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| | == Solution == | | == Solution == |
| − | | + | First, turn <math>\frac34 + \frac34i</math> into polar form as <math>\frac{3\sqrt{2}}{4}e^{\frac{\pi}{4}i}</math>. Restated using geometric probabilities, we are trying to find the portion of a square enlarged by a factor of <math>\frac{3\sqrt{2}}{4}</math> and rotated <math>45</math> degrees that lies within the original square. This skips all the absolute values required before. Finish with the symmetry method stated above. |
| − | We can directly compute <math>\left(\frac34 + \frac34i\right)z = \left(\frac34 + \frac34i\right)(x + iy) = \frac{3(x-y)}4 + \frac{3(x+y)}4 \cdot i</math>.
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| − | This number is in <math>S</math> if and only if <math>-1 \leq \frac{3(x-y)}4 \leq 1</math> and at the same time <math>-1 \leq \frac{3(x+y)}4 \leq 1</math>. This simplifies to <math>|x-y|\leq\frac 43</math> and <math>|x+y|\leq\frac 43</math>.
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| − | Let <math>T = \{ x + iy : |x-y|\leq\frac 43 ~\land~ |x+y|\leq\frac 43 \}</math>, and let <math>[X]</math> denote the area of the region <math>X</math>. Then obviously the probability we seek is <math>\frac {[S\cap T]}{[S]} = \frac{[S\cap T]}4</math>. All we need to do is to compute the area of the intersection of <math>S</math> and <math>T</math>. It is easiest to do this graphically:
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| − | <asy>
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| − | unitsize(2cm);
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| − | defaultpen(0.8);
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| − | path s = (-1,-1) -- (-1,1) -- (1,1) -- (1,-1) -- cycle;
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| − | path t = (4/3,0) -- (0,4/3) -- (-4/3,0) -- (0,-4/3) -- cycle;
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| − | path s_cap_t = (1/3,1) -- (1,1/3) -- (1,-1/3) -- (1/3,-1) -- (-1/3,-1) -- (-1,-1/3) -- (-1,1/3) -- (-1/3,1) -- cycle;
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| − | filldraw(s, lightred, black);
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| − | filldraw(t, lightgreen, black);
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| − | filldraw(s_cap_t, lightyellow, black);
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| − | draw( (-5/3,0) -- (5/3,0), dashed );
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| − | draw( (0,-5/3) -- (0,5/3), dashed );
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| − | </asy>
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| − | Coordinate axes are dashed, <math>S</math> is shown in red, <math>T</math> in green and their intersection is yellow. The intersections of the boundary of <math>S</math> and <math>T</math> are obviously at <math>(\pm 1,\pm 1/3)</math> and at <math>(\pm 1/3,\pm 1)</math>.
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| − | Hence each of the four red triangles is an isosceles right triangle with legs long <math>\frac 23</math>, and hence the area of a single red triangle is <math>\frac 12 \cdot \left( \frac 23 \right)^2 = \frac 29</math>. Then the area of all four is <math>\frac 89</math>, and therefore the area of <math>S\cap T</math> is <math>4 - \frac 89</math>. Then the probability we seek is <math>\frac{ [S\cap T]}4 = \frac{ 4 - \frac 89 }4 = 1 - \frac 29 = \boxed{\frac 79}</math>.
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| − | (Alternately, when we got to the point that we know that a single red triangle is <math>\frac 29</math>, we can directly note that the picture is symmetric, hence we can just consider the first quadrant and there the probability is <math>1 - \frac 29 = \frac 79</math>. This saves us the work of first multiplying and then dividing by <math>4</math>.)
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| − | == Solution 2 (Same idea) ==
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| − | The solution proposed above is good, but there is a more straightforward method. First, turn <math>\frac34 + \frac34i</math> into polar form as <math>\frac{3\sqrt{2}}{4}e^{\frac{\pi}{4}i}</math>. Restated using geometric probabilities, we are trying to find the portion of a square enlarged by a factor of <math>\frac{3\sqrt{2}}{4}</math> and rotated <math>45</math> degrees that lies within the original square. This skips all the absolute values required before. Finish with the symmetry method stated above.
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| | -asdf334 | | -asdf334 |
in the complex plane is defined by
![\[S = \{x + iy: - 1\le x\le1, - 1\le y\le1\}.\]](http://latex.artofproblemsolving.com/1/f/1/1f183c6f855fb40cc0df24c6d24b6a99b1218b0d.png)
A complex number 
is chosen uniformly at random from 
is also in 
into polar form as 
. Restated using geometric probabilities, we are trying to find the portion of a square enlarged by a factor of 
and rotated 
degrees that lies within the original square. This skips all the absolute values required before. Finish with the symmetry method stated above.
