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

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== Problem ==
 
== Problem ==
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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>.
+
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 ==
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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 &ndash; 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 &ndash; 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 75 \cdot (20 - 5\sqrt{2} - 5\sqrt{6})^2
+
:<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}}</math>
+
{{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 $\triangle ABC$, $A$ is located at the origin and $B$ is located at (20,0). Point $C$ is in the first quadrant with $\displaystyle AC = BC$ and angle $BAC = 75^{\circ}$. If triangle $ABC$ is rotated counterclockwise about point $A$ until the image of $C$ lies on the positive $y$-axis, the area of the region common to the original and the rotated triangle is in the form $p\sqrt{2} + q\sqrt{3} + r\sqrt{6} + s$, where $\displaystyle p,q,r,s$ are integers. Find $\frac{p-q+r-s}2$.

Solution

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Call the vertices of the new triangle $AB'C'$ ($A$, the origin, is a vertex of both triangles). $B'C'$ and $AB$ intersect at a single point, $D$. $BC$ intersect at two points; the one with the higher y-coordinate will be $E$, and the other $F$. The intersection of the two triangles is a quadrilateral $ADEF$. Notice that we can find this area by subtracting $[\triangle ADB'] - [\triangle EFB']$.

Since $\displaystyle \angle B'AC'$ and $\displaystyle \angle BAC$ both have measures $75^{\circ}$, both of their complements are $15^{\circ}$, and $\angle DAC' = 90 - 2(15) = 60^{\circ}$. We know that $C'B'A = 75^{\circ}$, and since the angles of a triangle add up to $180^{\circ}$, we find that $ADB' = 180 - 60 - 75 = 45^{\circ}$.

So $ADB'$ is a $45 - 60 - 75 \triangle$. It can be solved by drawing an altitude splitting the $75^{\circ}$ angle into $30^{\circ}$ and $45^{\circ}$ angles – this forms a $\displaystyle 30-60-90$ right triangle and a $\displaystyle 45-45-90$ isosceles right triangle. Since we know that $DB' = 20$, the base of the $\displaystyle 30-60-90$ triangle is $10$, the height is $10\sqrt{3}$, and the base of the $\displaystyle 45-45-90$ is $10\sqrt{3}$. Thus, the total area of $[\triangle ADB'] = \frac{1}{2}(10\sqrt{3})(10\sqrt{3} + 10) = 150 + 50\sqrt{3}$.

Now, we need to find $[\triangle EFB']$, which is a $\displaystyle 15-75-90$ right triangle. We can find its base by subtracting $AF$ from $20$. $\triangle AFB$ is also a $\displaystyle 15-75-90$ triangle, so we find that $AF = 20\sin 75 = 20 \sin (30 + 45) = 20\frac{\sqrt{2} + \sqrt{6}}4 = 5\sqrt{2} + 5\sqrt{6}$. $FB' = 20 - AF = 20 - 5\sqrt{2} - 5\sqrt{6}$.

To solve $[\triangle EFB']$, note that $\displaystyle [\triangle EFB'] = \frac{1}{2} FB' \cdot EF = \frac{1}{2} FB' \cdot (\tan 75 FB')$. Through algebra, we can calculate $(FB')^2 \cdot \tan 75$:

$\displaystyle \frac{1}{2}\tan (30 + 45) \cdot (20 - 5\sqrt{2} - 5\sqrt{6})^2$
$\displaystyle = \frac{1}{2} \left(\frac{\frac{1}{\sqrt{3}} + 1}{1 - \frac{1}{\sqrt{3}}}\right) [400 - 100\sqrt{2} - 100\sqrt{6} - 100\sqrt{2} + 50 + 50\sqrt{3} - 100\sqrt{6} + 50\sqrt{3} + 150]$
$= \frac{1}{2}(2 + \sqrt{3})[600 - 200\sqrt{2} - 200\sqrt{6} + 100\sqrt{3}]$
$=- 500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} +750$

To finish, find $[ADEF] = [\triangle ADB'] - [\triangle EFB']$$= (150 + 50\sqrt{3}) - (-500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} + 750)$$=500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600$. The solution is $\frac{500 + 350 + 300 + 600}2 = \frac{1750}2 = 875$.

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
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