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AIME 2012 II · 第 6 题

AIME 2012 II — Problem 6

专题
Contest Math
难度
L4
来源
AIME

题目详情

Problem 6

Let z=a+biz=a+bi be the complex number with z=5\vert z \vert = 5 and b>0b > 0 such that the distance between (1+2i)z3(1+2i)z^3 and z5z^5 is maximized, and let z4=c+diz^4 = c+di. Find c+dc+d.

解析

Solution

Let's consider the maximization constraint first: we want to maximize the value of z5(1+2i)z3|z^5 - (1+2i)z^3| Simplifying, we have

z3z2(1+2i)|z^3| * |z^2 - (1+2i)| =z3z2(1+2i)=|z|^3 * |z^2 - (1+2i)| =125z2(1+2i)=125|z^2 - (1+2i)|

Thus we only need to maximize the value of z2(1+2i)|z^2 - (1+2i)|.

To maximize this value, we must have that z2z^2 is in the opposite direction of 1+2i1+2i. The unit vector in the complex plane in the desired direction is 15+25i\frac{-1}{\sqrt{5}} + \frac{-2}{\sqrt{5}} i. Furthermore, we know that the magnitude of z2z^2 is 2525, because the magnitude of zz is 55. From this information, we can find that z2=5(510i)z^2 = \sqrt{5} (-5 - 10i)

Squaring, we get z4=5(25100+100i)=375+500iz^4 = 5 (25 - 100 + 100i) = -375 + 500i. Finally, c+d=375+500=125c+d = -375 + 500 = \boxed{125}

Solution 2

WLOG, let z1=5(cosθ1+isinθ1)z_{1}=5(\cos{\theta_{1}}+i\sin{\theta_{1}}) and

z2=1+2i=5(cosθ2+isinθ2)z_{2}=1+2i=\sqrt{5}(\cos{\theta_{2}+i\sin{\theta_{2}}})

This means that

z13=125(cos3θ1+isin3θ1)z_{1}^3=125(\cos{3\theta_{1}}+i\sin{3\theta_{1}}) z14=625(cos4θ1+isin4θ1)z_{1}^4=625(\cos{4\theta_{1}}+i\sin{4\theta_{1}})

Hence, this means that

z2z13=1255(cos(θ2+3θ1)+isin(θ2+3θ1))z_{2}z_{1}^3=125\sqrt{5}(\cos({\theta_{2}+3\theta_{1}})+i\sin({\theta_{2}+3\theta_{1}}))

And

z15=3125(cos5θ1+isin5θ1)z_{1}^5=3125(\cos{5\theta_{1}}+i\sin{5\theta_{1}})

Now, common sense tells us that the distance between these two complex numbers is maxed when they both are points satisfying the equation of the line yi=mxyi=mx, or when they are each a 180180^{\circ} rotation away from each other.

Hence, we must have that 5θ1=3θ1+θ2+180    θ1=θ2+18025\theta_{1}=3\theta_{1}+\theta_{2}+180^{\circ}\implies\theta_{1}=\frac{\theta_{2}+180^{\circ}}{2}

Now, plug this back into z14z_{1}^4(if you want to know why, reread what we want in the problem!)

So now, we have that z14=625(cos2θ2+isin2θ2)z_{1}^4=625(\cos{2\theta_{2}}+i\sin{2\theta_{2}})

Notice that cosθ2=15\cos\theta_{2}=\frac{1}{\sqrt{5}} and sinθ2=25\sin\theta_{2}=\frac{2}{\sqrt{5}}

Then, we have that cos2θ2=cos2θ2sin2θ2=35\cos{2\theta_{2}}=\cos^2{\theta_{2}}-\sin^2{\theta_{2}}=-\frac{3}{5} and sin2θ2=2sinθ2cosθ2=45\sin{2\theta_{2}}=2\sin{\theta_{2}}\cos{\theta_{2}}=\frac{4}{5}

Finally, plugging back in, we find that z14=625(35+4i5)=375+500iz_{1}^4=625(-\frac{3}{5}+\frac{4i}{5})=-375+500i

375+500=125-375+500=\boxed{125}

Solution 3

Clearly, we want arg((1+2i)z3)=π+arg(z5)\arg((1+2i)z^3) = \pi + \arg(z^5). This is equivalent to arg(1+2i)=π+2arg(z)\arg(1+2i) = \pi + 2 \arg(z) by the additive property of the argument in complex multiplication. Because we want to find z4z^4, we want an expression for arg(z4)\arg(z^4). We now have 2arg(1+2i)=2π+4arg(z)arg(3+4i)=4arg(z)2\arg(1+2i) = 2\pi + 4 \arg(z) \rightarrow \arg(-3+4i) = 4\arg(z). Thus, z4z^4 is in the direction of 3+4i-3 + 4i. To achieve a magnitude of 5^4, we need to multiple 3+4i-3+4i by 125, so z4=125(3+4i)z^4 = 125(-3+4i) and the answer is thus 125\boxed{125}.