Difference between revisions of "2007 AIME I Problems/Problem 12"
m |
(fmtting) |
||
Line 9: | Line 9: | ||
== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
− | + | Let the new triangle be <math>\triangle AB'C'</math> (<math>A</math>, the origin, is a vertex of both triangles). Let <math>\overline{B'C'}</math> intersect with <math>\overline{AC}</math> at point <math>D</math>, <math>\overline{BC}</math> intersect with <math>\overline{B'C'}</math> at <math>E</math>, and <math>\overline{BC}</math> intersect with <math>\overline{AB'}</math> at <math>F</math>. The region common to both triangles is the [[quadrilateral]] <math>ADEF</math>. Notice that <math>[ADEF] = [\triangle ADB'] - [\triangle EFB']</math>, where we let <math>[\ldots]</math> denote area. | |
− | |||
− | + | <center>To find <math>[\triangle ADB']</math>:</center> | |
− | + | 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 DAB' = 90 - 2(15) = 60^{\circ}</math>. We know that <math>\angle DB'A = 75^{\circ}</math>, so <math>\angle ADB' = 180 - 60 - 75 = 45^{\circ}</math>. | |
− | + | Thus <math>\triangle 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, forming 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 base of the <math>45-45-90</math> is <math>10\sqrt{3}</math>, and their common height is <math>10\sqrt{3}</math>. Thus, the total area of <math>[\triangle ADB'] = \frac{1}{2}(10\sqrt{3})(10\sqrt{3} + 10) = \boxed{150 + 50\sqrt{3}}</math>. | |
− | |||
− | |||
− | |||
− | |||
− | To | + | |
+ | <center>To find <math>[\triangle EFB']</math>:</center> | ||
+ | |||
+ | Since <math>\triangle AFB</math> is also a <math>15-75-90</math> triangle, | ||
+ | <cmath>AF = 20\sin 75 = 20 \sin (30 + 45) = 20\left(\frac{\sqrt{2} + \sqrt{6}}4\right) = 5\sqrt{2} + 5\sqrt{6}</cmath> | ||
+ | |||
+ | and | ||
+ | |||
+ | <cmath>FB' = AB' - AF = 20 - 5\sqrt{2} - 5\sqrt{6}</cmath> | ||
+ | |||
+ | Since <math>[\triangle EFB'] = \frac{1}{2} (FB' \cdot EF) = \frac{1}{2} (FB') (FB' \tan 75^{\circ})</math>. With some horrendous [[algebra]], we can calculate | ||
+ | <cmath>\begin{align*} | ||
+ | [\triangle EFB'] & = \frac{1}{2}\tan (30 + 45) \cdot (20 - 5\sqrt{2} - 5\sqrt{6})^2 \\ | ||
+ | & = 25 \left(\frac{\frac{1}{\sqrt{3}} + 1}{1 - \frac{1}{\sqrt{3}}}\right) \left(8 - 2\sqrt{2} - 2\sqrt{6} - 2\sqrt{2} + 1 + \sqrt{3} - 2\sqrt{6} + \sqrt{3} + 3\right) \\ | ||
+ | & = 25(2 + \sqrt{3})(12 - 4\sqrt{2} - 4\sqrt{6} + 2\sqrt{3}) \\ | ||
+ | [\triangle EFB'] & = \boxed{- 500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} +750}. | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | |||
+ | To finish, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | [ADEF] &= [\triangle ADB'] - [\triangle EFB']\\ | ||
+ | &= \left(150 + 50\sqrt{3}\right) - \left(-500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} + 750\right)\\ | ||
+ | &=500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600\\ \end{align*}</cmath> | ||
+ | |||
+ | |||
+ | Hence <math>\frac{p-q+r-s}{2} = \frac{500 + 350 + 300 + 600}2 = \frac{1750}2 = \boxed{\boxed{875}}</math>. | ||
=== Solution 2 === | === Solution 2 === | ||
Line 30: | Line 52: | ||
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 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>. < | + | 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>. |
+ | |||
+ | <cmath>\begin{align*} | ||
+ | [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) \\ | ||
+ | &= \frac{(800 + 400\sqrt{3})}{(2 + \sqrt{3})}\cdot\frac{2 - \sqrt{3}}{2-\sqrt{3}} \\ | ||
+ | &= \frac{400\sqrt{3} + 400}8 = 50\sqrt{3} + 50 | ||
+ | \end{align*}</cmath> | ||
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 21:44, 17 January 2008
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 .
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.
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 .
Since is also a triangle,
and
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 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 .
We take these points and tie them together by shoelace, and the answer should come out the same as .
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 |