# Difference between revisions of "2007 AIME I Problems/Problem 12"

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== Problem == | == Problem == | ||

− | In [[isosceles triangle]] <math>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> | + | 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>\displaysytle p,q,r,s</math> are integers. Find <math>\frac{p-q+r-s}2</math>. |

== Solution == | == Solution == | ||

− | {{ | + | {{image}} |

+ | 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>\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>. | ||

+ | |||

+ | 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>: | ||

+ | :<math>\frac{1}{2}\tan 75 \cdot (20 - 5\sqrt{2} - 5\sqrt{6})^2 | ||

== 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}}</math> |

## Revision as of 13:23, 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 $\displaysytle p,q,r,s$ (Error compiling LaTeX. ! Undefined control sequence.) 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 :

- $\frac{1}{2}\tan 75 \cdot (20 - 5\sqrt{2} - 5\sqrt{6})^2

== See also == {{AIME box|year=2007|n=I|num-b=11|num-a=13}}$ (Error compiling LaTeX. ! Missing $ inserted.)