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

m (Solution: + 2nd solution, but I did something wrong somewhere)
(Solution 2: got it, complete solution)
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Redefine the points in the same manner as the last time (<math>\displaystyle \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>.
 
Redefine the points in the same manner as the last time (<math>\displaystyle \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>\frac 12 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>\displaystyle [\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 75}</math>. The [[trigonometric identity|sine addition rule]] shows that <math>\frac{10}{\sin 75} = \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} + \frac{\sqrt{2} + \sqrt{6}}{4})</math>. <math>[ADC'] = \frac 12 AC' \cdot h = \frac 12 (10\sqrt{6} - 10\sqrt{2})(\frac{10\sqrt{6} - 10\sqrt{2}}{\sqrt{3} + \frac{\sqrt{2} + \sqrt{6}}{4}})</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 75}</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})(\frac{10\sqrt{6} + 10\sqrt{2}}{2\sqrt{3} + 2}</math><math>= \frac{(800 + 400\sqrt{3})}{(2 + \sqrt{3})}\cdot\frac{2 - \sqrt{3}}{2-\sqrt{3}}</math><math>= \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>.  

Revision as of 14:52, 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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Solution 1

Call the vertices of the new triangle $\displaystyle AB'C'$ ($A$, the origin, is a vertex of both triangles). $\displaystyle B'C'$ and $\displaystyle AB$ intersect at a single point, $D$. $\displaystyle BC$ 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 $\displaystyle ADEF$. 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 $\displaystyle ADB'$ 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 $\displaystyle DB' = 20$, 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 $(FB')^2 \cdot \tan 75$:

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

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

Solution 2

Redefine the points in the same manner as the last time ($\displaystyle \triangle AB'C'$, intersect at $D$, $E$, and $F$). This time, notice that $[ADEF] = [\triangle AB'C'] - ([\triangle ADC'] + [\triangle EFB'])$.

The area of $[\triangle AB'C'] = [\triangle ABC]$. The altitude of $\triangle ABC$ is clearly $10 \tan 75 = 10 \tan (30 + 45)$. The tangent addition rule yields $10(2 + \sqrt{3})$ (see above). Thus, $\displaystyle [\triangle ABC] = \frac{1}{2} 20 \cdot (20 + 10\sqrt{3}) = 200 + 100\sqrt{3}$.

The area of $[\triangle ADC']$ (with a side on the y-axis) can be found by splitting it into two triangles, $30-60-90$ and $15-75-90$ right triangles. $AC' = AC = \frac{10}{\sin 75}$. The sine subtraction rule shows that $\frac{10}{\sin 15} = \frac{10}{\frac{\sqrt{6} - \sqrt{2}}4} = \frac{40}{\sqrt{6} - \sqrt{2}} = 10(\sqrt{6} + \sqrt{2})$. $AC'$, in terms of the height of $\triangle ADC'$, is equal to $h(\sqrt{3} + \tan 75) = h(\sqrt{3} + 2 + \sqrt{3})$. $[ADC'] = \frac 12 AC' \cdot h = \frac 12 (10\sqrt{6} + 10\sqrt{2})(\frac{10\sqrt{6} + 10\sqrt{2}}{2\sqrt{3} + 2}$$= \frac{(800 + 400\sqrt{3})}{(2 + \sqrt{3})}\cdot\frac{2 - \sqrt{3}}{2-\sqrt{3}}$$= \frac{400\sqrt{3} + 400}8 = 50\sqrt{3} + 50$.

The area of $[\triangle EFB']$ was found in the previous solution to be $- 500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} +750$.

Therefore, $[ADEF] = (200 + 100\sqrt{3}) - \left((50 + 50\sqrt{3}) + (-500\sqrt{2} + 400\sqrt{3} - 300\sqrt{6} +750)\right) = 500\sqrt{2} - 350\sqrt{3} + 300\sqrt{6} - 600$, and our answer is $875$.

See also

2007 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AIME Problems and Solutions
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