https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Engarde&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T14:40:41ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_23&diff=443552005 AMC 10A Problems/Problem 232012-01-25T03:56:24Z<p>Engarde: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
Let <math>AB</math> be a diameter of a circle and let <math>C</math> be a point on <math>AB</math> with <math>2\cdot AC=BC</math>. Let <math>D</math> and <math>E</math> be points on the circle such that <math>DC \perp AB</math> and <math>DE</math> is a second diameter. What is the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math>?<br />
<br />
<math> \mathrm{(A) \ } \frac{1}{6}\qquad \mathrm{(B) \ } \frac{1}{4}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{2}{3} </math><br />
<br />
==Solution 1==<br />
[[File:Circlenc1.png]]<br />
<br />
Let us assume that the diameter is of length <math>1</math>. <br />
<br />
<math>AC</math> is <math>\frac{1}{3}</math> of diameter and <math>CO</math> is <math>\frac{1}{2}-\frac{1}{3} = \frac{1}{6}</math>. <br />
<br />
<math>OD</math> is the radius of the circle, so using the Pythagorean theorem height <math>CD</math> of <math>\triangle AOC</math> is <math>\sqrt{(\frac{1}{2})^2-(\frac{1}{6})^2} = \frac{\sqrt{2}}{3}</math>. This is also the height of the <math>\triangle ABD</math>.<br />
<br />
Area of the <math>\triangle DCO</math> is <math>\frac{1}{2}\cdot\frac{1}{6}\cdot\frac{\sqrt{2}}{3}</math> = <math>\frac{\sqrt{2}}{36}</math>.<br />
<br />
The height of <math>\triangle DCE</math> can be found using the area of <math>\triangle DCO</math> and <math>DO</math> as base. <br />
<br />
Hence the height of <math>\triangle DCE</math> is <math>\frac{\frac{\sqrt{2}}{36}}{\frac{1}{2}\cdot\frac{1}{2}}</math> = <math>\frac{\sqrt{2}}{9}</math>. <br />
<br />
The diameter is the base for both the triangles <math>\triangle DCE</math> and <math>\triangle ABD</math>. <br />
<br />
Hence, the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math> is<br />
<math>\frac{\frac{\sqrt{2}}{9}}{\frac{\sqrt{2}}{3}}</math> = <math>\frac{1}{3} \Rightarrow C</math><br />
<br />
==Solution 2==<br />
<br />
Since <math>\triangle DCE</math> and <math>\triangle ABD</math> share a base, the ratio of their areas is the ratio of their altitudes. Draw the altitude from <math>C</math> to <math>DE</math>. <br />
<br />
<asy><br />
import graph;<br />
import olympiad;<br />
pair O,A,B,C,D,E,F;<br />
O=(0,0);A=(15,0);B=(-15,0);C=(5,0);D=(5,14.142135623730950488016887242097);E=(-5,-14.142135623730950488016887242097);F=(0.5555555555555555,1.5713484026367722764463208046774);<br />
draw(Circle((0,0),15)); <br />
draw(A--B);draw(D--E);draw(C--D);draw(C--E);draw(C--F);draw(A--D);draw(D--B);<br />
label("A",A,NE);label("B",B,W);label("C",C,SE);label("D",D,NE);label("E",E,SW);label("O",O,SW);label("F",F,NW);<br />
markscalefactor=0.2;<br />
draw(anglemark(C,F,D),blue);draw(anglemark(D,C,B),blue);<br />
</asy><br />
<math>OD=r, OC=\frac{1}{3}r</math>.<br />
<br />
Since <math>m\angle DCO=m\angle DFC=90^\circ</math>, then <math>\triangle DCO\cong \triangle DFC</math>. So the ratio of the two altitudes is <math>\frac{CF}{DC}=\frac{OC}{DO}=\frac{1}{3}\Rightarrow \text{(C)}</math><br />
<br />
==See also==<br />
{{AMC10 box|year=2005|ab=A|num-b=22|num-a=24}}<br />
<br />
[[Category:Introductory Geometry Problems]]</div>Engardehttps://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_5&diff=404522001 AIME I Problems/Problem 52011-07-14T19:07:54Z<p>Engarde: /* Solution */</p>
<hr />
<div>== Problem ==<br />
An [[equilateral triangle]] is inscribed in the [[ellipse]] whose equation is <math>x^2+4y^2=4</math>. One vertex of the triangle is <math>(0,1)</math>, one altitude is contained in the y-axis, and the length of each side is <math>\sqrt{\frac mn}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
<center><asy><br />
pointpen = black; pathpen = black + linewidth(0.7);<br />
path e = xscale(2)*unitcircle; real x = -8/13*3^.5;<br />
D((-3,0)--(3,0)); D((0,-2)--(0,2)); /* axes */<br />
D(e); D(D((0,1))--(x,x*3^.5+1)--(-x,x*3^.5+1)--cycle);<br />
</asy></center><br />
=== Solution 1 ===<br />
Denote the vertices of the triangle <math>A,B,</math> and <math>C,</math> where <math>B</math> is in [[quadrant]] 4 and <math>C</math> is in quadrant <math>3.</math><br />
<br />
Note that the slope of <math>\overline{AC}</math> is <math>\tan 60^\circ = \sqrt {3}.</math> Hence, the equation of the line containing <math>\overline{AC}</math> is<br />
<cmath><br />
y = x\sqrt {3} + 1.<br />
</cmath><br />
This will intersect the ellipse when<br />
<cmath><br />
\begin{eqnarray*}4 = x^{2} + 4y^{2} & = & x^{2} + 4(x\sqrt {3} + 1)^{2} \\<br />
& = & x^{2} + 4(3x^{2} + 2x\sqrt {3} + 1) \implies x = \frac { - 8\sqrt {3}}{13}. \end{eqnarray*}<br />
</cmath><br />
Since the triangle is symmetric with respect to the y-axis, the coordinates of <math>B</math> and <math>C</math> are now <math>\left(\frac {8\sqrt {3}}{13},y_{0}\right)</math> and <math>\left(\frac { - 8\sqrt {3}}{13},y_{0}\right),</math> respectively, for some value of <math>y_{0}.</math><br />
<br />
Since we're going to use the distance formula, the value of <math>y_{0}</math> is irrelevant. Our answer is<br />
<cmath><br />
BC = \sqrt {2\left(\frac {8\sqrt {3}}{13}\right)^{2}} = \sqrt {\frac {768}{169}}\implies m + n = \boxed{937}.<br />
</cmath><br />
<br />
=== Solution 2 ===<br />
Solving for <math>y</math> in terms of <math>x</math> gives <math>y=\sqrt{4-x^2}/2</math>, so the two other points of the triangle are <math>(x,\sqrt{4-x^2}/2)</math> and <math>(-x,\sqrt{4-x^2}/2)</math>, which are a distance of <math>2x</math> apart. Thus <math>2x</math> equals the distance between <math>(x,\sqrt{4-x^2}/2)</math> and <math>(0,1)</math>, so by the distance formula we have <br />
<br />
<cmath>2x=\sqrt{x^2+(1-\sqrt{4-x^2}/2)^2}.</cmath> <br />
<br />
Squaring both sides and simplifying through algebra yields <math>x^2=192/169</math>, so <math>2x=\sqrt{768/169}</math> and the answer is <math>\boxed{937}</math>.<br />
<br />
=== Solution 3 ===<br />
Since the altitude goes along the <math>y</math> axis, this means that the base is a horizontal line, which means that the endpoints of the base are <math>(x,y)</math> and <math>(-x,y)</math>, and WLOG, we can say that <math>x</math> is positive.<br />
<br />
Now, since all sides of an equilateral triangle are the same, we can do this (distance from one of the endpoints of the base to the vertex and the length of the base):<br />
<br />
<math>\sqrt{x^2 + (y-1)^2} = 2x</math><br />
<br />
Square both sides,<br />
<br />
<math>x^2 + (y-1)^2 = 4x^2</math><br />
<math>(y-1)^2 = 3x^2</math><br />
<br />
Now, with the equation of the ellipse:<br />
<math>x^2 + 4y^2 = 4</math><br />
<math>x^2 = 4-4y^2</math><br />
<math>3x^2 = 12-12y^2</math><br />
<br />
Substituting, <br />
<br />
<math>12-12y^2 = y^2 - 2y +1</math><br />
<br />
Moving stuff around and solving:<br />
<br />
<math>y = \frac{-11}{3}, 1</math><br />
<br />
The second is found to be extraneous, so, when we go back and figure out <math>x</math> and then <math>2x</math> (which is the side length), we find it to be:<br />
<br />
<math>\sqrt{\frac{768}{169}}</math> <br />
<br />
and so we get the desired answer of <math>\boxed{937}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2001|n=I|num-b=4|num-a=6}}<br />
<br />
[[Category:Intermediate Geometry Problems]]</div>Engardehttps://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_5&diff=388361992 AIME Problems/Problem 52011-05-31T19:02:07Z<p>Engarde: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Let <math>S^{}_{}</math> be the set of all rational numbers <math>r^{}_{}</math>, <math>0^{}_{}<r<1</math>, that have a repeating decimal expansion in the form <math>0.abcabcabc\ldots=0.\overline{abc}</math>, where the digits <math>a^{}_{}</math>, <math>b^{}_{}</math>, and <math>c^{}_{}</math> are not necessarily distinct. To write the elements of <math>S^{}_{}</math> as fractions in lowest terms, how many different numerators are required?<br />
<br />
== Solution ==<br />
We consider the method in which repeating decimals are normally converted to fractions with an example:<br />
<br />
<math>x=0.\overline{176}</math><br />
<br />
<math>\Rightarrow 1000x=176.\overline{176}</math><br />
<br />
<math>\Rightarrow 999x=1000x-x=176</math><br />
<br />
<math>\Rightarrow x=\frac{176}{999}</math><br />
<br />
Thus, let <math>x=0.\overline{abc}</math><br />
<br />
<math>\Rightarrow 1000x=abc.\overline{abc}</math><br />
<br />
<math>\Rightarrow 999x=1000x-x=abc</math><br />
<br />
<math>\Rightarrow x=\frac{abc}{999}</math><br />
<br />
If <math>abc</math> is not divisible by <math>3</math> or <math>37</math>, then this is in lowest terms. Let us consider the other multiples: <math>333</math> multiples of <math>3</math>, <math>27</math> of <math>37</math>, and <math>9</math> of <math>3</math> and <math>37</math>, so <math>999-333-27+9 = 648</math>, which is the amount that are neither. The <math>12</math> numbers that are multiples of <math>81</math> reduce to multiples of <math>3</math>. We have to count these since it will reduce to a multiple of <math>3</math> which we have removed from <math>999</math>, but, this cannot be removed since the numerator cannot cancel the <math>3</math>.There aren't any numbers which are multiples of <math>37^2</math>, so we can't get numerators which are multiples of <math>37</math>. Therefore <math>648 + 12 = \boxed{660}</math>.<br />
<br />
{{AIME box|year=1992|num-b=4|num-a=6}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]</div>Engardehttps://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_3&diff=378461988 AIME Problems/Problem 32011-03-30T00:59:03Z<p>Engarde: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Find <math>(\log_2 x)^2</math> if <math>\log_2 (\log_8 x) = \log_8 (\log_2 x)</math>.<br />
<br />
== Solution ==<br />
Raise both as [[exponent]]s with base 8:<br />
<br />
<div style="text-align:center;"><br />
<math><br />
\begin{eqnarray*}<br />
8^{\log_2 (\log_8 x)} &=& 8^{\log_8 (\log_2 x)}\\<br />
2^{3 \log_2(\log_8x)}} &=& \log_2x\\<br />
(\log_8x)^3 &=& \log_2x\\<br />
\left(\frac{\log_2x}{\log_28}\right)^3 &=& \log_2x\\<br />
(\log_2x)^2 &=& (\log_28)^3 = \boxed{27}\\<br />
\end{eqnarray*}<br />
</math></div><br />
<br />
----<br />
<br />
A quick explanation of the steps: On the 1st step, we use the property of [[logarithm]]s that <math>a^{\log_a x} = x</math>. On the 2nd step, we use the fact that <math>k \log_a x = \log_a x^k</math>. On the 3rd step, we use the [[change of base formula]], which states <math>\log_a b = \frac{\log_k b}{\log_k a}</math> for arbitrary <math>k</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1988|num-b=2|num-a=4}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Engardehttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_5&diff=371631983 AIME Problems/Problem 52011-02-27T16:03:18Z<p>Engarde: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
Suppose that the sum of the squares of two complex numbers <math>x</math> and <math>y</math> is <math>7</math> and the sum of the cubes is <math>10</math>. What is the largest real value that <math>x + y</math> can have?<br />
<br />
== Solution 1==<br />
One way to solve this problem seems to be by [[substitution]].<br />
<br />
<math>x^2+y^2=(x+y)^2-2xy=7</math> and<br />
<math>x^3+y^3=(x+y)(x^2-xy+y^2)=(7-xy)(x+y)=10</math><br />
<br />
Because we are only left with <math>x+y</math> and <math>xy</math>, substitution won't be too bad. Let <math>w=x+y</math> and <math>z=xy</math>.<br />
<br />
We get <math>w^2-2z=7</math> and<br />
<math>w(7-z)=10</math><br />
<br />
Because we want the largest possible <math>w</math>, let's find an expression for <math>z</math> in terms of <math>w</math>.<br />
<br />
<math>w^2-7=2z \implies z=\frac{w^2-7}{2}</math>.<br />
<br />
Substituting, <math>w^3-21w+20=0</math>. Factored, <math>(w-1)(w+5)(w-4)=0</math> (the Rational Root Theorem may be used here, along with synthetic division)<br />
<br />
The largest possible solution is therefore <math>x+y=w=\boxed{004}</math>.<br />
<br />
== Solution 2==<br />
An alternate way to solve this is to let <math>x=a+bi</math> and <math>y=c+di</math>.<br />
<br />
Because we are looking for a value of <math>x+y</math> that is real, we know that <math>d=-b</math>, and thus <math>y=c-bi</math>.<br />
<br />
Expanding <math>x^2+y^2=7+0i</math> will give two equations, since the real and imaginary parts must match up.<br />
<br />
<math>(a+bi)^2+(c-bi)^2=7+0i</math><br />
<br />
<math>(a^2+c^2-2b^2)+(2ab-2cb)i=7+0i</math><br />
<br />
Looking at the imaginary part of that equation, <math>2ab-2cb=0</math>, so <math>a=c</math>, and <math>x</math> and <math>y</math> are actually complex conjugates.<br />
<br />
Looking at the real part of the equation and plugging in <math>a=c</math>, <math>2a^2-2b^2=7</math>, or <math>2b^2=2a^2-7</math>.<br />
<br />
Now, evaluating the real part of <math>(a+bi)^3+(a-bi)^3</math>, which equals <math>10</math> (ignoring the odd powers of <math>i</math>, since they would not result in something in the form of <math>10+0i</math>):<br />
<br />
<math>a^3+3a(bi)^2+a^3+3a(-bi)^2=10</math><br />
<br />
<math>2a^3-6ab^2=10</math><br />
<br />
Since we know that <math>2b^2=2a^2-7</math>, it can be plugged in for <math>b^2</math> in the above equataion to yield:<br />
<br />
<math>2a^3-3a(2a^2-7)=10</math><br />
<br />
<math>-4a^3+21a=10</math><br />
<br />
<math>4a^3-21a+10=0</math><br />
<br />
Since the problem is looking for <math>x+y=2a</math> to be a positive integer, only positive half-integers (and whole-integers) need to be tested. From the Rational Roots theorem, <math>a=10, a=5, a=\frac{5}{2}</math> all fail, but <math>a=2</math> does work. Thus, the real part of both numbers is <math>2</math>, and their sum is <math>\boxed{004}</math><br />
<br />
== See also ==<br />
{{AIME box|year=1983|num-b=4|num-a=6}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Engardehttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_5&diff=371621983 AIME Problems/Problem 52011-02-27T15:58:41Z<p>Engarde: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
Suppose that the sum of the squares of two complex numbers <math>x</math> and <math>y</math> is <math>7</math> and the sum of the cubes is <math>10</math>. What is the largest real value that <math>x + y</math> can have?<br />
<br />
== Solution 1==<br />
One way to solve this problem seems to be by [[substitution]].<br />
<br />
<math>x^2+y^2=(x+y)^2-2xy=7</math> and<br />
<math>x^3+y^3=(x+y)(x^2-xy+y^2)=(7-xy)(x+y)=10</math><br />
<br />
Because we are only left with <math>x+y</math> and <math>xy</math>, substitution won't be too bad. Let <math>w=x+y</math> and <math>z=xy</math>.<br />
<br />
We get <math>w^2-2z=7</math> and<br />
<math>w(7-z)=10</math><br />
<br />
Because we want the largest possible <math>w</math>, let's find an expression for <math>z</math> in terms of <math>w</math>.<br />
<br />
<math>w^2-7=2z \implies z=\frac{w^2-7}{2}</math>.<br />
<br />
Substituting, <math>w^3-21w+20=0</math>. Factored, <math>(w-1)(w+5)(w-4)=0</math> (the Rational Root Theorem may be used here, along with synthetic division)<br />
<br />
The largest possible solution is therefore <math>x+y=w=\boxed{004}</math>.<br />
<br />
== Solution 2==<br />
An alternate way to solve this is to let <math>x=a+bi</math> and <math>y=c+di</math>.<br />
<br />
Because we are looking for a value of <math>x+y</math> that is real, we know that <math>d=-b</math>, and thus <math>y=c-bi</math>.<br />
<br />
Expanding <math>x^2+y^2=7+0i</math> will give two equations, since the real and imaginary parts must match up.<br />
<br />
<math>(a+bi)^2+(c-bi)^2=7+0i</math><br />
<br />
<math>(a^2+c^2-2b^2)+(2ab-2cb)i=7+0i</math><br />
<br />
Looking at the imaginary part of that equation, <math>2ab-2cb=0</math>, so <math>a=c</math>, and <math>x</math> and <math>y</math> are actually complex conjugates.<br />
<br />
Looking at the real part of the equation and plugging in <math>a=c</math>, <math>2a^2-2b^2=7</math>, or <math>2b^2=2a^2-7</math>.<br />
<br />
Now, evaluating the real part of <math>(a+bi)^3+(a-bi)^3</math>, which equals <math>10</math> (ignoring the odd powers of <math>i</math>):<br />
<br />
<math>a^3+3a(bi)^2+a^3+3a(-bi)^2=10</math><br />
<br />
<math>2a^3-6ab^2=10</math><br />
<br />
Since we know that <math>2b^2=2a^2-7</math>, it can be plugged in for <math>b^2</math> in the above equataion to yield:<br />
<br />
<math>2a^3-3a(2a^2-7)=10</math><br />
<br />
<math>-4a^3+21a=10</math><br />
<br />
<math>4a^3-21a+10=0</math><br />
<br />
Since the problem is looking for <math>x+y=2a</math> to be a positive integer, only positive half-integers (and whole-integers) need to be tested. From the Rational Roots theorem, <math>a=10, a=5, a=\frac{5}{2}</math> all fail, but <math>a=2</math> does work. Thus, the real part of both numbers is <math>2</math>, and their sum is <math>\boxed{004}</math><br />
<br />
== See also ==<br />
{{AIME box|year=1983|num-b=4|num-a=6}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Engardehttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_5&diff=371611983 AIME Problems/Problem 52011-02-27T15:58:04Z<p>Engarde: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
Suppose that the sum of the squares of two complex numbers <math>x</math> and <math>y</math> is <math>7</math> and the sum of the cubes is <math>10</math>. What is the largest real value that <math>x + y</math> can have?<br />
<br />
== Solution 1==<br />
One way to solve this problem seems to be by [[substitution]].<br />
<br />
<math>x^2+y^2=(x+y)^2-2xy=7</math> and<br />
<math>x^3+y^3=(x+y)(x^2-xy+y^2)=(7-xy)(x+y)=10</math><br />
<br />
Because we are only left with <math>x+y</math> and <math>xy</math>, substitution won't be too bad. Let <math>w=x+y</math> and <math>z=xy</math>.<br />
<br />
We get <math>w^2-2z=7</math> and<br />
<math>w(7-z)=10</math><br />
<br />
Because we want the largest possible <math>w</math>, let's find an expression for <math>z</math> in terms of <math>w</math>.<br />
<br />
<math>w^2-7=2z \implies z=\frac{w^2-7}{2}</math>.<br />
<br />
Substituting, <math>w^3-21w+20=0</math>. Factored, <math>(w-1)(w+5)(w-4)=0</math> (the Rational Root Theorem may be used here, along with synthetic division)<br />
<br />
The largest possible solution is therefore <math>x+y=w=4</math>.<br />
<br />
== Solution 2==<br />
An alternate way to solve this is to let <math>x=a+bi</math> and <math>y=c+di</math>.<br />
<br />
Because we are looking for a value of <math>x+y</math> that is real, we know that <math>d=-b</math>, and thus <math>y=c-bi</math>.<br />
<br />
Expanding <math>x^2+y^2=7+0i</math> will give two equations, since the real and imaginary parts must match up.<br />
<br />
<math>(a+bi)^2+(c-bi)^2=7+0i</math><br />
<br />
<math>(a^2+c^2-2b^2)+(2ab-2cb)i=7+0i</math><br />
<br />
Looking at the imaginary part of that equation, <math>2ab-2cb=0</math>, so <math>a=c</math>, and <math>x</math> and <math>y</math> are actually complex conjugates.<br />
<br />
Looking at the real part of the equation and plugging in <math>a=c</math>, <math>2a^2-2b^2=7</math>, or <math>2b^2=2a^2-7</math>.<br />
<br />
Now, evaluating the real part of <math>(a+bi)^3+(a-bi)^3</math>, which equals <math>10</math> (ignoring the odd powers of <math>i</math>):<br />
<br />
<math>a^3+3a(bi)^2+a^3+3a(-bi)^2=10</math><br />
<br />
<math>2a^3-6ab^2=10</math><br />
<br />
Since we know that <math>2b^2=2a^2-7</math>, it can be plugged in for <math>b^2</math> in the above equataion to yield:<br />
<br />
<math>2a^3-3a(2a^2-7)=10</math><br />
<br />
<math>-4a^3+21a=10</math><br />
<br />
<math>4a^3-21a+10=0</math><br />
<br />
Since the problem is looking for <math>x+y=2a</math> to be a positive integer, only positive half-integers (and whole-integers) need to be tested. From the Rational Roots theorem, <math>a=10, a=5, a=\frac{5}{2}</math> all fail, but <math>a=2</math> does work. Thus, the real part of both numbers is <math>2</math>, and their sum is <math>\boxed{004}</math><br />
<br />
== See also ==<br />
{{AIME box|year=1983|num-b=4|num-a=6}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Engardehttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_5&diff=371601983 AIME Problems/Problem 52011-02-27T15:56:48Z<p>Engarde: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
Suppose that the sum of the squares of two complex numbers <math>x</math> and <math>y</math> is <math>7</math> and the sum of the cubes is <math>10</math>. What is the largest real value that <math>x + y</math> can have?<br />
<br />
== Solution 1==<br />
One way to solve this problem seems to be by [[substitution]].<br />
<br />
<math>x^2+y^2=(x+y)^2-2xy=7</math> and<br />
<math>x^3+y^3=(x+y)(x^2-xy+y^2)=(7-xy)(x+y)=10</math><br />
<br />
Because we are only left with <math>x+y</math> and <math>xy</math>, substitution won't be too bad. Let <math>w=x+y</math> and <math>z=xy</math>.<br />
<br />
We get <math>w^2-2z=7</math> and<br />
<math>w(7-z)=10</math><br />
<br />
Because we want the largest possible <math>w</math>, let's find an expression for <math>z</math> in terms of <math>w</math>.<br />
<br />
<math>w^2-7=2z \implies z=\frac{w^2-7}{2}</math>.<br />
<br />
Substituting, <math>w^3-21w+20=0</math>. Factored, <math>(w-1)(w+5)(w-4)=0</math><br />
<br />
The largest possible solution is therefore <math>x+y=w=4</math>.<br />
<br />
== Solution 2==<br />
An alternate way to solve this is to let <math>x=a+bi</math> and <math>y=c+di</math>.<br />
<br />
Because we are looking for a value of <math>x+y</math> that is real, we know that <math>d=-b</math>, and thus <math>y=c-bi</math>.<br />
<br />
Expanding <math>x^2+y^2=7+0i</math> will give two equations, since the real and imaginary parts must match up.<br />
<br />
<math>(a+bi)^2+(c-bi)^2=7+0i</math><br />
<br />
<math>(a^2+c^2-2b^2)+(2ab-2cb)i=7+0i</math><br />
<br />
Looking at the imaginary part of that equation, <math>2ab-2cb=0</math>, so <math>a=c</math>, and <math>x</math> and <math>y</math> are actually complex conjugates.<br />
<br />
Looking at the real part of the equation and plugging in <math>a=c</math>, <math>2a^2-2b^2=7</math>, or <math>2b^2=2a^2-7</math>.<br />
<br />
Now, evaluating the real part of <math>(a+bi)^3+(a-bi)^3</math>, which equals <math>10</math> (ignoring the odd powers of <math>i</math>):<br />
<br />
<math>a^3+3a(bi)^2+a^3+3a(-bi)^2=10</math><br />
<br />
<math>2a^3-6ab^2=10</math><br />
<br />
Since we know that <math>2b^2=2a^2-7</math>, it can be plugged in for <math>b^2</math> in the above equataion to yield:<br />
<br />
<math>2a^3-3a(2a^2-7)=10</math><br />
<br />
<math>-4a^3+21a=10</math><br />
<br />
<math>4a^3-21a+10=0</math><br />
<br />
Since the problem is looking for <math>x+y=2a</math> to be a positive integer, only positive half-integers (and whole-integers) need to be tested. From the Rational Roots theorem, <math>a=10, a=5, a=\frac{5}{2}</math> all fail, but <math>a=2</math> does work. Thus, the real part of both numbers is <math>2</math>, and their sum is <math>\boxed{004}</math><br />
<br />
== See also ==<br />
{{AIME box|year=1983|num-b=4|num-a=6}}<br />
<br />
[[Category:Intermediate Algebra Problems]]</div>Engarde