeated application of the [[trigonometric identity|half-angle formula]], we can find that <math>\tan \frac{15}{2} = \sqrt{2} - \sqrt{3} + \sqrt{6} - 2</math>. The area of <math>[AED] = \frac 12 \left(20 \cos 15\right)^2 \left(\tan \frac{15}{2}\right)</math> Thus, <math>[AED] + [AE'D] = 2\left(\frac 12((5\sqrt{6} + 5\sqrt{2})^2 \cdot (\sqrt{2} - \sqrt{3} + \sqrt{6} - 2))\right)</math>. This eventually simplifies to <math>500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600</math>. |
Adding them together, we find that the solution is <math>[AEDF] = [AE'F] + [AED] + [AE'D] = 100 + 50\sqrt{3} + 500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600 = 500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600</math>. | Adding them together, we find that the solution is <math>[AEDF] = [AE'F] + [AED] + [AE'D] = 100 + 50\sqrt{3} + 500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600 = 500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600</math>. |
Revision as of 22:51, 14 November 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
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 .
Solution 3
Call the points of the intersections of the triangles as , , and as noted in the above 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.
. Since , we find that .
Now, we need to find . The acute angles of the triangles are . By repeated application of the half-angle formula, we can find that . The area of Thus, . This eventually simplifies to .
Adding them together, we find that the solution 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, , and .
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 |