https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mathboy11&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T10:26:06ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_12&diff=1048851997 AIME Problems/Problem 122019-03-23T18:32:24Z<p>Mathboy11: /* Solution 7 */</p>
<hr />
<div>== Problem ==<br />
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to<br />
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath><br />
In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.<br />
<br />
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so<br />
<cmath>\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath><br />
<br />
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.<br />
<br />
=== Solution 2 ===<br />
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>. <br />
<br />
(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get <br />
<br />
<cmath>\begin{align*}<br />
19 &= \frac{b - \frac da}{19 - e} + e\\<br />
97 &= \frac{b - \frac da}{97 - e} + e\\<br />
b - \frac da &= (19 - e)^2 = (97 - e)^2\\<br />
19 - e &= \pm (97 - e)<br />
\end{align*}</cmath><br />
<br />
Clearly we can discard the positive root, so <math>e = 58</math>.<br />
<br />
=== Solution 3 ===<br />
<!-- some linear algebra --><br />
We first note (as before) that the number not in the range of<br />
<cmath> f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} </cmath><br />
is <math>a/c</math>, as <math>\frac{b-ad/c}{cx+d}</math> is evidently never 0 (otherwise, <math>f</math><br />
would be a constant function, violating the condition <math>f(19) \neq f(97)</math>).<br />
<br />
We may represent the real number <math>x/y</math> as<br />
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]<br />
considered equivalent if they are scalar multiples of each other. Similarly,<br />
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix<br />
<math>\begin{pmatrix} A & B\\ C& D \end{pmatrix}</math>. Function composition and<br />
evaluation then become matrix multiplication.<br />
<br />
Now in general,<br />
<cmath> f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} =<br />
\frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .</cmath><br />
In our problem <math>f^2(x) = x</math>. It follows that<br />
<cmath> \begin{pmatrix} a & b \\ c& d \end{pmatrix} = K<br />
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , </cmath><br />
for some nonzero real <math>K</math>. Since<br />
<cmath> \frac{a}{d} = \frac{b}{-b} = K, </cmath><br />
it follows that <math>a = -d</math>. (In fact, this condition condition is equivalent<br />
to the condition that <math>f(f(x)) = x</math> for all <math>x</math> in the domain of <math>f</math>.)<br />
<br />
We next note that the function<br />
<cmath> g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} </cmath><br />
evaluates to 0 when <math>x</math> equals 19 and 97. Therefore<br />
<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath><br />
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>,<br />
our answer.<br />
<br />
=== Solution 4 ===<br />
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.<br />
<br />
<br />
Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath><br />
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath><br />
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that <math>a = -d</math>.<br />
<br />
This solution follows in the same manner as the last paragraph of the first solution.<br />
<br />
=== Solution 5 ===<br />
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.<br />
<br />
=== Solution 6===<br />
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy<br />
<br />
=== Solution 7===<br />
Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^{-1}(x)=\frac{19d-b}{a-19c}</math>. Since <math>f(f(x))=x</math>, <math>f(x)=f^{-1}(x)</math>, so substituting 19 and 97 yields the system, <math>\begin{array}{lcl} \frac{19a+b}{19c+d} & = & \frac{19d-b}{a-19c} \\ \frac{97a+b}{97c+d} & = & \frac{97d-b}{a-97c} \end{array}</math>, and after multiplying each equation out and subtracting equation 1 from 2, and after simplifying, you will get <math>116c=a-d</math>. Coincidentally, then <math>116c+d=a</math>, which is familiar because <math>f(116)=\frac{116a+b}{116c+d}</math>, and since <math>116c+d=a</math>, <math>f(116)=\frac{116a+b}{a}</math>. Also, <math>f(f(116))=\frac{a(\frac{116a+b}{a})+b}{c(\frac{116a+b}{a})+d}=116</math>, due to <math>f(f(x))=x</math>. This simplifies to <math>\frac{116a+2b}{c(\frac{116a+b}{a})+d}=116</math>, <math>116a+2b=116(c(\frac{116a+b}{a})+d)</math>, <math>116a+2b=116(c(116+\frac{b}{a})+d)</math>, <math>116a+2b=116c(116+\frac{b}{a})+116d</math>, and substituting <math>116c+d=a</math> and simplifying, you get <math>2b=116c(\frac{b}{a})</math>, then <math>\frac{a}{c}=58</math>. Looking at <math>116c=a-d</math> one more time, we get <math>116=\frac{a}{c}+\frac{-d}{c}</math>, and substituting, we get <math>\frac{-d}{c}=58</math>, and we are done. (Answer is 58)<br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_12&diff=1048841997 AIME Problems/Problem 122019-03-23T18:31:40Z<p>Mathboy11: /* Solution 7 */</p>
<hr />
<div>== Problem ==<br />
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to<br />
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath><br />
In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.<br />
<br />
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so<br />
<cmath>\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath><br />
<br />
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.<br />
<br />
=== Solution 2 ===<br />
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>. <br />
<br />
(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get <br />
<br />
<cmath>\begin{align*}<br />
19 &= \frac{b - \frac da}{19 - e} + e\\<br />
97 &= \frac{b - \frac da}{97 - e} + e\\<br />
b - \frac da &= (19 - e)^2 = (97 - e)^2\\<br />
19 - e &= \pm (97 - e)<br />
\end{align*}</cmath><br />
<br />
Clearly we can discard the positive root, so <math>e = 58</math>.<br />
<br />
=== Solution 3 ===<br />
<!-- some linear algebra --><br />
We first note (as before) that the number not in the range of<br />
<cmath> f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} </cmath><br />
is <math>a/c</math>, as <math>\frac{b-ad/c}{cx+d}</math> is evidently never 0 (otherwise, <math>f</math><br />
would be a constant function, violating the condition <math>f(19) \neq f(97)</math>).<br />
<br />
We may represent the real number <math>x/y</math> as<br />
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]<br />
considered equivalent if they are scalar multiples of each other. Similarly,<br />
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix<br />
<math>\begin{pmatrix} A & B\\ C& D \end{pmatrix}</math>. Function composition and<br />
evaluation then become matrix multiplication.<br />
<br />
Now in general,<br />
<cmath> f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} =<br />
\frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .</cmath><br />
In our problem <math>f^2(x) = x</math>. It follows that<br />
<cmath> \begin{pmatrix} a & b \\ c& d \end{pmatrix} = K<br />
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , </cmath><br />
for some nonzero real <math>K</math>. Since<br />
<cmath> \frac{a}{d} = \frac{b}{-b} = K, </cmath><br />
it follows that <math>a = -d</math>. (In fact, this condition condition is equivalent<br />
to the condition that <math>f(f(x)) = x</math> for all <math>x</math> in the domain of <math>f</math>.)<br />
<br />
We next note that the function<br />
<cmath> g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} </cmath><br />
evaluates to 0 when <math>x</math> equals 19 and 97. Therefore<br />
<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath><br />
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>,<br />
our answer.<br />
<br />
=== Solution 4 ===<br />
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.<br />
<br />
<br />
Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath><br />
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath><br />
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that <math>a = -d</math>.<br />
<br />
This solution follows in the same manner as the last paragraph of the first solution.<br />
<br />
=== Solution 5 ===<br />
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.<br />
<br />
=== Solution 6===<br />
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy<br />
<br />
=== Solution 7===<br />
Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^{-1}(x)=\frac{19d-b}{a-19c}</math>. Since <math>f(f(x))=x</math>, <math>f(x)=f^{-1}(x)</math>, so substituting 19 and 97 yields the system, <math>\begin{array}{lcl} \frac{19a+b}{19c+d} & = & \frac{19d-b}{a-19c} \\ \frac{97a+b}{97c+d} & = & \frac{97d-b}{a-97c} \end{array}</math>, and after multiplying each equation out and subtracting equation 1 from 2, and after simplifying, you will get <math>116c=a-d</math>. Coincidentally, then <math>116c+d=a</math>, which is familiar because <math>f(116)=\frac{116a+b}{116c+d}</math>, and since <math>116c+d=a</math>, <math>f(116)=\frac{116a+b}{a}</math>. Also, <math>f(f(116))=\frac{a(\frac{116a+b}{a})+b}{c(\frac{116a+b}{a})+d}=116</math>, due to <math>f(f(x))=x</math>. This simplifies to <math>\frac{116a+2b}{c(\frac{116a+b}{a})+d}=116</math>, <math>116a+2b=116(c(\frac{116a+b}{a})+d)</math>, <math>116a+2b=116(c(116+\frac{b}{a})+d)</math>, <math>116a+2b=116c(116+\frac{b}{a})+116d</math>, and substituting <math>116c+d=a</math> and simplifying, you get <math>2b=116c(\frac{b}{a})</math>, then <math>\frac{a}{c}=58</math>. Looking at <math>116c=a-d</math> one more time, we get $116=\frac{a}{c}+\frac{-d}{c}<br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_12&diff=1048831997 AIME Problems/Problem 122019-03-23T18:29:54Z<p>Mathboy11: /* Solution 7 */</p>
<hr />
<div>== Problem ==<br />
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to<br />
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath><br />
In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.<br />
<br />
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so<br />
<cmath>\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath><br />
<br />
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.<br />
<br />
=== Solution 2 ===<br />
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>. <br />
<br />
(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get <br />
<br />
<cmath>\begin{align*}<br />
19 &= \frac{b - \frac da}{19 - e} + e\\<br />
97 &= \frac{b - \frac da}{97 - e} + e\\<br />
b - \frac da &= (19 - e)^2 = (97 - e)^2\\<br />
19 - e &= \pm (97 - e)<br />
\end{align*}</cmath><br />
<br />
Clearly we can discard the positive root, so <math>e = 58</math>.<br />
<br />
=== Solution 3 ===<br />
<!-- some linear algebra --><br />
We first note (as before) that the number not in the range of<br />
<cmath> f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} </cmath><br />
is <math>a/c</math>, as <math>\frac{b-ad/c}{cx+d}</math> is evidently never 0 (otherwise, <math>f</math><br />
would be a constant function, violating the condition <math>f(19) \neq f(97)</math>).<br />
<br />
We may represent the real number <math>x/y</math> as<br />
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]<br />
considered equivalent if they are scalar multiples of each other. Similarly,<br />
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix<br />
<math>\begin{pmatrix} A & B\\ C& D \end{pmatrix}</math>. Function composition and<br />
evaluation then become matrix multiplication.<br />
<br />
Now in general,<br />
<cmath> f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} =<br />
\frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .</cmath><br />
In our problem <math>f^2(x) = x</math>. It follows that<br />
<cmath> \begin{pmatrix} a & b \\ c& d \end{pmatrix} = K<br />
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , </cmath><br />
for some nonzero real <math>K</math>. Since<br />
<cmath> \frac{a}{d} = \frac{b}{-b} = K, </cmath><br />
it follows that <math>a = -d</math>. (In fact, this condition condition is equivalent<br />
to the condition that <math>f(f(x)) = x</math> for all <math>x</math> in the domain of <math>f</math>.)<br />
<br />
We next note that the function<br />
<cmath> g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} </cmath><br />
evaluates to 0 when <math>x</math> equals 19 and 97. Therefore<br />
<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath><br />
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>,<br />
our answer.<br />
<br />
=== Solution 4 ===<br />
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.<br />
<br />
<br />
Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath><br />
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath><br />
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that <math>a = -d</math>.<br />
<br />
This solution follows in the same manner as the last paragraph of the first solution.<br />
<br />
=== Solution 5 ===<br />
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.<br />
<br />
=== Solution 6===<br />
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy<br />
<br />
=== Solution 7===<br />
Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^{-1}(x)=\frac{19d-b}{a-19c}</math>. Since <math>f(f(x))=x</math>, <math>f(x)=f^{-1}(x)</math>, so substituting 19 and 97 yields the system, <math>\begin{array}{lcl} \frac{19a+b}{19c+d} & = & \frac{19d-b}{a-19c} \\ \frac{97a+b}{97c+d} & = & \frac{97d-b}{a-97c} \end{array}</math>, and after multiplying each equation out and subtracting equation 1 from 2, and after simplifying, you will get <math>116c=a-d</math>. Coincidentally, then <math>116c+d=a</math>, which is familiar because <math>f(116)=\frac{116a+b}{116c+d}</math>, and since <math>116c+d=a</math>, <math>f(116)=\frac{116a+b}{a}</math>. Also, <math>f(f(116))=\frac{a(\frac{116a+b}{a})+b}{c(\frac{116a+b}{a})+d}=116</math>, due to <math>f(f(x))=x</math>. This simplifies to <math>\frac{116a+2b}{c(\frac{116a+b}{a})+d}=116</math>, <math>116a+2b=116(c(\frac{116a+b}{a})+d)</math>, <math>116a+2b=116(c(116+\frac{b}{a})+d)</math>, <math>116a+2b=116c(116+\frac{b}{a})+116d</math>, and substituting <math>116c+d=a</math> and simplifying, you get <math>2b=116c(\frac{b}{a})</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_12&diff=1048821997 AIME Problems/Problem 122019-03-23T18:29:40Z<p>Mathboy11: /* Solution 7 */</p>
<hr />
<div>== Problem ==<br />
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to<br />
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath><br />
In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.<br />
<br />
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so<br />
<cmath>\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath><br />
<br />
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.<br />
<br />
=== Solution 2 ===<br />
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>. <br />
<br />
(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get <br />
<br />
<cmath>\begin{align*}<br />
19 &= \frac{b - \frac da}{19 - e} + e\\<br />
97 &= \frac{b - \frac da}{97 - e} + e\\<br />
b - \frac da &= (19 - e)^2 = (97 - e)^2\\<br />
19 - e &= \pm (97 - e)<br />
\end{align*}</cmath><br />
<br />
Clearly we can discard the positive root, so <math>e = 58</math>.<br />
<br />
=== Solution 3 ===<br />
<!-- some linear algebra --><br />
We first note (as before) that the number not in the range of<br />
<cmath> f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} </cmath><br />
is <math>a/c</math>, as <math>\frac{b-ad/c}{cx+d}</math> is evidently never 0 (otherwise, <math>f</math><br />
would be a constant function, violating the condition <math>f(19) \neq f(97)</math>).<br />
<br />
We may represent the real number <math>x/y</math> as<br />
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]<br />
considered equivalent if they are scalar multiples of each other. Similarly,<br />
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix<br />
<math>\begin{pmatrix} A & B\\ C& D \end{pmatrix}</math>. Function composition and<br />
evaluation then become matrix multiplication.<br />
<br />
Now in general,<br />
<cmath> f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} =<br />
\frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .</cmath><br />
In our problem <math>f^2(x) = x</math>. It follows that<br />
<cmath> \begin{pmatrix} a & b \\ c& d \end{pmatrix} = K<br />
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , </cmath><br />
for some nonzero real <math>K</math>. Since<br />
<cmath> \frac{a}{d} = \frac{b}{-b} = K, </cmath><br />
it follows that <math>a = -d</math>. (In fact, this condition condition is equivalent<br />
to the condition that <math>f(f(x)) = x</math> for all <math>x</math> in the domain of <math>f</math>.)<br />
<br />
We next note that the function<br />
<cmath> g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} </cmath><br />
evaluates to 0 when <math>x</math> equals 19 and 97. Therefore<br />
<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath><br />
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>,<br />
our answer.<br />
<br />
=== Solution 4 ===<br />
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.<br />
<br />
<br />
Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath><br />
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath><br />
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that <math>a = -d</math>.<br />
<br />
This solution follows in the same manner as the last paragraph of the first solution.<br />
<br />
=== Solution 5 ===<br />
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.<br />
<br />
=== Solution 6===<br />
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy<br />
<br />
=== Solution 7===<br />
Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^{-1}(x)=\frac{19d-b}{a-19c}</math>. Since <math>f(f(x))=x</math>, <math>f(x)=f^{-1}(x)</math>, so substituting 19 and 97 yields the system, <math>\begin{array}{lcl} \frac{19a+b}{19c+d} & = & \frac{19d-b}{a-19c} \\ \frac{97a+b}{97c+d} & = & \frac{97d-b}{a-97c} \end{array}</math>, and after multiplying each equation out and subtracting equation 1 from 2, and after simplifying, you will get <math>116c=a-d</math>. Coincidentally, then <math>116c+d=a</math>, which is familiar because <math>f(116)=\frac{116a+b}{116c+d}</math>, and since <math>116c+d=a</math>, <math>f(116)=\frac{116a+b}{a}</math>. Also, <math>f(f(116))=\frac{a(\frac{116a+b}{a})+b}{c(\frac{116a+b}{a})+d}=116</math>, due to <math>f(f(x))=x</math>. This simplifies to <math>\frac{116a+2b}{c(\frac{116a+b}{a})+d}=116</math>, <math>116a+2b=116(c(\frac{116a+b}{a})+d)</math>, <math>116a+2b=116(c(116+\frac{b}{a})+d)</math>, <math>116a+2b=116c(116+\frac{b}{a})+116d</math>, and substituting <math>116c+d=a</math> and simplifying, you get <math>2b=116c(/frac{b}{a})</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_12&diff=1048811997 AIME Problems/Problem 122019-03-23T18:26:45Z<p>Mathboy11: /* Solution 7 */</p>
<hr />
<div>== Problem ==<br />
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to<br />
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath><br />
In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.<br />
<br />
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so<br />
<cmath>\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath><br />
<br />
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.<br />
<br />
=== Solution 2 ===<br />
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>. <br />
<br />
(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get <br />
<br />
<cmath>\begin{align*}<br />
19 &= \frac{b - \frac da}{19 - e} + e\\<br />
97 &= \frac{b - \frac da}{97 - e} + e\\<br />
b - \frac da &= (19 - e)^2 = (97 - e)^2\\<br />
19 - e &= \pm (97 - e)<br />
\end{align*}</cmath><br />
<br />
Clearly we can discard the positive root, so <math>e = 58</math>.<br />
<br />
=== Solution 3 ===<br />
<!-- some linear algebra --><br />
We first note (as before) that the number not in the range of<br />
<cmath> f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} </cmath><br />
is <math>a/c</math>, as <math>\frac{b-ad/c}{cx+d}</math> is evidently never 0 (otherwise, <math>f</math><br />
would be a constant function, violating the condition <math>f(19) \neq f(97)</math>).<br />
<br />
We may represent the real number <math>x/y</math> as<br />
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]<br />
considered equivalent if they are scalar multiples of each other. Similarly,<br />
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix<br />
<math>\begin{pmatrix} A & B\\ C& D \end{pmatrix}</math>. Function composition and<br />
evaluation then become matrix multiplication.<br />
<br />
Now in general,<br />
<cmath> f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} =<br />
\frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .</cmath><br />
In our problem <math>f^2(x) = x</math>. It follows that<br />
<cmath> \begin{pmatrix} a & b \\ c& d \end{pmatrix} = K<br />
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , </cmath><br />
for some nonzero real <math>K</math>. Since<br />
<cmath> \frac{a}{d} = \frac{b}{-b} = K, </cmath><br />
it follows that <math>a = -d</math>. (In fact, this condition condition is equivalent<br />
to the condition that <math>f(f(x)) = x</math> for all <math>x</math> in the domain of <math>f</math>.)<br />
<br />
We next note that the function<br />
<cmath> g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} </cmath><br />
evaluates to 0 when <math>x</math> equals 19 and 97. Therefore<br />
<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath><br />
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>,<br />
our answer.<br />
<br />
=== Solution 4 ===<br />
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.<br />
<br />
<br />
Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath><br />
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath><br />
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that <math>a = -d</math>.<br />
<br />
This solution follows in the same manner as the last paragraph of the first solution.<br />
<br />
=== Solution 5 ===<br />
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.<br />
<br />
=== Solution 6===<br />
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy<br />
<br />
=== Solution 7===<br />
Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^{-1}(x)=\frac{19d-b}{a-19c}</math>. Since <math>f(f(x))=x</math>, <math>f(x)=f^{-1}(x)</math>, so substituting 19 and 97 yields the system, <math>\begin{array}{lcl} \frac{19a+b}{19c+d} & = & \frac{19d-b}{a-19c} \\ \frac{97a+b}{97c+d} & = & \frac{97d-b}{a-97c} \end{array}</math>, and after multiplying each equation out and subtracting equation 1 from 2, and after simplifying, you will get <math>116c=a-d</math>. Coincidentally, then <math>116c+d=a</math>, which is familiar because <math>f(116)=\frac{116a+b}{116c+d}</math>, and since <math>116c+d=a</math>, <math>f(116)=\frac{116a+b}{a}</math>. Also, <math>f(f(116))=\frac{a(\frac{116a+b}{a})+b}{c(\frac{116a+b}{a})+d}=116</math>, due to <math>f(f(x))=x</math>. This simplifies to <math>\frac{116a+2b}{c(\frac{116a+b}{a})+d}=116</math>, <math>116a+2b=116(c(\frac{116a+b}{a})+d)</math>, <math>116a+2b=116(c(116+\frac{b}{a})+d)</math>, <math>116a+2b=116c(116+\frac{b}{a})+116d)</math><br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_12&diff=1048801997 AIME Problems/Problem 122019-03-23T18:25:49Z<p>Mathboy11: /* Solution 7 */</p>
<hr />
<div>== Problem ==<br />
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to<br />
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath><br />
In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.<br />
<br />
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so<br />
<cmath>\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath><br />
<br />
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.<br />
<br />
=== Solution 2 ===<br />
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>. <br />
<br />
(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get <br />
<br />
<cmath>\begin{align*}<br />
19 &= \frac{b - \frac da}{19 - e} + e\\<br />
97 &= \frac{b - \frac da}{97 - e} + e\\<br />
b - \frac da &= (19 - e)^2 = (97 - e)^2\\<br />
19 - e &= \pm (97 - e)<br />
\end{align*}</cmath><br />
<br />
Clearly we can discard the positive root, so <math>e = 58</math>.<br />
<br />
=== Solution 3 ===<br />
<!-- some linear algebra --><br />
We first note (as before) that the number not in the range of<br />
<cmath> f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} </cmath><br />
is <math>a/c</math>, as <math>\frac{b-ad/c}{cx+d}</math> is evidently never 0 (otherwise, <math>f</math><br />
would be a constant function, violating the condition <math>f(19) \neq f(97)</math>).<br />
<br />
We may represent the real number <math>x/y</math> as<br />
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]<br />
considered equivalent if they are scalar multiples of each other. Similarly,<br />
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix<br />
<math>\begin{pmatrix} A & B\\ C& D \end{pmatrix}</math>. Function composition and<br />
evaluation then become matrix multiplication.<br />
<br />
Now in general,<br />
<cmath> f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} =<br />
\frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .</cmath><br />
In our problem <math>f^2(x) = x</math>. It follows that<br />
<cmath> \begin{pmatrix} a & b \\ c& d \end{pmatrix} = K<br />
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , </cmath><br />
for some nonzero real <math>K</math>. Since<br />
<cmath> \frac{a}{d} = \frac{b}{-b} = K, </cmath><br />
it follows that <math>a = -d</math>. (In fact, this condition condition is equivalent<br />
to the condition that <math>f(f(x)) = x</math> for all <math>x</math> in the domain of <math>f</math>.)<br />
<br />
We next note that the function<br />
<cmath> g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} </cmath><br />
evaluates to 0 when <math>x</math> equals 19 and 97. Therefore<br />
<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath><br />
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>,<br />
our answer.<br />
<br />
=== Solution 4 ===<br />
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.<br />
<br />
<br />
Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath><br />
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath><br />
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that <math>a = -d</math>.<br />
<br />
This solution follows in the same manner as the last paragraph of the first solution.<br />
<br />
=== Solution 5 ===<br />
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.<br />
<br />
=== Solution 6===<br />
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy<br />
<br />
=== Solution 7===<br />
Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^{-1}(x)=\frac{19d-b}{a-19c}</math>. Since <math>f(f(x))=x</math>, <math>f(x)=f^{-1}(x)</math>, so substituting 19 and 97 yields the system, <math>\begin{array}{lcl} \frac{19a+b}{19c+d} & = & \frac{19d-b}{a-19c} \\ \frac{97a+b}{97c+d} & = & \frac{97d-b}{a-97c} \end{array}</math>, and after multiplying each equation out and subtracting equation 1 from 2, and after simplifying, you will get <math>116c=a-d</math>. Coincidentally, then <math>116c+d=a</math>, which is familiar because <math>f(116)=\frac{116a+b}{116c+d}</math>, and since <math>116c+d=a</math>, <math>f(116)=\frac{116a+b}{a}</math>. Also, <math>f(f(116))=\frac{a(\frac{116a+b}{a})+b}{c(\frac{116a+b}{a})+d}=116</math>, due to <math>f(f(x))=x</math>. This simplifies to <math>\frac{116a+2b}{c(\frac{116a+b}{a})+d}=116</math>, <math>116a+2b=116(c(\frac{116a+b}{a})+d)</math>, <math>116a+2b=116(c(116+\frac{b}{a})+d)</math>, <math>116a+2b=116c(\frac{116a+b}{a})+116d)</math><br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_12&diff=1048791997 AIME Problems/Problem 122019-03-23T18:17:08Z<p>Mathboy11: /* Solution 7 */</p>
<hr />
<div>== Problem ==<br />
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to<br />
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath><br />
In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.<br />
<br />
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so<br />
<cmath>\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath><br />
<br />
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.<br />
<br />
=== Solution 2 ===<br />
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>. <br />
<br />
(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get <br />
<br />
<cmath>\begin{align*}<br />
19 &= \frac{b - \frac da}{19 - e} + e\\<br />
97 &= \frac{b - \frac da}{97 - e} + e\\<br />
b - \frac da &= (19 - e)^2 = (97 - e)^2\\<br />
19 - e &= \pm (97 - e)<br />
\end{align*}</cmath><br />
<br />
Clearly we can discard the positive root, so <math>e = 58</math>.<br />
<br />
=== Solution 3 ===<br />
<!-- some linear algebra --><br />
We first note (as before) that the number not in the range of<br />
<cmath> f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} </cmath><br />
is <math>a/c</math>, as <math>\frac{b-ad/c}{cx+d}</math> is evidently never 0 (otherwise, <math>f</math><br />
would be a constant function, violating the condition <math>f(19) \neq f(97)</math>).<br />
<br />
We may represent the real number <math>x/y</math> as<br />
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]<br />
considered equivalent if they are scalar multiples of each other. Similarly,<br />
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix<br />
<math>\begin{pmatrix} A & B\\ C& D \end{pmatrix}</math>. Function composition and<br />
evaluation then become matrix multiplication.<br />
<br />
Now in general,<br />
<cmath> f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} =<br />
\frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .</cmath><br />
In our problem <math>f^2(x) = x</math>. It follows that<br />
<cmath> \begin{pmatrix} a & b \\ c& d \end{pmatrix} = K<br />
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , </cmath><br />
for some nonzero real <math>K</math>. Since<br />
<cmath> \frac{a}{d} = \frac{b}{-b} = K, </cmath><br />
it follows that <math>a = -d</math>. (In fact, this condition condition is equivalent<br />
to the condition that <math>f(f(x)) = x</math> for all <math>x</math> in the domain of <math>f</math>.)<br />
<br />
We next note that the function<br />
<cmath> g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} </cmath><br />
evaluates to 0 when <math>x</math> equals 19 and 97. Therefore<br />
<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath><br />
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>,<br />
our answer.<br />
<br />
=== Solution 4 ===<br />
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.<br />
<br />
<br />
Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath><br />
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath><br />
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that <math>a = -d</math>.<br />
<br />
This solution follows in the same manner as the last paragraph of the first solution.<br />
<br />
=== Solution 5 ===<br />
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.<br />
<br />
=== Solution 6===<br />
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy<br />
<br />
=== Solution 7===<br />
Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^{-1}(x)=\frac{19d-b}{a-19c}</math>. Since <math>f(f(x))=x</math>, <math>f(x)=f^{-1}(x)</math>, so substituting 19 and 97 yields the system, <math>\begin{array}{lcl} \frac{19a+b}{19c+d} & = & \frac{19d-b}{a-19c} \\ \frac{97a+b}{97c+d} & = & \frac{97d-b}{a-97c} \end{array}</math>, and after multiplying each equation out and subtracting equation 1 from 2, and after simplifying, you will get <math>116c=a-d</math>. Coincidentally, then <math>116c+d=a</math>, which is familiar because <math>f(116)=\frac{116a+b}{116c+d}</math>, and since <math>116c+d=a</math>, <math>f(116)=\frac{116a+b}{a}</math>. Also, <math>f(f(116))=\frac{a(\frac{116a+b}{a})+b}{c(\frac{116a+b}{a})+d}=116</math>, due to <math>f(f(x))=x</math>. This simplifies to <math>\frac{116a+2b}{c(\frac{116a+b}{a})+d}=116</math>, <math>116a+2b=116(c(\frac{116a+b}{a})+d)</math>, <math>116a+2b=116(c(116+\frac{b}{a})+d)</math><br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_12&diff=1048781997 AIME Problems/Problem 122019-03-23T18:14:01Z<p>Mathboy11: /* Solution 7 */</p>
<hr />
<div>== Problem ==<br />
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to<br />
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath><br />
In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.<br />
<br />
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so<br />
<cmath>\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath><br />
<br />
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.<br />
<br />
=== Solution 2 ===<br />
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>. <br />
<br />
(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get <br />
<br />
<cmath>\begin{align*}<br />
19 &= \frac{b - \frac da}{19 - e} + e\\<br />
97 &= \frac{b - \frac da}{97 - e} + e\\<br />
b - \frac da &= (19 - e)^2 = (97 - e)^2\\<br />
19 - e &= \pm (97 - e)<br />
\end{align*}</cmath><br />
<br />
Clearly we can discard the positive root, so <math>e = 58</math>.<br />
<br />
=== Solution 3 ===<br />
<!-- some linear algebra --><br />
We first note (as before) that the number not in the range of<br />
<cmath> f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} </cmath><br />
is <math>a/c</math>, as <math>\frac{b-ad/c}{cx+d}</math> is evidently never 0 (otherwise, <math>f</math><br />
would be a constant function, violating the condition <math>f(19) \neq f(97)</math>).<br />
<br />
We may represent the real number <math>x/y</math> as<br />
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]<br />
considered equivalent if they are scalar multiples of each other. Similarly,<br />
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix<br />
<math>\begin{pmatrix} A & B\\ C& D \end{pmatrix}</math>. Function composition and<br />
evaluation then become matrix multiplication.<br />
<br />
Now in general,<br />
<cmath> f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} =<br />
\frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .</cmath><br />
In our problem <math>f^2(x) = x</math>. It follows that<br />
<cmath> \begin{pmatrix} a & b \\ c& d \end{pmatrix} = K<br />
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , </cmath><br />
for some nonzero real <math>K</math>. Since<br />
<cmath> \frac{a}{d} = \frac{b}{-b} = K, </cmath><br />
it follows that <math>a = -d</math>. (In fact, this condition condition is equivalent<br />
to the condition that <math>f(f(x)) = x</math> for all <math>x</math> in the domain of <math>f</math>.)<br />
<br />
We next note that the function<br />
<cmath> g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} </cmath><br />
evaluates to 0 when <math>x</math> equals 19 and 97. Therefore<br />
<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath><br />
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>,<br />
our answer.<br />
<br />
=== Solution 4 ===<br />
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.<br />
<br />
<br />
Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath><br />
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath><br />
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that <math>a = -d</math>.<br />
<br />
This solution follows in the same manner as the last paragraph of the first solution.<br />
<br />
=== Solution 5 ===<br />
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.<br />
<br />
=== Solution 6===<br />
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy<br />
<br />
=== Solution 7===<br />
Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^{-1}(x)=\frac{19d-b}{a-19c}</math>. Since <math>f(f(x))=x</math>, <math>f(x)=f^{-1}(x)</math>, so substituting 19 and 97 yields the system, <math>\begin{array}{lcl} \frac{19a+b}{19c+d} & = & \frac{19d-b}{a-19c} \\ \frac{97a+b}{97c+d} & = & \frac{97d-b}{a-97c} \end{array}</math>, and after multiplying each equation out and subtracting equation 1 from 2, and after simplifying, you will get <math>116c=a-d</math>. Coincidentally, then <math>116c+d=a</math>, which is familiar because <math>f(116)=\frac{116a+b}{116c+d}</math>, and since <math>116c+d=a</math>, <math>f(116)=\frac{116a+b}{a}</math>. Also, <math>f(f(116))=\frac{a(\frac{116a+b}{a})+b)}{c(\frac{116a+b}{a})+d)}=116</math>, due to <math>f(f(x))=x</math><br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_12&diff=1048771997 AIME Problems/Problem 122019-03-23T18:12:30Z<p>Mathboy11: /* Solution 7 */</p>
<hr />
<div>== Problem ==<br />
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to<br />
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath><br />
In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.<br />
<br />
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so<br />
<cmath>\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath><br />
<br />
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.<br />
<br />
=== Solution 2 ===<br />
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>. <br />
<br />
(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get <br />
<br />
<cmath>\begin{align*}<br />
19 &= \frac{b - \frac da}{19 - e} + e\\<br />
97 &= \frac{b - \frac da}{97 - e} + e\\<br />
b - \frac da &= (19 - e)^2 = (97 - e)^2\\<br />
19 - e &= \pm (97 - e)<br />
\end{align*}</cmath><br />
<br />
Clearly we can discard the positive root, so <math>e = 58</math>.<br />
<br />
=== Solution 3 ===<br />
<!-- some linear algebra --><br />
We first note (as before) that the number not in the range of<br />
<cmath> f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} </cmath><br />
is <math>a/c</math>, as <math>\frac{b-ad/c}{cx+d}</math> is evidently never 0 (otherwise, <math>f</math><br />
would be a constant function, violating the condition <math>f(19) \neq f(97)</math>).<br />
<br />
We may represent the real number <math>x/y</math> as<br />
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]<br />
considered equivalent if they are scalar multiples of each other. Similarly,<br />
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix<br />
<math>\begin{pmatrix} A & B\\ C& D \end{pmatrix}</math>. Function composition and<br />
evaluation then become matrix multiplication.<br />
<br />
Now in general,<br />
<cmath> f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} =<br />
\frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .</cmath><br />
In our problem <math>f^2(x) = x</math>. It follows that<br />
<cmath> \begin{pmatrix} a & b \\ c& d \end{pmatrix} = K<br />
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , </cmath><br />
for some nonzero real <math>K</math>. Since<br />
<cmath> \frac{a}{d} = \frac{b}{-b} = K, </cmath><br />
it follows that <math>a = -d</math>. (In fact, this condition condition is equivalent<br />
to the condition that <math>f(f(x)) = x</math> for all <math>x</math> in the domain of <math>f</math>.)<br />
<br />
We next note that the function<br />
<cmath> g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} </cmath><br />
evaluates to 0 when <math>x</math> equals 19 and 97. Therefore<br />
<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath><br />
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>,<br />
our answer.<br />
<br />
=== Solution 4 ===<br />
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.<br />
<br />
<br />
Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath><br />
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath><br />
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that <math>a = -d</math>.<br />
<br />
This solution follows in the same manner as the last paragraph of the first solution.<br />
<br />
=== Solution 5 ===<br />
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.<br />
<br />
=== Solution 6===<br />
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy<br />
<br />
=== Solution 7===<br />
Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^{-1}(x)=\frac{19d-b}{a-19c}</math>. Since <math>f(f(x))=x</math>, <math>f(x)=f^{-1}(x)</math>, so substituting 19 and 97 yields the system, <math>\begin{array}{lcl} \frac{19a+b}{19c+d} & = & \frac{19d-b}{a-19c} \\ \frac{97a+b}{97c+d} & = & \frac{97d-b}{a-97c} \end{array}</math>, and after multiplying each equation out and subtracting equation 1 from 2, and after simplifying, you will get <math>116c=a-d</math>. Coincidentally, then <math>116c+d=a</math>, which is familiar because <math>f(116)=\frac{116a+b}{116c+d}</math>, and since <math>116c+d=a</math>, <math>f(116)=\frac{116a+b}{a}</math>. Also, <math>f(f(116))=\frac{116(\frac{116a+b}{a})+b)}{116(\frac{116a+b}{a})+b)}</math><br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_12&diff=1048761997 AIME Problems/Problem 122019-03-23T18:09:50Z<p>Mathboy11: /* Solution 7 */</p>
<hr />
<div>== Problem ==<br />
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to<br />
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath><br />
In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.<br />
<br />
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so<br />
<cmath>\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath><br />
<br />
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.<br />
<br />
=== Solution 2 ===<br />
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>. <br />
<br />
(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get <br />
<br />
<cmath>\begin{align*}<br />
19 &= \frac{b - \frac da}{19 - e} + e\\<br />
97 &= \frac{b - \frac da}{97 - e} + e\\<br />
b - \frac da &= (19 - e)^2 = (97 - e)^2\\<br />
19 - e &= \pm (97 - e)<br />
\end{align*}</cmath><br />
<br />
Clearly we can discard the positive root, so <math>e = 58</math>.<br />
<br />
=== Solution 3 ===<br />
<!-- some linear algebra --><br />
We first note (as before) that the number not in the range of<br />
<cmath> f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} </cmath><br />
is <math>a/c</math>, as <math>\frac{b-ad/c}{cx+d}</math> is evidently never 0 (otherwise, <math>f</math><br />
would be a constant function, violating the condition <math>f(19) \neq f(97)</math>).<br />
<br />
We may represent the real number <math>x/y</math> as<br />
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]<br />
considered equivalent if they are scalar multiples of each other. Similarly,<br />
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix<br />
<math>\begin{pmatrix} A & B\\ C& D \end{pmatrix}</math>. Function composition and<br />
evaluation then become matrix multiplication.<br />
<br />
Now in general,<br />
<cmath> f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} =<br />
\frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .</cmath><br />
In our problem <math>f^2(x) = x</math>. It follows that<br />
<cmath> \begin{pmatrix} a & b \\ c& d \end{pmatrix} = K<br />
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , </cmath><br />
for some nonzero real <math>K</math>. Since<br />
<cmath> \frac{a}{d} = \frac{b}{-b} = K, </cmath><br />
it follows that <math>a = -d</math>. (In fact, this condition condition is equivalent<br />
to the condition that <math>f(f(x)) = x</math> for all <math>x</math> in the domain of <math>f</math>.)<br />
<br />
We next note that the function<br />
<cmath> g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} </cmath><br />
evaluates to 0 when <math>x</math> equals 19 and 97. Therefore<br />
<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath><br />
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>,<br />
our answer.<br />
<br />
=== Solution 4 ===<br />
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.<br />
<br />
<br />
Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath><br />
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath><br />
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that <math>a = -d</math>.<br />
<br />
This solution follows in the same manner as the last paragraph of the first solution.<br />
<br />
=== Solution 5 ===<br />
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.<br />
<br />
=== Solution 6===<br />
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy<br />
<br />
=== Solution 7===<br />
Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^{-1}(x)=\frac{19d-b}{a-19c}</math>. Since <math>f(f(x))=x</math>, <math>f(x)=f^{-1}(x)</math>, so substituting 19 and 97 yields the system, <math>\begin{array}{lcl} \frac{19a+b}{19c+d} & = & \frac{19d-b}{a-19c} \\ \frac{97a+b}{97c+d} & = & \frac{97d-b}{a-97c} \end{array}</math>, and after multiplying each equation out and subtracting equation 1 from 2, and after simplifying, you will get <math>116c=a-d</math>. Coincidentally, then <math>116c+d=a</math>, which is familiar because <math>f(116)=\frac{116a+b}{116c+d}</math>, and since <math>116c+d=a</math>, <math>f(116)=\frac{116a+b}{a}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_12&diff=1048751997 AIME Problems/Problem 122019-03-23T18:06:22Z<p>Mathboy11: /* Solution 7 */</p>
<hr />
<div>== Problem ==<br />
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to<br />
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath><br />
In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.<br />
<br />
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so<br />
<cmath>\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath><br />
<br />
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.<br />
<br />
=== Solution 2 ===<br />
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>. <br />
<br />
(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get <br />
<br />
<cmath>\begin{align*}<br />
19 &= \frac{b - \frac da}{19 - e} + e\\<br />
97 &= \frac{b - \frac da}{97 - e} + e\\<br />
b - \frac da &= (19 - e)^2 = (97 - e)^2\\<br />
19 - e &= \pm (97 - e)<br />
\end{align*}</cmath><br />
<br />
Clearly we can discard the positive root, so <math>e = 58</math>.<br />
<br />
=== Solution 3 ===<br />
<!-- some linear algebra --><br />
We first note (as before) that the number not in the range of<br />
<cmath> f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} </cmath><br />
is <math>a/c</math>, as <math>\frac{b-ad/c}{cx+d}</math> is evidently never 0 (otherwise, <math>f</math><br />
would be a constant function, violating the condition <math>f(19) \neq f(97)</math>).<br />
<br />
We may represent the real number <math>x/y</math> as<br />
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]<br />
considered equivalent if they are scalar multiples of each other. Similarly,<br />
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix<br />
<math>\begin{pmatrix} A & B\\ C& D \end{pmatrix}</math>. Function composition and<br />
evaluation then become matrix multiplication.<br />
<br />
Now in general,<br />
<cmath> f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} =<br />
\frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .</cmath><br />
In our problem <math>f^2(x) = x</math>. It follows that<br />
<cmath> \begin{pmatrix} a & b \\ c& d \end{pmatrix} = K<br />
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , </cmath><br />
for some nonzero real <math>K</math>. Since<br />
<cmath> \frac{a}{d} = \frac{b}{-b} = K, </cmath><br />
it follows that <math>a = -d</math>. (In fact, this condition condition is equivalent<br />
to the condition that <math>f(f(x)) = x</math> for all <math>x</math> in the domain of <math>f</math>.)<br />
<br />
We next note that the function<br />
<cmath> g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} </cmath><br />
evaluates to 0 when <math>x</math> equals 19 and 97. Therefore<br />
<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath><br />
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>,<br />
our answer.<br />
<br />
=== Solution 4 ===<br />
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.<br />
<br />
<br />
Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath><br />
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath><br />
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that <math>a = -d</math>.<br />
<br />
This solution follows in the same manner as the last paragraph of the first solution.<br />
<br />
=== Solution 5 ===<br />
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.<br />
<br />
=== Solution 6===<br />
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy<br />
<br />
=== Solution 7===<br />
Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^{-1}(x)=\frac{19d-b}{a-19c}</math>. Since <math>f(f(x))=x</math>, <math>f(x)=f^{-1}(x)</math>, so substituting 19 and 97 yields the system, <math>\begin{array}{lcl} \frac{19a+b}{19c+d} & = & \frac{19d-b}{a-19c} \\ \frac{97a+b}{97c+d} & = & \frac{97d-b}{a-97c} \end{array}</math><br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_12&diff=1048741997 AIME Problems/Problem 122019-03-23T18:05:34Z<p>Mathboy11: /* Solution 7 */</p>
<hr />
<div>== Problem ==<br />
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to<br />
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath><br />
In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.<br />
<br />
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so<br />
<cmath>\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath><br />
<br />
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.<br />
<br />
=== Solution 2 ===<br />
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>. <br />
<br />
(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get <br />
<br />
<cmath>\begin{align*}<br />
19 &= \frac{b - \frac da}{19 - e} + e\\<br />
97 &= \frac{b - \frac da}{97 - e} + e\\<br />
b - \frac da &= (19 - e)^2 = (97 - e)^2\\<br />
19 - e &= \pm (97 - e)<br />
\end{align*}</cmath><br />
<br />
Clearly we can discard the positive root, so <math>e = 58</math>.<br />
<br />
=== Solution 3 ===<br />
<!-- some linear algebra --><br />
We first note (as before) that the number not in the range of<br />
<cmath> f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} </cmath><br />
is <math>a/c</math>, as <math>\frac{b-ad/c}{cx+d}</math> is evidently never 0 (otherwise, <math>f</math><br />
would be a constant function, violating the condition <math>f(19) \neq f(97)</math>).<br />
<br />
We may represent the real number <math>x/y</math> as<br />
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]<br />
considered equivalent if they are scalar multiples of each other. Similarly,<br />
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix<br />
<math>\begin{pmatrix} A & B\\ C& D \end{pmatrix}</math>. Function composition and<br />
evaluation then become matrix multiplication.<br />
<br />
Now in general,<br />
<cmath> f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} =<br />
\frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .</cmath><br />
In our problem <math>f^2(x) = x</math>. It follows that<br />
<cmath> \begin{pmatrix} a & b \\ c& d \end{pmatrix} = K<br />
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , </cmath><br />
for some nonzero real <math>K</math>. Since<br />
<cmath> \frac{a}{d} = \frac{b}{-b} = K, </cmath><br />
it follows that <math>a = -d</math>. (In fact, this condition condition is equivalent<br />
to the condition that <math>f(f(x)) = x</math> for all <math>x</math> in the domain of <math>f</math>.)<br />
<br />
We next note that the function<br />
<cmath> g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} </cmath><br />
evaluates to 0 when <math>x</math> equals 19 and 97. Therefore<br />
<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath><br />
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>,<br />
our answer.<br />
<br />
=== Solution 4 ===<br />
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.<br />
<br />
<br />
Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath><br />
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath><br />
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that <math>a = -d</math>.<br />
<br />
This solution follows in the same manner as the last paragraph of the first solution.<br />
<br />
=== Solution 5 ===<br />
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.<br />
<br />
=== Solution 6===<br />
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy<br />
<br />
=== Solution 7===<br />
Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^{-1}(x)=\frac{19d-b}{a-19c}</math>. Since <math>f(f(x))=x</math>, <math>f(x)=f^{-1}(x)</math>, so substituting 19 and 97 yields the system, <math>\begin{array}{lcl} \frac{19a+b}{19c+d} & = & \frac{19d-b}{a-19c} \\ f(x,y,z) & = & x + y + z \end{array}</math><br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_12&diff=1048731997 AIME Problems/Problem 122019-03-23T18:03:50Z<p>Mathboy11: /* Solution 7 */</p>
<hr />
<div>== Problem ==<br />
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to<br />
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath><br />
In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.<br />
<br />
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so<br />
<cmath>\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath><br />
<br />
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.<br />
<br />
=== Solution 2 ===<br />
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>. <br />
<br />
(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get <br />
<br />
<cmath>\begin{align*}<br />
19 &= \frac{b - \frac da}{19 - e} + e\\<br />
97 &= \frac{b - \frac da}{97 - e} + e\\<br />
b - \frac da &= (19 - e)^2 = (97 - e)^2\\<br />
19 - e &= \pm (97 - e)<br />
\end{align*}</cmath><br />
<br />
Clearly we can discard the positive root, so <math>e = 58</math>.<br />
<br />
=== Solution 3 ===<br />
<!-- some linear algebra --><br />
We first note (as before) that the number not in the range of<br />
<cmath> f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} </cmath><br />
is <math>a/c</math>, as <math>\frac{b-ad/c}{cx+d}</math> is evidently never 0 (otherwise, <math>f</math><br />
would be a constant function, violating the condition <math>f(19) \neq f(97)</math>).<br />
<br />
We may represent the real number <math>x/y</math> as<br />
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]<br />
considered equivalent if they are scalar multiples of each other. Similarly,<br />
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix<br />
<math>\begin{pmatrix} A & B\\ C& D \end{pmatrix}</math>. Function composition and<br />
evaluation then become matrix multiplication.<br />
<br />
Now in general,<br />
<cmath> f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} =<br />
\frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .</cmath><br />
In our problem <math>f^2(x) = x</math>. It follows that<br />
<cmath> \begin{pmatrix} a & b \\ c& d \end{pmatrix} = K<br />
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , </cmath><br />
for some nonzero real <math>K</math>. Since<br />
<cmath> \frac{a}{d} = \frac{b}{-b} = K, </cmath><br />
it follows that <math>a = -d</math>. (In fact, this condition condition is equivalent<br />
to the condition that <math>f(f(x)) = x</math> for all <math>x</math> in the domain of <math>f</math>.)<br />
<br />
We next note that the function<br />
<cmath> g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} </cmath><br />
evaluates to 0 when <math>x</math> equals 19 and 97. Therefore<br />
<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath><br />
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>,<br />
our answer.<br />
<br />
=== Solution 4 ===<br />
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.<br />
<br />
<br />
Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath><br />
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath><br />
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that <math>a = -d</math>.<br />
<br />
This solution follows in the same manner as the last paragraph of the first solution.<br />
<br />
=== Solution 5 ===<br />
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.<br />
<br />
=== Solution 6===<br />
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy<br />
<br />
=== Solution 7===<br />
Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^{-1}(x)=\frac{19d-b}{a-19c}</math>. Since <math>f(f(x))=x</math>, <math>f(x)=f^{-1}(x)</math>, so <math>\frac{19a+b}{19c+d}=\frac{19d-b}{a-19c}</math>, so substituting 19 and 97 yields the system <math>\begin{array}{lcl} z & = & a \\ f(x,y,z) & = & x + y + z \end{array}</math><br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_12&diff=1048721997 AIME Problems/Problem 122019-03-23T18:01:56Z<p>Mathboy11: /* Solution 7 */</p>
<hr />
<div>== Problem ==<br />
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to<br />
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath><br />
In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.<br />
<br />
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so<br />
<cmath>\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath><br />
<br />
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.<br />
<br />
=== Solution 2 ===<br />
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>. <br />
<br />
(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get <br />
<br />
<cmath>\begin{align*}<br />
19 &= \frac{b - \frac da}{19 - e} + e\\<br />
97 &= \frac{b - \frac da}{97 - e} + e\\<br />
b - \frac da &= (19 - e)^2 = (97 - e)^2\\<br />
19 - e &= \pm (97 - e)<br />
\end{align*}</cmath><br />
<br />
Clearly we can discard the positive root, so <math>e = 58</math>.<br />
<br />
=== Solution 3 ===<br />
<!-- some linear algebra --><br />
We first note (as before) that the number not in the range of<br />
<cmath> f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} </cmath><br />
is <math>a/c</math>, as <math>\frac{b-ad/c}{cx+d}</math> is evidently never 0 (otherwise, <math>f</math><br />
would be a constant function, violating the condition <math>f(19) \neq f(97)</math>).<br />
<br />
We may represent the real number <math>x/y</math> as<br />
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]<br />
considered equivalent if they are scalar multiples of each other. Similarly,<br />
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix<br />
<math>\begin{pmatrix} A & B\\ C& D \end{pmatrix}</math>. Function composition and<br />
evaluation then become matrix multiplication.<br />
<br />
Now in general,<br />
<cmath> f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} =<br />
\frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .</cmath><br />
In our problem <math>f^2(x) = x</math>. It follows that<br />
<cmath> \begin{pmatrix} a & b \\ c& d \end{pmatrix} = K<br />
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , </cmath><br />
for some nonzero real <math>K</math>. Since<br />
<cmath> \frac{a}{d} = \frac{b}{-b} = K, </cmath><br />
it follows that <math>a = -d</math>. (In fact, this condition condition is equivalent<br />
to the condition that <math>f(f(x)) = x</math> for all <math>x</math> in the domain of <math>f</math>.)<br />
<br />
We next note that the function<br />
<cmath> g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} </cmath><br />
evaluates to 0 when <math>x</math> equals 19 and 97. Therefore<br />
<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath><br />
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>,<br />
our answer.<br />
<br />
=== Solution 4 ===<br />
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.<br />
<br />
<br />
Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath><br />
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath><br />
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that <math>a = -d</math>.<br />
<br />
This solution follows in the same manner as the last paragraph of the first solution.<br />
<br />
=== Solution 5 ===<br />
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.<br />
<br />
=== Solution 6===<br />
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy<br />
<br />
=== Solution 7===<br />
Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^{-1}(x)=\frac{19d-b}{a-19c}</math>. Since <math>f(f(x))=x</math>, <math>f(x)=f^{-1}(x)</math>, so <math>\frac{19a+b}{19c+d}=\frac{19d-b}{a-19c}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_12&diff=1048711997 AIME Problems/Problem 122019-03-23T18:00:57Z<p>Mathboy11: /* Solution 7 */</p>
<hr />
<div>== Problem ==<br />
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to<br />
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath><br />
In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.<br />
<br />
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so<br />
<cmath>\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath><br />
<br />
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.<br />
<br />
=== Solution 2 ===<br />
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>. <br />
<br />
(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get <br />
<br />
<cmath>\begin{align*}<br />
19 &= \frac{b - \frac da}{19 - e} + e\\<br />
97 &= \frac{b - \frac da}{97 - e} + e\\<br />
b - \frac da &= (19 - e)^2 = (97 - e)^2\\<br />
19 - e &= \pm (97 - e)<br />
\end{align*}</cmath><br />
<br />
Clearly we can discard the positive root, so <math>e = 58</math>.<br />
<br />
=== Solution 3 ===<br />
<!-- some linear algebra --><br />
We first note (as before) that the number not in the range of<br />
<cmath> f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} </cmath><br />
is <math>a/c</math>, as <math>\frac{b-ad/c}{cx+d}</math> is evidently never 0 (otherwise, <math>f</math><br />
would be a constant function, violating the condition <math>f(19) \neq f(97)</math>).<br />
<br />
We may represent the real number <math>x/y</math> as<br />
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]<br />
considered equivalent if they are scalar multiples of each other. Similarly,<br />
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix<br />
<math>\begin{pmatrix} A & B\\ C& D \end{pmatrix}</math>. Function composition and<br />
evaluation then become matrix multiplication.<br />
<br />
Now in general,<br />
<cmath> f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} =<br />
\frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .</cmath><br />
In our problem <math>f^2(x) = x</math>. It follows that<br />
<cmath> \begin{pmatrix} a & b \\ c& d \end{pmatrix} = K<br />
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , </cmath><br />
for some nonzero real <math>K</math>. Since<br />
<cmath> \frac{a}{d} = \frac{b}{-b} = K, </cmath><br />
it follows that <math>a = -d</math>. (In fact, this condition condition is equivalent<br />
to the condition that <math>f(f(x)) = x</math> for all <math>x</math> in the domain of <math>f</math>.)<br />
<br />
We next note that the function<br />
<cmath> g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} </cmath><br />
evaluates to 0 when <math>x</math> equals 19 and 97. Therefore<br />
<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath><br />
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>,<br />
our answer.<br />
<br />
=== Solution 4 ===<br />
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.<br />
<br />
<br />
Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath><br />
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath><br />
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that <math>a = -d</math>.<br />
<br />
This solution follows in the same manner as the last paragraph of the first solution.<br />
<br />
=== Solution 5 ===<br />
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.<br />
<br />
=== Solution 6===<br />
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy<br />
<br />
=== Solution 7===<br />
Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^{-1}(x)=/frac{19d-b}{a-19c}</math>. Since <math>f(f(x))=x</math>, <math>f(x)=f^{-1}(x)</math>, so $<br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_12&diff=1048701997 AIME Problems/Problem 122019-03-23T17:59:43Z<p>Mathboy11: /* Solution 7 */</p>
<hr />
<div>== Problem ==<br />
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to<br />
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath><br />
In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.<br />
<br />
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so<br />
<cmath>\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath><br />
<br />
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.<br />
<br />
=== Solution 2 ===<br />
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>. <br />
<br />
(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get <br />
<br />
<cmath>\begin{align*}<br />
19 &= \frac{b - \frac da}{19 - e} + e\\<br />
97 &= \frac{b - \frac da}{97 - e} + e\\<br />
b - \frac da &= (19 - e)^2 = (97 - e)^2\\<br />
19 - e &= \pm (97 - e)<br />
\end{align*}</cmath><br />
<br />
Clearly we can discard the positive root, so <math>e = 58</math>.<br />
<br />
=== Solution 3 ===<br />
<!-- some linear algebra --><br />
We first note (as before) that the number not in the range of<br />
<cmath> f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} </cmath><br />
is <math>a/c</math>, as <math>\frac{b-ad/c}{cx+d}</math> is evidently never 0 (otherwise, <math>f</math><br />
would be a constant function, violating the condition <math>f(19) \neq f(97)</math>).<br />
<br />
We may represent the real number <math>x/y</math> as<br />
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]<br />
considered equivalent if they are scalar multiples of each other. Similarly,<br />
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix<br />
<math>\begin{pmatrix} A & B\\ C& D \end{pmatrix}</math>. Function composition and<br />
evaluation then become matrix multiplication.<br />
<br />
Now in general,<br />
<cmath> f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} =<br />
\frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .</cmath><br />
In our problem <math>f^2(x) = x</math>. It follows that<br />
<cmath> \begin{pmatrix} a & b \\ c& d \end{pmatrix} = K<br />
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , </cmath><br />
for some nonzero real <math>K</math>. Since<br />
<cmath> \frac{a}{d} = \frac{b}{-b} = K, </cmath><br />
it follows that <math>a = -d</math>. (In fact, this condition condition is equivalent<br />
to the condition that <math>f(f(x)) = x</math> for all <math>x</math> in the domain of <math>f</math>.)<br />
<br />
We next note that the function<br />
<cmath> g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} </cmath><br />
evaluates to 0 when <math>x</math> equals 19 and 97. Therefore<br />
<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath><br />
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>,<br />
our answer.<br />
<br />
=== Solution 4 ===<br />
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.<br />
<br />
<br />
Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath><br />
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath><br />
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that <math>a = -d</math>.<br />
<br />
This solution follows in the same manner as the last paragraph of the first solution.<br />
<br />
=== Solution 5 ===<br />
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.<br />
<br />
=== Solution 6===<br />
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy<br />
<br />
=== Solution 7===<br />
Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^{-1}(x)=\frac{19d-b}{a-19c}</math>. Since <math>f(f(x))=x</math>, <math>f(x)=f^{-1}(x)</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_12&diff=1048691997 AIME Problems/Problem 122019-03-23T17:58:40Z<p>Mathboy11: /* Solution 7 */</p>
<hr />
<div>== Problem ==<br />
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to<br />
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath><br />
In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.<br />
<br />
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so<br />
<cmath>\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath><br />
<br />
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.<br />
<br />
=== Solution 2 ===<br />
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>. <br />
<br />
(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get <br />
<br />
<cmath>\begin{align*}<br />
19 &= \frac{b - \frac da}{19 - e} + e\\<br />
97 &= \frac{b - \frac da}{97 - e} + e\\<br />
b - \frac da &= (19 - e)^2 = (97 - e)^2\\<br />
19 - e &= \pm (97 - e)<br />
\end{align*}</cmath><br />
<br />
Clearly we can discard the positive root, so <math>e = 58</math>.<br />
<br />
=== Solution 3 ===<br />
<!-- some linear algebra --><br />
We first note (as before) that the number not in the range of<br />
<cmath> f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} </cmath><br />
is <math>a/c</math>, as <math>\frac{b-ad/c}{cx+d}</math> is evidently never 0 (otherwise, <math>f</math><br />
would be a constant function, violating the condition <math>f(19) \neq f(97)</math>).<br />
<br />
We may represent the real number <math>x/y</math> as<br />
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]<br />
considered equivalent if they are scalar multiples of each other. Similarly,<br />
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix<br />
<math>\begin{pmatrix} A & B\\ C& D \end{pmatrix}</math>. Function composition and<br />
evaluation then become matrix multiplication.<br />
<br />
Now in general,<br />
<cmath> f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} =<br />
\frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .</cmath><br />
In our problem <math>f^2(x) = x</math>. It follows that<br />
<cmath> \begin{pmatrix} a & b \\ c& d \end{pmatrix} = K<br />
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , </cmath><br />
for some nonzero real <math>K</math>. Since<br />
<cmath> \frac{a}{d} = \frac{b}{-b} = K, </cmath><br />
it follows that <math>a = -d</math>. (In fact, this condition condition is equivalent<br />
to the condition that <math>f(f(x)) = x</math> for all <math>x</math> in the domain of <math>f</math>.)<br />
<br />
We next note that the function<br />
<cmath> g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} </cmath><br />
evaluates to 0 when <math>x</math> equals 19 and 97. Therefore<br />
<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath><br />
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>,<br />
our answer.<br />
<br />
=== Solution 4 ===<br />
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.<br />
<br />
<br />
Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath><br />
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath><br />
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that <math>a = -d</math>.<br />
<br />
This solution follows in the same manner as the last paragraph of the first solution.<br />
<br />
=== Solution 5 ===<br />
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.<br />
<br />
=== Solution 6===<br />
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy<br />
<br />
=== Solution 7===<br />
Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^{-1}(x)=\frac{19d-b}{a-19c}</math><br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_12&diff=1048681997 AIME Problems/Problem 122019-03-23T17:58:13Z<p>Mathboy11: /* Solution 7 */</p>
<hr />
<div>== Problem ==<br />
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to<br />
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath><br />
In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.<br />
<br />
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so<br />
<cmath>\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath><br />
<br />
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.<br />
<br />
=== Solution 2 ===<br />
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>. <br />
<br />
(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get <br />
<br />
<cmath>\begin{align*}<br />
19 &= \frac{b - \frac da}{19 - e} + e\\<br />
97 &= \frac{b - \frac da}{97 - e} + e\\<br />
b - \frac da &= (19 - e)^2 = (97 - e)^2\\<br />
19 - e &= \pm (97 - e)<br />
\end{align*}</cmath><br />
<br />
Clearly we can discard the positive root, so <math>e = 58</math>.<br />
<br />
=== Solution 3 ===<br />
<!-- some linear algebra --><br />
We first note (as before) that the number not in the range of<br />
<cmath> f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} </cmath><br />
is <math>a/c</math>, as <math>\frac{b-ad/c}{cx+d}</math> is evidently never 0 (otherwise, <math>f</math><br />
would be a constant function, violating the condition <math>f(19) \neq f(97)</math>).<br />
<br />
We may represent the real number <math>x/y</math> as<br />
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]<br />
considered equivalent if they are scalar multiples of each other. Similarly,<br />
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix<br />
<math>\begin{pmatrix} A & B\\ C& D \end{pmatrix}</math>. Function composition and<br />
evaluation then become matrix multiplication.<br />
<br />
Now in general,<br />
<cmath> f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} =<br />
\frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .</cmath><br />
In our problem <math>f^2(x) = x</math>. It follows that<br />
<cmath> \begin{pmatrix} a & b \\ c& d \end{pmatrix} = K<br />
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , </cmath><br />
for some nonzero real <math>K</math>. Since<br />
<cmath> \frac{a}{d} = \frac{b}{-b} = K, </cmath><br />
it follows that <math>a = -d</math>. (In fact, this condition condition is equivalent<br />
to the condition that <math>f(f(x)) = x</math> for all <math>x</math> in the domain of <math>f</math>.)<br />
<br />
We next note that the function<br />
<cmath> g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} </cmath><br />
evaluates to 0 when <math>x</math> equals 19 and 97. Therefore<br />
<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath><br />
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>,<br />
our answer.<br />
<br />
=== Solution 4 ===<br />
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.<br />
<br />
<br />
Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath><br />
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath><br />
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that <math>a = -d</math>.<br />
<br />
This solution follows in the same manner as the last paragraph of the first solution.<br />
<br />
=== Solution 5 ===<br />
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.<br />
<br />
=== Solution 6===<br />
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy<br />
<br />
=== Solution 7===<br />
Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^{-1}(x)=frac{19d-b}{a-19c}</math><br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_12&diff=1048671997 AIME Problems/Problem 122019-03-23T17:57:23Z<p>Mathboy11: /* Solution 7 */</p>
<hr />
<div>== Problem ==<br />
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to<br />
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath><br />
In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.<br />
<br />
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so<br />
<cmath>\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath><br />
<br />
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.<br />
<br />
=== Solution 2 ===<br />
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>. <br />
<br />
(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get <br />
<br />
<cmath>\begin{align*}<br />
19 &= \frac{b - \frac da}{19 - e} + e\\<br />
97 &= \frac{b - \frac da}{97 - e} + e\\<br />
b - \frac da &= (19 - e)^2 = (97 - e)^2\\<br />
19 - e &= \pm (97 - e)<br />
\end{align*}</cmath><br />
<br />
Clearly we can discard the positive root, so <math>e = 58</math>.<br />
<br />
=== Solution 3 ===<br />
<!-- some linear algebra --><br />
We first note (as before) that the number not in the range of<br />
<cmath> f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} </cmath><br />
is <math>a/c</math>, as <math>\frac{b-ad/c}{cx+d}</math> is evidently never 0 (otherwise, <math>f</math><br />
would be a constant function, violating the condition <math>f(19) \neq f(97)</math>).<br />
<br />
We may represent the real number <math>x/y</math> as<br />
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]<br />
considered equivalent if they are scalar multiples of each other. Similarly,<br />
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix<br />
<math>\begin{pmatrix} A & B\\ C& D \end{pmatrix}</math>. Function composition and<br />
evaluation then become matrix multiplication.<br />
<br />
Now in general,<br />
<cmath> f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} =<br />
\frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .</cmath><br />
In our problem <math>f^2(x) = x</math>. It follows that<br />
<cmath> \begin{pmatrix} a & b \\ c& d \end{pmatrix} = K<br />
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , </cmath><br />
for some nonzero real <math>K</math>. Since<br />
<cmath> \frac{a}{d} = \frac{b}{-b} = K, </cmath><br />
it follows that <math>a = -d</math>. (In fact, this condition condition is equivalent<br />
to the condition that <math>f(f(x)) = x</math> for all <math>x</math> in the domain of <math>f</math>.)<br />
<br />
We next note that the function<br />
<cmath> g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} </cmath><br />
evaluates to 0 when <math>x</math> equals 19 and 97. Therefore<br />
<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath><br />
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>,<br />
our answer.<br />
<br />
=== Solution 4 ===<br />
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.<br />
<br />
<br />
Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath><br />
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath><br />
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that <math>a = -d</math>.<br />
<br />
This solution follows in the same manner as the last paragraph of the first solution.<br />
<br />
=== Solution 5 ===<br />
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.<br />
<br />
=== Solution 6===<br />
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy<br />
<br />
=== Solution 7===<br />
Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^{-1}(x)=frac{(19d-b)/(a-19c)}</math><br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_12&diff=1048661997 AIME Problems/Problem 122019-03-23T17:56:22Z<p>Mathboy11: /* Solution 7 */</p>
<hr />
<div>== Problem ==<br />
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to<br />
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath><br />
In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.<br />
<br />
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so<br />
<cmath>\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath><br />
<br />
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.<br />
<br />
=== Solution 2 ===<br />
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>. <br />
<br />
(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get <br />
<br />
<cmath>\begin{align*}<br />
19 &= \frac{b - \frac da}{19 - e} + e\\<br />
97 &= \frac{b - \frac da}{97 - e} + e\\<br />
b - \frac da &= (19 - e)^2 = (97 - e)^2\\<br />
19 - e &= \pm (97 - e)<br />
\end{align*}</cmath><br />
<br />
Clearly we can discard the positive root, so <math>e = 58</math>.<br />
<br />
=== Solution 3 ===<br />
<!-- some linear algebra --><br />
We first note (as before) that the number not in the range of<br />
<cmath> f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} </cmath><br />
is <math>a/c</math>, as <math>\frac{b-ad/c}{cx+d}</math> is evidently never 0 (otherwise, <math>f</math><br />
would be a constant function, violating the condition <math>f(19) \neq f(97)</math>).<br />
<br />
We may represent the real number <math>x/y</math> as<br />
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]<br />
considered equivalent if they are scalar multiples of each other. Similarly,<br />
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix<br />
<math>\begin{pmatrix} A & B\\ C& D \end{pmatrix}</math>. Function composition and<br />
evaluation then become matrix multiplication.<br />
<br />
Now in general,<br />
<cmath> f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} =<br />
\frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .</cmath><br />
In our problem <math>f^2(x) = x</math>. It follows that<br />
<cmath> \begin{pmatrix} a & b \\ c& d \end{pmatrix} = K<br />
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , </cmath><br />
for some nonzero real <math>K</math>. Since<br />
<cmath> \frac{a}{d} = \frac{b}{-b} = K, </cmath><br />
it follows that <math>a = -d</math>. (In fact, this condition condition is equivalent<br />
to the condition that <math>f(f(x)) = x</math> for all <math>x</math> in the domain of <math>f</math>.)<br />
<br />
We next note that the function<br />
<cmath> g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} </cmath><br />
evaluates to 0 when <math>x</math> equals 19 and 97. Therefore<br />
<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath><br />
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>,<br />
our answer.<br />
<br />
=== Solution 4 ===<br />
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.<br />
<br />
<br />
Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath><br />
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath><br />
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that <math>a = -d</math>.<br />
<br />
This solution follows in the same manner as the last paragraph of the first solution.<br />
<br />
=== Solution 5 ===<br />
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.<br />
<br />
=== Solution 6===<br />
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy<br />
<br />
=== Solution 7===<br />
Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^{-1}(x)</math><br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_12&diff=1048651997 AIME Problems/Problem 122019-03-23T17:56:07Z<p>Mathboy11: /* Solution 7 */</p>
<hr />
<div>== Problem ==<br />
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to<br />
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath><br />
In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.<br />
<br />
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so<br />
<cmath>\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath><br />
<br />
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.<br />
<br />
=== Solution 2 ===<br />
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>. <br />
<br />
(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get <br />
<br />
<cmath>\begin{align*}<br />
19 &= \frac{b - \frac da}{19 - e} + e\\<br />
97 &= \frac{b - \frac da}{97 - e} + e\\<br />
b - \frac da &= (19 - e)^2 = (97 - e)^2\\<br />
19 - e &= \pm (97 - e)<br />
\end{align*}</cmath><br />
<br />
Clearly we can discard the positive root, so <math>e = 58</math>.<br />
<br />
=== Solution 3 ===<br />
<!-- some linear algebra --><br />
We first note (as before) that the number not in the range of<br />
<cmath> f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} </cmath><br />
is <math>a/c</math>, as <math>\frac{b-ad/c}{cx+d}</math> is evidently never 0 (otherwise, <math>f</math><br />
would be a constant function, violating the condition <math>f(19) \neq f(97)</math>).<br />
<br />
We may represent the real number <math>x/y</math> as<br />
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]<br />
considered equivalent if they are scalar multiples of each other. Similarly,<br />
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix<br />
<math>\begin{pmatrix} A & B\\ C& D \end{pmatrix}</math>. Function composition and<br />
evaluation then become matrix multiplication.<br />
<br />
Now in general,<br />
<cmath> f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} =<br />
\frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .</cmath><br />
In our problem <math>f^2(x) = x</math>. It follows that<br />
<cmath> \begin{pmatrix} a & b \\ c& d \end{pmatrix} = K<br />
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , </cmath><br />
for some nonzero real <math>K</math>. Since<br />
<cmath> \frac{a}{d} = \frac{b}{-b} = K, </cmath><br />
it follows that <math>a = -d</math>. (In fact, this condition condition is equivalent<br />
to the condition that <math>f(f(x)) = x</math> for all <math>x</math> in the domain of <math>f</math>.)<br />
<br />
We next note that the function<br />
<cmath> g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} </cmath><br />
evaluates to 0 when <math>x</math> equals 19 and 97. Therefore<br />
<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath><br />
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>,<br />
our answer.<br />
<br />
=== Solution 4 ===<br />
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.<br />
<br />
<br />
Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath><br />
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath><br />
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that <math>a = -d</math>.<br />
<br />
This solution follows in the same manner as the last paragraph of the first solution.<br />
<br />
=== Solution 5 ===<br />
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.<br />
<br />
=== Solution 6===<br />
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy<br />
<br />
=== Solution 7===<br />
Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^[-1](x)</math><br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_12&diff=1048641997 AIME Problems/Problem 122019-03-23T17:55:52Z<p>Mathboy11: /* Solution 7 */</p>
<hr />
<div>== Problem ==<br />
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to<br />
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath><br />
In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.<br />
<br />
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so<br />
<cmath>\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath><br />
<br />
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.<br />
<br />
=== Solution 2 ===<br />
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>. <br />
<br />
(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get <br />
<br />
<cmath>\begin{align*}<br />
19 &= \frac{b - \frac da}{19 - e} + e\\<br />
97 &= \frac{b - \frac da}{97 - e} + e\\<br />
b - \frac da &= (19 - e)^2 = (97 - e)^2\\<br />
19 - e &= \pm (97 - e)<br />
\end{align*}</cmath><br />
<br />
Clearly we can discard the positive root, so <math>e = 58</math>.<br />
<br />
=== Solution 3 ===<br />
<!-- some linear algebra --><br />
We first note (as before) that the number not in the range of<br />
<cmath> f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} </cmath><br />
is <math>a/c</math>, as <math>\frac{b-ad/c}{cx+d}</math> is evidently never 0 (otherwise, <math>f</math><br />
would be a constant function, violating the condition <math>f(19) \neq f(97)</math>).<br />
<br />
We may represent the real number <math>x/y</math> as<br />
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]<br />
considered equivalent if they are scalar multiples of each other. Similarly,<br />
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix<br />
<math>\begin{pmatrix} A & B\\ C& D \end{pmatrix}</math>. Function composition and<br />
evaluation then become matrix multiplication.<br />
<br />
Now in general,<br />
<cmath> f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} =<br />
\frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .</cmath><br />
In our problem <math>f^2(x) = x</math>. It follows that<br />
<cmath> \begin{pmatrix} a & b \\ c& d \end{pmatrix} = K<br />
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , </cmath><br />
for some nonzero real <math>K</math>. Since<br />
<cmath> \frac{a}{d} = \frac{b}{-b} = K, </cmath><br />
it follows that <math>a = -d</math>. (In fact, this condition condition is equivalent<br />
to the condition that <math>f(f(x)) = x</math> for all <math>x</math> in the domain of <math>f</math>.)<br />
<br />
We next note that the function<br />
<cmath> g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} </cmath><br />
evaluates to 0 when <math>x</math> equals 19 and 97. Therefore<br />
<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath><br />
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>,<br />
our answer.<br />
<br />
=== Solution 4 ===<br />
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.<br />
<br />
<br />
Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath><br />
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath><br />
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that <math>a = -d</math>.<br />
<br />
This solution follows in the same manner as the last paragraph of the first solution.<br />
<br />
=== Solution 5 ===<br />
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.<br />
<br />
=== Solution 6===<br />
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy<br />
<br />
=== Solution 7===<br />
Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^-1(x)</math><br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_12&diff=1048631997 AIME Problems/Problem 122019-03-23T17:55:01Z<p>Mathboy11: /* Solution 6 */</p>
<hr />
<div>== Problem ==<br />
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to<br />
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath><br />
In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.<br />
<br />
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so<br />
<cmath>\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath><br />
<br />
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.<br />
<br />
=== Solution 2 ===<br />
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>. <br />
<br />
(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get <br />
<br />
<cmath>\begin{align*}<br />
19 &= \frac{b - \frac da}{19 - e} + e\\<br />
97 &= \frac{b - \frac da}{97 - e} + e\\<br />
b - \frac da &= (19 - e)^2 = (97 - e)^2\\<br />
19 - e &= \pm (97 - e)<br />
\end{align*}</cmath><br />
<br />
Clearly we can discard the positive root, so <math>e = 58</math>.<br />
<br />
=== Solution 3 ===<br />
<!-- some linear algebra --><br />
We first note (as before) that the number not in the range of<br />
<cmath> f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} </cmath><br />
is <math>a/c</math>, as <math>\frac{b-ad/c}{cx+d}</math> is evidently never 0 (otherwise, <math>f</math><br />
would be a constant function, violating the condition <math>f(19) \neq f(97)</math>).<br />
<br />
We may represent the real number <math>x/y</math> as<br />
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]<br />
considered equivalent if they are scalar multiples of each other. Similarly,<br />
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix<br />
<math>\begin{pmatrix} A & B\\ C& D \end{pmatrix}</math>. Function composition and<br />
evaluation then become matrix multiplication.<br />
<br />
Now in general,<br />
<cmath> f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} =<br />
\frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .</cmath><br />
In our problem <math>f^2(x) = x</math>. It follows that<br />
<cmath> \begin{pmatrix} a & b \\ c& d \end{pmatrix} = K<br />
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , </cmath><br />
for some nonzero real <math>K</math>. Since<br />
<cmath> \frac{a}{d} = \frac{b}{-b} = K, </cmath><br />
it follows that <math>a = -d</math>. (In fact, this condition condition is equivalent<br />
to the condition that <math>f(f(x)) = x</math> for all <math>x</math> in the domain of <math>f</math>.)<br />
<br />
We next note that the function<br />
<cmath> g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} </cmath><br />
evaluates to 0 when <math>x</math> equals 19 and 97. Therefore<br />
<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath><br />
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>,<br />
our answer.<br />
<br />
=== Solution 4 ===<br />
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.<br />
<br />
<br />
Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath><br />
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath><br />
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that <math>a = -d</math>.<br />
<br />
This solution follows in the same manner as the last paragraph of the first solution.<br />
<br />
=== Solution 5 ===<br />
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.<br />
<br />
=== Solution 6===<br />
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy<br />
<br />
=== Solution 7===<br />
Begin by finding the inverse function of <math>f(X)</math><br />
<br />
== See also ==<br />
{{AIME box|year=1997|num-b=11|num-a=13}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_8&diff=1043412019 AIME I Problems/Problem 82019-03-14T23:44:14Z<p>Mathboy11: /* Solution */</p>
<hr />
<div>The 2019 AIME I takes place on March 13, 2019.<br />
<br />
==Problem 8==<br />
==Solution(BASH)==<br />
Remember <math>sin^2(x)+cos^2(x)=1</math>. This means <math>sin^{10}(x)+cos^{10}(x)=(sin^2(x)+cos^2(x))sin^{10}(x)+(sin^2(x)+cos^2(x))cos^{10})(x)=sin^{12}(x)+cos^{12}(x)+sin^2(x)cos^{10}(x)+sin^{10}(x)cos^{2}(x)=sin^{12}(x)+cos^{12}(x)+sin^2(x)cos^{2}(x)[sin^8(x)+cos^8(x)]</math>. Let us look at $sin^8(x)+cos^8(x)=(sin^2(x)+cos^2(x))sin^{8}(x)+(sin^2(x)+cos^2(x))cos^{8})(x)=sin^{10}(x)+cos^{10})(x)<br />
<br />
==See Also==<br />
{{AIME box|year=2019|n=I|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_8&diff=1043382019 AIME I Problems/Problem 82019-03-14T23:42:09Z<p>Mathboy11: /* Solution */</p>
<hr />
<div>The 2019 AIME I takes place on March 13, 2019.<br />
<br />
==Problem 8==<br />
==Solution==<br />
Remember <math>sin^2(x)+cos^2(x)=1</math>. This means <math>sin^{10}(x)+cos^{10}(x)=(sin^2(x)+cos^2(x))sin^{10}(x)+(sin^2(x)+cos^2(x))cos^{10})(x)=sin^{12}(x)+cos^{12}(x)+sin^2(x)cos^{10}(x)+sin^{10}(x)cos^{2}(x)=sin^{12}(x)+cos^{12}(x)+sin^2(x)cos^{2}[sin^8(x)+cos^8(x)</math>.<br />
<br />
==See Also==<br />
{{AIME box|year=2019|n=I|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_8&diff=1043372019 AIME I Problems/Problem 82019-03-14T23:41:02Z<p>Mathboy11: /* Solution */</p>
<hr />
<div>The 2019 AIME I takes place on March 13, 2019.<br />
<br />
==Problem 8==<br />
==Solution==<br />
Remember <math>sin^2(x)+cos^2(x)=1</math>. This means <math>sin^{10}(x)+cos^{10}(x)=(sin^2(x)+cos^2(x))sin^{10}(x)+(sin^2(x)+cos^2(x))cos^{10})(x)=sin^{12}(x)+cos^{12}(x)+sin^2(x)cos^{10}(x)+sin^{10}(x)cos^{2}(x)=11/36</math>.<br />
<br />
==See Also==<br />
{{AIME box|year=2019|n=I|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_8&diff=1043362019 AIME I Problems/Problem 82019-03-14T23:40:49Z<p>Mathboy11: /* Solution */</p>
<hr />
<div>The 2019 AIME I takes place on March 13, 2019.<br />
<br />
==Problem 8==<br />
==Solution==<br />
Remember <math>sin^2(x)+cos^2(x)=1</math>. This means <math>sin^{10}(x)+cos^{10}(x)=(sin^2(x)+cos^2(x))sin^{10}(x)+(sin^2(x)+cos^2(x))cos^{10})(x)=sin^{12}(x)+cos^{12}(x)+sin^2(x)cos^{10}(x)+sin^{10}(x)cos^{2}(x)=frac{11/36}</math>.<br />
<br />
==See Also==<br />
{{AIME box|year=2019|n=I|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_8&diff=1043342019 AIME I Problems/Problem 82019-03-14T23:40:10Z<p>Mathboy11: /* Solution */</p>
<hr />
<div>The 2019 AIME I takes place on March 13, 2019.<br />
<br />
==Problem 8==<br />
==Solution==<br />
Remember <math>sin^2(x)+cos^2(x)=1</math>. This means <math>sin^{10}(x)+cos^{10}(x)=(sin^2(x)+cos^2(x))sin^{10}(x)+(sin^2(x)+cos^2(x))cos^{10})(x)=sin^{12}(x)+cos^{12}(x)+<br />
sin^2(x)cos^{10}(x)+sin^{10}(x)cos^{2}(x)</math><br />
<br />
==See Also==<br />
{{AIME box|year=2019|n=I|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_8&diff=1043332019 AIME I Problems/Problem 82019-03-14T23:39:42Z<p>Mathboy11: /* Solution */</p>
<hr />
<div>The 2019 AIME I takes place on March 13, 2019.<br />
<br />
==Problem 8==<br />
==Solution==<br />
Remember <math>sin^2(x)+cos^2(x)=1</math>. This means <math>sin^{10}(x)+cos^{10}(x)=(sin^2(x)+cos^2(x))sin^{10}(x)+(sin^2(x)+cos^2(x))cos^{10})(x)=sin^{12}(x)+cos^{12}(x)+sin^2(x)cos^{10}(x)+sin^{10}(x)cos^{2}(x)</math><br />
<br />
==See Also==<br />
{{AIME box|year=2019|n=I|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_8&diff=1043312019 AIME I Problems/Problem 82019-03-14T23:38:29Z<p>Mathboy11: /* Solution */</p>
<hr />
<div>The 2019 AIME I takes place on March 13, 2019.<br />
<br />
==Problem 8==<br />
==Solution==<br />
Remember <math>sin^2(x)+cos^2(x)=1</math>. This means <math>sin^{10}(x)+cos^{10}(x)=(sin^2(x)+cos^2(x))sin^{10}(x)+(sin^2(x)+cos^2(x))cos^{10})(x)</math><br />
<br />
==See Also==<br />
{{AIME box|year=2019|n=I|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_8&diff=1043302019 AIME I Problems/Problem 82019-03-14T23:36:46Z<p>Mathboy11: /* Solution */</p>
<hr />
<div>The 2019 AIME I takes place on March 13, 2019.<br />
<br />
==Problem 8==<br />
==Solution==<br />
Remember <math>sin^2(x)+cos^2(x)=1</math>. This means <math>sin^(10)(x)+cos^(10)(x)=(sin^2(x)+cos^2(x))sin^(10)(x)+(sin^2(x)+cos^2(x))cos^(10)(x)</math><br />
<br />
==See Also==<br />
{{AIME box|year=2019|n=I|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_8&diff=1043292019 AIME I Problems/Problem 82019-03-14T23:35:57Z<p>Mathboy11: /* Solution */</p>
<hr />
<div>The 2019 AIME I takes place on March 13, 2019.<br />
<br />
==Problem 8==<br />
==Solution==<br />
Remember <math>sin^2(x)+cos^2(x)=1</math>. This means <math>sin^10(x)+cos^10(x)=(sin^2(x)+cos^2(x))sin^10(x)+(sin^2(x)+cos^2(x))cos^10(x)</math><br />
<br />
==See Also==<br />
{{AIME box|year=2019|n=I|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_8&diff=1043282019 AIME I Problems/Problem 82019-03-14T23:34:30Z<p>Mathboy11: /* Solution */</p>
<hr />
<div>The 2019 AIME I takes place on March 13, 2019.<br />
<br />
==Problem 8==<br />
==Solution==<br />
Remember <math>sin^2(x)+cos^2(x)=1</math>.<br />
<br />
==See Also==<br />
{{AIME box|year=2019|n=I|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_19&diff=978492018 AMC 12A Problems/Problem 192018-09-18T00:54:06Z<p>Mathboy11: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
<br />
Let <math>A</math> be the set of positive integers that have no prime factors other than <math>2</math>, <math>3</math>, or <math>5</math>. The infinite sum <cmath>\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \cdots</cmath>of the reciprocals of the elements of <math>A</math> can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?<br />
<br />
<math>\textbf{(A)} \text{ 16} \qquad \textbf{(B)} \text{ 17} \qquad \textbf{(C)} \text{ 19} \qquad \textbf{(D)} \text{ 23} \qquad \textbf{(E)} \text{ 36}</math><br />
<br />
== Solution ==<br />
It's just <cmath><br />
\sum_{a\ge 0}\frac1{2^a}\sum_{b\ge 0}\frac1{3^b}\sum_{c\ge 0}\frac{1}{5^c} = 2 \cdot \frac32 \cdot \frac54 = \frac{15}{4}\Rightarrow\textbf{(C)}.<br />
</cmath> since this represents all the numbers in the denominator.<br />
(ayushk)<br />
== Solution 2==<br />
Separate into 7 separate infinite series's so we can calculate each and find the original sum. The first infinite sequence shall be all the reciprocals of the powers of 2, the second shall be reciprocals of the powers of 3, and the third is reciprocals of the powers of 5. We can easily calculate these to be <math>1, 1/2, 1/4</math> respectively. The fourth infinite series shall be all real numbers in the form <math> 1/(2^a3^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The fifth is all real numbers in the form <math> 1/(2^a5^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The sixth is all real numbers in the form <math> 1/(3^a5^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The seventh infinite series is all real numbers in the form <math> 1/(2^a3^b5^c)</math>, where <math>a</math> and <math>b</math> and <math>c</math> are greater than or equal to 1. Let us denote the first sequence as <math>a_{1}</math>, the second as <math>a_{2}</math>, etc. We know <math>a_{1}=1</math>, <math>a_{2}=1/2</math>, <math>a_{3}=1/4</math>, let us find <math>a_{4}</math>. factoring out <math>1/6</math> from the terms in this subsequence, we would get <math>a_{4}=1/6(1+a_{1}+a_{2}+a_{4})</math>. Knowing <math>a_{1}</math> and <math>a_{2}</math>, we can substitute and solve for <math>a_{4}</math>, and we get <math>1/2</math>. If we do the similar procedures for the fifth and sixth sequences, we can solve for them too, and we get after solving them <math>1/4</math> and <math>1/8</math>. Finally, for the seventh sequence, we see <math>a_{7}=1/30(a_{8})</math>, where <math>a_{8}</math> is the infinite series the problem is asking us to solve for. The sum of all seven subsequences will equal the one we are looking for, so solving, we get <math>1+1/2+1/4+1/2+1/4+1/8+1/30(a_{8})=a_{8}</math>, but when we separated the sequence into its parts, we ignored the <math>1/1</math>, so adding in the <math>1</math>, we get <math>1+1+1/2+1/4+1/2+1/4+1/8+1/30(a_{8})=a_{8}</math>, which when we solve for, we get <math>29/8=29/30(a_{8})</math>, <math>1/8=1/30(a_{8})</math>, <math>30/8=(a_{8})</math>, <math>15/4=(a_{8})</math>. So our answer is 15/4, but we are asked to add the numerator and denominator, which sums up to 19, which is the answer.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2018|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_19&diff=978482018 AMC 12A Problems/Problem 192018-09-18T00:53:04Z<p>Mathboy11: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
<br />
Let <math>A</math> be the set of positive integers that have no prime factors other than <math>2</math>, <math>3</math>, or <math>5</math>. The infinite sum <cmath>\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \cdots</cmath>of the reciprocals of the elements of <math>A</math> can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?<br />
<br />
<math>\textbf{(A)} \text{ 16} \qquad \textbf{(B)} \text{ 17} \qquad \textbf{(C)} \text{ 19} \qquad \textbf{(D)} \text{ 23} \qquad \textbf{(E)} \text{ 36}</math><br />
<br />
== Solution ==<br />
It's just <cmath><br />
\sum_{a\ge 0}\frac1{2^a}\sum_{b\ge 0}\frac1{3^b}\sum_{c\ge 0}\frac{1}{5^c} = 2 \cdot \frac32 \cdot \frac54 = \frac{15}{4}\Rightarrow\textbf{(C)}.<br />
</cmath> since this represents all the numbers in the denominator.<br />
(ayushk)<br />
== Solution 2==<br />
Separate into 7 separate infinite series's so we can calculate each and find the original sum. The first infinite sequence shall be all the reciprocals of the powers of 2, the second shall be reciprocals of the powers of 3, and the third is reciprocals of the powers of 5. We can easily calculate these to be <math>1, 1/2, 1/4</math> respectively. The fourth infinite series shall be all real numbers in the form <math> 1/(2^a3^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The fifth is all real numbers in the form <math> 1/(2^a5^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The sixth is all real numbers in the form <math> 1/(3^a5^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The seventh infinite series is all real numbers in the form <math> 1/(2^a3^b5^c)</math>, where <math>a</math> and <math>b</math> and <math>c</math> are greater than or equal to 1. Let us denote the first sequence as <math>a_{1}</math>, the second as <math>a_{2}</math>, etc. We know <math>a_{1}=1</math>, <math>a_{2}=1/2</math>, <math>a_{3}=1/4</math>, let us find <math>a_{4}</math>. factoring out <math>1/6</math> from the terms in this subsequence, we would get <math>a_{4}=1/6(1+a_{1}+a_{2}+a_{4})</math>. Knowing <math>a_{1}</math> and <math>a_{2}</math>, we can substitute and solve for <math>a_{4}</math>, and we get <math>1/2</math>. If we do the similar procedures for the fifth and sixth sequences, we can solve for them too, and we get after solving them <math>1/4</math> and <math>1/8</math>. Finally, for the seventh sequence, we see <math>a_{7}=1/30(a_{8})</math>, where <math>a_{8}</math> is the infinite series the problem is asking us to solve for. The sum of all seven subsequences will equal the one we are looking for, so solving, we get <math>1+1/2+1/4+1/2+1/4+1/8+1/30(a_{8})=a_{8}</math>, but when we separated the sequence into its parts, we ignored the <math>1/1</math>, so adding in the <math>1</math>, we get <math>1+1+1/2+1/4+1/2+1/4+1/8+1/30(a_{8})=a_{8}</math>, which when we solve for, we get <math>29/8=29/30(a_{8})</math>, <math>1/8=1/30(a_{8})</math>, <math>30/8=(a_{8})</math>, <math>15/4=(a_{8})</math>. So our answer is 15/4, but we are asked to add the numerator and denominator, which sums up to 19.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2018|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_19&diff=978472018 AMC 12A Problems/Problem 192018-09-18T00:51:45Z<p>Mathboy11: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
<br />
Let <math>A</math> be the set of positive integers that have no prime factors other than <math>2</math>, <math>3</math>, or <math>5</math>. The infinite sum <cmath>\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \cdots</cmath>of the reciprocals of the elements of <math>A</math> can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?<br />
<br />
<math>\textbf{(A)} \text{ 16} \qquad \textbf{(B)} \text{ 17} \qquad \textbf{(C)} \text{ 19} \qquad \textbf{(D)} \text{ 23} \qquad \textbf{(E)} \text{ 36}</math><br />
<br />
== Solution ==<br />
It's just <cmath><br />
\sum_{a\ge 0}\frac1{2^a}\sum_{b\ge 0}\frac1{3^b}\sum_{c\ge 0}\frac{1}{5^c} = 2 \cdot \frac32 \cdot \frac54 = \frac{15}{4}\Rightarrow\textbf{(C)}.<br />
</cmath> since this represents all the numbers in the denominator.<br />
(ayushk)<br />
== Solution 2==<br />
Separate into 7 separate infinite series's so we can calculate each and find the original sum. The first infinite sequence shall be all the reciprocals of the powers of 2, the second shall be reciprocals of the powers of 3, and the third is reciprocals of the powers of 5. We can easily calculate these to be <math>1, 1/2, 1/4</math> respectively. The fourth infinite series shall be all real numbers in the form <math> 1/(2^a3^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The fifth is all real numbers in the form <math> 1/(2^a5^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The sixth is all real numbers in the form <math> 1/(3^a5^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The seventh infinite series is all real numbers in the form <math> 1/(2^a3^b5^c)</math>, where <math>a</math> and <math>b</math> and <math>c</math> are greater than or equal to 1. Let us denote the first sequence as <math>a_{1}</math>, the second as <math>a_{2}</math>, etc. We know <math>a_{1}=1</math>, <math>a_{2}=1/2</math>, <math>a_{3}=1/4</math>, let us find <math>a_{4}</math>. factoring out <math>1/6 from the terms in this subsequence, we would get </math>a_{4}=1/6(1+a_{1}+a_{2}+a_{4})<math>. Knowing </math>a_{1}<math> and </math>a_{2}<math>, we can substitute and solve for </math>a_{4}<math>, and we get </math>1/2<math>. If we do the similar procedures for the fifth and sixth sequences, we can solve for them too, and we get after solving them </math>1/4<math> and </math>1/8<math>. Finally, for the seventh sequence, we see </math>a_{7}=1/30(a_{8})<math>, where </math>a_{8}<math> is the infinite series the problem is asking us to solve for. The sum of all seven subsequences will equal the one we are looking for, so solving, we get </math>1+1/2+1/4+1/2+1/4+1/8+1/30(a_{8})=a_{8}<math>, but when we separated the sequence into its parts, we ignored the </math>1/1<math>, so adding in the </math>1<math>, we get </math>1+1+1/2+1/4+1/2+1/4+1/8+1/30(a_{8})=a_{8}<math>, which when we solve for, we get </math>29/8=29/30(a_{8})<math>, </math>1/8=1/30(a_{8})<math>, </math>30/8=(a_{8})<math>, </math>15/4=(a_{8})$. So our answer is 15/4, but we are asked to add the numerator and denominator, which sums up to 19.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2018|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_19&diff=978462018 AMC 12A Problems/Problem 192018-09-18T00:35:36Z<p>Mathboy11: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
<br />
Let <math>A</math> be the set of positive integers that have no prime factors other than <math>2</math>, <math>3</math>, or <math>5</math>. The infinite sum <cmath>\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \cdots</cmath>of the reciprocals of the elements of <math>A</math> can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?<br />
<br />
<math>\textbf{(A)} \text{ 16} \qquad \textbf{(B)} \text{ 17} \qquad \textbf{(C)} \text{ 19} \qquad \textbf{(D)} \text{ 23} \qquad \textbf{(E)} \text{ 36}</math><br />
<br />
== Solution ==<br />
It's just <cmath><br />
\sum_{a\ge 0}\frac1{2^a}\sum_{b\ge 0}\frac1{3^b}\sum_{c\ge 0}\frac{1}{5^c} = 2 \cdot \frac32 \cdot \frac54 = \frac{15}{4}\Rightarrow\textbf{(C)}.<br />
</cmath> since this represents all the numbers in the denominator.<br />
(ayushk)<br />
== Solution 2==<br />
Separate into 7 separate infinite series's so we can calculate each and find the original sum. The first infinite sequence shall be all the reciprocals of the powers of 2, the second shall be reciprocals of the powers of 3, and the third is reciprocals of the powers of 5. We can easily calculate these to be <math>1, 1/2, 1/4</math> respectively. The fourth infinite series shall be all real numbers in the form <math> 1/(2^a3^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The fifth is all real numbers in the form <math> 1/(2^a5^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The sixth is all real numbers in the form <math> 1/(3^a5^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The seventh infinite series is all real numbers in the form <math> 1/(2^a3^b5^c)</math>, where <math>a</math> and <math>b</math> and <math>c</math> are greater than or equal to 1. Let us denote the first sequence as <math>a_{exp}</math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2018|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_19&diff=978452018 AMC 12A Problems/Problem 192018-09-18T00:34:51Z<p>Mathboy11: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
<br />
Let <math>A</math> be the set of positive integers that have no prime factors other than <math>2</math>, <math>3</math>, or <math>5</math>. The infinite sum <cmath>\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \cdots</cmath>of the reciprocals of the elements of <math>A</math> can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?<br />
<br />
<math>\textbf{(A)} \text{ 16} \qquad \textbf{(B)} \text{ 17} \qquad \textbf{(C)} \text{ 19} \qquad \textbf{(D)} \text{ 23} \qquad \textbf{(E)} \text{ 36}</math><br />
<br />
== Solution ==<br />
It's just <cmath><br />
\sum_{a\ge 0}\frac1{2^a}\sum_{b\ge 0}\frac1{3^b}\sum_{c\ge 0}\frac{1}{5^c} = 2 \cdot \frac32 \cdot \frac54 = \frac{15}{4}\Rightarrow\textbf{(C)}.<br />
</cmath> since this represents all the numbers in the denominator.<br />
(ayushk)<br />
== Solution 2==<br />
Separate into 7 separate infinite series's so we can calculate each and find the original sum. The first infinite sequence shall be all the reciprocals of the powers of 2, the second shall be reciprocals of the powers of 3, and the third is reciprocals of the powers of 5. We can easily calculate these to be <math>1, 1/2, 1/4</math> respectively. The fourth infinite series shall be all real numbers in the form <math> 1/(2^a3^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The fifth is all real numbers in the form <math> 1/(2^a5^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The sixth is all real numbers in the form <math> 1/(3^a5^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The seventh infinite series is all real numbers in the form <math> 1/(2^a3^b5^c)</math>, where <math>a</math> and <math>b</math> and <math>c</math> are greater than or equal to 1. Let us denote the first sequence as <math>asub1</math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2018|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_19&diff=978442018 AMC 12A Problems/Problem 192018-09-18T00:31:05Z<p>Mathboy11: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Let <math>A</math> be the set of positive integers that have no prime factors other than <math>2</math>, <math>3</math>, or <math>5</math>. The infinite sum <cmath>\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \cdots</cmath>of the reciprocals of the elements of <math>A</math> can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?<br />
<br />
<math>\textbf{(A)} \text{ 16} \qquad \textbf{(B)} \text{ 17} \qquad \textbf{(C)} \text{ 19} \qquad \textbf{(D)} \text{ 23} \qquad \textbf{(E)} \text{ 36}</math><br />
<br />
== Solution ==<br />
It's just <cmath><br />
\sum_{a\ge 0}\frac1{2^a}\sum_{b\ge 0}\frac1{3^b}\sum_{c\ge 0}\frac{1}{5^c} = 2 \cdot \frac32 \cdot \frac54 = \frac{15}{4}\Rightarrow\textbf{(C)}.<br />
</cmath> since this represents all the numbers in the denominator.<br />
(ayushk)<br />
== Solution 2==<br />
Separate into 7 separate infinite series's so we can calculate each and find the original sum. The first infinite sequence shall be all the reciprocals of the powers of 2, the second shall be reciprocals of the powers of 3, and the third is reciprocals of the powers of 5. We can easily calculate these to be <math>1, 1/2, 1/4</math> respectively. The fourth infinite series shall be all real numbers in the form <math> 1/(2^a3^b)</math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2018|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_12&diff=972652016 AMC 12A Problems/Problem 122018-08-16T15:09:56Z<p>Mathboy11: /* Solution 4 */</p>
<hr />
<div>==Problem 12==<br />
<br />
In <math>\triangle ABC</math>, <math>AB = 6</math>, <math>BC = 7</math>, and <math>CA = 8</math>. Point <math>D</math> lies on <math>\overline{BC}</math>, and <math>\overline{AD}</math> bisects <math>\angle BAC</math>. Point <math>E</math> lies on <math>\overline{AC}</math>, and <math>\overline{BE}</math> bisects <math>\angle ABC</math>. The bisectors intersect at <math>F</math>. What is the ratio <math>AF</math> : <math>FD</math>?<br />
<br />
<asy> pair A = (0,0), B=(6,0), C=intersectionpoints(Circle(A,8),Circle(B,7))[0], F=incenter(A,B,C), D=extension(A,F,B,C),E=extension(B,F,A,C); draw(A--B--C--A--D^^B--E); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,NE); label("$E$",E,NW); label("$F$",F,1.5*N); </asy><br />
<br />
<math>\textbf{(A)}\ 3:2\qquad\textbf{(B)}\ 5:3\qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 7:3\qquad\textbf{(E)}\ 5:2</math><br />
<br />
== Solution 1==<br />
<br />
Applying the angle bisector theorem to <math>\triangle ABC</math> with <math>\angle CAB</math> being bisected by <math>AD</math>, we have<br />
<br />
<cmath>\frac{CD}{AC}=\frac{BD}{AB}.</cmath><br />
<br />
Thus, we have <br />
<br />
<cmath>\frac{CD}{8}=\frac{BD}{6},</cmath><br />
<br />
and cross multiplying and dividing by <math>2</math> gives us<br />
<br />
<cmath>3\cdot CD=4\cdot BD.</cmath><br />
<br />
<br />
Since <math>CD+BD=BC=7</math>, we can substitute <math>CD=7-BD</math> into the former equation. Therefore, we get <math>3(7-BD)=4BD</math>, so <math>BD=3</math>.<br />
<br />
<br />
Apply the angle bisector theorem again to <math>\triangle ABD</math> with <math>\angle ABC</math> being bisected. This gives us<br />
<br />
<cmath>\frac{AB}{AF}=\frac{BD}{FD},</cmath><br />
<br />
and since <math>AB=6</math> and <math>BD=3</math>, we have<br />
<br />
<cmath>\frac{6}{AF}=\frac{3}{FD}.</cmath><br />
<br />
Cross multiplying and dividing by <math>3</math> gives us <br />
<br />
<cmath>AF=2\cdot FD,</cmath><br />
<br />
and dividing by <math>FD</math> gives us<br />
<br />
<cmath>\frac{AF}{FD}=\frac{2}{1}.</cmath><br />
<br />
Therefore,<br />
<br />
<cmath>AF:FD=\frac{AF}{FD}=\frac{2}{1}=\boxed{\textbf{(C)}\; 2 : 1}.</cmath><br />
<br />
== Solution 2==<br />
<br />
By the angle bisector theorem, <math>\frac{AB}{AE} = \frac{CB}{CE}</math><br />
<br />
<math>\frac{6}{AE} = \frac{7}{8 - AE}</math> so <math>AE = \frac{48}{13}</math><br />
<br />
Similarly, <math>CD = 4</math>.<br />
<br />
Now, we use mass points. Assign point <math>C</math> a mass of <math>1</math>.<br />
<br />
<math>mC \cdot CD = mB \cdot DB</math> , so <math>mB = \frac{4}{3}</math><br />
<br />
Similarly, <math>A</math> will have a mass of <math>\frac{7}{6}</math><br />
<br />
<math>mD = mC + mB = 1 + \frac{4}{3} = \frac{7}{3}</math><br />
<br />
So <math>\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}</math><br />
<br />
== Solution 3==<br />
<br />
Denote <math>[\triangle{ABC}]</math> as the area of triangle ABC and let <math>r</math> be the inradius. Also, as above, use the angle bisector theorem to find that <math>BD = 3</math>. There are two ways to continue from here:<br />
<br />
<math>1.</math> Note that <math>F</math> is the incenter. Then, <math>\frac{AF}{FD} = \frac{[\triangle{AFB}]}{[\triangle{BFD}]} = \frac{AB * \frac{r}{2}}{BD * \frac{r}{2}} = \frac{AB}{BD} = \boxed{\textbf{(C)}\; 2 : 1}</math><br />
<br />
<math>2.</math> Apply the angle bisector theorem on <math>\triangle{ABD}</math> to get <math>\frac{AF}{FD} = \frac{AB}{BD} = \frac{6}{3} = \boxed{\textbf{(C)}\; 2 : 1}</math><br />
<br />
==Solution 4==<br />
Draw the third angle bisector, and denote the point where this bisector intersects AB as P. Using angle bisector theorem, we see <math>AE=48/13 , EC=56/13, AP=16/5, PB=14/5</math>. Applying Van Aubel's theorem, <math>AF/FD=(48/13)/(56/13) + (16/5)/(14/5)=(6/7)+(8/7)=14/7=2/1</math>, and so the answer is <math>\boxed{\textbf{(C)}\; 2 : 1}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=A|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_12&diff=972642016 AMC 12A Problems/Problem 122018-08-16T15:09:04Z<p>Mathboy11: /* Solution 3 */</p>
<hr />
<div>==Problem 12==<br />
<br />
In <math>\triangle ABC</math>, <math>AB = 6</math>, <math>BC = 7</math>, and <math>CA = 8</math>. Point <math>D</math> lies on <math>\overline{BC}</math>, and <math>\overline{AD}</math> bisects <math>\angle BAC</math>. Point <math>E</math> lies on <math>\overline{AC}</math>, and <math>\overline{BE}</math> bisects <math>\angle ABC</math>. The bisectors intersect at <math>F</math>. What is the ratio <math>AF</math> : <math>FD</math>?<br />
<br />
<asy> pair A = (0,0), B=(6,0), C=intersectionpoints(Circle(A,8),Circle(B,7))[0], F=incenter(A,B,C), D=extension(A,F,B,C),E=extension(B,F,A,C); draw(A--B--C--A--D^^B--E); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,NE); label("$E$",E,NW); label("$F$",F,1.5*N); </asy><br />
<br />
<math>\textbf{(A)}\ 3:2\qquad\textbf{(B)}\ 5:3\qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 7:3\qquad\textbf{(E)}\ 5:2</math><br />
<br />
== Solution 1==<br />
<br />
Applying the angle bisector theorem to <math>\triangle ABC</math> with <math>\angle CAB</math> being bisected by <math>AD</math>, we have<br />
<br />
<cmath>\frac{CD}{AC}=\frac{BD}{AB}.</cmath><br />
<br />
Thus, we have <br />
<br />
<cmath>\frac{CD}{8}=\frac{BD}{6},</cmath><br />
<br />
and cross multiplying and dividing by <math>2</math> gives us<br />
<br />
<cmath>3\cdot CD=4\cdot BD.</cmath><br />
<br />
<br />
Since <math>CD+BD=BC=7</math>, we can substitute <math>CD=7-BD</math> into the former equation. Therefore, we get <math>3(7-BD)=4BD</math>, so <math>BD=3</math>.<br />
<br />
<br />
Apply the angle bisector theorem again to <math>\triangle ABD</math> with <math>\angle ABC</math> being bisected. This gives us<br />
<br />
<cmath>\frac{AB}{AF}=\frac{BD}{FD},</cmath><br />
<br />
and since <math>AB=6</math> and <math>BD=3</math>, we have<br />
<br />
<cmath>\frac{6}{AF}=\frac{3}{FD}.</cmath><br />
<br />
Cross multiplying and dividing by <math>3</math> gives us <br />
<br />
<cmath>AF=2\cdot FD,</cmath><br />
<br />
and dividing by <math>FD</math> gives us<br />
<br />
<cmath>\frac{AF}{FD}=\frac{2}{1}.</cmath><br />
<br />
Therefore,<br />
<br />
<cmath>AF:FD=\frac{AF}{FD}=\frac{2}{1}=\boxed{\textbf{(C)}\; 2 : 1}.</cmath><br />
<br />
== Solution 2==<br />
<br />
By the angle bisector theorem, <math>\frac{AB}{AE} = \frac{CB}{CE}</math><br />
<br />
<math>\frac{6}{AE} = \frac{7}{8 - AE}</math> so <math>AE = \frac{48}{13}</math><br />
<br />
Similarly, <math>CD = 4</math>.<br />
<br />
Now, we use mass points. Assign point <math>C</math> a mass of <math>1</math>.<br />
<br />
<math>mC \cdot CD = mB \cdot DB</math> , so <math>mB = \frac{4}{3}</math><br />
<br />
Similarly, <math>A</math> will have a mass of <math>\frac{7}{6}</math><br />
<br />
<math>mD = mC + mB = 1 + \frac{4}{3} = \frac{7}{3}</math><br />
<br />
So <math>\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}</math><br />
<br />
== Solution 3==<br />
<br />
Denote <math>[\triangle{ABC}]</math> as the area of triangle ABC and let <math>r</math> be the inradius. Also, as above, use the angle bisector theorem to find that <math>BD = 3</math>. There are two ways to continue from here:<br />
<br />
<math>1.</math> Note that <math>F</math> is the incenter. Then, <math>\frac{AF}{FD} = \frac{[\triangle{AFB}]}{[\triangle{BFD}]} = \frac{AB * \frac{r}{2}}{BD * \frac{r}{2}} = \frac{AB}{BD} = \boxed{\textbf{(C)}\; 2 : 1}</math><br />
<br />
<math>2.</math> Apply the angle bisector theorem on <math>\triangle{ABD}</math> to get <math>\frac{AF}{FD} = \frac{AB}{BD} = \frac{6}{3} = \boxed{\textbf{(C)}\; 2 : 1}</math><br />
<br />
==Solution 4==<br />
Draw the third angle bisector, and denote the point where this bisector intersects AB as P. Using angle bisector theorem, we see <math>AE=48/13 , EC=56/13, AP=16/5, PB=14/5</math>. Applying Van Aubel's theorem, <math>AF/FD=(48/13)/(56/13) + (16/5)/(14/5)=(6/7)+(8/7)=14/7=2/1</math>, and so the answer is \boxed{\textbf{(C)}\; 2 : 1}.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=A|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_12&diff=972632016 AMC 12A Problems/Problem 122018-08-16T15:04:52Z<p>Mathboy11: /* Solution 3 */</p>
<hr />
<div>==Problem 12==<br />
<br />
In <math>\triangle ABC</math>, <math>AB = 6</math>, <math>BC = 7</math>, and <math>CA = 8</math>. Point <math>D</math> lies on <math>\overline{BC}</math>, and <math>\overline{AD}</math> bisects <math>\angle BAC</math>. Point <math>E</math> lies on <math>\overline{AC}</math>, and <math>\overline{BE}</math> bisects <math>\angle ABC</math>. The bisectors intersect at <math>F</math>. What is the ratio <math>AF</math> : <math>FD</math>?<br />
<br />
<asy> pair A = (0,0), B=(6,0), C=intersectionpoints(Circle(A,8),Circle(B,7))[0], F=incenter(A,B,C), D=extension(A,F,B,C),E=extension(B,F,A,C); draw(A--B--C--A--D^^B--E); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,NE); label("$E$",E,NW); label("$F$",F,1.5*N); </asy><br />
<br />
<math>\textbf{(A)}\ 3:2\qquad\textbf{(B)}\ 5:3\qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 7:3\qquad\textbf{(E)}\ 5:2</math><br />
<br />
== Solution 1==<br />
<br />
Applying the angle bisector theorem to <math>\triangle ABC</math> with <math>\angle CAB</math> being bisected by <math>AD</math>, we have<br />
<br />
<cmath>\frac{CD}{AC}=\frac{BD}{AB}.</cmath><br />
<br />
Thus, we have <br />
<br />
<cmath>\frac{CD}{8}=\frac{BD}{6},</cmath><br />
<br />
and cross multiplying and dividing by <math>2</math> gives us<br />
<br />
<cmath>3\cdot CD=4\cdot BD.</cmath><br />
<br />
<br />
Since <math>CD+BD=BC=7</math>, we can substitute <math>CD=7-BD</math> into the former equation. Therefore, we get <math>3(7-BD)=4BD</math>, so <math>BD=3</math>.<br />
<br />
<br />
Apply the angle bisector theorem again to <math>\triangle ABD</math> with <math>\angle ABC</math> being bisected. This gives us<br />
<br />
<cmath>\frac{AB}{AF}=\frac{BD}{FD},</cmath><br />
<br />
and since <math>AB=6</math> and <math>BD=3</math>, we have<br />
<br />
<cmath>\frac{6}{AF}=\frac{3}{FD}.</cmath><br />
<br />
Cross multiplying and dividing by <math>3</math> gives us <br />
<br />
<cmath>AF=2\cdot FD,</cmath><br />
<br />
and dividing by <math>FD</math> gives us<br />
<br />
<cmath>\frac{AF}{FD}=\frac{2}{1}.</cmath><br />
<br />
Therefore,<br />
<br />
<cmath>AF:FD=\frac{AF}{FD}=\frac{2}{1}=\boxed{\textbf{(C)}\; 2 : 1}.</cmath><br />
<br />
== Solution 2==<br />
<br />
By the angle bisector theorem, <math>\frac{AB}{AE} = \frac{CB}{CE}</math><br />
<br />
<math>\frac{6}{AE} = \frac{7}{8 - AE}</math> so <math>AE = \frac{48}{13}</math><br />
<br />
Similarly, <math>CD = 4</math>.<br />
<br />
Now, we use mass points. Assign point <math>C</math> a mass of <math>1</math>.<br />
<br />
<math>mC \cdot CD = mB \cdot DB</math> , so <math>mB = \frac{4}{3}</math><br />
<br />
Similarly, <math>A</math> will have a mass of <math>\frac{7}{6}</math><br />
<br />
<math>mD = mC + mB = 1 + \frac{4}{3} = \frac{7}{3}</math><br />
<br />
So <math>\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}</math><br />
<br />
== Solution 3==<br />
<br />
Denote <math>[\triangle{ABC}]</math> as the area of triangle ABC and let <math>r</math> be the inradius. Also, as above, use the angle bisector theorem to find that <math>BD = 3</math>. There are two ways to continue from here:<br />
<br />
<math>1.</math> Note that <math>F</math> is the incenter. Then, <math>\frac{AF}{FD} = \frac{[\triangle{AFB}]}{[\triangle{BFD}]} = \frac{AB * \frac{r}{2}}{BD * \frac{r}{2}} = \frac{AB}{BD} = \boxed{\textbf{(C)}\; 2 : 1}</math><br />
<br />
<math>2.</math> Apply the angle bisector theorem on <math>\triangle{ABD}</math> to get <math>\frac{AF}{FD} = \frac{AB}{BD} = \frac{6}{3} = \boxed{\textbf{(C)}\; 2 : 1}</math><br />
<br />
Solution 4: <br />
Draw the third angle bisector, and denote the point where this bisector intersects AB as P. Using angle bisector theorem, we see <math>AE=48/13 , EC=56/13</math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2016|ab=A|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_12A_Problems/Problem_17&diff=968912014 AMC 12A Problems/Problem 172018-08-09T15:35:02Z<p>Mathboy11: /* Solution */</p>
<hr />
<div>==Problem==<br />
A <math>4\times 4\times h</math> rectangular box contains a sphere of radius <math>2</math> and eight smaller spheres of radius <math>1</math>. The smaller spheres are each tangent to three sides of the box, and the larger sphere is tangent to each of the smaller spheres. What is <math>h</math>?<br />
<br />
<center><asy><br />
import graph3;<br />
import solids;<br />
real h=2+2*sqrt(7);<br />
currentprojection=orthographic((0.75,-5,h/2+1),target=(2,2,h/2));<br />
currentlight=light(4,-4,4);<br />
draw((0,0,0)--(4,0,0)--(4,4,0)--(0,4,0)--(0,0,0)^^(4,0,0)--(4,0,h)--(4,4,h)--(0,4,h)--(0,4,0));<br />
draw(shift((1,3,1))*unitsphere,gray(0.85));<br />
draw(shift((3,3,1))*unitsphere,gray(0.85));<br />
draw(shift((3,1,1))*unitsphere,gray(0.85));<br />
draw(shift((1,1,1))*unitsphere,gray(0.85));<br />
draw(shift((2,2,h/2))*scale(2,2,2)*unitsphere,gray(0.85));<br />
draw(shift((1,3,h-1))*unitsphere,gray(0.85));<br />
draw(shift((3,3,h-1))*unitsphere,gray(0.85));<br />
draw(shift((3,1,h-1))*unitsphere,gray(0.85));<br />
draw(shift((1,1,h-1))*unitsphere,gray(0.85));<br />
draw((0,0,0)--(0,0,h)--(4,0,h)^^(0,0,h)--(0,4,h));<br />
</asy></center><br />
<br />
<math>\textbf{(A) }2+2\sqrt 7\qquad<br />
\textbf{(B) }3+2\sqrt 5\qquad<br />
\textbf{(C) }4+2\sqrt 7\qquad<br />
\textbf{(D) }4\sqrt 5\qquad<br />
\textbf{(E) }4\sqrt 7\qquad</math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
==Solution==<br />
<br />
Let <math>A</math> be the point in the same plane as the centers of the top spheres equidistant from said centers. Let <math>B</math> be the analogous point for the bottom spheres, and let <math>C</math> be the midpoint of <math>\overline{AB}</math> and the center of the large sphere. Let <math>D</math> and <math>E</math> be the points at which line <math>AB</math> intersects the top of the box and the bottom, respectively.<br />
<br />
Let <math>O</math> be the center of any of the top spheres (you choose!). We have <math>AO=1\cdot\sqrt{2}</math>, and <math>CO=3</math>, so <math>AC=\sqrt{3^2-\sqrt2^2}=\sqrt{7}</math>. Similarly, <math>BC=\sqrt{7}</math>. <math>\overline{AD}</math> and <math>\overline{BE}</math> are clearly equal to the radius of the small spheres, <math>1</math>. Thus the total height is <math>AD+AC+BC+BE=2+2\sqrt7</math>, or <math>\boxed{\textbf{(A)}}</math>.<br />
<br />
(Solution by AwesomeToad)<br />
<br />
==Solution 2==<br />
<br />
Use the 4 bottom spheres. (In progress)<br />
<br />
==See Also==<br />
{{AMC12 box|year=2014|ab=A|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_23&diff=968272017 AMC 10B Problems/Problem 232018-08-07T14:42:20Z<p>Mathboy11: /* See Also */</p>
<hr />
<div>==Problem 23==<br />
Let <math>N=123456789101112\dots4344</math> be the <math>79</math>-digit number that is formed by writing the integers from <math>1</math> to <math>44</math> in order, one after the other. What is the remainder when <math>N</math> is divided by <math>45</math>?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44</math><br />
<br />
==Solution==<br />
We only need to find the remainders of N when divided by 5 and 9 to determine the answer.<br />
By inspection, <math>N \equiv 4 \text{ (mod 5)}</math>.<br />
The remainder when <math>N</math> is divided by <math>9</math> is <math>1+2+3+4+ \cdots +1+0+1+1 +1+2 +\cdots + 4+3+4+4</math>, but since <math>10 \equiv 1 \text{ (mod 9)}</math>, we can also write this as <math>1+2+3 +\cdots +10+11+12+ \cdots 43 + 44 = \frac{44 \cdot 45}2 = 22 \cdot 45</math>, which has a remainder of 0 mod 9. Therefore, by inspection, the answer is <math>\boxed{\textbf{(C) } 9}</math>.<br />
<br />
Note: the sum of the digits of <math>N</math> is <math>270</math>.<br />
<br />
==Solution 2==<br />
Noting the solution above, we try to find the sum of the digits to figure out its remainder when divided by <math>9</math>. From <math>1</math> thru <math>9</math>, the sum is <math>45</math>. <math>10</math> thru <math>19</math>, the sum is <math>55</math>, <math>20</math> thru <math>29</math> is <math>65</math>, and <math>30</math> thru <math>39</math> is <math>75</math>. Thus the sum of the digits is <math>45+55+65+75+4+5+6+7+8 = 240+30 = 270</math>, and thus <math>N</math> is divisible by <math>9</math>. Now, refer to the above solution. <math>N \equiv 4 \text{ (mod 5)}</math> and <math>N \equiv 0 \text{ (mod 9)}</math>. From this information, we can conclude that <math>N \equiv 54 \text{ (mod 5)}</math> and <math>N \equiv 54 \text{ (mod 9)}</math>. Therefore, <math>N \equiv 54 \text{ (mod 45)}</math> and <math>N \equiv 9 \text{ (mod 45)}</math> so the remainder is <math>\boxed{\textbf{(C) }9}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=B|num-b=22|num-a=24}}<br />
{{MAA Notice}}<br />
yeet</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_21&diff=936212011 AMC 10B Problems/Problem 212018-03-31T01:20:06Z<p>Mathboy11: /* Solution 4 */</p>
<hr />
<div>== Problem ==<br />
<br />
Brian writes down four integers <math>w > x > y > z</math> whose sum is <math>44</math>. The pairwise positive differences of these numbers are <math>1, 3, 4, 5, 6,</math> and <math>9</math>. What is the sum of the possible values for <math>w</math>?<br />
<br />
<math> \textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93</math><br />
<br />
== Solution 1 ==<br />
The largest difference, <math>9,</math> must be between <math>w</math> and <math>z.</math><br />
<br />
The smallest difference, <math>1,</math> must be directly between two integers. This also means the differences directly between the other two should add up to <math>8.</math> The only remaining differences that would make this possible are <math>3</math> and <math>5.</math> However, those two differences can't be right next to each other because they would make a difference of <math>8.</math> This means <math>1</math> must be the difference between <math>y</math> and <math>x.</math> We can express the possible configurations as the lines.<br />
<br />
<br />
<center><asy><br />
unitsize(14mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
pair Z1=(0,1), Y1=(1,1), X1=(2,1), W1=(3,1);<br />
pair Z4=(4,1), Y4=(5,1), X4=(6,1), W4=(7,1);<br />
<br />
draw(Z1--W1); draw(Z4--W4);<br />
<br />
pair[] ps={W1,W4,X1,X4,Y1,Y4,Z1,Z4};<br />
dot(ps);<br />
label("$z$",Z1,N); label("$y$",Y1,N); label("$x$",X1,N); label("$w$",W1,N);<br />
label("$z$",Z4,N); label("$y$",Y4,N); label("$x$",X4,N); label("$w$",W4,N);<br />
<br />
label("$1$",(X1--Y1),N); label("$1$",(X4--Y4),N);<br />
label("$3$",(Y1--Z1),N); label("$3$",(W4--X4),N); label("$5$",(X1--W1),N); label("$5$",(Y4--Z4),N);<br />
<br />
</asy><br />
</center><br />
<br />
If we look at the first number line, you can express <math>x</math> as <math>w-5,</math> <math>y</math> as <math>w-6,</math> and <math>z</math> as <math>w-9.</math> Since the sum of all these integers equal <math>44</math>,<br />
<cmath>\begin{align*}<br />
w+w-5+w-6+w-9&=44\\<br />
4w&=64\\<br />
w&=16 \end{align*}</cmath><br />
You can do something similar to this with the second number line to find the other possible value of <math>w.</math><br />
<cmath>\begin{align*}<br />
w+w-3+w-4+w-9&=44\\<br />
4w&=60\\<br />
w&=15 \end{align*}</cmath><br />
The sum of the possible values of <math>w</math> is <math>16+15 = \boxed{\textbf{(B) }31}</math><br />
<br />
== Solution 2 ==<br />
<br />
First, like Solution 1, we know that <math>w-z=9 \ \text{(1)}</math>, because no sum could be smaller. Next, we find the sum of all the differences; since <math>w</math> is in the positive part of a difference 3 times, and has no differences where it contributes as the negative part, the sum of the differences includes <math>3w</math>. Continuing in this way, we find that <cmath>3w+x-y-3z=28 \ \text{(2)}</cmath>. Now, we can subtract <math>3w-3z=27</math> from (2) to get <math>x-y=1 \ \text{(3)}</math>. Also, adding (2) with <math>w+x+y+z=44</math> gives <math>4w+2x-2z=72</math>, or <math>2w+x-z=36</math>. Subtracting (1) from this gives <math>w+x=27</math>. Since we know <math>w-z</math> and <math>x-y</math>, we find that <cmath>(w-z)+(x-y)=(w-y)+(x-z)=9+1=10</cmath>. This means that <math>w-y</math> and <math>x-z</math> must be 4 and 6, in some order. If <math>w-y=6</math>, then subtracting this from (3) gives <math>(w-y)-(x-y)=6-1=5</math>, so <math>w-x=5</math>. This means that <math>(w-x)+(w+x)=2w=27+5=32</math>, so <math>w=16</math>. Similarly, <math>w</math> can also equal <math>15</math>.<br />
<br />
Now if you are in a rush, you would have just answered <math>16+15=\boxed{\textbf{(B) }31}</math>. But we do have to check if these work. In fact, they do, giving solutions <math>\boxed{16, 11, 10, 7}</math> and <math>\boxed{15, 12, 11, 6}</math>.<br />
<br />
<br />
== Solution 3 ==<br />
<br />
Let <math>w - x = a</math>, <math>w - y = b</math>, <math>w - z = c</math>. As above, we know that <math>c = 9</math>. Thus, <math>a < b < c</math>. <br />
So, we have <math>w + x + y + z = w + (w - a) + (w - b) + (w - 9) = 4w - a - b - 9 = 44</math>. This means <math>a + b + 9</math> is a multiple of <math>4</math>. Testing values of <math>a</math> and <math>b</math>, we find <math>(a, b, c) = (1, 6, 9), (3, 4, 9),</math> and <math>(5, 6, 9)</math> all satisfy this relation. The corresponding <math>(w, x, y, z)</math> sets are <math>(15, 14, 9, 6), (15, 12, 11, 6),</math> and <math>(16, 11, 10, 7)</math>. The first set does not satisfy the given conditions, but the other two do. Thus, <math>w = 15</math> and <math>w = 16</math> are both possible solutions so the answer is <math>15 + 16 = 31</math>.<br />
<br />
~~ solution by ccx09 Roy Short<br />
~~latexed by jk23541<br />
== Solution 4 ==<br />
From the problem, we know that <math>w+x+y+z=44</math>. Since it is said that the pairwise positive differences between numbers are 1, 3, 4, 5, 6, 9, and we can figure that the pairwise positive differences are <math>(w-x)</math>, <math>(w-y)</math>, <math>(w-z)</math>, <math>(x-y)</math>, <math>(x-z)</math>, <math>(y-z)</math>, the sum of <math>(w-x), (w-y), (w-z), (x-y), (x-z), (y-z)</math> is equal to the sum of 1, 3, 4, 5, 6, 9, so <math>(w-x)+(w-y)+(w-z)+(x-y)+(x-z)+(y-z)=28</math>. Simplifying, we get <math>3w-3z+x-y=28</math>. Adding <math>w+x+y+z=44</math> and <math>(w-x)+(w-y)+(w-z)+(x-y)+(x-z)+(y-z)=28</math>, we get <math>4w+2x-2z=72</math>, and simplifying we get <math>2w+x-z=36</math>. Since <math>x-z</math> is one of our positive differences, we can start guessing values for <math>w</math>, and if the equation simplifies to one of our numerical positive differences, that value of <math>w</math> should work. We can start at <math>w=18</math> and keep going down, because our sum has to be positive. For <math>w=18</math>, <math>x-z=0</math>, which is not one of our sums. For <math>w=17</math>, <math>x-z=2</math>,which is not one of our sums. For <math>w=16</math>, <math>x-z=4</math>, which is one of our sums, so 16 works. For <math>w=15</math>, <math>x-z=6</math>, which is one of our sums, so 15 works. For <math>w=14</math>, <math>x-z=8</math>, which is not one of our sums. If we keep going, <math>x-z</math> will soon exceed 10 and exceed all our sums, so any value below <math>w=14</math> will not work. Therefore, our only solutions for <math>w</math> are 15 and 16, which means our sum is <math>\boxed{31}</math>. You can check that 15 and 16 work by forming a string of 4 numbers as shown above.<br />
<br />
== See Also==<br />
<br />
{{AMC10 box|year=2011|ab=B|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_21&diff=936202011 AMC 10B Problems/Problem 212018-03-31T01:19:28Z<p>Mathboy11: /* Solution 4 */</p>
<hr />
<div>== Problem ==<br />
<br />
Brian writes down four integers <math>w > x > y > z</math> whose sum is <math>44</math>. The pairwise positive differences of these numbers are <math>1, 3, 4, 5, 6,</math> and <math>9</math>. What is the sum of the possible values for <math>w</math>?<br />
<br />
<math> \textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93</math><br />
<br />
== Solution 1 ==<br />
The largest difference, <math>9,</math> must be between <math>w</math> and <math>z.</math><br />
<br />
The smallest difference, <math>1,</math> must be directly between two integers. This also means the differences directly between the other two should add up to <math>8.</math> The only remaining differences that would make this possible are <math>3</math> and <math>5.</math> However, those two differences can't be right next to each other because they would make a difference of <math>8.</math> This means <math>1</math> must be the difference between <math>y</math> and <math>x.</math> We can express the possible configurations as the lines.<br />
<br />
<br />
<center><asy><br />
unitsize(14mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
pair Z1=(0,1), Y1=(1,1), X1=(2,1), W1=(3,1);<br />
pair Z4=(4,1), Y4=(5,1), X4=(6,1), W4=(7,1);<br />
<br />
draw(Z1--W1); draw(Z4--W4);<br />
<br />
pair[] ps={W1,W4,X1,X4,Y1,Y4,Z1,Z4};<br />
dot(ps);<br />
label("$z$",Z1,N); label("$y$",Y1,N); label("$x$",X1,N); label("$w$",W1,N);<br />
label("$z$",Z4,N); label("$y$",Y4,N); label("$x$",X4,N); label("$w$",W4,N);<br />
<br />
label("$1$",(X1--Y1),N); label("$1$",(X4--Y4),N);<br />
label("$3$",(Y1--Z1),N); label("$3$",(W4--X4),N); label("$5$",(X1--W1),N); label("$5$",(Y4--Z4),N);<br />
<br />
</asy><br />
</center><br />
<br />
If we look at the first number line, you can express <math>x</math> as <math>w-5,</math> <math>y</math> as <math>w-6,</math> and <math>z</math> as <math>w-9.</math> Since the sum of all these integers equal <math>44</math>,<br />
<cmath>\begin{align*}<br />
w+w-5+w-6+w-9&=44\\<br />
4w&=64\\<br />
w&=16 \end{align*}</cmath><br />
You can do something similar to this with the second number line to find the other possible value of <math>w.</math><br />
<cmath>\begin{align*}<br />
w+w-3+w-4+w-9&=44\\<br />
4w&=60\\<br />
w&=15 \end{align*}</cmath><br />
The sum of the possible values of <math>w</math> is <math>16+15 = \boxed{\textbf{(B) }31}</math><br />
<br />
== Solution 2 ==<br />
<br />
First, like Solution 1, we know that <math>w-z=9 \ \text{(1)}</math>, because no sum could be smaller. Next, we find the sum of all the differences; since <math>w</math> is in the positive part of a difference 3 times, and has no differences where it contributes as the negative part, the sum of the differences includes <math>3w</math>. Continuing in this way, we find that <cmath>3w+x-y-3z=28 \ \text{(2)}</cmath>. Now, we can subtract <math>3w-3z=27</math> from (2) to get <math>x-y=1 \ \text{(3)}</math>. Also, adding (2) with <math>w+x+y+z=44</math> gives <math>4w+2x-2z=72</math>, or <math>2w+x-z=36</math>. Subtracting (1) from this gives <math>w+x=27</math>. Since we know <math>w-z</math> and <math>x-y</math>, we find that <cmath>(w-z)+(x-y)=(w-y)+(x-z)=9+1=10</cmath>. This means that <math>w-y</math> and <math>x-z</math> must be 4 and 6, in some order. If <math>w-y=6</math>, then subtracting this from (3) gives <math>(w-y)-(x-y)=6-1=5</math>, so <math>w-x=5</math>. This means that <math>(w-x)+(w+x)=2w=27+5=32</math>, so <math>w=16</math>. Similarly, <math>w</math> can also equal <math>15</math>.<br />
<br />
Now if you are in a rush, you would have just answered <math>16+15=\boxed{\textbf{(B) }31}</math>. But we do have to check if these work. In fact, they do, giving solutions <math>\boxed{16, 11, 10, 7}</math> and <math>\boxed{15, 12, 11, 6}</math>.<br />
<br />
<br />
== Solution 3 ==<br />
<br />
Let <math>w - x = a</math>, <math>w - y = b</math>, <math>w - z = c</math>. As above, we know that <math>c = 9</math>. Thus, <math>a < b < c</math>. <br />
So, we have <math>w + x + y + z = w + (w - a) + (w - b) + (w - 9) = 4w - a - b - 9 = 44</math>. This means <math>a + b + 9</math> is a multiple of <math>4</math>. Testing values of <math>a</math> and <math>b</math>, we find <math>(a, b, c) = (1, 6, 9), (3, 4, 9),</math> and <math>(5, 6, 9)</math> all satisfy this relation. The corresponding <math>(w, x, y, z)</math> sets are <math>(15, 14, 9, 6), (15, 12, 11, 6),</math> and <math>(16, 11, 10, 7)</math>. The first set does not satisfy the given conditions, but the other two do. Thus, <math>w = 15</math> and <math>w = 16</math> are both possible solutions so the answer is <math>15 + 16 = 31</math>.<br />
<br />
~~ solution by ccx09 Roy Short<br />
~~latexed by jk23541<br />
== Solution 4 ==<br />
From the problem, we know that <math>w+x+y+z=44</math>. Since it is said that the pairwise positive differences between numbers are 1, 3, 4, 5, 6, 9, and we can figure that the pairwise positive differences are <math>(w-x)</math>, <math>(w-y)</math>, <math>(w-z)</math>, <math>(x-y)</math>, <math>(x-z)</math>, <math>(y-z)</math>, the sum of <math>(w-x), (w-y), (w-z), (x-y), (x-z), (y-z)</math> is equal to the sum of 1, 3, 4, 5, 6, 9, so <math>(w-x)+(w-y)+(w-z)+(x-y)+(x-z)+(y-z)=28</math>. Simplifying, we get <math>3w-3z+x-y=28</math>. Adding <math>w+x+y+z=44</math> and <math>(w-x)+(w-y)+(w-z)+(x-y)+(x-z)+(y-z)=28</math>, we get <math>4w+2x-2z=72</math>, and simplifying we get <math>2w+x-z=36</math>. Since <math>x-z</math> is one of our positive differences, we can start guessing values for <math>w</math>, and if the equation simplifies to one of our numerical positive differences, that value of <math>w</math> should work. We can start at <math>w=18</math> and keep going down, because our sum has to be positive. For <math>w=18</math>, <math>x-z=0</math>, which is not one of our sums. For <math>w=17</math>, <math>x-z=2</math>,which is not one of our sums. For <math>w=16</math>, <math>x-z=4</math>, which is one of our sums, so 16 works. For <math>w=15</math>, <math>x-z=6</math>, which is one of our sums, so 15 works. For <math>w=14</math>, <math>x-z=8</math>, which is not one of our sums. If we keep going, <math>x-z</math> will soon exceed 10 and exceed all our sums, so any value below <math>w=14</math> will not work. Therefore, our only solutions for <math>w</math> are 15 and 16, which means our sum is <math>\boxed{31}</math>. You can check that 15 and 16 work by forming a string of 4 numbers as shown above. <br />
/pi<br />
<br />
== See Also==<br />
<br />
{{AMC10 box|year=2011|ab=B|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_21&diff=936192011 AMC 10B Problems/Problem 212018-03-31T01:19:18Z<p>Mathboy11: /* Solution 4 */</p>
<hr />
<div>== Problem ==<br />
<br />
Brian writes down four integers <math>w > x > y > z</math> whose sum is <math>44</math>. The pairwise positive differences of these numbers are <math>1, 3, 4, 5, 6,</math> and <math>9</math>. What is the sum of the possible values for <math>w</math>?<br />
<br />
<math> \textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93</math><br />
<br />
== Solution 1 ==<br />
The largest difference, <math>9,</math> must be between <math>w</math> and <math>z.</math><br />
<br />
The smallest difference, <math>1,</math> must be directly between two integers. This also means the differences directly between the other two should add up to <math>8.</math> The only remaining differences that would make this possible are <math>3</math> and <math>5.</math> However, those two differences can't be right next to each other because they would make a difference of <math>8.</math> This means <math>1</math> must be the difference between <math>y</math> and <math>x.</math> We can express the possible configurations as the lines.<br />
<br />
<br />
<center><asy><br />
unitsize(14mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
pair Z1=(0,1), Y1=(1,1), X1=(2,1), W1=(3,1);<br />
pair Z4=(4,1), Y4=(5,1), X4=(6,1), W4=(7,1);<br />
<br />
draw(Z1--W1); draw(Z4--W4);<br />
<br />
pair[] ps={W1,W4,X1,X4,Y1,Y4,Z1,Z4};<br />
dot(ps);<br />
label("$z$",Z1,N); label("$y$",Y1,N); label("$x$",X1,N); label("$w$",W1,N);<br />
label("$z$",Z4,N); label("$y$",Y4,N); label("$x$",X4,N); label("$w$",W4,N);<br />
<br />
label("$1$",(X1--Y1),N); label("$1$",(X4--Y4),N);<br />
label("$3$",(Y1--Z1),N); label("$3$",(W4--X4),N); label("$5$",(X1--W1),N); label("$5$",(Y4--Z4),N);<br />
<br />
</asy><br />
</center><br />
<br />
If we look at the first number line, you can express <math>x</math> as <math>w-5,</math> <math>y</math> as <math>w-6,</math> and <math>z</math> as <math>w-9.</math> Since the sum of all these integers equal <math>44</math>,<br />
<cmath>\begin{align*}<br />
w+w-5+w-6+w-9&=44\\<br />
4w&=64\\<br />
w&=16 \end{align*}</cmath><br />
You can do something similar to this with the second number line to find the other possible value of <math>w.</math><br />
<cmath>\begin{align*}<br />
w+w-3+w-4+w-9&=44\\<br />
4w&=60\\<br />
w&=15 \end{align*}</cmath><br />
The sum of the possible values of <math>w</math> is <math>16+15 = \boxed{\textbf{(B) }31}</math><br />
<br />
== Solution 2 ==<br />
<br />
First, like Solution 1, we know that <math>w-z=9 \ \text{(1)}</math>, because no sum could be smaller. Next, we find the sum of all the differences; since <math>w</math> is in the positive part of a difference 3 times, and has no differences where it contributes as the negative part, the sum of the differences includes <math>3w</math>. Continuing in this way, we find that <cmath>3w+x-y-3z=28 \ \text{(2)}</cmath>. Now, we can subtract <math>3w-3z=27</math> from (2) to get <math>x-y=1 \ \text{(3)}</math>. Also, adding (2) with <math>w+x+y+z=44</math> gives <math>4w+2x-2z=72</math>, or <math>2w+x-z=36</math>. Subtracting (1) from this gives <math>w+x=27</math>. Since we know <math>w-z</math> and <math>x-y</math>, we find that <cmath>(w-z)+(x-y)=(w-y)+(x-z)=9+1=10</cmath>. This means that <math>w-y</math> and <math>x-z</math> must be 4 and 6, in some order. If <math>w-y=6</math>, then subtracting this from (3) gives <math>(w-y)-(x-y)=6-1=5</math>, so <math>w-x=5</math>. This means that <math>(w-x)+(w+x)=2w=27+5=32</math>, so <math>w=16</math>. Similarly, <math>w</math> can also equal <math>15</math>.<br />
<br />
Now if you are in a rush, you would have just answered <math>16+15=\boxed{\textbf{(B) }31}</math>. But we do have to check if these work. In fact, they do, giving solutions <math>\boxed{16, 11, 10, 7}</math> and <math>\boxed{15, 12, 11, 6}</math>.<br />
<br />
<br />
== Solution 3 ==<br />
<br />
Let <math>w - x = a</math>, <math>w - y = b</math>, <math>w - z = c</math>. As above, we know that <math>c = 9</math>. Thus, <math>a < b < c</math>. <br />
So, we have <math>w + x + y + z = w + (w - a) + (w - b) + (w - 9) = 4w - a - b - 9 = 44</math>. This means <math>a + b + 9</math> is a multiple of <math>4</math>. Testing values of <math>a</math> and <math>b</math>, we find <math>(a, b, c) = (1, 6, 9), (3, 4, 9),</math> and <math>(5, 6, 9)</math> all satisfy this relation. The corresponding <math>(w, x, y, z)</math> sets are <math>(15, 14, 9, 6), (15, 12, 11, 6),</math> and <math>(16, 11, 10, 7)</math>. The first set does not satisfy the given conditions, but the other two do. Thus, <math>w = 15</math> and <math>w = 16</math> are both possible solutions so the answer is <math>15 + 16 = 31</math>.<br />
<br />
~~ solution by ccx09 Roy Short<br />
~~latexed by jk23541<br />
== Solution 4 ==<br />
From the problem, we know that <math>w+x+y+z=44</math>. Since it is said that the pairwise positive differences between numbers are 1, 3, 4, 5, 6, 9, and we can figure that the pairwise positive differences are <math>(w-x)</math>, <math>(w-y)</math>, <math>(w-z)</math>, <math>(x-y)</math>, <math>(x-z)</math>, <math>(y-z)</math>, the sum of <math>(w-x), (w-y), (w-z), (x-y), (x-z), (y-z)</math> is equal to the sum of 1, 3, 4, 5, 6, 9, so <math>(w-x)+(w-y)+(w-z)+(x-y)+(x-z)+(y-z)=28</math>. Simplifying, we get <math>3w-3z+x-y=28</math>. Adding <math>w+x+y+z=44</math> and <math>(w-x)+(w-y)+(w-z)+(x-y)+(x-z)+(y-z)=28</math>, we get <math>4w+2x-2z=72</math>, and simplifying we get <math>2w+x-z=36</math>. Since <math>x-z</math> is one of our positive differences, we can start guessing values for <math>w</math>, and if the equation simplifies to one of our numerical positive differences, that value of <math>w</math> should work. We can start at <math>w=18</math> and keep going down, because our sum has to be positive. For <math>w=18</math>, <math>x-z=0</math>, which is not one of our sums. For <math>w=17</math>, <math>x-z=2</math>,which is not one of our sums. For <math>w=16</math>, <math>x-z=4</math>, which is one of our sums, so 16 works. For <math>w=15</math>, <math>x-z=6</math>, which is one of our sums, so 15 works. For <math>w=14</math>, <math>x-z=8</math>, which is not one of our sums. If we keep going, <math>x-z</math> will soon exceed 10 and exceed all our sums, so any value below <math>w=14</math> will not work. Therefore, our only solutions for <math>w</math> are 15 and 16, which means our sum is <math>\boxed{31}</math>. You can check that 15 and 16 work by forming a string of 4 numbers as shown above. <br />
<math>/pi</math><br />
<br />
== See Also==<br />
<br />
{{AMC10 box|year=2011|ab=B|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_21&diff=936182011 AMC 10B Problems/Problem 212018-03-31T01:18:48Z<p>Mathboy11: /* Solution 4 */</p>
<hr />
<div>== Problem ==<br />
<br />
Brian writes down four integers <math>w > x > y > z</math> whose sum is <math>44</math>. The pairwise positive differences of these numbers are <math>1, 3, 4, 5, 6,</math> and <math>9</math>. What is the sum of the possible values for <math>w</math>?<br />
<br />
<math> \textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93</math><br />
<br />
== Solution 1 ==<br />
The largest difference, <math>9,</math> must be between <math>w</math> and <math>z.</math><br />
<br />
The smallest difference, <math>1,</math> must be directly between two integers. This also means the differences directly between the other two should add up to <math>8.</math> The only remaining differences that would make this possible are <math>3</math> and <math>5.</math> However, those two differences can't be right next to each other because they would make a difference of <math>8.</math> This means <math>1</math> must be the difference between <math>y</math> and <math>x.</math> We can express the possible configurations as the lines.<br />
<br />
<br />
<center><asy><br />
unitsize(14mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
pair Z1=(0,1), Y1=(1,1), X1=(2,1), W1=(3,1);<br />
pair Z4=(4,1), Y4=(5,1), X4=(6,1), W4=(7,1);<br />
<br />
draw(Z1--W1); draw(Z4--W4);<br />
<br />
pair[] ps={W1,W4,X1,X4,Y1,Y4,Z1,Z4};<br />
dot(ps);<br />
label("$z$",Z1,N); label("$y$",Y1,N); label("$x$",X1,N); label("$w$",W1,N);<br />
label("$z$",Z4,N); label("$y$",Y4,N); label("$x$",X4,N); label("$w$",W4,N);<br />
<br />
label("$1$",(X1--Y1),N); label("$1$",(X4--Y4),N);<br />
label("$3$",(Y1--Z1),N); label("$3$",(W4--X4),N); label("$5$",(X1--W1),N); label("$5$",(Y4--Z4),N);<br />
<br />
</asy><br />
</center><br />
<br />
If we look at the first number line, you can express <math>x</math> as <math>w-5,</math> <math>y</math> as <math>w-6,</math> and <math>z</math> as <math>w-9.</math> Since the sum of all these integers equal <math>44</math>,<br />
<cmath>\begin{align*}<br />
w+w-5+w-6+w-9&=44\\<br />
4w&=64\\<br />
w&=16 \end{align*}</cmath><br />
You can do something similar to this with the second number line to find the other possible value of <math>w.</math><br />
<cmath>\begin{align*}<br />
w+w-3+w-4+w-9&=44\\<br />
4w&=60\\<br />
w&=15 \end{align*}</cmath><br />
The sum of the possible values of <math>w</math> is <math>16+15 = \boxed{\textbf{(B) }31}</math><br />
<br />
== Solution 2 ==<br />
<br />
First, like Solution 1, we know that <math>w-z=9 \ \text{(1)}</math>, because no sum could be smaller. Next, we find the sum of all the differences; since <math>w</math> is in the positive part of a difference 3 times, and has no differences where it contributes as the negative part, the sum of the differences includes <math>3w</math>. Continuing in this way, we find that <cmath>3w+x-y-3z=28 \ \text{(2)}</cmath>. Now, we can subtract <math>3w-3z=27</math> from (2) to get <math>x-y=1 \ \text{(3)}</math>. Also, adding (2) with <math>w+x+y+z=44</math> gives <math>4w+2x-2z=72</math>, or <math>2w+x-z=36</math>. Subtracting (1) from this gives <math>w+x=27</math>. Since we know <math>w-z</math> and <math>x-y</math>, we find that <cmath>(w-z)+(x-y)=(w-y)+(x-z)=9+1=10</cmath>. This means that <math>w-y</math> and <math>x-z</math> must be 4 and 6, in some order. If <math>w-y=6</math>, then subtracting this from (3) gives <math>(w-y)-(x-y)=6-1=5</math>, so <math>w-x=5</math>. This means that <math>(w-x)+(w+x)=2w=27+5=32</math>, so <math>w=16</math>. Similarly, <math>w</math> can also equal <math>15</math>.<br />
<br />
Now if you are in a rush, you would have just answered <math>16+15=\boxed{\textbf{(B) }31}</math>. But we do have to check if these work. In fact, they do, giving solutions <math>\boxed{16, 11, 10, 7}</math> and <math>\boxed{15, 12, 11, 6}</math>.<br />
<br />
<br />
== Solution 3 ==<br />
<br />
Let <math>w - x = a</math>, <math>w - y = b</math>, <math>w - z = c</math>. As above, we know that <math>c = 9</math>. Thus, <math>a < b < c</math>. <br />
So, we have <math>w + x + y + z = w + (w - a) + (w - b) + (w - 9) = 4w - a - b - 9 = 44</math>. This means <math>a + b + 9</math> is a multiple of <math>4</math>. Testing values of <math>a</math> and <math>b</math>, we find <math>(a, b, c) = (1, 6, 9), (3, 4, 9),</math> and <math>(5, 6, 9)</math> all satisfy this relation. The corresponding <math>(w, x, y, z)</math> sets are <math>(15, 14, 9, 6), (15, 12, 11, 6),</math> and <math>(16, 11, 10, 7)</math>. The first set does not satisfy the given conditions, but the other two do. Thus, <math>w = 15</math> and <math>w = 16</math> are both possible solutions so the answer is <math>15 + 16 = 31</math>.<br />
<br />
~~ solution by ccx09 Roy Short<br />
~~latexed by jk23541<br />
== Solution 4 ==<br />
From the problem, we know that <math>w+x+y+z=44</math>. Since it is said that the pairwise positive differences between numbers are 1, 3, 4, 5, 6, 9, and we can figure that the pairwise positive differences are <math>(w-x)</math>, <math>(w-y)</math>, <math>(w-z)</math>, <math>(x-y)</math>, <math>(x-z)</math>, <math>(y-z)</math>, the sum of <math>(w-x), (w-y), (w-z), (x-y), (x-z), (y-z)</math> is equal to the sum of 1, 3, 4, 5, 6, 9, so <math>(w-x)+(w-y)+(w-z)+(x-y)+(x-z)+(y-z)=28</math>. Simplifying, we get <math>3w-3z+x-y=28</math>. Adding <math>w+x+y+z=44</math> and <math>(w-x)+(w-y)+(w-z)+(x-y)+(x-z)+(y-z)=28</math>, we get <math>4w+2x-2z=72</math>, and simplifying we get <math>2w+x-z=36</math>. Since <math>x-z</math> is one of our positive differences, we can start guessing values for <math>w</math>, and if the equation simplifies to one of our numerical positive differences, that value of <math>w</math> should work. We can start at <math>w=18</math> and keep going down, because our sum has to be positive. For <math>w=18</math>, <math>x-z=0</math>, which is not one of our sums. For <math>w=17</math>, <math>x-z=2</math>,which is not one of our sums. For <math>w=16</math>, <math>x-z=4</math>, which is one of our sums, so 16 works. For <math>w=15</math>, <math>x-z=6</math>, which is one of our sums, so 15 works. For <math>w=14</math>, <math>x-z=8</math>, which is not one of our sums. If we keep going, <math>x-z</math> will soon exceed 10 and exceed all our sums, so any value below <math>w=14</math> will not work. Therefore, our only solutions for <math>w</math> are 15 and 16, which means our sum is <math>\boxed{31}</math>. You can check that 15 and 16 work by forming a string of 4 numbers as shown above. <br />
<math>/pi/</math><br />
<br />
== See Also==<br />
<br />
{{AMC10 box|year=2011|ab=B|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_21&diff=936172011 AMC 10B Problems/Problem 212018-03-31T01:18:34Z<p>Mathboy11: /* Solution 4 */</p>
<hr />
<div>== Problem ==<br />
<br />
Brian writes down four integers <math>w > x > y > z</math> whose sum is <math>44</math>. The pairwise positive differences of these numbers are <math>1, 3, 4, 5, 6,</math> and <math>9</math>. What is the sum of the possible values for <math>w</math>?<br />
<br />
<math> \textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93</math><br />
<br />
== Solution 1 ==<br />
The largest difference, <math>9,</math> must be between <math>w</math> and <math>z.</math><br />
<br />
The smallest difference, <math>1,</math> must be directly between two integers. This also means the differences directly between the other two should add up to <math>8.</math> The only remaining differences that would make this possible are <math>3</math> and <math>5.</math> However, those two differences can't be right next to each other because they would make a difference of <math>8.</math> This means <math>1</math> must be the difference between <math>y</math> and <math>x.</math> We can express the possible configurations as the lines.<br />
<br />
<br />
<center><asy><br />
unitsize(14mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
pair Z1=(0,1), Y1=(1,1), X1=(2,1), W1=(3,1);<br />
pair Z4=(4,1), Y4=(5,1), X4=(6,1), W4=(7,1);<br />
<br />
draw(Z1--W1); draw(Z4--W4);<br />
<br />
pair[] ps={W1,W4,X1,X4,Y1,Y4,Z1,Z4};<br />
dot(ps);<br />
label("$z$",Z1,N); label("$y$",Y1,N); label("$x$",X1,N); label("$w$",W1,N);<br />
label("$z$",Z4,N); label("$y$",Y4,N); label("$x$",X4,N); label("$w$",W4,N);<br />
<br />
label("$1$",(X1--Y1),N); label("$1$",(X4--Y4),N);<br />
label("$3$",(Y1--Z1),N); label("$3$",(W4--X4),N); label("$5$",(X1--W1),N); label("$5$",(Y4--Z4),N);<br />
<br />
</asy><br />
</center><br />
<br />
If we look at the first number line, you can express <math>x</math> as <math>w-5,</math> <math>y</math> as <math>w-6,</math> and <math>z</math> as <math>w-9.</math> Since the sum of all these integers equal <math>44</math>,<br />
<cmath>\begin{align*}<br />
w+w-5+w-6+w-9&=44\\<br />
4w&=64\\<br />
w&=16 \end{align*}</cmath><br />
You can do something similar to this with the second number line to find the other possible value of <math>w.</math><br />
<cmath>\begin{align*}<br />
w+w-3+w-4+w-9&=44\\<br />
4w&=60\\<br />
w&=15 \end{align*}</cmath><br />
The sum of the possible values of <math>w</math> is <math>16+15 = \boxed{\textbf{(B) }31}</math><br />
<br />
== Solution 2 ==<br />
<br />
First, like Solution 1, we know that <math>w-z=9 \ \text{(1)}</math>, because no sum could be smaller. Next, we find the sum of all the differences; since <math>w</math> is in the positive part of a difference 3 times, and has no differences where it contributes as the negative part, the sum of the differences includes <math>3w</math>. Continuing in this way, we find that <cmath>3w+x-y-3z=28 \ \text{(2)}</cmath>. Now, we can subtract <math>3w-3z=27</math> from (2) to get <math>x-y=1 \ \text{(3)}</math>. Also, adding (2) with <math>w+x+y+z=44</math> gives <math>4w+2x-2z=72</math>, or <math>2w+x-z=36</math>. Subtracting (1) from this gives <math>w+x=27</math>. Since we know <math>w-z</math> and <math>x-y</math>, we find that <cmath>(w-z)+(x-y)=(w-y)+(x-z)=9+1=10</cmath>. This means that <math>w-y</math> and <math>x-z</math> must be 4 and 6, in some order. If <math>w-y=6</math>, then subtracting this from (3) gives <math>(w-y)-(x-y)=6-1=5</math>, so <math>w-x=5</math>. This means that <math>(w-x)+(w+x)=2w=27+5=32</math>, so <math>w=16</math>. Similarly, <math>w</math> can also equal <math>15</math>.<br />
<br />
Now if you are in a rush, you would have just answered <math>16+15=\boxed{\textbf{(B) }31}</math>. But we do have to check if these work. In fact, they do, giving solutions <math>\boxed{16, 11, 10, 7}</math> and <math>\boxed{15, 12, 11, 6}</math>.<br />
<br />
<br />
== Solution 3 ==<br />
<br />
Let <math>w - x = a</math>, <math>w - y = b</math>, <math>w - z = c</math>. As above, we know that <math>c = 9</math>. Thus, <math>a < b < c</math>. <br />
So, we have <math>w + x + y + z = w + (w - a) + (w - b) + (w - 9) = 4w - a - b - 9 = 44</math>. This means <math>a + b + 9</math> is a multiple of <math>4</math>. Testing values of <math>a</math> and <math>b</math>, we find <math>(a, b, c) = (1, 6, 9), (3, 4, 9),</math> and <math>(5, 6, 9)</math> all satisfy this relation. The corresponding <math>(w, x, y, z)</math> sets are <math>(15, 14, 9, 6), (15, 12, 11, 6),</math> and <math>(16, 11, 10, 7)</math>. The first set does not satisfy the given conditions, but the other two do. Thus, <math>w = 15</math> and <math>w = 16</math> are both possible solutions so the answer is <math>15 + 16 = 31</math>.<br />
<br />
~~ solution by ccx09 Roy Short<br />
~~latexed by jk23541<br />
== Solution 4 ==<br />
From the problem, we know that <math>w+x+y+z=44</math>. Since it is said that the pairwise positive differences between numbers are 1, 3, 4, 5, 6, 9, and we can figure that the pairwise positive differences are <math>(w-x)</math>, <math>(w-y)</math>, <math>(w-z)</math>, <math>(x-y)</math>, <math>(x-z)</math>, <math>(y-z)</math>, the sum of <math>(w-x), (w-y), (w-z), (x-y), (x-z), (y-z)</math> is equal to the sum of 1, 3, 4, 5, 6, 9, so <math>(w-x)+(w-y)+(w-z)+(x-y)+(x-z)+(y-z)=28</math>. Simplifying, we get <math>3w-3z+x-y=28</math>. Adding <math>w+x+y+z=44</math> and <math>(w-x)+(w-y)+(w-z)+(x-y)+(x-z)+(y-z)=28</math>, we get <math>4w+2x-2z=72</math>, and simplifying we get <math>2w+x-z=36</math>. Since <math>x-z</math> is one of our positive differences, we can start guessing values for <math>w</math>, and if the equation simplifies to one of our numerical positive differences, that value of <math>w</math> should work. We can start at <math>w=18</math> and keep going down, because our sum has to be positive. For <math>w=18</math>, <math>x-z=0</math>, which is not one of our sums. For <math>w=17</math>, <math>x-z=2</math>,which is not one of our sums. For <math>w=16</math>, <math>x-z=4</math>, which is one of our sums, so 16 works. For <math>w=15</math>, <math>x-z=6</math>, which is one of our sums, so 15 works. For <math>w=14</math>, <math>x-z=8</math>, which is not one of our sums. If we keep going, <math>x-z</math> will soon exceed 10 and exceed all our sums, so any value below <math>w=14</math> will not work. Therefore, our only solutions for <math>w</math> are 15 and 16, which means our sum is <math>\boxed{31}</math>. You can check that 15 and 16 work by forming a string of 4 numbers as shown above. <br />
<math>pi</math><br />
<br />
== See Also==<br />
<br />
{{AMC10 box|year=2011|ab=B|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Mathboy11https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_21&diff=935212011 AMC 10B Problems/Problem 212018-03-26T22:03:34Z<p>Mathboy11: /* Solution 4 */</p>
<hr />
<div>== Problem ==<br />
<br />
Brian writes down four integers <math>w > x > y > z</math> whose sum is <math>44</math>. The pairwise positive differences of these numbers are <math>1, 3, 4, 5, 6,</math> and <math>9</math>. What is the sum of the possible values for <math>w</math>?<br />
<br />
<math> \textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93</math><br />
<br />
== Solution 1 ==<br />
The largest difference, <math>9,</math> must be between <math>w</math> and <math>z.</math><br />
<br />
The smallest difference, <math>1,</math> must be directly between two integers. This also means the differences directly between the other two should add up to <math>8.</math> The only remaining differences that would make this possible are <math>3</math> and <math>5.</math> However, those two differences can't be right next to each other because they would make a difference of <math>8.</math> This means <math>1</math> must be the difference between <math>y</math> and <math>x.</math> We can express the possible configurations as the lines.<br />
<br />
<br />
<center><asy><br />
unitsize(14mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
pair Z1=(0,1), Y1=(1,1), X1=(2,1), W1=(3,1);<br />
pair Z4=(4,1), Y4=(5,1), X4=(6,1), W4=(7,1);<br />
<br />
draw(Z1--W1); draw(Z4--W4);<br />
<br />
pair[] ps={W1,W4,X1,X4,Y1,Y4,Z1,Z4};<br />
dot(ps);<br />
label("$z$",Z1,N); label("$y$",Y1,N); label("$x$",X1,N); label("$w$",W1,N);<br />
label("$z$",Z4,N); label("$y$",Y4,N); label("$x$",X4,N); label("$w$",W4,N);<br />
<br />
label("$1$",(X1--Y1),N); label("$1$",(X4--Y4),N);<br />
label("$3$",(Y1--Z1),N); label("$3$",(W4--X4),N); label("$5$",(X1--W1),N); label("$5$",(Y4--Z4),N);<br />
<br />
</asy><br />
</center><br />
<br />
If we look at the first number line, you can express <math>x</math> as <math>w-5,</math> <math>y</math> as <math>w-6,</math> and <math>z</math> as <math>w-9.</math> Since the sum of all these integers equal <math>44</math>,<br />
<cmath>\begin{align*}<br />
w+w-5+w-6+w-9&=44\\<br />
4w&=64\\<br />
w&=16 \end{align*}</cmath><br />
You can do something similar to this with the second number line to find the other possible value of <math>w.</math><br />
<cmath>\begin{align*}<br />
w+w-3+w-4+w-9&=44\\<br />
4w&=60\\<br />
w&=15 \end{align*}</cmath><br />
The sum of the possible values of <math>w</math> is <math>16+15 = \boxed{\textbf{(B) }31}</math><br />
<br />
== Solution 2 ==<br />
<br />
First, like Solution 1, we know that <math>w-z=9 \ \text{(1)}</math>, because no sum could be smaller. Next, we find the sum of all the differences; since <math>w</math> is in the positive part of a difference 3 times, and has no differences where it contributes as the negative part, the sum of the differences includes <math>3w</math>. Continuing in this way, we find that <cmath>3w+x-y-3z=28 \ \text{(2)}</cmath>. Now, we can subtract <math>3w-3z=27</math> from (2) to get <math>x-y=1 \ \text{(3)}</math>. Also, adding (2) with <math>w+x+y+z=44</math> gives <math>4w+2x-2z=72</math>, or <math>2w+x-z=36</math>. Subtracting (1) from this gives <math>w+x=27</math>. Since we know <math>w-z</math> and <math>x-y</math>, we find that <cmath>(w-z)+(x-y)=(w-y)+(x-z)=9+1=10</cmath>. This means that <math>w-y</math> and <math>x-z</math> must be 4 and 6, in some order. If <math>w-y=6</math>, then subtracting this from (3) gives <math>(w-y)-(x-y)=6-1=5</math>, so <math>w-x=5</math>. This means that <math>(w-x)+(w+x)=2w=27+5=32</math>, so <math>w=16</math>. Similarly, <math>w</math> can also equal <math>15</math>.<br />
<br />
Now if you are in a rush, you would have just answered <math>16+15=\boxed{\textbf{(B) }31}</math>. But we do have to check if these work. In fact, they do, giving solutions <math>\boxed{16, 11, 10, 7}</math> and <math>\boxed{15, 12, 11, 6}</math>.<br />
<br />
<br />
== Solution 3 ==<br />
<br />
Let <math>w - x = a</math>, <math>w - y = b</math>, <math>w - z = c</math>. As above, we know that <math>c = 9</math>. Thus, <math>a < b < c</math>. <br />
So, we have <math>w + x + y + z = w + (w - a) + (w - b) + (w - 9) = 4w - a - b - 9 = 44</math>. This means <math>a + b + 9</math> is a multiple of <math>4</math>. Testing values of <math>a</math> and <math>b</math>, we find <math>(a, b, c) = (1, 6, 9), (3, 4, 9),</math> and <math>(5, 6, 9)</math> all satisfy this relation. The corresponding <math>(w, x, y, z)</math> sets are <math>(15, 14, 9, 6), (15, 12, 11, 6),</math> and <math>(16, 11, 10, 7)</math>. The first set does not satisfy the given conditions, but the other two do. Thus, <math>w = 15</math> and <math>w = 16</math> are both possible solutions so the answer is <math>15 + 16 = 31</math>.<br />
<br />
~~ solution by ccx09 Roy Short<br />
~~latexed by jk23541<br />
== Solution 4 ==<br />
From the problem, we know that <math>w+x+y+z=44</math>. Since it is said that the pairwise positive differences between numbers are 1, 3, 4, 5, 6, 9, and we can figure that the pairwise positive differences are <math>(w-x)</math>, <math>(w-y)</math>, <math>(w-z)</math>, <math>(x-y)</math>, <math>(x-z)</math>, <math>(y-z)</math>, the sum of <math>(w-x), (w-y), (w-z), (x-y), (x-z), (y-z)</math> is equal to the sum of 1, 3, 4, 5, 6, 9, so <math>(w-x)+(w-y)+(w-z)+(x-y)+(x-z)+(y-z)=28</math>. Simplifying, we get <math>3w-3z+x-y=28</math>. Adding <math>w+x+y+z=44</math> and <math>(w-x)+(w-y)+(w-z)+(x-y)+(x-z)+(y-z)=28</math>, we get <math>4w+2x-2z=72</math>, and simplifying we get <math>2w+x-z=36</math>. Since <math>x-z</math> is one of our positive differences, we can start guessing values for <math>w</math>, and if the equation simplifies to one of our numerical positive differences, that value of <math>w</math> should work. We can start at <math>w=18</math> and keep going down, because our sum has to be positive. For <math>w=18</math>, <math>x-z=0</math>, which is not one of our sums. For <math>w=17</math>, <math>x-z=2</math>,which is not one of our sums. For <math>w=16</math>, <math>x-z=4</math>, which is one of our sums, so 16 works. For <math>w=15</math>, <math>x-z=6</math>, which is one of our sums, so 15 works. For <math>w=14</math>, <math>x-z=8</math>, which is not one of our sums. If we keep going, <math>x-z</math> will soon exceed 10 and exceed all our sums, so any value below <math>w=14</math> will not work. Therefore, our only solutions for <math>w</math> are 15 and 16, which means our sum is <math>\boxed{31}</math>. You can check that 15 and 16 work by forming a string of 4 numbers as shown above.<br />
<br />
== See Also==<br />
<br />
{{AMC10 box|year=2011|ab=B|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Mathboy11