Difference between revisions of "2009 AIME I Problems/Problem 2"
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<cmath>n = \boxed{697}.</cmath> | <cmath>n = \boxed{697}.</cmath> | ||
+ | |||
+ | ==Solution 3 (Highly not recommended)== | ||
+ | Below is an image of the complex plane. Let <math>\operatorname{Im}(z)</math> denote the imaginary part of a complex number <math>z</math>. | ||
+ | <asy> | ||
+ | unitsize(1cm); | ||
+ | |||
+ | |||
+ | xaxis("Re",Arrows); | ||
+ | yaxis("Im",Arrows); | ||
+ | |||
+ | real f(real x) {return 164;} | ||
+ | pair F(real x) {return (x,f(x));} | ||
+ | |||
+ | draw(graph(f,-700,100),red,Arrows); | ||
+ | |||
+ | label("Im$(z)=164$",F(-330),N); | ||
+ | |||
+ | |||
+ | pair z = (-656,164); | ||
+ | dot(Label("$z$",align=N),z); | ||
+ | dot(Label("$z+n$",align=N),z+(697,0)); | ||
+ | |||
+ | draw(Label("$4x$"),z--(0,0)); | ||
+ | draw(Label("$x$"),(0,0)--z+(697,0)); | ||
+ | |||
+ | markscalefactor=2; | ||
+ | draw(rightanglemark(z,(0,0),z+(697,0))); | ||
+ | </asy> | ||
+ | <math>z</math> must lie on the line <math>\operatorname{Im}(z)=164</math>. <math>z+n</math> must also lie on the same line, since <math>n</math> is real and does not affect the imaginary part of <math>z</math>. | ||
+ | |||
+ | Consider <math>z</math> and <math>z+n</math> in terms of their magnitude (distance from the origin) and phase (angle formed by the point, the origin, and the positive real axis, measured counterclockwise from said axis). When multiplying/dividing two complex numbers, you can multiply/divide their magnitudes and add/subtract their phases to get the magnitude and phase of the product/quotient. Expressed in a formula, we have <math>z_1z_2 = r_1\angle\theta_1 \cdot r_2\angle\theta_2 = r_1r_2\angle(\theta_1+\theta_2)</math> and <math>\frac{z_1}{z_2} = \frac{r_1\angle\theta_1}{r_2\angle\theta_2} = \frac{r_1}{r_2}\angle(\theta_1-\theta_2)</math>, where <math>r</math> is the magnitude and <math>\theta</math> is the phase, and <math>z_n=r_n\angle\theta_n</math>. | ||
+ | |||
+ | Since <math>4i</math> has magnitude <math>4</math> and phase <math>90^\circ</math> (since the positive imaginary axis points in a direction <math>90^\circ</math> counterclockwise from the positive real axis), <math>z</math> must have a magnitude <math>4</math> times that of <math>z+n</math>. We denote the length from the origin to <math>z+n</math> with the value <math>x</math> and the length from the origin to <math>z</math> with the value <math>4x</math>. Additionally, <math>z</math>, the origin, and <math>z+n</math> must form a right angle, with <math>z</math> counterclockwise from <math>z+n</math>. | ||
+ | |||
+ | This means that <math>z</math>, the origin, and <math>z+n</math> form a right triangle. The hypotenuse is the length from <math>z</math> to <math>z+n</math> and has length <math>n</math>, since <math>n</math> is defined to be a positive integer. The area of the triangle can be expressed using the two legs, as <math>\frac{x \cdot 4x}{2}</math>, or using the hypotenuse and its corresponding altitude, as <math>\frac{164n}{2}</math>, so <math>\frac{x \cdot 4x}{2} = \frac{164n}{2} \implies x^2 = 41n</math>. By Pythagorean Theorem, <math>x^2+(4x)^2 = n^2 \implies 17x^2 = n^2</math>. Substituting out <math>x^2</math> using the earlier equation, we get <math>17\cdot41n = n^2 \implies n = \boxed{697}</math>. ~[[User:emerald_block|emerald_block]] | ||
==Video Solution== | ==Video Solution== |
Revision as of 20:52, 8 June 2020
Contents
Problem
There is a complex number with imaginary part and a positive integer such that
Find .
Solution 1
Let .
Then and
By comparing coefficients, equating the real terms on the leftmost and rightmost side of the equation,
we conclude that
By equating the imaginary terms on each side of the equation,
we conclude that
We now have an equation for :
and this equation shows that
Solution 2
Since their imaginary part has to be equal,
Solution 3 (Highly not recommended)
Below is an image of the complex plane. Let denote the imaginary part of a complex number . must lie on the line . must also lie on the same line, since is real and does not affect the imaginary part of .
Consider and in terms of their magnitude (distance from the origin) and phase (angle formed by the point, the origin, and the positive real axis, measured counterclockwise from said axis). When multiplying/dividing two complex numbers, you can multiply/divide their magnitudes and add/subtract their phases to get the magnitude and phase of the product/quotient. Expressed in a formula, we have and , where is the magnitude and is the phase, and .
Since has magnitude and phase (since the positive imaginary axis points in a direction counterclockwise from the positive real axis), must have a magnitude times that of . We denote the length from the origin to with the value and the length from the origin to with the value . Additionally, , the origin, and must form a right angle, with counterclockwise from .
This means that , the origin, and form a right triangle. The hypotenuse is the length from to and has length , since is defined to be a positive integer. The area of the triangle can be expressed using the two legs, as , or using the hypotenuse and its corresponding altitude, as , so . By Pythagorean Theorem, . Substituting out using the earlier equation, we get . ~emerald_block
Video Solution
~IceMatrix
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.